Understanding PWM of Motor (current problem)

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i think that part you are missing is the CATCH DIODE. When the switch switches off, the current continues to flow through the diode. So with the right inducantce, and a 50% DF, the current in the load flows all the time, the current through the battery flows 1/2 time so the average output current is 2x the input current and the output voltage is 1/2 the input voltage. Input power =3D output power. This is simplifed and ignores losses of course. Mark

No, I realize it's there and is passing the current caused by the collapsing magnetic field in the motor.

When you say "the current" I'm assuming you mean the current from the collapsing field?

What is DF? Your saying the same thing as the EV group. Now I need to understand where the (non battery supplied) current is coming from. Mike

Perhaps it'll be helpful to think about power flow. If a dc-dc converter consumes no power, and it delivers power at half the voltage of an input, it must consume half the current from the input. Consider, it it consumes the same current, on the average, then where would the unused half of the power go?

Lately I've been working with the "bus converter" concept. Use PWM converters with a fixed duty cycle, like 63 or 75%, and transform DC voltages around, without power loss, to speak of anyway.

For example, in two stages I go from a 24V 3.3A source to a supply delivering +4V, -12V at 5A. Both source and destination are 80W. OK, the source may have to provide 85 watts, but who's worrying about the extra 5 watts?

When you're thinking about motors, the really interesting stuff happens at start-up, when you need high torque =3D high current, delivered to the motor, but not taken from the power supply. Where does it come from? The supply designer knows the high-current buck stops at the low- resistance low-ac-loss inductor and the low esr bypass capacitors. 50A at a 10% duty cycle is only 5A average -- if you have a capacitor that can deliver 50A for the other 90% of the time.

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Your on the right track. Fifty percent duty cycle means the voltage is 50 % and also the average current is 50% as you mentioned. So, the power is 1/4 or 25%. That is to say both the motor speed and the motor torque are reduced by 1/2.

It's wrong to say that the motor current is 50% of the battery current. The average currents of the battery and the motor are the same. No current is gained or lost unless some form of dynamic breaking is in the mix. The 50% number is relative to the Maximum current for a given load. The situation is made more complicated by acceleration, torque required, back emf, speed and breaking and other issues.

In general, think of the PWM acting like a simple analog rheostat dropping the voltage available to the motor. Both the voltage and the current are reduced.

Ok so far.

That is what I think, the waveform may be different but the average or effective value is the same. But the EVers are saying if you use two shunts, one to measure battery current and one to measure motor current the numbers will be quite different.

That's what I think.

Now how do you explain what Mark had to say?

I can't make this work, but I see others give the same logic.

Still searching, Mike

Perhaps it'll be helpful to think about power flow. If a dc-dc converter consumes no power, and it delivers power at half the voltage of an input, it must consume half the current from the input. Consider, it it consumes the same current, on the average, then where would the unused half of the power go?

Lately I've been working with the "bus converter" concept. Use PWM converters with a fixed duty cycle, like 63 or 75%, and transform DC voltages around, without power loss, to speak of anyway.

For example, in two stages I go from a 24V 3.3A source to a supply delivering +4V, -12V at 5A. Both source and destination are 80W. OK, the source may have to provide 85 watts, but who's worrying about the extra 5 watts?

When you're thinking about motors, the really interesting stuff happens at start-up, when you need high torque = high current, delivered to the motor, but not taken from the power supply. Where does it come from? The supply designer knows the high-current buck stops at the low- resistance low-ac-loss inductor and the low esr bypass capacitors. 50A at a 10% duty cycle is only 5A average -- if you have a capacitor that can deliver 50A for the other 90% of the time.

Win, go to bed, get a good nights rest, get up and have your coffee. Sorry, I'm picking on you a little cause I know you can explain it better than you have so far. After your coffee, explain a PWM motor speed controller driving a motor with inductance. I'm using a 300 amp 48 volt motor control, so I don't know if it has capacitors large enough to deliver that 300 amps for the other 90% of the time. Mike PS. If you see a guy on an electric gokart holding an oscilloscope dragging an extension cord, That's me :-)

No, the peak currents are the same. With a PWM assuming - the presence of a freewheel diode or equivalent - 50% PWM - switching frequency is high

then the average motor current will be twice the average battery current.

The PWM turns on and current is drawn from the controller/battery through the motor. Call this I1

The PWM turns off the current continues to flow through the motor winding and freewheel diode.

Over several PWM cycles this will ramp up the current until it reaches a steady state. Where the average current drawn during the on time matches that in the freewheel diode during the off time. So the current in the motor winding is I1*ton/tduty + I1*toff/tduty. The current from the battery is I1*ton/tduty (1/2 that of the motor).

The effect is commonly referred to as current multiplication.

Another way of looking at is the energy from the battery is Ibat*Vbat, the energy in the motor side is Imot*Vmot. Since Vmot is Vbat/2 then Imot must be about Ibat*2 or something is getting awfully hot.

You must run the PWM fast enough to avoid significant decay in the motor inductance but that's easier than it used to be. The bigger subtle point is that the wire inductance from the power source (I keep calling it a battery) must be compensated for in the controller in order to avoid power spikes in the power devices.

Robert

Mostly. The caps only have to deliver 45A and only for the 5% on time.

For the 5% on time the capacitors provide 45A, the power supply provides

During the 90% off time the current in the motor is simply re- circulating through the freewheel diode. For a sero'th oder approximation consider an ideal diode with a zero forward drop and a motor with zero winding resistance. Once that concept clicks in add the forward drop and winding resistance.

They key to this is a fast PWM. Slow PWMs like the EV-1 used do not have this behaviour and show significant current ripple in the motor.

Robert

"Bob Eld"

In a correctly functioning PWM, DC motor controller - the motor has an almost steady current flow ( due to the free-wheeling diode) while the battery experiences a PWM current draw.

During " on " periods, motor and battery currents are the same.

So with a 50% duty cycle, the average battery current is one half that of the motor.

..... Phil

Is this current from the collapse of the magnetic field in the motor?

Where is the extra current coming from?

I think mine runs at 16Khz, Power source, mine is a battery or 4-12volt car batteries. Mike

Just as in any other inductor. THe same thing happens in contactor and relay coils as well. Also in DC-DC converters such as a standard PWM buck converter.

With most EV motors anything over 5K IS good enough. Modern controllers tend to run from 10kHz to 20kHz, dropping to 5KHz during plugging for series wound motors.

The GE EV-1 and similar SCR based controllers used to run at a few hundred Hz. You could always tell since they made a pleasant tune as they accelerated.

Robert

The extreme case is a superconductive inductor, like the main field electromagnet in an MRI machine. This is a big inductor with literally zero resistance. They connect a power supply to it for an hour maybe, build up the current, then they short the inductor leads and remove the power supply, literally load it back on the truck. The current keeps circulating in the coil for years.

The motor has inductance. The power supply is applied to it part of the time, and some current I flows from the supply into to the inductor when they're connected. Then some fets are switched around to disconnect the power supply and short the motor/inductor. Current I continues to flow in the shorted inductor (the motor windings) because that's what inductors do; power supply current is zero in this state.

Ignoring resistive losses, the ratio of average motor current to power supply current can be very large. The motor of course has resistance, and if it's spinning and driving a load it's doing work, so those "losses" keep the current multiplication ratio down. Motor current will, in practice, droop a bit while the motor is shorted, due to various losses.

John

You are right, I misunderstood the question. The PWM is simply a "DC transformer." The Power is constant (excluding losses). So if the voltage is cut in half, the current has to double, Duh! If the input voltage is twice the output voltage, the input current has to be half of the output current. Was that the question?

Thanks John, I guess the problem I have is that energy stored in the Magnetic field had to come from the battery. Right? Please look at

Also, assuming 50% duty cycle; If I put a shunt in the battery circuit and a shunt in the motor circuit and use analog meters for their averaging effect, is it your opinion that the battery meter will read 1/2 of the current the motor meter reads? Thanks, Mike

Sure. But it's *stored* in the magnetic field, so once it's there, the battery isn't needed to keep it there.

If the frequency is low, like in the 500 Hz case, the current ramp up and ramp down are big slopes. The rampup is the power supply charging the inductor, and the rampdown is caused by losses in the inductor/motor, caused by resistance in the windings and especially any real work being done.

At higher frequencies, the slopes *appear* to be flatter because we're operating in shorter time slices. When the battery is disconnected, the circulating current decays only a little, so when we switch back on to top off the current, we only need a little up-slope. The higher the frequency, the less ripple in the inductor current. Less ripple means that the power supply "on" current is very close to the inductor circulating "off" current.

If the frequency is reasonably high, so that motor inductive ripple is low, then yes, close to 1/2.

John

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Yes. The energy is stored in the mag field and the current tries to continnue to flow...think of it as if the electronis in an inductor have intertia...

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DF =3D Duty Factor =3D Duty Cycle...same thing

Mark

Yes. The energy is stored in the mag field and the current tries to continnue to flow...think of it as if the electronis in an inductor have intertia...

DF = Duty Factor = Duty Cycle...same thing

Mark

Thanks, Mark Mike

Ok, I think I have it, when I said to myself, the voltage across the motor should reverse polarity during the off time, why don't I see it in the graph. Then I realized I do see it, but it's clamped to the diode drop. I see why responses kept going back to power in = power out. With a 50% duty cycle the voltage is effectively 1/2 so the current has to be double. Intuitively, It seems like the current should be higher during the on time (in comparison to the off time) to run the motor and store the energy for off time. And looking at the the graph referenced before, the current is higher, but not as much as I would think, especially for this 30% duty cycle. Comments, Thanks, Mike

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A PWM controller behaves like a variable transformer: The input power has to match the output power (ignoring losses), so the voltage and current on the output can be different from those present on the input. The product of voltage and current will be the same.

I think you lot are getting mixed up over the pulsed current in the versus the rms-current flowing in the motor