# RPM, Torque, and Battery Amps



## Arniem (Apr 24, 2008)

I'm grappling with calculating with what the *battery* amps draw will be when using a DC controller (eg Curtis 1231C) to control a DC motor. 

I understand that DC controllers typically use a PWM technique to pulse the full voltage available to it from the battery pack so rapidly (by changing the "duty cycle") that the motor at the other end "sees" an average of the pulse peaks, thereby presenting a varying voltage level, from 0 to pack voltage. 

The golden rules of DC motors (eg Warp9) are that rising and falling voltage determines DC motor RPM (so doubling the voltage presented to the motor will theoretically double the motor RPM - unless it flies apart from the rotation stresses...), and that amps drawn by the motor are proportional to the torque required to be delivered *at the voltage applied to the motor*, and that the current draw is the same irrespective of the voltage applied (ie even if double the voltage is applied the current draw will be the same for the same torque load - only the RPM at which the torque load can be sustained is higher). 

So, using the torque curve charts for a given DC motor, if you know the RPM that the motor should rotate at for a given point you can work out the voltage required to be applied. And if you know the torque required at that same point you can work out the current that will be drawn by the motor. The controller will PWM the full voltage of the battery pack (say 144v) down to the voltage required for the RPM needed at the motor, but what will the current draw at the batteries be? 

Am I right to use first principles, that W = Amps x Voltage, and that Amps(Batt) x Voltage(Batt) = Amps(Motor) x Voltage(Motor) - (some losses)? 

As an example (ignore the other losses for now): Amps(Motor: 100A) x Voltage(Motor: 72v) = Amps(Batt: ?A) x Voltage(Batt: 144v), so Amps(Batt) = (100A x 72v)/144v = 50 Battery Amps. 

Can it be as simple as that?


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## frodus (Apr 12, 2008)

Arniem said:


> I'm grappling with calculating with what the *battery* amps draw will be when using a DC controller (eg Curtis 1231C) to control a DC motor.
> 
> Am I right to use first principles, that W = Amps x Voltage, and that Amps(Batt) x Voltage(Batt) = Amps(Motor) x Voltage(Motor) - (some losses)?
> 
> ...



Pretty much 7200W on one side is pretty much 7200 on the other side - losses.

It can be easier to consider the controller as a power converter.


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## Arniem (Apr 24, 2008)

Thanks Travis. I have seen elsewhere on this forum the idea of treating the controller as an "energy converter", but still wasn't clear about the smoke-and-mirrors behind something like that working...

So applying the rule of watts-in-(almost)-equals-watts-out, you would have the situation where at low RPM the voltage applied would be very low (say 20v), but the current draw at the motor is extremely high (say 400A), as when taking off slowly up an incline. This would make for a warm motor and controller, but the battery amps would be a very reasonable 55A at 144v. 

Any idea of the losses that should be expected from a controller - I am guesstimating around 90% efficiency?

AM


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## frodus (Apr 12, 2008)

Depends on the controller and the voltage its being used at. The controller Synkromotive is working on is around 95% when running above 120V... I don't know for sure the efficiencies of the other controllers out there, but around 90% should be ballpark. 

the heat lost in the controller and motor is due to the efficiency. 7200W and a 10% loss of controller is 720W, pretty warm.

But yes, at 144V, 55A, the 20V and 396A is approximately correct, but to get a real number, adjust for inefficiency of the controller.... the motor efficiency means not all of the electrical energy will be converted to mechanical.


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