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Old 09-10-2012, 01:43 PM   #11 (permalink)
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Quote:
Originally Posted by DJBecker View Post
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.
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?

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Old 09-10-2012, 02:29 PM   #12 (permalink)
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Originally Posted by Daox View Post
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.

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
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.

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Old 09-10-2012, 02:49 PM   #13 (permalink)
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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.
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Old 09-10-2012, 03:12 PM   #14 (permalink)
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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)?
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Old 09-10-2012, 03:21 PM   #15 (permalink)
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HMMM...... I wish I had access to a system, I may have mixed some of the stuff up a bit.
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Old 09-11-2012, 02:18 PM   #16 (permalink)
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Quote:
Originally Posted by Daox View Post
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)?
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)
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Old 09-21-2012, 05:35 AM   #17 (permalink)
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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...?
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Old 09-28-2012, 02:47 PM   #18 (permalink)
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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.
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Old 09-28-2012, 03:34 PM   #19 (permalink)
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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.
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Old 09-28-2012, 03:42 PM   #20 (permalink)
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Quote:
Originally Posted by Daox View Post
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.

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