# 3 phase motor drive current sanity check

• posted

Hi

I have a client using a three phase brushless motor driven by H-bridges. I know little about it except that it uses high frequency PWM to generate sinusoidal excitation.

As I see it (and there's no possibility to look on a scope, it's on a different continent) each motor winding must take current as half a sine from one diagonal pair of H-bridge switches, then the next half from the other pair, which means the /supply/ current looks like an upside-down full wave rectified sine.

Add the three phases together and the supply current is the same shape, but three times the frequency and about a quarter the p-p amplitude.

The client thinks there should be no supply ripple.

I think I'm right. Am I?

Cheers

```--
Clive```
• posted

Perhaps.

No supply ripple is definitely wrong, but there shouldn't be all that much.

The idea of a three phase motor is that each winding s driven by a sine wav e of current, and the three sine waves involved are staggered by 120 degree s.

sin (? + ?) = (sin ? cos ? + cos ? sin ? )

Sine(120 degrees) = 0.866 sine (240 degrees)= -0.866

Cosine (120 degrees) = -0.5 Cosine(240 degrees) = - 0.5

so you can work it out for yourself.

If I remember rightly, there's not a lot of ripple with an ideal drive, and nothing below three times the operating frequency.

High frequency pulse-width modulated approximations to these sinusoidal cur rents could well have quite a bit of high frequency content.

The motor windings will have quite a bit of inductance, but also quite a bi t of parallel capacitance, and magnetic fields involved tend to go through metal pole pieces, so any high-frequency content in the current going throu gh the motor windings is likely to warm up any metal pole pieces, which wou ldn't help performance.

It's a good idea to filter out as much as possible of the high frequency sw itching spikes before they get to the motor windings.

```--
Bill Sloman, Sydney```
• posted

• posted

If indeed the excitation of the bridges aims to simulate a three-phase supply, then the overall supply current is constant, while the currents in the individual windings are sinusoids.

Forget about half a sine being applied to any windings. Each motor terminal gets a full sine, but with 120 degrees of shift from one to the next.

Jeroen Belleman

• posted

It depends what you (or your client) mean with ripple.

There may be a low frequency more or less periodic ripple due to cogging, o r gear box if the motor is speed regulated.

You may see some kind of ripple on the supply current with the same frequen cy of the PWM, this is because at the end (to put it simply) the PWM is a o n/off switch: switch on -> current increase, switch off -> current decrease s (the speed of increase/decrease) depends on the inductance of the coils.

Bye Jack

• posted

Maybe I'm missing something...

Looking at a single winding, let's say it's a linear drive, let's call it a resistive load driven by two single-supply op-amps in bridge mode.

You drive a bi-polar sinusoidal current through that load by having one op-amp go up and down sinusoidally while the other goes down and up, both with respect to half supply.

Surely in this case the supply current looks like a full-wave rectified sine? If so, then adding in two more at 120' won't make it smooth.

Or is it because it's PWM and it's arranged such that there's /always/ only one of three coils on at any instant? Yes, that might work...

Cheers

```--
Clive```
• posted

No. You forget about winding inductance. Only _part_ of winding current comes from supply. Part is "free wheeling" in inductance. In other words H-brige is acting as buck power supply (but with variable setpoint). Ideally _energy_ comes from supply and this is _square_ of sine, that is 1/2 + (1/2)*sin(2*omega*t). Three sine terms cancel due to phase shift, so in three winding together there is steady flow of energy. Of course, in real circuit there will be losses. Resitive losses does not change much. Nonlinar losses will give some ripple. But much smaller than what you suggest.

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Waldek Hebisch```
• posted

You are correct! But PWM outputs are not like opamp outputs. They will happily accept a current flowing in the opposite direction, restoring it to the supply or drawing it from GND with only little loss. An opamp cannot do that.

Jeroen Belleman

• posted

Thanks. Gottit.

Cheers

```--
Clive```
• posted

Hi,

Just in case it is useful, shows the 6 state BLDC phase commutation:

cheers, Jamie

• posted

uency

this might be a good application for a pspice sim.

m
• posted

I don't think so.

you've got instantaneous power into each phase that's constant plus a sine-wave. the three are offset by 240 degrees (not 120 because the frequency is doubled for power) but those three power loads still add to DC + nothing.

probably needs better filtering or some "magic" tuning of the PWM

```--
When I tried casting out nines I made a hash of it.```
• posted

Shouldn't be an H-bridge. Should be three windings linked (Delta or Wye) and three half-bridges. I like to call it a Zhe bridge (for the Cyrillic character).

Tim

```--
Seven Transistor Labs, LLC
Electrical Engineering Consultation and Design ```
• posted

ency

But it will make it a lot smoother.

It might work as an idea, but it wouldn't be a good way to run a motor.

If the current through the coils looked like trapezoids, rather than raised half sines, you could set up a situation where the total current drawn was constant, but the torque generated wouldn't be constant, and you'd see so me ripple on the angular velocity of the motor shaft (though probably not a lot).

The current waveform would rise and fall linearly and there'd be a flat pea k current - you could cut back the flat top far enough to end up with a tri angular wave current drive, but this would be sub-optimal.

```--
Bill Sloman, Sydney```
• posted

Are you sure there are three full H-bridges ?

This would require that both ends of each winding is separately available. In the old days large motors (multiple kW) used wye/delta "soft" starting, first connecting the windings in wye (e.g. 230 V) and when the motor is spinning, connect the windings in delta (e.g. 400 V) to get full power. In such motors both ends of each winding are available in the terminal block.Without a soft stater, the induction motor starting current would be up to 6 times the nominal current.

If only three terminals are available, the windings are internally connected to either wye or delta.The VFD consists of three half bridges not full H-bridges.

If the motor is connected directly to the mains without a VFD, what is the motor nominal voltage ? How much DC bus voltage is available ? Based on this data, it is possible to determine, if full H-bridges and independently available winding ends are needed. .

• posted

That explains why another client states that their motors use trapezoidal excitation. I hadn't paid much attention because he was a salesman and it didn't matter to me anyway.

Cheers

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Clive```
• posted

Ok thanks, didn't know that, just made assumptions.

Cheers

```--
Clive```
• posted

There's several VFD gurus that hang out in Mike Holt's forum, engineering section. You might ask them.

```--
A host is a host from coast to coast.................wb8foz@nrk.com
& no one will talk to a host that's close.......................... ```

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