Back-EMF speed detection in motors

How calculated from PWM?

Flyback diode? Or full H-bridge? ...Jim Thompson

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I assume you are using a PWM drive scheme? Can you arrange to sample the back EMF during the PWM off-time? That would get rid of the R factor to a large extent, depending on filtering.

Jon

What about the inductance? ...Jim Thompson

As JL alluded to, if you PWM slowly enough to allow that to work, the current in the motor has to be discontinuous, which means high RMS current ( => large I^2*R losses).

Best regards, Spehro Pefhany

In our 'heavy current' lectures we were taught to allow 2v drop per brush, regardless of current. No explanation was given for the rule-of-thumb.

Le Fri, 22 Mar 2013 12:51:29 -0500, Tim Wescott a écrit:

How do you supply the motor? Is that a MOSFET bridge or do you have some free wheeling diode somewhere?

With the latter you have Vavg = D Vs - (1-D) Vd = D(Vs-Vd) - Vd which indeed has about a 1V offset per diode.

Le Sat, 23 Mar 2013 15:37:54 +0000, Fred Bartoli a écrit:

Oops, obviously make that Vavg = D Vs - (1-D) Vd = D(Vs+Vd) - Vd

Works just as well the other way around...

This is based on a half-H-bridge drive, thus no diode forward drop to confuse the issue.

But I suspect the diode drop could be likewise easily subtracted out. ...Jim Thompson

Pondering my navel, I don't think it matters, half-H-bridge or simply a flyback diode... the average of the voltage at the upper motor terminal is still equal to Vm + 2*Im*Rm ...Jim Thompson

Tim, Suggestion... Instal a fudge factor to take out the offset or gain error or whatever you think you are seeing and re test the system. Use this as a troubleshooting step, not as a fix. It is possible that the fudge factor will not behave as you expect and this will give you additional insight to the true problem. For example you may install a linear gain fudge factor term and discover that the error is actually second order.... Etc. Be sure to come back and let us know the answer. Mark

Full H-bridge. There'd be no way of knowing with flyback diodes -- you'd have to measure.

I'm using an external resistor. It's stored in a variable named MotorR.

It's H bridge, with very little dead time.

On a sunny day (Sat, 23 Mar 2013 21:32:50 +0100) it happened Fabio_78 wrote in :

mm, maybe I should order a lipo-safety bag... :-) But with all those modifications I think Boeing made, may as well use lipo! Would claim back at least some weight !

Le Sat, 23 Mar 2013 16:39:14 -0500, Tim Wescott a écrit:

Yes, but which techno, and how are the switches driven? Is it MOSFET, or BJT/IGBT, in which case you have to have free wheeling diodes across the transistors. Do you have some "clever" low current ripple mode, or just plain diagonal switching?

You say the offset is circa 20/30RPM. What are your typical figures? (Vrail, motor constant or nominal speed at Vrail, typical current)

The motor needs to reverse?

See my other posts, I'm no longer certain that a flyback diode hurts you calculation in the slightest. ...Jim Thompson

;-) ...Jim Thompson

Ignoring my added compensating R for the moment, what does the average terminal voltage (at the motor) have to be?

Take Vm = speed voltage, Im = motor current, Rm = winding resistance

The _average_ voltage has to be Vavg = Vm + Im*Rm

Irrespective of how it's driven, flyback diode or not.

Why is that? Return to fundamentals to see how easy the solution is

...Jim Thompson

Many (most?) DC motors have brushes that span more than one commutator segment.