Motor amps vs battery amps

[Daox]

I was talking with Darin the other day about this and I think we got it figured out. But, if someone is sure or can explain it better it would be great.

Lets say you have a 100V battery pack. You apply that 100V to your motor and it spins up to 4000 rpm at some load. Now, you reduce the amperage with a pwm controler to the motor until the rpms are at 2000. Lets say you are now applying 100A to the motor. Since your only at 2000 rpm its only requiring ~50V (half of the 100V @ 4000 rpm) to achieve that rpm. The motor is using 50V * 100A = 5000W. But, your battery pack is 100V, not 50V, yet it is still putting out 5000W. So, your battery amps are 5000W / 100V = 50A. So the motor is seeing 100A but the battery pack is only seeing 50A.

Is this correct?

edit: added pwm controller info

[TheEnemy]

No it is not correct. You are still applying @100v to the motor.

When a motor spins slower it uses more current the 4000 rpm current should be much lower than the 2000 rpm current.

You will have a lower voltage at 2000 rpm though due to the batteries internal resistance, resistance in your speed controller, and the resistance in the cables supplying the power, so it is possible that at 2000 rpm you would have half of the voltage but that is also a sign that you need to fix something.

[Daox]

Sorry, that doesn't seem to make sense. I've read that motor amps can be significantly higher (multiple times higher) than battery amps in an electric car. You're saying that can't happen unless there is something very wrong. I require further elaboration.

[TheEnemy]

I guess I should ask if you are also using a pulse width modulation type speed controller, that would complicate measurements a bit.

[Daox]

I'm not using anything personally. But yes, this theoretical example would be using a PWM controller. I should have added that to the 1st post. I shall revise my example.

[MPaulHolmes]

Power in = power out, So


BatteryVoltage*BatteryAmps = MotorVoltage*MotorAmps

Because MotorVoltage = BatteryVoltage*PWM_DUTY, where PWM_DUTY is in [0,1],

BatteryVoltage*BatteryAmps = BatteryVoltage*PWM_DUTY*MotorAmps

So,

BatteryAmps = PWM_DUTY*MotorAmps

I'm not sure if that answers any questions, since I never read any of the previous posts, since my son is yelling at me to go outside with him. haha.

[TheEnemy]

A pwm might use a stabilizing capacitor on the input, and output and there would be two ways of measuring the current and voltage. The results would depend on how exactly the pwm is built. Ill give multiple examples and their results for simplicity I will ignore voltage losses through the system.

With no capacitors on either side.
The average and instatanious (single pulse) current would be the same, during the on phase the voltage would be the same as the battery voltage, and drop to zero during the off phase. The average voltage on the motor (multiple pulses) which is what a typical DVM would see would be the battery voltage times the duty cycle. ex. 100v * 50% duty cycle would be 50v. I'm not entirely sure about the voltage part because of the back EMF from the motor.

Capacitor on the input, not on the output.
The average and instatanious current on the battery pack should be about the same. There will always be some ripple. The current and voltage on the ouput would be the same as above.

Capacitors on both the input and output.
The current on both sides will be pretty steady, and approxametly the same. The voltage on the motor side will be lower than the battery side, and this is not a sign of a problem.

[TheEnemy]

Now if you are using an AC motor and an inverter then the calculations are a bit beyond me.

[DJBecker]

Originally Posted by TheEnemy

A pwm might use a stabilizing capacitor on the input, and output and there would be two ways of measuring the current and voltage. The results would depend on how exactly the pwm is built.

The results don't depend on the capacitors.

At 50% pulse width the motor sees 100V half the time, and approximately 0V half the time. It averages out to 50V, although the insulation and commutator still has to withstand 100V.

During the time the battery is being applied (the on pulse), the current comes from the battery. When the battery is not being applied the current flows through a "freewheel", either a diode or diode emulation. Thus the current is "multiplied" by 2x. In reality the current isn't multiplied, but with a high frequency PWM, some capacitors (see below) and a slow current meter you measure the average current from the battery rather than the pulsed current.

The capacitors are vital to avoid voltage spikes during the switching transitions, and help even out the current flow from the battery, but don't change the gross calculations.

[TheEnemy]

Yes and no, it depends on the size of the capacitor in relation to the impedance of the motor.