Motor amps vs battery amps
 

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

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.

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. :)

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

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.

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.

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.

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

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.

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

Could this be elaborated upon?

I understand that the average voltage would be 50V, and therefore your current must increase to have the same power transmitted as was taken. I don't understand how the current increases though. From my basic understanding of electronics, you apply X voltage to Y resistance and you get Z current (ohms law). The resistance of the load isn't changing and the voltage is actually going down. How does current increase?

PWM causes a lot of strangeness in the real world so we will ignore it.

On my 48v car I can typically get up to 30mph @ 65amps or so (summer)

If I were to peg the car to make the motor have 24v (whatever that takes as PWM percentages don't really match directly to a neat voltage realationship on most motors)
I would end up driving about 16-18mph at about half the amps.

Real world this is what I observe battery side, my guess (and its probably right) is in my circumstance the motor amps are also slightly reduced because the aero loss is less.

Cheers
Ryan

Its basically a trick of the meter and the way it samples the measurement. The part he is talking about with the diodes, is about the protection diodes that protect the circuit from the pulse from switching off the current to the coils of the motor.

The coil on a motor is essentually an inductor. An inductor stors current the same way a capacitor stores voltage. When you apply a voltage to it the current slowly increases until it either burns up or you max out the current due to resistance. However when you turn off the supply the inductor will spike the voltage to keep the current going, that spike tends to arc through things and burn them up. The diodes give a path for this stored current to flow without blowing out parts. It is short lived and is current that was originally supplied by the battery.

The best way to measure each side is with a scope that way you can see the voltage and current waveforms. Most modern scopes can even calculate power which will also appear as a waveform.

Ok, so the controller shuts off its PWM, then you get voltage spike (I'm a little familiar with this due to some reading on switching mode power supplies) which is increasing the current to get rid of the stored inductance of the motor? So this means that even though the PWM pulse is over with, you're actually still getting some motive power (for a very very short time)?

HMMM...... I wish I had access to a system, I may have mixed some of the stuff up a bit.

Yep,

Because the motor is a huge inductor once current starts to flow it wants to continue to flow even after the battery is disconnected.

This is why PWM even works at all, if there were no inductance you would get effectively a short circuit and blow up (inductance opposes current flow and opposes current shut off as the field slowly callapses)
Also because every motor has a different amount of inductance it means that a PWM percentage does not directly coorelate to a voltage.

In the case of my little 48v motor 50% PWM on roughly 52v gives me about 36 volts give or take, not the expected 26v, obviously there are a few other things going on with my meter but a motor is an analog device with a curve much like a gasoline engine so each one is unique and behaves differently when exposed to a given voltage and PWM percentage.

PWM speed also affects this, since the decay time is different on slow switching speeds (which also increase friction/resistance losses)

I have to admit I honestly thought it was a lot simpler than this. Appreciating that there are some complexities introduced by the controller PWM, running the motor at a lower RPM and therefore V means that for a given battery power the current delivered can be considerably higher than the batt pack given that the controller is capable...?

For motor versus battery amps and volts, convert both to watts. For example, 100 battery volts at 50 A would be 5,000 watts. At a low RPM, the motor might only see an average of 25 volts, but would be at 200A; 25V * 200A = 5,000 watts. The motor controller determines how much voltage is needed based on throttle position, and the amount of current is based on how much energy the motor needs to reach the RPM indicated by the motor voltage. So, 5000 watts from the battery equals 5000 watts to the motor.

At the motor, volts = rpm and amps = torque.

The duty cycle of the controller output (i.e PWM) complicates the picture, but in simplest terms, watts in = watts out.

I understand the equation and know that the energy out of the pack must equal energy into the motor (minus heat losses). I guess I was hoping for a more indepth reason and explination of what is actually happening when a motor is seeing higher amps than the battery pack draw.

It is an issue of averages. The motor current can't ever be more than the battery current, it is just the motor inductance averages the current.
So if the motor current is 100amps, the battery current is 100amps, but only half the time, which AVERAGES to 50amps.

How is it that my 115V arc welder running off a 20A outlet can put 90A to a stick?

Or, the DC to DC converter in my car which takes the 120V pack and converts it down to 13.5V to keep the 12V car battery charged. When the car's 12V system is drawing a steady 20A, the DC-DC is drawing only about 2A from the 120V pack, with or without the 12V battery in the circuit.

In a PWM controller the meters are probably averaging to some degree, but here is a picture of the controller showing V & A for both motor and battery. The car was just barely rolling during the shot.

(hopefully the picture attaches this time...)

A welder is using a transformer, two coils with different windings.