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EVcycle
01 December 2010, 1541
Is it better to

1) Have 72 volts at 50 AH on a 72 volt Controller

or

2) 90 volts at 40 AH to a controller that will reduce the output to the motor at 72 volts?

DaveAK
01 December 2010, 1544
If the controller is rated by voltage, then you probably won't want to exceed that rating. Is 90V within the maximum controller limit?

ETA: OK, I misunderstood, you're not talking about the same controller in both examples. So my personal opinion would be to lessen the stress on the system. If a 72V supply to a 72V controller requires less work on the part of the controller than chopping 90V down to 72V then I'd go with 1.

cycleguy
01 December 2010, 1551
I would recommend that your batteries do not exceed the nominal voltage rating of the controller and that they have enough Ah capacity to supply maximum amperage to the controller with enough headroom as not to overstress the batteries.

What is the maximum voltage and amp rating of your controller?

EVcycle
01 December 2010, 1559
The controller I have is a 90 volt max controller. The batteries will charge to 94 but rest at about 88 so no issues there.

This is not for max speed, but max distance and good speed. I have the present 300 amp controller turned down to 250 amps and can still get almost 70 MPH.

DaveAK
01 December 2010, 1656
I was planning to run my 80V Sevcon at 84V. It's max is 100V I think and my batteries were going to be under that fully charged. I wasn't too concerned at this because people have done it successfully before.

Gut feeling tells me that if you're not going use the volts for speed then go for 72V with a 72V controller if you have one, but I'd use the 90V one if that's what I already had. Personally I can't offer any more than this because I have far less experience than you and I am no kind of electrical engineer.

But my advice is always free and worth every penny you pay for it! :D

EVcycle
01 December 2010, 1710
Thanks Dave!

cycleguy
01 December 2010, 1726
I guess it depends on what you are after, better acceleration or a higher top speed. Lower voltage with a higher Ah capacity would allow you to turn the controller back up to 350A and give you better low end acceleration, but torque will fall off earlier which will limit your top speed.

It looks to me like your battery pack is underpowered for the controller. At 40Ah, and 350Amps you would be drawing 8.75C from the batteries, which is too high unless you are using high discharge batteries. Ideally, you want the battery pack to be sufficient enough to allow the controller to operate to it's maximum, otherwise you are just trading between acceleration and speed. I would suggest at least 70ah to take full advantage of your controller.

It doesn't really matter whether you lower the voltage and raise the Ah or raise the voltage and lower the Ah, the motor will consume the same amount of watts depending on your throttle input, up to the limits of your controller or battery pack.

harlan
01 December 2010, 1823
Higher voltage on the battery side will be better since there will be less Peukert.

For instance, to get 3600 Watts to the motor (assuming all things are equal with the controller) will require:

72V @ 50A from choice A

or

90V @ 40A from choice B

Either way you're pulling 1C out of the batteries, but less amps means less Peukert, means less loss, right?

Tony Coiro
01 December 2010, 1910
Is it better to

1) Have 72 volts at 50 AH on a 72 volt Controller

or

2) 90 volts at 40 AH to a controller that will reduce the output to the motor at 72 volts?

You sure the output will be reduced to 72V? An easy way to test would be to jack up the rear tire, full throttle and check voltage across the motor terminals. If I understand controllers correctly, you should see 90V motor side. I guess my justification for this is when my pack is REALLY sagging, my top speed loss is proportional to voltage sag. I think the controller nominal voltage actually describes what the FET's or IGBT's in the controller are rated for.

cycleguy
01 December 2010, 1958
Either way you're pulling 1C out of the batteries, but less amps means less Peukert, means less loss, right?

I believe you're right Harlan, Peukert's equation would indicate higher amperage draw would increase losses.
So, if given a choice between 96V and 300 amps, or 72V and 400 amps, The higher voltage would be more efficient.

Nuts & Volts
01 December 2010, 2013
90V all the way. For these reasons-
1.) Less peukert on the batteries
2.) Less Current in the systems = less resistive heat losses in the cntrl, wires, motor
3.) It wont change your max speed, because it sounds like the controller will be turning the volts in into amps out.
4.) Full charge will be quicker less total AH to put in (ie charges at a higher voltage)
5.) Potentially easier wiring (less parallel cells if any)

-Kyle

edit:balancing would not be easier sorry

chef
01 December 2010, 2110
5.) Potentially easier wiring and balancing (less parallel cells if any)
Curious why you would say that. I was under the impression parallel cells make balancing easier (the ones in parallel are naturally balanced). Whereas series cells are prone to falling out of balance voltage-wise.

Tony Coiro
01 December 2010, 2121
Maybe I misunderstood, is your controller stepping 90VDC to 72VDC?

lugnut
01 December 2010, 2145
Is it better to

1) Have 72 volts at 50 AH on a 72 volt Controller

or

2) 90 volts at 40 AH to a controller that will reduce the output to the motor at 72 volts?

If I understand your question right, I say it is the same either way. You have 3600 Wh. Example:
1) 24 cells in series where each is 3 volts and 50 Ah.

2) 30 cells in series where each is 3 volts and 40 Ah.

Both systems use the same motor with a max of 72 volts. So at equal motor output, both battery systems would be discharging at the same respective C-rate. So discharge times would be equal.

Assuming cell chemistry is the same for both cells, the amount of active material will be the same, but you will have slightly more mass and volume with choice #2 due to additional cell packaging, but you may find greater packaging flexibility for your installation with the smaller cells.

In the real world there are a couple of things to consider. If you're the type who believes in BMS, choice #2 will require 6 additional cell cards. Also, 6 additional intercell connections. However the intercell connections and battery cables could be 20% smaller for choice #2.

harlan
01 December 2010, 2145
Maybe I misunderstood, is your controller stepping 90VDC to 72VDC?

Isn't that what a DC motor controller is? A voltage controller. RPMs are proportional to Voltage.

harlan
01 December 2010, 2147
If I understand your question right, I say it is the same either way. You have 3600 Wh.

In an ideal world, I would agree with you. Due to inefficiencies and other losses, I stand by my theory that higher voltage/less amps is the better choice.

DaveAK
01 December 2010, 2158
Maybe I misunderstood, is your controller stepping 90VDC to 72VDC?
What you said earlier that putting 90V in to a 72V rated controller will get you 90V out at full throttle makes sense to me. So if he's running a 72V motor, and doesn't want to run it higher than that he could limit the throttle. This goes with what I was originally thinking in that the controller will be doing extra work in this scenario - limiting 90V to 72V.

EVcycle
01 December 2010, 2353
The controller (Altrax) can be programmed to limit the output voltage or Amps to the motor. The controller limits/adds volts all the time as apart of its normal function as
amps and throttle are needed/wanted by the rider.
In this case I am Using Headways 10AH 3.32 V batteries. Balancing is not problem either way as there will be at least 4 in parallel.

Each pack of 4 or 5 batteries will be charged individually. A balanced pack every charge.

BaldBruce
02 December 2010, 0704
A DC motor controller is simply one particular type of switch mode power supply. It takes the input voltage and PWMs it to obtain the desired output voltage and current. It takes advantage of the fact that a DC motor is a very large inductor, but other than that it is similar to any other variable power supply. Always more efficient in the power supply world to use the highest voltage that is practical due to efficiency. It is simply I^2R losses. At any given power level, As voltage halves, the current doubles (P=IE), but the power loss goes up by 4!! (P=I^2R) Now there are always several other issues to be taken into account like the fact that the difference between the input and output voltages also drives efficiency (but nowhere near as much as the I change) and the practical limits of the controller input. I have a friend with a small EV car that has to bleed off the surface charge of his LA batteries with his headlights before he can leave after a float charge because ot the overvoltage limit on his controller.........
Bruce

DaveAK
02 December 2010, 0927
All those saying highest voltage wins can you explain if that's input voltage or output voltage, or if it doesn't matter, 'cause I'm not following. I understand that a controller PWMs input V into output V, but if input and output are the same, no PWMing, right? Or is that wrong because I'm ignoring current?

ETA: I actually understand that higher voltage/lower current is better, I just can't seem to apply it to this situation where the max output V isn't the same as the input V.

lugnut
02 December 2010, 1245
I understand that a controller PWMs input V into output V, but if input and output are the same, no PWMing, right? Or is that wrong because I'm ignoring current?

I think you are correct. No PWMing??? The more correct way to describe that condition would be 100% duty cycle, or dc for short, not to be confused with DC as in Direct Current. So the dc (duty cycle) is simply the on-time / period. The period is the time for one complete cycle of PWM, or on-time plus off-time. Let's use a 10 millisecond (mS) period for example.

If you have a 90V battery and need 72 volts to the motor, then the controller would have an 80% dc, on for 8mS and off for 2mS. 90V * 0.8 = 72V. And the motor current would be 1.25 times the battery current (1 / 0.8).

If you have a 72V battery and need 72 volts to the motor, then the controller would have a 100% dc, on for 10mS, off for 0mS. 72V * 1.0 = 72V. And the motor current would equal the battery current.

frodus
02 December 2010, 1248
Higher voltage on the battery side will be better since there will be less Peukert.

For instance, to get 3600 Watts to the motor (assuming all things are equal with the controller) will require:

72V @ 50A from choice A

or

90V @ 40A from choice B

Either way you're pulling 1C out of the batteries, but less amps means less Peukert, means less loss, right?

It all depends on if the 50Ah batteries and the 40Ah batteries are the same style of batteries. If they have the same discharge curve at 1C, then the peukert remains the same.

Now if you draw 50A out of one, and 50A out of the other, the C ratings change, and your peukert on the 40Ah battery at 50A discharge worsens.

frodus
02 December 2010, 1301
90V all the way. For these reasons-
1.) Less peukert on the batteries
2.) Less Current in the systems = less resistive heat losses in the cntrl, wires, motor
3.) It wont change your max speed, because it sounds like the controller will be turning the volts in into amps out.
4.) Full charge will be quicker less total AH to put in (ie charges at a higher voltage)
5.) Potentially easier wiring (less parallel cells if any)

-Kyle

edit:balancing would not be easier sorry
1) nope, its the same if the discharge power is the same for both, the C rate will be the same, and if they're the same chemistry and design (i.e. ts 40Ah cells and ts 60Ah cells should have similar discharge curves at 1C and should be the same if they're made the same). If we're talking about a TS 40Ah cell and 5 10Ah headways, they have different peukert, but that is comparing apples to oranges. they should have the same discharge rate at 1C, which would be 1 hour. Peukert is the same if we're talking 5-10Ah headways and 4 10Ah headways. The 1C discharge curve is exactly the same.

2) Current is less in the system, YES, exactly. This reduces connection loss, cable loss and here's one thing that no one has mentioned.... Switching loss. When you run FET's at a higher voltage, to get the same power, they run less current through them..... which means that they're going to have less LOSS, which means less heat and higher controller efficiency. Now this all depends on the FET's in the controller (lower voltage controllers aren't IGBT), but if its the same controller running either 90V or 72V, then choose higher voltage. It'l run cooler and more efficient because of lower current and lower switching losses.

3) why is he limiting his motor to 72V? Why not just run 0-100% in either case? A 72v motor can run at 1-2 times its rated voltage as long as it doesn't exceed mechanical RPM limits. Why not get the extra RPM out of the motor?

4) Charging time is the exact same (although charger efficiency might NOT be the same). The total Wh of the battery pack, if discharged the same DOD, is exactly the same. To put ~3600Wh back into the batteries, you would still need to put the same amount of Wh back in. If you charge at 72V, you charge at ~13.8A (1000W charger), it takes ~3.5hours. If you charge at 90V and 1000W, or 11.1A, it would take the same amount of time. There is no change here, but you are limited to the charge current limit of the batteries.

5) Yes, definately. putting 5 headways in parallel compared to 4 is a big difference in bussbar and connection points.

Nuts & Volts
02 December 2010, 1351
Thanks for the corrections travis!
1.) Good to know about the peukert thing, which makes sense.
2.) yep
3.) I was just going with the 72V output that Ed said the controller had in his first post (but I agree with your correction
4.) My bad I was think V up but forgot that a 0.5C charge would be different for each case.
5.) yep

Kyle

DaveAK
02 December 2010, 1408
We don't know the specs of the motor do we? If it will take 90V then I'd run it at 90V, but if it's a 36V motor that he wants to run at 72V, then pushing it to 90V might not be a good idea.

Ed? Enlighten us. :)

frodus
02 December 2010, 1425
Thanks Dave, thats a really good point. If its REALLY 72V, then its probably going to be just fine running at 90V. If its a 36 or 48V motor, it might be marginally pushing it, but even then, if he can set the controller to limit PWM output, then he's set.

Although, I'm wondering why he would think about doing 90V. Thats 26 cells in series, which is going to be a PITA to find a charger for.

DaveAK
02 December 2010, 1440
Thanks Dave, thats a really good point. If its REALLY 72V, then its probably going to be just fine running at 90V. If its a 36 or 48V motor, it might be marginally pushing it, but even then, if he can set the controller to limit PWM output, then he's set.

Although, I'm wondering why he would think about doing 90V. Thats 26 cells in series, which is going to be a PITA to find a charger for.
I'm doing 26 cells, well I was but I'm probably going to do 24 now. I built my own 15A charger, where as Ed already uses a single charger per cell, so I don't think he'll have any problems there.

frodus
02 December 2010, 1458
oh, but its going to take a while to charge isn't it?

DaveAK
02 December 2010, 1505
oh, but its going to take a while to charge isn't it?
Yup, I would think so, but he's happy with it I believe. I think his current chargers are 2A but I don't know the Ah of his current pack. Maybe he has a different plan though. Maybe he isn't planning anything excpet to get us talking about a hypothetical situation. :)

Skeezmour
02 December 2010, 1515
Suckers :)

Good one Ed!

EVcycle
02 December 2010, 1754
It is the RT Mars motor rated for 72 volts. At 90 volts it gets a bit warm, but doable.

I was keeping it at 72 via the controller as it gives me good speed and keeps the motor happy.

EVcycle
02 December 2010, 1758
Yup, I would think so, but he's happy with it I believe. I think his current chargers are 2A but I don't know the Ah of his current pack. Maybe he has a different plan though. Maybe he isn't planning anything excpet to get us talking about a hypothetical situation. :)

Single chargers at 2A

No need to rush the charge for me as I plug it in when I get home from work and it is charged by the time go the next day. It is a commuter bike.

2A for each 40AH pack.

frodus
02 December 2010, 1758
Great discussion topic though.....

run higher volts, less Ah.

Same energy, same power, but higher efficiency and lower losses.

EVcycle
02 December 2010, 1759
then run higher volts, less Ah.

Same energy, same power, but higher efficiency and lower losses.

That was my thought. Just wanted it bounce it off the experts!


Thanks.



Now for the next project....... :)

EVcycle
02 December 2010, 1803
Yup, I would think so, but he's happy with it I believe. I think his current chargers are 2A but I don't know the Ah of his current pack. Maybe he has a different plan though. Maybe he isn't planning anything excpet to get us talking about a hypothetical situation. :)

Not hypothetical, Just upgrading the Juiced bike with a simpler setup and in this case, better.

To go to 24S5P would have been a major redo.
To go to 26S4P (or maybe even 27S4P) Just add batteries in the string and a few more chargers. It will take the same time to charge as before. :)

lugnut
02 December 2010, 1821
Great discussion topic though.....

run higher volts, less Ah.

Same energy, same power, but higher efficiency and lower losses.

For the difference, 72 to 90, not a big deal. But I fail to see where more losses/less efficiency comes into play. Yeah, for more battery current, he needs slightly larger cable. But on the other hand, for higher voltage, he needs more series cells and therefore more chargers and less reliability.

Yeah, not a big deal one way or the other, 72 to 90 volts. But higher voltage is not automatically assurance of a better system. And since he is doing an upgrade where additional series cells are easier to implement, I agree. But every system and application needs to be looked at on its own merits. Higher voltage isn't always better. It can bring with it higher stresses and more expense.

EVcycle
02 December 2010, 1825
less reliability?

harlan
02 December 2010, 1841
For the difference, 72 to 90, not a big deal. But I fail to see where more losses/less efficiency comes into play. Yeah, for more battery current, he needs slightly larger cable. But on the other hand, for higher voltage, he needs more series cells and therefore more chargers and less reliability.

Yeah, not a big deal one way or the other, 72 to 90 volts. But higher voltage is not automatically assurance of a better system. And since he is doing an upgrade where additional series cells are easier to implement, I agree. But every system and application needs to be looked at on its own merits. Higher voltage isn't always better. It can bring with it higher stresses and more expense.

I2R losses are not negligible when you're talking about so many amps.

lugnut
02 December 2010, 1841
less reliability?

Sure. Anytime you increase the number of components in a system, you decrease the reliability, don't you? More things to go wrong. So to go from 72 to 90 volts with your charging system, you have to increase the number of chargers. Right? It increases the chance of a single charger failure shutting you down. Hopefully you will detect this if it occurs and correct it. But nevertheless, whenever you increase the components in a system, like # of chargers, connections or whatever, you need to factor in the reliability. Yeah, maybe you won't notice it off the bat. But a few years from now, when one of the modular chargers has quit, and you fry a parallel bank of cells, you'll think of me :-)

lugnut
02 December 2010, 1845
I2R losses are not negligible when you're talking about so many amps.

All you have to do is increase the cable size instead of making it longer to accommodate the additional cells. Same amount of copper, just in a different place. No increase in resistive loss.

BaldBruce
02 December 2010, 2201
All you have to do is increase the cable size instead of making it longer to accommodate the additional cells. Same amount of copper, just in a different place. No increase in resistive loss.

For the same controller, you can't modify anything internal (easily) and therfore Harlan and Travis point still stands. Losses go down if you operate at a higher voltage. I^2 R is a powerful law. (Pun intended....):eek:

Your cable concept is also flawed. Losses go up or down as a squared fuction of current. Therefore to keep cable losses equal, the resistance also needs to change as a function of the squared term to equate losses.

Really big difference example of 100A at 10V versus 25A at 40V. (Both 1kW) Lets say I have a 10 mohm cable in the original 100A example. This is a loss of 0.01 X 100 X 100 = 100watts. Changing to 25A example we get way lower losses with the same cable! 0.01 X 25 X 25 =6.25 watts. The losses on the same cable when we dropped the voltage by 4 equate to a 16 times decrease in power loss. Or conversely, I could use a cable 16 times smaller to get equal losses. (100W/25^2 V=0.16 Ohms)

I know this is an extreme example, but it clearly shows the advantages of higher voltages. Is it a coincidence that all EVs being commercially made use very high voltage packs????

lugnut
03 December 2010, 0633
For the same controller, you can't modify anything internal.....

Yes, I know. So let's call it a 100V max controller and use the previous example. For equal conditions on the motor, that controller will likely operate a little more efficient at 72 volts than at 90. Equal conditions mean that the controller output to the motor is the same. Therefore at 90 volts, the switching duty cycle is less causing more ripple in the caps and more loss. And more ripple implies higher peaks for the same RMS, so the IČ works against you. In this example, the efficiency difference would be so small in the controller it would be difficult to measure, except for the time where it operates at 72V to the motor. There it would be full-on for the low voltage system and switching at 80% for the high voltage. Full-on implies no switching loss, therefore higher efficiency.

As far as the loss in the cables (and connectors) on the battery side, I fully agree that IČR loss favors higher voltage systems. However in the example, as I stated, if he were to use the copper needed to extend the length of the battery circuit to connect additional cells to make the shorter cable length larger in cross section, he would have reduced the resistance for the higher current lower voltage system and kept the IČR losses about the same. Yeah, a pretty academic argument and likely not practical in his case.

And I agree that higher voltage systems are preferable as a general rule. But not necessarily better for every situation. Especially when it comes to batteries due to the increase in cell count. I think this is evidenced by Toyota's design decision in the Prius. They used high voltage motors (around 500V) but stuck with a lower voltage battery (around 200).

Again, this example of 72 and 90V, no big deal either way. He should do what's easiest. But higher voltage in general brings with it some concerns. Higher voltage presents greater hazard, like being able to arc further distance and sustain the arc. Higher dielectric stress meaning thicker more expensive insulation systems. More expensive fusing and switching devices, both mechanical and solid state. Remember a few years ago when there was the push towards 42V automotive auxiliary systems. Hear much about that lately?

For equal power systems, that IČR law does favor higher voltage systems WRT to conductor, cable and wiring mass. I agree. But you want a real shocker? Compare the cost of equal power systems (like in the 150 kW range) between 350 and 700 volts DC. You'd wish the cost increase was only a squared function!

frodus
03 December 2010, 0905
All you have to do is increase the cable size instead of making it longer to accommodate the additional cells. Same amount of copper, just in a different place. No increase in resistive loss.

Please re-read this


2) Current is less in the system, YES, exactly. This reduces connection loss, cable loss and here's one thing that no one has mentioned.... Switching loss. When you run FET's at a higher voltage, to get the same power, they run less current through them..... which means that they're going to have less LOSS, which means less heat and higher controller efficiency. Now this all depends on the FET's in the controller (lower voltage controllers aren't IGBT), but if its the same controller running either 90V or 72V, then choose higher voltage. It'l run cooler and more efficient because of lower current and lower switching losses.

and I read your comment, but how often do you run at 100% throttle? or even 80%?

You're almost never there, so switching losses are almost always an issue.

And the argument of 72V and 90V with respect to arcing and danger...... its a difference of 18V. Yes it'l be slightly different, but your CURRENT decreases. Current, voltage and gap distance is an issue, decrease one, increase the other and arcing is not going to change much for 18V.

lugnut
03 December 2010, 0953
but how often do you run at 100% throttle? or even 80%?

Who me? A lot :-)


Please re-read this

If you discount the 100% duty cycle condition, you're probably correct. However it seems you just consider the main switch in the controller. Since we would want to compare identical conditions, being the same motor current and voltage, then the output of the controller would see the same current in both cases. And since the main switch is seeing less current, the diode must see more, correct? And the diode is part of the controller. So the diode would have increased loss which would to some degree offset the savings in the switch (mosfet or igbt). And then the higher voltage and ripple in the bus caps, which are also a controller loss.

Between 72 and 90 volts, discounting the 100% condition, I think it would be pretty much a wash.

frodus
03 December 2010, 1030
I doubt you would even know if you were at 100% duty cycle unless you've got a probe looking at the gate signal on the FET's. Turing the throttle to 100% doesn't mean much. Most controllers are torque control. Pegging the throttle doesn't neccessarily mean that its going to 100% duty.

I think you're probably right that its nearly the same in both cases, but higher voltage is better for losses in general, but as you point out, it can be worse for charging/BMS. One more thing to go wrong, but if he installs some sort of monitoring (like a cell-log) he should be good.




And BTW, I haven't seen your motorcycle, do you have a link? I'd like to read up about your project. You haven't posted much information about your build.

lugnut
03 December 2010, 1048
I doubt you would even know if you were at 100% duty cycle unless you've got a probe looking at the gate signal on the FET's. Turing the throttle to 100% doesn't mean much. Most controllers are torque control. Pegging the throttle doesn't neccessarily mean that its going to 100% duty.

Let's see. If I'm at full throttle and the battery current peaks and starts to decline, I'm at 100%, right?

Yeah, I'll have to get crackin' on that 2-wheeler EV project. Still have a ways to go. Guess I'll shut this stupid computer off and go out, drag it out of the shed and into the workshop. You guys are a real inspiration :-)

BaldBruce
03 December 2010, 1207
Let's put this in perspective by studying the actual problem given.

What Ed is asking IMO is how to best use his extra cells in an existing bike. So let's just make some assumptions and see where we land.

I'll assume a 15KW motor with a 300A current limit to the motor. That means at the current limit we need an RMS voltage of 50V output. Assuming a constant loss for the controller of 2KW we will compare the losses of just the input cables and connectors. (I show below why they are not exactly equal, but let's just make the best case assumption that they are equal in losses.)

17KW/72V= 236A input current and a D of 69.4%
17KW/90V=188A input current and a D of 55.5%

If we have 4AWG cable of 5 feet long with 4 connectors in our system we are in the 5 mohm range. Therefore:

236^2 *.005 = 278.48W loss
188^2 *.005 = 176.72W loss

100watts less loss in the same system! You'll note that the assumed resistance is a linear scaler so it really doesn't matter what gauge wire you analyze. It will always come out to be 36.5% less loss if I use the higher voltage in an existing system.

Now getting back to the controller it should be quite clear that given MOSFETs with a given Rds On, operating at a higher voltage and lower current will always result in lower losses. These controllers are constant frequency and are just varying the duty cycle of the PWM and therfore the switching loss is relatively constant between the two examples. Losses in the capacitors ESR is also almost constant. The same logic applies to IGBTs or transistors because they have a constant voltage drop and therefore the drop is a smaller percentage when operating at higher voltages.

Now it is clearly possible to lower these losses by changing controllers or dropping the AWG wire size down, but I believe the root question is answered as to which is better for his particular case.

Whether 100W is large or small is a up to the builder..........

lugnut
03 December 2010, 1257
Let's put this in perspective by studying the actual problem given.

What Ed is asking IMO is how to best use his extra cells in an existing bike. So let's just make some assumptions and see where we land......

If we have 4AWG cable of 5 feet long with 4 connectors in our system we are in the 5 mohm range. Therefore:

236^2 *.005 = 278.48W loss
188^2 *.005 = 176.72W loss
.......

Yes, this is the case if the battery wiring was the same for the higher voltage pack. But my point was, he would have to add wire when he added the cells to increase pack voltage. So let's continue your example. Say the extra cells have to go under the seat and required an additional 2 feet of #4 cable. Then the R goes up to 0.007Ω and comes to 247W loss, up from the 177W.

Instead of using that copper for 2 ft longer cable adding the cells, let him use it to increase the size of the existing 5 ft of cable. Then that R goes down to 0.0036Ω and the loss is 200W for the low voltage system, down from 278W and even lower than the high voltage system.

And in your controller loss explanation, you do not include the increase loss in the diode. And the bus cap ESR loss will increase due to the higher voltage and shorter duty cycle causing increased ripple.

In this particular case, 72 vs 90V, I do not think the argument can be made on system efficiency. Maybe comparing a 24V system to a 90, yes. He should do what is best for his installation considering the work, cost and reliability.

frodus
03 December 2010, 1305
Let's see. If I'm at full throttle and the battery current peaks and starts to decline, I'm at 100%, right?

Yeah, I'll have to get crackin' on that 2-wheeler EV project. Still have a ways to go. Guess I'll shut this stupid computer off and go out, drag it out of the shed and into the workshop. You guys are a real inspiration :-)

Oh, you haven't even built anything yet.......

Just because current is at max, and then starts to decrease doesn't mean its at 100% PWM.

Once current to the motor equals current from the batteries you are at 100%. Then and only then.

lugnut
03 December 2010, 1328
Oh, you haven't even built anything yet....... I'll just stop arguing with you, its like talking to a wall. Lots of "research" and little experience.

I didn't say I had not built anything yet. In fact I have a number of EVs under my belt, so to speak. Just that my EV 2-wheeler project is lagging. Excuse me, but I did not see a resume required to participate on this forum. Why not address what I say instead of what or who you think I am?


Just because current is at max, and then starts to decrease doesn't mean its at 100% PWM.

Once current to the motor equals current from the batteries you are at 100%. Then and only then.

So what you're saying is that if I have a full throttle acceleration and my battery ammeter peaks and starts to decline, I have not reached the point where battery current equals motor current, ie 100% duty cycle?

Tony Coiro
03 December 2010, 1329
I am in between classes, so I apologize I am not up to date on this thread, gotta type fast. Why is high voltage/lower current more efficient? Sure, I^2*R is power, but so it V^2/R. Also, higher efficiency doesn't make sense to me on a quantum level either. Everything seems to cancel out when I think about it..

frodus
03 December 2010, 1332
because your FET's are switching less battery current. Less current, less loss.

Tony Coiro
03 December 2010, 1429
because your FET's are switching less battery current. Less current, less loss.

But fundamentally, why is less current more efficient?

Tony Coiro
03 December 2010, 1454
Really big difference example of 100A at 10V versus 25A at 40V. (Both 1kW) Lets say I have a 10 mohm cable in the original 100A example. This is a loss of 0.01 X 100 X 100 = 100watts. Changing to 25A example we get way lower losses with the same cable! 0.01 X 25 X 25 =6.25 watts. The losses on the same cable when we dropped the voltage by 4 equate to a 16 times decrease in power loss. Or conversely, I could use a cable 16 times smaller to get equal losses. (100W/25^2 V=0.16 Ohms)

I know this is an extreme example, but it clearly shows the advantages of higher voltages. Is it a coincidence that all EVs being commercially made use very high voltage packs????

This is a misleading example, sure they are both 1kW, however, you cannot apply 40V to something with 0.01 Ohm and only get 25A, not even close. The only way this could happen is if the wire had 1.6 Ohms of resistance, a number 160 times bigger than the resistance in the 10V example. I am about to get tar and feather'ed off the forum but I cannot believe at this point that higher voltage is more efficient, at least in terms of electrical power transmission. All I^2*R is is a substitution of Ohm's law (V=I*R) into power (Power = V*I). It is just as fair and correct to say V^2 / R is power.

EDIT

Also, the current is wrong in the first example as well. If you apply 10V to a 0.01 ohm wire, you will measure 1,000A, not 100.

frodus
03 December 2010, 1515
Had to dig out my engineering book....

Its not the switching losses that is the issue, I appologize. I knew what I meant, just not how to explain it. The losses while you're switching ON and switching OFFremain the same (just did some quick calculations on an IRLR3110 FET to confirm).

The issue at hand are the losses while the device is on. P(on) = Duty cycle* I^2 * R(ds). So lets say the output voltage on both is fixed at 72V with an R(ds) of 14milliohms. The P(on) for 72V at 3600W input would be P(on) - 1.00 * 50*50 * 0.014 Ohm = 35W. Now, with 90V, the duty cycle to output 72V would be 80%. P(on) = 0.8 * 40 * 40 *.014 = 17.92W

Thats almost half the loss just by going up in voltage and decreasing amps by 10A.

Efficiency = [power(in) - p(total)]/ power(in) * 100
lets ignore the switching loss, and just go with conduction loss.

(3600 - 35)/3600 = 99.027%

(3600 - 17.92)/3600 = 99.502%

0.5% Not a ton, but its still a little bit of a difference. that efficiency gives you more range.

For a 5kwh pack

90V would be 4975Wh
72V would be 4950Wh

Thats 25Wh, and could get you what, 1/4 mile?

lugnut
03 December 2010, 1519
But fundamentally, why is less current more efficient?

You have two state variables in the electric system. Voltage (V) is the across variable and current (I) is the thru variable. Voltage can exists as a potential difference without current, hence no power and no work being done. Set aside superconducting systems. Current cannot exist without a potential difference. And when you have current flow, the power is V*I. And there is always a loss mechanism associated with current, called resistance, or sometimes a voltage drop as in a semiconductor. As in most systems, the loss is associated with the thru variable, not the across variable. Like in a hydrualic system, the loss due to a restriction is associated with the flow, not the pressure.

So it seems logical to associate loss in an electric circuit to current and not voltage. However, we know that V and I are always related by Ohm's Law.

So what they are saying is that for equal power systems, a higher voltage system will have less current and therefore less loss and be more efficient. What they fail to see is that the loss mechanisms could be altered to mitigate this. As with the FETs, one would just have to add parallel FETs to bring down the Rds on equivalent to get back to the efficiency level in the low voltage system. Or with resistance in conductors, just increase the copper size to lower the resistance and loss.

They then say, yeah, but that makes it heavier and more expensive. But they tend to discount the fact that adding cells to get higher voltage also has a downside.

Personally, like both low and high voltage systems and have done vehicles from 6V to 800V. There are some real world practicalities when getting into higher power systems. It just would be out of the question to attempt a 200kW system at 12V. But in the mid range power levels of elmoto, it becomes a trade off of V and I for the power needed where the particular application is going to dictate the best choice. Most of this centers around constraints such as available components, envelop and budget.

Back to your question, probably the biggest fundamental reasons less current is more efficient is the fixed semiconductor voltage drop and the increased mass of conductors to maintain reasonable resistance.

Tony Coiro
03 December 2010, 1526
Had to dig out my engineering book....

Its not the switching losses that is the issue, I appologize. I knew what I meant, just not how to explain it. The losses while you're switching ON and switching OFFremain the same (just did some quick calculations on an IRLR3110 FET to confirm).

The issue at hand are the losses while the device is on. P(on) = Duty cycle* I^2 * R(ds). So lets say the output voltage on both is fixed at 72V with an R(ds) of 14milliohms. The P(on) for 72V at 3600W would be P(on) - 1.00 * 50*50 * 0.014 Ohm = 35W. Now, with 90V, the duty cycle to output 72V would be 80%. P(on) = 0.8 * 40 * 40 *.014 = 17.92W

Thats almost half the loss just by going up in voltage and decreasing amps by 10A.

This is wrong, the duty cycle is correctly scaled, however, assuming they are connected to identical loads (which they should be for a fair comparison) they will both be drawing the exact same current, since both mean voltages are equal.

frodus
03 December 2010, 1529
Lugnut, you're right, you could mitigate, but I am assuming that he is using the same controller, same motor, just altering the configuration of batteries. Keeping those things the same, efficiency is higher. Its not a ton, but it adds up.

frodus
03 December 2010, 1530
thinking....

I forgot to take into account the caps. The caps are in parallel with the input bus and, so its still switching the 50A..... you're right, sorry about that.


I'm trying to remember what it was though, because we did bench tests on the synkromotive controller and maybe I'm just forgetting why the higher voltage helped efficiency... it was a combination of things.

lugnut
03 December 2010, 1537
Lugnut, you're right, you could mitigate, but I am assuming that he is using the same controller, same motor, just altering the configuration of batteries. Keeping those things the same, efficiency is higher. Its not a ton, but it adds up.

Thanks for recognizing that. I think we all get into confusion and see some try to draw general assumptions from specific examples.

BTW, in your FET example, I think the freewheeling diode loss would be 12 or 13 watts which narrow your gain considerably.

Tony Coiro
03 December 2010, 1537
You have two state variables in the electric system. Voltage (V) is the across variable and current (I) is the thru variable. Voltage can exists as a potential difference without current, hence no power and no work being done. Set aside superconducting systems.

This is still true in superconducting systems, some work must be done initially to increase the kinetic energy of electrons.


Current cannot exist without a potential difference. And when you have current flow, the power is V*I. And there is always a loss mechanism associated with current, called resistance, or sometimes a voltage drop as in a semiconductor. As in most systems, the loss is associated with the thru variable, not the across variable. Like in a hydrualic system, the loss due to a restriction is associated with the flow, not the pressure.

So it seems logical to associate loss in an electric circuit to current and not voltage. However, we know that V and I are always related by Ohm's Law.

So what they are saying is that for equal power systems, a higher voltage system will have less current and therefore less loss and be more efficient.

This is where I disagree, if the two quantities are related, like you said (Ohm's Law), raising one and lowering another is the exact same as lowering one and raising the other.



What they fail to see is that the loss mechanisms could be altered to mitigate this. As with the FETs, one would just have to add parallel FETs to bring down the Rds on equivalent to get back to the efficiency level in the low voltage system. Or with resistance in conductors, just increase the copper size to lower the resistance and loss.

Correct, the only way that I see to increase efficiency is to lower resistance. Resistance is independent of current and voltage.

frodus
03 December 2010, 1546
ahhh, I remember. They're more efficient at 100% duty, which doesn't happen a ton, so I guess its kinda moot point.

DaveAK
03 December 2010, 1557
So Ed, did you decide on a color yet?

Tony Coiro
03 December 2010, 1558
Back to your question, probably the biggest fundamental reasons less current is more efficient is the fixed semiconductor voltage drop and the increased mass of conductors to maintain reasonable resistance.

I had to research this a bit to double check but this is wrong too, at least in the context on the controller MOSFETs. There isn't a band gap (or set voltage drop) between the drain and source of a MOSFET. When "turned on" in conducts across either an entirely N-channel or P-channel depending on the construction of the MOSFET. There is only a voltage (set) drop when going from positively to negatively (or vice versa) doped semiconductor material, like in a diode.

lugnut
03 December 2010, 1608
There is only a voltage (set) drop when going from positively to negatively (or vice versa) doped semiconductor material, like in a diode.

That is what I was referring to, not FETs. Diodes and junction semiconductor devices like transistors (IGBTs, bipolar and such).

EVcycle
03 December 2010, 1653
So Ed, did you decide on a color yet?

Still Orange!

Thanks for asking? :):)




New on ABC Television.

"Battle of the EEs!"

Fun for .... um ...only a few of us since most people have no idea what the hell they are talking about!!!!! :)

Tony Coiro
03 December 2010, 1740
Haha, who are you calling a EE? As if I'd associate with those guys. ;) But seriously, the idea that higher voltage, lower current is higher in efficiency has been bothering me for ages. I can't figure out why/if it's true.

lugnut
03 December 2010, 1924
But seriously, the idea that higher voltage, lower current is higher in efficiency has been bothering me for ages. I can't figure out why/if it's true.

One thing I like to do is take it to the limits. Was that an Eagles tune? Anyway, take my previous example. 200 kW at 12 V. Then calculate how much 20 ft of suitable conductor would weigh. For kicks, compare it to a 350 V system.

Now, when these elmoto guys talk about efficiency, it isn't Po/Pi. It is Wh/mile. And that is on a duty cycle which includes multiple velocity changes. In most cases, without regeneration. So every extra kg of mass is penalizing them in K.E. Making the Wh/m efficiency worse. And not only that, especially on a bike, that extra mass can be a handful to handle.

And then you can take it to the other extreme. Say a 12,000 volt system. Just the warning signs would be enough to outweigh the bike :-(

So we look for some reasonable compromises in between :-)

BaldBruce
03 December 2010, 1924
Haha, who are you calling a EE? As if I'd associate with those guys. ;) But seriously, the idea that higher voltage, lower current is higher in efficiency has been bothering me for ages. I can't figure out why/if it's true.

I smell a physicist versus engineer discussion coming on.....:eek:

The resistance in the physics world is a number that can be arbitarily set.

The resistance in the engineering world has practical limits like cost, size, weight and manufacturability.

So, in the practical world, engineers almost always use the highest practical voltage to minimize losses in the resistive elements. (MOSFETS, resistors, wire, etc)

As a practical example, run the numbers of what the amperage would be to deliver the 150,000 watts to run your street with a fairly typical 14,400 volt system versus just using 240V that ends up at your house anyway. Then calcuate what wire size you need to run that amperage with only a 5% power loss from 5000 ft away. You will very quickly understand why they chose 115KV and higher for long distance and 9600 or 14400V for local distribution.

Clear now why higher voltage is more efficient? (given practical limitations....)

Sorry Ed, I'll shut up now.

EVcycle
03 December 2010, 2328
Sorry Ed, I'll shut up now.


No Problem! Great stuff! :)

Tony Coiro
04 December 2010, 1950
So, in the practical world, engineers almost always use the highest practical voltage to minimize losses in the resistive elements. (MOSFETS, resistors, wire, etc)


This is what I am disagreeing with. I do not see a reason why higher voltage is more efficient. If you know why, please explain.



As a practical example, run the numbers of what the amperage would be to deliver the 150,000 watts to run your street with a fairly typical 14,400 volt system versus just using 240V that ends up at your house anyway. Then calcuate what wire size you need to run that amperage with only a 5% power loss from 5000 ft away. You will very quickly understand why they chose 115KV and higher for long distance and 9600 or 14400V for local distribution.

Clear now why higher voltage is more efficient? (given practical limitations....)


Yes, I agree it is more cost effective, I have never disagreed with that. What I am disagreeing with is how everyone has been quoting wattage efficiency gains but so far, every example, including yours, has violated Ohm's Law. Do they teach that one in practical world or is it only for physicists? :p

lugnut
04 December 2010, 2041
I do not see a reason why higher voltage is more efficient. If you know why, please explain.


elmoto guys talk about efficiency, it isn't Po/Pi. It is Wh/mile.

You have to consider the context of how the word efficiency is used.

BaldBruce
04 December 2010, 2135
Tony,
For any given resistance of the tansmission media (wire, power supply, connectors, ect.), you always choose the highest practical voltage to get the lowest current. We are not violating ohms law, we are using it to our advantage. Note that the resistance of the load is not what we are talking about here, but the resistance of the transmission media. It is this power loss that we are trying to lower in delivering a desired power. P=IE is not being violated either, but this form of the equation does not reflect the fixed resistance of a given transmission media. I understand your confusion when all variables are infinetly available. In the practical world they are not. They are subjected to very real constraints.

Again, a simple practical example of delivering 100 watts using a 1 ohm resistance to connect my source to my load. In this example I am the designer, so I get to pick all the souce and load resistances but I am stuck with a given transmission resistance. If I configure my source as 100 V, I need 1 amp to deliver my input power. I now have 99 watts deliver to the load.

But what if I had choosen a 50V input? Then Mr. Ohm tells me I have 2 Amps. Therefore using the same exact transmission media, I only get 96 watts out. I lost 3 more Watts in my transmission media, so we claim an efficiency gain!

Now, as Monsiour Lugnut has suggested, if you vary the reisitance, you can clearly simply change that variable and get right back to the same losses. Problem is that this is not often a practical solution. For instance I can buy MOSFETs from the single digit mOhm to several Ohm range, but no practical devices outside this range. Wire limitations are obvious. Any clearer??? Hope this helps.

Tony Coiro
05 December 2010, 0023
You have to consider the context of how the word efficiency is used.

Yes, I am only considering when the units of Watts are used. I have never debated the merits of weight savings.


Tony,
For any given resistance of the tansmission media (wire, power supply, connectors, ect.), you always choose the highest practical voltage to get the lowest current. We are not violating ohms law, we are using it to our advantage. Note that the resistance of the load is not what we are talking about here, but the resistance of the transmission media.

You're joking right? You cannot determine the voltage drop across a wire without knowing the resistance of the load. Using Kirchhoff's voltage law, the most fundamental law of EE, the voltage drop across the wire is:

Delta V across wire = Vapplied * (Resistance of wire / (resistance of wire + resistance of load))

Please explain how you solved for this without knowing the resistance of the load.

Tony Coiro
05 December 2010, 0031
Again, a simple practical example of delivering 100 watts using a 1 ohm resistance to connect my source to my load. In this example I am the designer, so I get to pick all the souce and load resistances but I am stuck with a given transmission resistance. If I configure my source as 100 V, I need 1 amp to deliver my input power. I now have 99 watts deliver to the load.

But what if I had choosen a 50V input? Then Mr. Ohm tells me I have 2 Amps. Therefore using the same exact transmission media, I only get 96 watts out. I lost 3 more Watts in my transmission media, so we claim an efficiency gain!

First example:
100W = 100 V * 1 A
100V = (1 + x)ohms * 1A
x = 99 Ohms

Second example
100W = 50V * 2 A
100V = (1+x)ohms * 2A
x = 49 Ohms

Hmmm, that's weird, it's like your examples used two very different loads. (99 Ohms vs. 49 Ohms) Changing more than one variable at a time can give misleading results so I understand your confusion. Let's keep the load constant this time (a Thevenin resistance of 10 Ohms, we'll keep the wire at 1 Ohm) and see what happens:

First example:
100V = x A * 10 Ohms
x = 10 A
(10 A)^2 * 10 Ohms = (100 V) ^2 / 10 Ohms
1000W = 1000W

Power lost in wire:
10 A ^2 * 1 Ohms = 100 W

90% efficient

Second Example
x = 5 A
(5 A)^2 * 10 = (50 V) ^ 2 / 10 Ohms
250 W = 250 W

Power lost in wire:
5 A ^ 2 * 1 Ohm = 25 W

90% efficient

OMG, it's the same thing! It's like I've been saying that all along. The only time it would appear to be different is if you vary the value of the load in the examples, which you claimed was earlier was irrelevant.


Any clearer??? Hope this helps.

:D

magicsmoke
05 December 2010, 0337
OMG, it's the same thing! It's like I've been saying that all along. The only time it would appear to be different is if you vary the value of the load in the examples, which you claimed was earlier was irrelevant.

Tony, you have a fundamental misunderstanding of what LOAD is.
It is the POWER demanded from a source and not simply the value of the resistor(s).

A physicisty sort of analogy could be mass and weight.
A mass of say 1Kg only has weight under the influence of gravity. Wherever in the universe I move this 1Kg to it will always be 1Kg mass but it's weight will vary depending on the local gravity.
Similarly, a resistor of say 1ohm will always have a resistance of 1ohm wherever I put it in a circuit but the load it represents will depend upon the voltage applied across it.

As you say, power = V^2/R, therefore, if you alter V, you MUST also alter R to maintain a fixed load.

Now, this discussion is referring to a 2 part load, namely the cable and the user.

Because these two parts are in series, the same current flows through both and so clearly if the cable and user elements of the load were equal in resistance then the total power draw I^2(Rc+Ru) would be split equally between them and the system could be referred to as 50% efficient in that only half of the developed power is being developed in the user load.

If the cable resistance were greater than the user resistance then the majority of the power would be developed in the cable and vice versa.



First example:
100V = x A * 10 Ohms
x = 10 A
(10 A)^2 * 10 Ohms = (100 V) ^2 / 10 Ohms
1000W = 1000W

Power lost in wire:
10 A ^2 * 1 Ohms = 100 W

90% efficient

Second Example
x = 5 A
(5 A)^2 * 10 = (50 V) ^ 2 / 10 Ohms
250 W = 250 W

Power lost in wire:
5 A ^ 2 * 1 Ohm = 25 W

90% efficient


So, to redo your second example properly, i.e. with the same load (power), gives ..


Second Example
1000W @ 50V requires 20A
total resistance = 2.5 ohms
x = 20 A
(20 A)^2 * 2.5 = (50 V) ^ 2 / 2.5 Ohms
1000 W = 1000 W

Power lost in wire:
20 A ^ 2 * 1 Ohm = 400 W

60% efficient

Rob

teddillard
05 December 2010, 0439
An engineering student is walking along when a fellow student arrives on a new bicycle. Impressed, he asks, "Where did you got this beautiful bicycle?"

"Well," the second engineering student says, "A couple of days ago I was just walking along when this gorgeous blonde pulls up, hops off the bike, rips off all her clothes, and says 'take what you want'."

The other engineering student nods and says "Good choice. The clothes probably wouldn't have fit."

(Like I said, Danger gets more chicks...)

As far as Physicists go:
A farmer has problems with his chickens: all of the sudden, they are all getting very sick. After trying all conventional means, he calls a physist to see if they can figure out what is wrong. The physist trys. He stands there and looks at the chickens for a long time without touching them or anything. Then all of the sudden he starts scribbling away in a notebook. Finally, after several gruesome calculations, he exclaims, "I've got it! But it only works for spherical chickens in a vaccum."

As far as Ted goes:
Relativity: Family get-togethers at Christmas.
Gravity: Strength of a glass of beer.
Time travel: Throwing the alarm clock at the wall.
Black holes: What you get in black socks.
Critical mass: A gaggle of film reviewers.
Hyperspace: Where you park at the superstore.

Philosophers:
"M. Descartes... would you care for another cup of coffee?" "I think not."
>POOF<
...he disappears.

lugnut
05 December 2010, 0638
First example:
100W = 100 V * 1 A
100V = (1 + x)ohms * 1A
x = 99 Ohms

Second example
100W = 50V * 2 A
100V = (1+x)ohms * 2A
x = 49 Ohms

Hmmm, that's weird, it's like your examples used two very different loads. (99 Ohms vs. 49 Ohms) Changing more than one variable at a time can give misleading results

All loads are not purely resistive. And in the elmoto context, consider the load a motor in each case. Example 1 is a 100W motor wound for 100V, 1A. Example 2 is a 100W motor wound for 50V, 2A. In both cases, the wires from the source to motor are 1Ω. Then example 1 would have 1W transmission loss and example 2 would have 4W loss. Here both examples have a 100W load and a 1Ω wire. We have just changed from a 101V source to a 52V source and increased the system loss. Everybody (I think) understands that you can increase the wire size fourfold and get back to the same system efficiency.

Tony Coiro
05 December 2010, 0816
Tony, you have a fundamental misunderstanding of what LOAD is.
It is the POWER demanded from a source and not simply the value of the resistor(s).


But the power IS determined by the voltage and the entire series of the resistance. This is why, when you changed the value of the resistance in my example, you now have a new efficiency. My example was correct, you changed a value, recalculated and then said I was wrong. Hardly seems fair. Unfortunately, there's gonna be a break in the action for me, I'm going snowboarding with the girlfriend all day, won't be back til late and my Manday is totally full til like 6PM. Cya then, I'm sure there will be plenty more to read by then. :D

Skeezmour
05 December 2010, 0921
First example:
100W = 100 V * 1 A
100V = (1 + x)ohms * 1A
x = 99 Ohms

Second example
100W = 50V * 2 A
100V = (1+x)ohms * 2A
x = 49 Ohms

Hmmm, that's weird, it's like your examples used two very different loads. (99 Ohms vs. 49 Ohms) Changing more than one variable at a time can give misleading results so I understand your confusion. Let's keep the load constant this time (a Thevenin resistance of 10 Ohms, we'll keep the wire at 1 Ohm) and see what happens:

First example:
100V = x A * 10 Ohms
x = 10 A
(10 A)^2 * 10 Ohms = (100 V) ^2 / 10 Ohms
1000W = 1000W

Power lost in wire:
10 A ^2 * 1 Ohms = 100 W

90% efficient

Second Example
x = 5 A
(5 A)^2 * 10 = (50 V) ^ 2 / 10 Ohms
250 W = 250 W

Power lost in wire:
5 A ^ 2 * 1 Ohm = 25 W

90% efficient

OMG, it's the same thing! It's like I've been saying that all along. The only time it would appear to be different is if you vary the value of the load in the examples, which you claimed was earlier was irrelevant.



:D

So it does look in your 50v example here your only delivering 1/4 the power to get the same 90% effiency. Which would be what others are talking about.

Remotecontact
05 December 2010, 0949
One other thing I have yet to see mentioned is that on a given motor higher voltage means more power given an unlimited current source. An etek at 12v is going to be puny, even on a zilla with a massive A123 pack, it doesnt matter. You can't apply more current, only voltage.

Increasing voltage does increase RPMs but it also increases the stall amperage. KT the same, higher voltage gives the motor the ability to draw more current. With an unlimited current source, if you double the voltage of a motor you quadruple the horsepower. RPM increases twofold and so does the stall amperage (stall torque).

Of course with all of our fancy doodads we can increase voltage but current limit to keep power the same.

Switching losses are an issue....So you should always just go WOT. :)

In reality, Ed's alltrax will work harder because the 90v is causing the etek to draw more current overall. This can be mitigated some by gearing down further, something he can afford due to his increased RPM. The etek and the alltrax can probably handle it fine though. Just reminding people that when you increase voltage you are generally increasing power too.

The subject is Volts or AH. If you want a reliable commuter vehicle with cheap parts, go AH (golf cart). if you want any kind of power you have to increase voltage simply because there are no readily available motors that can push my ninja and me to 80mph on 36v. Not only that, smaller wire is just easier to deal with. I run 4-6 awg on my bike on 120v and wiring was a breeze.

Nuts & Volts
05 December 2010, 1101
Alright here is my 2 cents. We are talking about power loss through 3 elements here. Motor, Controller, and wiring.

Lets do wiring first. The Voltage is always the same (created by batteries). Thus the power loss will be P=I^2*R. This is always the power lost in a resistive element correct. So according to V=I*R, R is constant in wiring, V goes up so I goes down. Thus the Power lost will go down.
So less resistive power loss in the wires! higher voltage is more efficient.

Ok lets try the motor next. (I am no EE or physicist, I am a ME in training so this is my interpretation)
Power from the motor is P=RPM*Torque or P=V*I. R of motor is a constant. Throttle is WOT for now. Motor is at full speed.
At
70V motor spins at 700RPM
and the torque needed is 30 (units irrelevant)
gear for 50 MPH on rear wheel
draws I1

90V motor spins at 900RPM
and the torque needed is (700*30/900) = 23.333
gear for 50MPH on rear wheel
draws I2

Current is proportionally less in 90V system I2=.777*I1, less current means less resistive losses. Less power lost in a (DC) motor running on higher voltage.
**May have inductive or magnetic losses which is beyond me right now**

Controller-
I am lost here. Sounds like the controller will be roughly just as efficient at either voltage (thou you still have less resistive losses if running at higher V)?

What i think Tony is not considering is that in our application, resistance is a set parameter and the voltage is a set parameter.

A motorcycle built with an Etek, Alltrax, and Thundersky and with same wiring and geared for same top speed, a 90V system should run more efficient than a 70 or 72V system if driven with the same driving habits (same person).

Please let me know where I am wrong
Kyle

DaveAK
05 December 2010, 1111
I only know the first rule of Fight Club, but I've already said too much.

Remotecontact
05 December 2010, 1135
Unless you're current limiting or have some other current bottleneck (TS), the 90v system will put out more power than the 72v system. So, for regular riding you would use more wh/m.

Nuts & Volts
05 December 2010, 1234
Unless you're current limiting or have some other current bottleneck (TS), the 90v system will put out more power than the 72v system. So, for regular riding you would use more wh/m.

You may use more energy if you accelerate faster, but that is why i said the same person will be driving both systems. If the same driving habit is taken, the torque at the rear wheel will be the same in both systems (one will being using Volts to make RPM to be geared into torque, the other will be using Amps to make torque directly). Thus both systems will have the same power usage over a time period, which is in turn energy used. However the 90V system will use less energy because it is a little bit more efficient.

Tony Coiro
06 December 2010, 0849
So it does look in your 50v example here your only delivering 1/4 the power to get the same 90% effiency. Which would be what others are talking about.

Oh, yea, I have only been saying the efficiency would be the same, I agree, the power changes. Have we all be talking about the same thing?



Lets do wiring first. The Voltage is always the same (created by batteries). Thus the power loss will be P=I^2*R. This is always the power lost in a resistive element correct. So according to V=I*R, R is constant in wiring, V goes up so I goes down. Thus the Power lost will go down.
So less resistive power loss in the wires! higher voltage is more efficient.


V= I*R
If R is held constant and you double voltage, amperage must be doubled as well, which doubles I in I^2 * R, resulting in the same efficiency regardless of voltage but four times the power (but four times the losses as well.)

Tony Coiro
06 December 2010, 0903
Unless you're current limiting or have some other current bottleneck (TS), the 90v system will put out more power than the 72v system. So, for regular riding you would use more wh/m.

Wouldn't the PWM just run the 90V system at a lower duty cycle? If 90V is outputting more power, you are either turning something into a heater or pushing more air out of the way (going faster) or it's violating conservation of energy, unless it's PWM'ed to a lower duty cycle.

Nuts & Volts
06 December 2010, 0937
Then does a controller works by varying its resistance (I guess as a function of the back emf from the motor maybe)? because the controller input always has a 90V input from the batteries, but it varies the current draw from the batteries.

Also then the wires from the battery to the controller have a set resistance (as a function of length and area) and a set Voltage (90V). So they dont obey V=I*R because current changes freely??

Your last explanation above makes sense, but I am trying to understand all of it
Thanks
Kyle

Tony Coiro
06 December 2010, 0945
If controllers changed resistance, we'd all need much bigger heatsinks. :D They use something called pulse width modulation or PWM. This video will help a lot:

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The wires do obey Ohm's law, what V = I * R really is is a voltage drop. Imagine we have two resistors of equal value connected by a magical superconducting wire. ( I am a physicist, I can do that.) If we apply some voltage to the circuit and measure the voltage across one of those resistor, we will measure half of the applied voltage. The same drop will be measured across the other resistor as well. By the end of the circuit, it is "out of voltage". This is a requirement for conservation of energy and is called Kirchoff's voltage law. Basically, the applied voltage minus the sum of the voltage drops across the circuit must equal zero. The voltage drop is calculated by this:

Voltage drop at X = Applied voltage * (Resistance of X) / (Resistance of X + Resistance of everything else in the circuit.)

So yes, there will be some voltage drop in any wire (and the voltage drop times the current is the power going to heating the wire [V*I or I^2*R]) but what that voltage drop is will be determined by the resistance of the entire circuit, called the Thevenin equivalent. Also, this tends to come up, but superconductors also obey Ohm's Law, they just have zero voltage drop, so there is no opposition to current, allowing all of the power to be used at whatever you are powering. Here's an interesting one to think about, what would happen if you wound a motor with superconducting wire? There's one more cool analogy for how resistance works on a quantum level but I've already written a novel. If anyone wants to hear it, let me know.

Nuts & Volts
06 December 2010, 0955
I can understand the video and PWM pretty well, but how does that effect the battery side current draw? The video only shows the V, how does the current behave on both sides of the PWM circuit?

Tony Coiro
06 December 2010, 1020
You can solve for the current which varies based on the load. (BTW, magicsmoke, I was using the term load incorrectly, I'm self-taught in electronic stuff, so I screw up terminology quite a bit.) Battery side power and motor side power need to be equal (well, minus inefficiencies), or it violates conservation of energy. The simplest version is: Battery side voltage * current x = PWM mean voltage * current y, where the mean voltage is directly proportional to motor speed. When Travis was talking about the capacitors in the controller a few pages back, those act as a voltage filter, they help make a time average power draw, which makes the other components a little happier. BTW Kyle, I edited my last post with some more info.

Skeezmour
06 December 2010, 1030
First example:
100V = x A * 10 Ohms
x = 10 A
(10 A)^2 * 10 Ohms = (100 V) ^2 / 10 Ohms
1000W = 1000W

Power lost in wire:
10 A ^2 * 1 Ohms = 100 W

90% efficient

Second Example
x = 5 A
(5 A)^2 * 10 = (50 V) ^ 2 / 10 Ohms
250 W = 250 W

Power lost in wire:
5 A ^ 2 * 1 Ohm = 25 W

90% efficient

OMG, it's the same thing! It's like I've been saying that all along. The only time it would appear to be different is if you vary the value of the load in the examples, which you claimed was earlier was irrelevant.



:D[/QUOTE]

Hey Tony could you do your 2 examples up here again but with both being a 1000w total power output and tell me what the effiency is then? Having one 1000w example and one 250w example and claiming that since they both are 90% some how makes them equal is beyond me.

frodus
06 December 2010, 1034
The caps also help the controller amplify the current.

When the FET is OFF, the batteries and capacitors on the supply side of the FET are in parallel. The capacitors are charged fully.

When the FET is on, the batteries and capacitors discharge through the FET, so you get I(battery) + I(capacitor) = I(load).

When the FET is off, the current still flows because of a freewheeling diode, that allows the current to keep going in the motor loop.

Remotecontact
06 December 2010, 1057
Wouldn't the PWM just run the 90V system at a lower duty cycle? If 90V is outputting more power, you are either turning something into a heater or pushing more air out of the way (going faster) or it's violating conservation of energy, unless it's PWM'ed to a lower duty cycle.

Increasing voltage increases the motors ability to draw current. The motor would be heating up because it's drawing more current and outputting more power. Thats it.

I understand the theoretical debates here but this is the real world answer to ed's question. Your system will be slightly more efficient and your power out will go up. And depending on current limiting (programming or driver) your wh/m may or may not suffer. Because he's adding energy to his pack I think the additional wh/m is suitable for the performance bump.

Nuts & Volts
06 December 2010, 1336
Tony, you are the man! I just learned all that stuff in my circuits stuff this past quarter. It makes sense now. Even if i thin out the wiring to increase the resistance the resistance of the controller (ie the load on the motor) will compensate for the differences and normalize the system efficiencies (ok there is more detail, but this is how my head gets it).

Travis, thanks for the cap explaination. Thats what I thought they were used for, just like putting a pack of supercaps in parallel with your batteries to take the load of the peak current bursts).

Remote, the current doesnt have to increase because the load on the motor shouldnt change, thus the power needed doesnt change. The current will increase if you want to acceleration faster which requires more energy and power.

-Kyle

Tony Coiro
06 December 2010, 1451
(Just read the next post, managed to double post somehow while editing for clarity. Pretty much every time I post, when I go back and read it, I find sentences where I've like omitted verbs, misspellings and silly things like that. Feel free to delete mods.)

Tony Coiro
06 December 2010, 1456
Hey Tony could you do your 2 examples up here again but with both being a 1000w total power output and tell me what the efficiency is then? Having one 1000w example and one 250w example and claiming that since they both are 90% some how makes them equal is beyond me.

My point with all of this was showing using the same circuit and only changing the voltage applied to it, percentage of the losses is voltage independent and is just a function of R. Higher voltage will increase power, as you'd expect, but you will see the same percentage of your power going to heat. When magicsmoke did the 50V example and got 60% efficiency, it was correctly done and 60% is less than 90% so it must sound like I'm contradicting myself. But, to get 1000W out of 50V, you need 2.5 Ohms of resistance in the entire circuit as his example did, my example uses 10 Ohms which was based on getting 1000W from a 100V source. This is consistent, the voltage is half, so the resistance must be four times less for equal power. Here is the reason the efficiency changed: the wire was set in both examples to have 1 Ohm resistance, voltage drop is power used. (Vdrop = Vapplied * (R of something/R of everthing)) Vdrop and Vapplied are directly proportional, so they increase relative to one another. The R value is the real culprit in the efficiency change:

1 Ohm / 10 Ohm = .1, 1-.1 = .9 = 90% efficiency
1 Ohm / 2.5 Ohm = . 4, 1-.4 = .6 = 60% efficiency

Regardless of increasing or decreasing voltage, using these resistances, you will always get these percentages with these resistance unless you add a transmission. (That last part was a joke. :D )

seanece
06 December 2010, 1502
I have been dying to chime in on this subject. Just finished 2 finals and here I go.....
KISS = Keep It Simple Stupid


First example:
100V = x A * 10 Ohms
x = 10 A
KISS P = 100V * 10A
1000W

Power lost in wire:
10 A ^2 * 1 Ohms = 100 W HEAT

90% efficient




Second Example
'50V = x A * 10 ohm' Edit Note: Disregard due to regular idiocy

KISS P = 50 * 20A
'x = 20 A' Edit Note: Moved for clarification
1000W

Power lost in wire:
20 A ^ 2 * 1 Ohm = 400 W HEAT

60% efficient


On a completely different topic, here is an oldy but goody tribute to Skeez (via my idiocy). For a 1million cell battery pack charging at 2Ah, you are discharging a cell at 2Ah; what are the other batteries in the pack charging at (relatively speaking)?

Oh..... and for the very top example, please note copper increases its resistance as it gets warmer, which causes more losses in heat, which causes increased resistance, which causes more losses in heat, etc, etc, etc

Remotecontact
06 December 2010, 1521
Remote, the current doesnt have to increase because the load on the motor shouldnt change, thus the power needed doesnt change. The current will increase if you want to acceleration faster which requires more energy and power.

-Kyle

That's obvious. All I'm saying is that we're talking about Ed's bike. If there's more power available, my money is on him using it. That's all.

Tony Coiro
06 December 2010, 1521
Second Example
50V = x A * 10 ohm
x = 20 A
KISS P = 50 * 20A
1000W

Power lost in wire:
20 A ^ 2 * 1 Ohm = 400 W HEAT

60% efficient


V = I * R
50V = 20A * 10R?

It's easy to forget R is a connected number. V and I are joined at the hip by R.

Skeezmour
06 December 2010, 1525
Ok you made me laugh

Nuts & Volts
06 December 2010, 1530
That's obvious. All I'm saying is that we're talking about Ed's bike. If there's more power available, my money is on him using it. That's all.

Haha yes I would take that bet as well

seanece
06 December 2010, 1542
LOL

V = I * R
50V = 20A * 10R?

It's easy to forget R is a connected number. V and I are joined at the hip by R.



I forgot my KISS rule already! Please disregard and refer to Watts appropriately..... given a constant R

frodus
06 December 2010, 1544
This is a fun thread.

Anyway, Tony, I did prove that the same power output at 100% PWM for 72V and 90V is different.


from before:
P(on) = Duty cycle* I^2 * R(ds).

72V @ 3600W:
P(on) - 1.00 * 50*50 * 0.014 Ohm = 35W
(3600W - 35W)/3600W = 99.027%

90V @3600W
P(on) - 1.00 * 40*40* 0.014 Ohm = 22.5W
(3600W - 22.5W)/3600W= 99.375%

0.35% difference Not a ton at only 50A, but when you move up to 20kW....

72V @ 20kW:
P(on) - 1.00 * 277*277 0.014 Ohm = 1074W
(2000W - 1074W)/20000W = 94.63%

90V @3600W
P(on) - 1.00 * 222*222* 0.014 Ohm = 689W
(20000W - 689W)/20000W= 96.56%

That's ~2% difference in just conduction losses in the controller..... and a difference of ~400W worth of heat.

I'd personally want that extra 400W going to my motor.

seanece
06 December 2010, 1551
Ohhhhhh, you wanna use REAL resistance measurements. I see how it is Travis. Hehe,
http://messenger.msn.com/MMM2006-04-19_17.00/Resource/emoticons/thumbs_up.gif
http://messenger.msn.com/MMM2006-04-19_17.00/Resource/emoticons/beer_mug.gif

frodus
06 December 2010, 1606
Sean, yup

I looked at the datasheet for a standard FET.... 14mOhm.

seanece
06 December 2010, 1626
Gotcha, cool. So that is on a very efficient, simple motor controller...... and then we have to take the magnetic reluctance and return feed resistance into effect (sooooo sorry, too much engineering. Oh dear geekery....... I need a beer!)

I have got a class to teach soon and I may actually use your values rather than mine, Travis. Were there any other questions on this thread? I must get caught up. This is great!

EVcycle
06 December 2010, 2346
That's obvious. All I'm saying is that we're talking about Ed's bike. If there's more power available, my money is on him using it. That's all.

Nope! I want distance.... Gots plenty of power now ( Never say never) :D

Also another factor is that the weight of adding 10 AH is 24 more cells and wiring.
Going from 72 to 90 is only 12 more cells.

teddillard
07 December 2010, 0501
Nope! I want distance.... Gots plenty of power now ( Never say never) :D

Also another factor is that the weight of adding 10 AH is 24 more cells and wiring.
Going from 72 to 90 is only 12 more cells.

Well, Ed, I think it's obvious. There's only one thing to do. Build it both ways and see which really is better! :D

lugnut
07 December 2010, 0546
Also another factor is that the weight of adding 10 AH is 24 more cells and wiring.
Going from 72 to 90 is only 12 more cells.

I ran out of fingers and toes, but wouldn't that be 24 more cells either way? But I guess you could add 4, 8, 12, 16 or 20. Up to you.

frodus
07 December 2010, 0910
to keep the energy the same, and to make this whole entire discussion even applicable, you'd need to add the same number of cells either way. Either 40Ah of 90V worth of cells, or 50Ah of 72V worth of cells.

EVcycle
07 December 2010, 0944
to keep the energy the same, and to make this whole entire discussion even applicable, you'd need to add the same number of cells either way. Either 40Ah of 90V worth of cells, or 50Ah of 72V worth of cells.

But I don't wanna....... :):)

EVcycle
07 December 2010, 0947
I ran out of fingers and toes, but wouldn't that be 24 more cells either way? But I guess you could add 4, 8, 12, 16 or 20. Up to you.

Here is another wrench into the works....

72 volt 40 AH = 24S4P = 96 cells

72 volt 50 AH = 24S5P = 120 cells (24 cell difference)

90 volt 40 AH = 27S4P = 108 cells (12 cell difference)

frodus
07 December 2010, 1012
27 cells? thats not 90V.... thats like 86.5V. Normal rule of thumb when people talk about voltage in the EV world, we say 4 cells for every 12V jump in voltage. Remember we're not talking voltages here, we're talking voltage ranges in increments of 12V. 72V is 24 cells and 90 would be 90V/12V = 7.5 then multiply by 4 to get 30 cells. Yes, Yes I know, its nominal voltage is 102.4V (just like my motorcycle), but thats just how you count with lithium.

I think we all thought you were talking the same exact pack, reconfigured.

You want more range? then use the lower voltage 50Ah pack because its obviously got the most energy.



Devil is in the details ..... Maybe you should talk in configuration instead of voltage, i.e. 24s5p and 27s4p. They're not even close to equivalent. the 24s5p pack has 3840Wh and the 27s4p pack has 3456Wh. Thats a 400Wh difference, or about 4 miles.

EVcycle
08 December 2010, 0423
Why do have to talk about 12 volts group when Lifepo4 come in 3.2V cells.

You are thinking to much inside the box.

When I charge my 27 groups of cells they stop at 98.5 volts. After a period of time after the charge is complete, the cells "rest" at 89. Well within the range of my present controller. Same charge time as before but now 12 more cells worth of energy. (and easier to add to a already installed system.)

frodus
08 December 2010, 0949
Ed,

So you're saying that 72V and 24 cells is calculated using the 12V rule, and then you're saying the 90V isn't calculated that way, you're contradicting yourself and thinking inside the box for one example but not the other? Com on Ed....

72V/24 = 3V
so
90V/3V = 30 cells.
but you are using a different way to calculate pack size for 90V than you are for 72V. I'm still not sure what nominal value for these cells you're using for the 90V calculation (3.33V?), but then you're using 3V for the 72V example. If you did 3.33V for the 72V example, you'd get ~80V. Now the answer is simple.

So, back to why I thought this way:

Go read page 2, Post 14 from lugnut:

If I understand your question right, I say it is the same either way. You have 3600 Wh. Example:
1) 24 cells in series where each is 3 volts and 50 Ah.

2) 30 cells in series where each is 3 volts and 40 Ah.


You never commented either way, so I assumed because of your lack of response, that this is what you were doing. Not sure why you would ask if 108 batteries compared to 120 batteries would yeild a difference in range. 120 batteries would give more range and more power.

But you thought this way for 72V, but not for 90V and are telling me I'm thinking too much inside the box..... give us more information next time :)

EVcycle
08 December 2010, 1125
" So you're saying that 72V and 24 cells is calculated using the 12V rule, and then you're saying the 90V isn't calculated that way, you're contradicting yourself and thinking inside the box for one example but not the other? Com on Ed...."

Nope, was not saying that at all. Well maybe .....(Although it looks that way.) :) Ok, yes.

The "Box" comment was relating to the "everything in groups of 12v rule" . But I digress....or regress....or egress.....

I agree, I was using the old way of thinking of what 24s was in the eariler posts. For some reason most of the folks on this page associated 24 cells or groups of that with 72 volt. I have 24 groups of cells because that is all I could fit. Actually the original 24S pack sits at 78 volts at rest. (3.3v X 24)
The new bigger pack of 27s rests at 89.

I did not comment to lugnuts post as I have been working 16 hour days and have not gotten on here as often as I should to read everything.
I will try to do better on that. :)

Ed
P.S. I own a box but never use it! I will on occasion trip over it though.

frodus
08 December 2010, 1405
Well, we're all on the same page now. I hope work lets up a bit so you can come visit us more, we mis you.... lol


So, whats controller are you using? Does it have a wide voltage swing? or is it 90V max?

The reason I ask, is I'm thinking maybe you should go higher voltage 30 cells at 40Ah (I'm doing 30s6p with the 10Ah headways if I can fit, but its tight).

EVcycle
08 December 2010, 1528
I have the standard Altrax 7245. 90 volts max. I may upgrade soon. Time and money will tell.

I am on salary so no extra income from that. :(


Please keep us updated on your project.

Stay cool!

Ed

frodus
08 December 2010, 1644
So yeah, either 26 or 27 cells is about the max you can do. You might be able to squeeze 28 in there and just turn the headlights on first (will drop pack voltage a little), but that is pushing it.