Back-EMF speed detection in motors



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.
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replying to Tim Wescott , passerby wrote:

Measuring the zero crossing point too close to the commutation time while commutation noise is still affecting the reading (and/or your cables are on the long side) or there's enough capacitance in the leads / circuit for the initial jump during commutation to push the BEMF zero crossing time forward for just the right amount of time?
Wait - are you reading it sooner than anticipated or later? trying to wrap my head around the description you posted while caffeine hasn't fully kicked in. If it's sooner, then you're probably reading it too soon and there may actually be yet another (real) zero crossing in the same period .
Take anything I say on the subject with a huge grain of salt - I've never actually dealt with anything brushed that did not have an encoder attached to it. Do you still do it in a way similar to what you would have done for a sensorless brushless? I would imagine brushed commutation would be *way* noisier than brushless so would require long[er] ZC hold-off times, perhaps even too long for high RPMs.
Sorry, not offering a solution here, just trying to think along ...
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On Fri, 22 Mar 2013 18:49:46 +0000, passerby wrote:

DC _brushed_ motors. One doesn't have access to the zero crossing time (if I did, I'd know the motor speed). One measures current and voltage, then one takes the (theoretically known) resistance, and backs out a number for the back EMF. Then one uses that to determine motor speed.
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On Fri, 22 Mar 2013 12:51:29 -0500, Tim Wescott wrote:

My WAG is that your "armature resistance" is not really a constant. There will be times when (depending on the commutator) the brushes will likely be connected to two separate windings; other times only one. Given sufficient rpm you could probably use some time-averaged value; is that what you've done?
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Ah, that's a good one too...
Tim, you say it's zero at zero (locked rotor), but is it always zero? Does the offset change with position..?
Tim
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On Fri, 22 Mar 2013 21:49:55 +0000, Frank Miles wrote:

Many (most?) DC motors have brushes that span more than one commutator segment.
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On Sun, 24 Mar 2013 13:49:10 -0700, Fred Abse

There

likely

that

And my mind went ding. I have seen an awful lot of two brush DC motors that have an odd number of commutator segments. Thus most of the time one of the brushes is spanning two segments. It is a thought.
?-)
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Tim Wescott wrote:

You've received a number of hints already. One more: Could it be that this motor is arcing? Then you would not see any effect when at standstill.
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wrote:

In the past I've done really good with just using an external resistor equal to the winding resistance, so I'm not quite sure what setup Tim is using.          ...Jim Thompson
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On Fri, 22 Mar 2013 17:58:57 -0700, Jim Thompson wrote:

I'm using an external resistor. It's stored in a variable named MotorR.
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On Sat, 23 Mar 2013 16:38:04 -0500, Tim Wescott

;-)          ...Jim Thompson
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wrote:

Here's the way I've done it in the past...
http://www.analog-innovations.com/SED/DC_Motor_Control.pdf          ...Jim Thompson
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On Fri, 22 Mar 2013 18:53:13 -0700, Jim Thompson

Cool. That exactly doubles the copper loss.
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On Fri, 22 Mar 2013 18:53:13 -0700, Jim Thompson
[snip]

Works just as well the other way around...
http://www.analog-innovations.com/SED/MotorDriver_PWM.pdf
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
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On Sat, 23 Mar 2013 13:53:45 -0700, 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
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Le Fri, 22 Mar 2013 12:51:29 -0500, Tim Wescott a crit:

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.
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Le Sat, 23 Mar 2013 15:37:54 +0000, Fred Bartoli a crit:
<...>

Oops, obviously make that Vavg = D Vs - (1-D) Vd = D(Vs+Vd) - Vd
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On Sat, 23 Mar 2013 15:37:54 +0000, Fred Bartoli wrote:

It's H bridge, with very little dead time.
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Le Sat, 23 Mar 2013 16:39:14 -0500, Tim Wescott a crit:

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)
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On 23 Mar 2013 23:09:20 GMT, Fred Bartoli

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
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