# Voltage to Ah Relationship Question



## Aaron Oz (Sep 15, 2009)

Assuming a vehicle is driven the same for both examples, I was curious about something...

Would each battery pack described below provide the same running time (again, given that both scenario's are run exactly the same way):

1) X Volts @ Y Ah
2) X*2 Volts @ Y/2 Ah

I understand the significant performance differences between the two, just looking to compare two exact same driving scenarios and see if I understand the situation.

Oz


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## Jan (Oct 5, 2009)

Hi Aaron, You will probably notice no difference in running time. They have the same energy.

In theory the current can make a difference. The higher the current, the higher the resistance losses. And to reduce current you have to increase the voltage.


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## Aaron Oz (Sep 15, 2009)

Thanks Jan.

I'm beginning to grasp the 'theory' behind some of this. The higher resistance with higher current is an interesting consideration.

My first thought was to the efficiency of the controller, pulsing the higher voltages off for longer durations... I think your point would contribute just as much consideration as controller efficiency, perhaps even more.

Yet, in real world testing... I would think the measured differences would be relatively insignificant.

Oz


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## Jan (Oct 5, 2009)

Aaron Oz said:


> Yet, in real world testing... I would think the measured differences would be relatively insignificant.


I think that no smart or expensive efficiency solution can beat the man driving the car. I mean, a driver with range in his mind and knowing how to get the most out of his battery pack will win it from a more efficient setup with someone who doesn't care.

But if you care, I think the most important thing to consider is peukert. If you're considering lead accid. If you're thinking of Lithium it is less important. But with lead it is crusial. A bad sized pack with fo example 50Ah batteries will not have half the range of a 100Ah pack, but maybe only 25%. If you pull to much Amps from a lead accid battery it's potential energy will be reduced significantly.

So calculate your power need. And choose batteries that can deliver those amps easy.

Again Lithium doesn't care much for Peukert.


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## mszhao (Oct 17, 2009)

Aaron Oz said:


> Assuming a vehicle is driven the same for both examples, I was curious about something...
> 
> Would each battery pack described below provide the same running time (again, given that both scenario's are run exactly the same way):
> 
> ...


this is a hard question.
their prices are the same.
but running time is not neccessary the same.
it depends on the specification of the motor,how many Amp of continuous and peak discharge current.double the voltage can not double the driving distance,personally i think.how many voltage of battery pack you should use depends on the motor.


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## GerhardRP (Nov 17, 2009)

Aaron Oz said:


> Assuming a vehicle is driven the same for both examples, I was curious about something...
> 
> Would each battery pack described below provide the same running time (again, given that both scenario's are run exactly the same way):
> 
> ...


Hi Oz, here is how I look at it.

#1. A modern controller is a DC power transformer with a variable voltage stepdown. The motor voltage equals the battery voltage times PWM%. The motor current equals the battery current divided by the same PWM%. The power out is equal to the power in minus maybe 2% inefficiency.

#2. Battery power loss due to internal resistance and loss of capacity from the Peukert effect are dependant only on the the total power delivered by the pack. 

#3. It is possible to set up "two exact same driving scenarios" by setting the motor voltage limit in the controller to the value of X in your question. With this setup, everything is the same in both senarios with regard to battery power and motor power, so the two ranges and performances will be identical. The only difference is that resistive losses in the cableing, fuse(s) and contactor between the battery and the controller will be smaller in the high voltage case.

#4. The usual reason to go to higher voltages is to allow power to be delivered at higher RPM. Look at the second attachment in this post http://www.diyelectriccar.com/forums/showthread.php/dc-motor-theory-and-model-39931.html. Compare particularly the 75 and 150 volt curves. You can see that the power available at high RPMs is doubled giving seriously better "performance", but it comes at the expence of double the stress on the batteries resulting in greater internal power loss and Peukert capacity loss. 

#5 Bottom line is you can trade range for performance, mainly through battery losses. Same way you trade operating speed for range through aero losses.

Gerhard


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## Qer (May 7, 2008)

I second Gerhards comment (and is grateful for not having to write it myself  ) and just want to fill in a minor details.



GerhardRP said:


> The only difference is that resistive losses in the cableing, fuse(s) and contactor between the battery and the controller will be smaller in the high voltage case.


With really high pack voltage the losses will start to increase, so there's a "sweet spot" that depends on the controller and the cable areas. The increasing losses at high voltage is the reason why the Soliton cut down maximum motor current from 1000 to 900 Ampere above 310 Volt (iirc).

But then, for ordinary commuting cars there's probably no point in having that high pack voltage anyway, it just makes it harder (or at least more expensive) to find a good charger for the pack...


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