Is this the correct way to measure the Back EMF Voltage?

[fishandchips]

Hello, I measured the Back EMF Voltage of a DC motor by connecting the shaft of two motors together. Then, applied 11.1V to the driving motor (called Motor A). While the motors were spinning, I measured the voltage across Motor A and also the voltage across Motor B (the one being driven) at the same time. Meanwhile, I measured the spinning speed. I obtained the following data: Voltage across Motor A: 9.37V. Voltage across Motor B: 8.15V. Am I correct that the Back EMF Voltage is: 8.15V? How come the voltage across Motor A was 9.37V rather than the applied voltage 11.1V?

[neko efecktz]

NO...
You are using motor A to drive motor B.
DC motors are also generators.
You are simply measuring the voltage that motor 2 is producing.
this has nothing to do with back EMF.
To learn more on Back EMF there is an enless supply of information on line .
wiki pedia has some interesting information

A motor has coils turning inside magnetic fields, and a coil turning inside a magnetic field induces an emf. This emf, known as the back emf, acts against the applied voltage that's causing the motor to spin in the first place, and reduces the current flowing through the coils of the motor.

[fishandchips]

I used the method from a book recommended by a forum user. See 2-77 Under Voltage Constant and Fig. 2.10.13. I think the Eg mentioned there is the Back EMF voltage. Perhaps I read the method incorrectly? Could anybody please double check?

https://www.elsevier.com/books/dc-motors-speed-controls-servo-systems/zhou/978-0-08-021714-7

[rstofer]

re: link - TLDR;

You are making this harder than it is.  You apply a voltage to the terminals and, given the measured resistance, you know what current should flow.  But you measure the current of an unloaded motor and you find it is far less than what you calculated for the winding resistance and applied voltage.  The difference is the back EMF.  What you really have is the applied voltage minus the back EMF (producing a net lower voltage) divided by the measured resistance.  Ohm's Law...

As you load the motor, the back EMF is reduced and the current increases accordingly.

So, (Applied Voltage - Back EMF) / Measured Resistance => Input Current

Back EMF = Applied Voltage minus (Input Current times Measured Resistance)    <= this is the equation you want!

[Ian.M]

Identical DC motors with the same field current (or identical field magnets) spinning in the same direction at the same speed will have the same back-EMF.   However if you couple the two motors shaft to shaft, they are spinning in opposite directions and asymmetries in the brush positions will result in non-equal back-EMFs.   Many DC motors have asymmetric brush positions for improved high speed performance in one direction at the expense of performance in reverse.

If you need to actually measure^* the back-EMF of a permanent magnet motor, the easiest is to PWM the motor supply with a high voltage MOSFET (to withstand the inductive kickback with only a minimal snubber) and measure the generated voltage during the off time (discarding the initial inductive transient).  As the duty cycle approaches 100% the measured back-EMF will approach its non-PWMed limit for that supply voltage.   Shunt wound motors can be handled similarly, but you need to maintain power to the field circuit.   Series or compound wound motors are not amenable to such methods of measurement.

* Directly measure, rather than calculate it from other measurements.

[Zero999]

You are using motor A to drive motor B.
DC motors are also generators.
You are simply measuring the voltage that motor 2 is producing.
this has nothing to do with back EMF.

The voltage generated by motor 2 spinning is the back EMF.

How come the voltage across Motor A was 9.37V rather than the applied voltage 11.1V?

It could be the resistance of the wiring, power supply or PSU current limit.

re: link - TLDR;

You are making this harder than it is.  You apply a voltage to the terminals and, given the measured resistance, you know what current should flow.  But you measure the current of an unloaded motor and you find it is far less than what you calculated for the winding resistance and applied voltage.  The difference is the back EMF.  What you really have is the applied voltage minus the back EMF (producing a net lower voltage) divided by the measured resistance.  Ohm's Law...

As you load the motor, the back EMF is reduced and the current increases accordingly.

So, (Applied Voltage - Back EMF) / Measured Resistance => Input Current

Back EMF = Applied Voltage minus (Input Current times Measured Resistance)    <= this is the equation you want!

Yes, that's the easiest way to do it:
BackEMF = V_IN - R×I

[fishandchips]

About "Back EMF = Applied Voltage minus (Input Current times Measured Resistance)"

I conducted another set of 10 experiments. I set the applied voltage from a power supply to 11.1V and measured the voltage across the two terminals of the DC motor as well as the current in series with the motor. Meanwhile, I measured the turning speed using a tachometer.

The averaged voltage measured across the motor was 10.18V.
The averaged current measured in series with the motor was 1.45A

Note that the measured voltages across the motor of all 10 runs were about 1V less than the supplied voltage of 11.1V.

I calculated the resistance by stalling the motor and measured the voltage across the motor and the current in series with the motor. Then, calculated the resistance as averaged voltage of 10 runs divided by averaged current of 10 runs. I obtained 0.1 Ohm.

Is the Back EMF = 11.1V - (1.45A * 0.1Ohm) = 10.96V?

[max_torque]

The driving motor is doing useful work (spinning itself and the "load" motor) and this takes power, that means a non zero phase current is flowing in the driving motor. And, as the driving motor cannot have a zero phase resistance, than means there is a voltage drop across the phases.


The "Load" motor is, in this case, effectively unconnected (assuming you were measuring the voltage with a high impedance multimeter!) and hence has zero phase current, and so no voltage drop. 

So, the Back EMF measured from the second motor is of course lower than the voltage supplied to the first motor!

For example, lets say at speed X, with both motors being identical (and ignoring any unsymmetrical brush effects) the KE (the motors back emf vs rotational speed) of each motor generates a back emf of 10 volts.  In the driving motor, there is a positive phase current, of say 1 amp, and if the winding impedance were 1 ohm, then 1 volt would be required to push that current through the phase windings, and so you'd have to apply 11 volts to the driving motor, yet get just 10V out of the load motor.

[fishandchips]

I am using the following instruments. Are they OK for the kind of experiments I am doing to characterize the motor?

To measure the voltage across the DC motor:
http://overseas.sanwa-meter.co.jp/items/detail.php?id=30

To measure the current in series with the motor:
Mastercraft 052-0052-2 (the version that could measure 20A)
https://www.eevblog.com/forum/testgear/canadian-tire-mastercraft-dmm-new-and-old-revision-teardown/

Power Supply (I used the front outputs to supply voltage to the motor)
http://www.alinco.com/Products/ps/DM-330/

[Ian.M]

Max_torque's explanation has a further implication: Because the driving motor has to provide power to overcome twice the friction and windage losses of an unloaded motor, you will need more voltage to match the speed of an unloaded motor.  You therefore cant get accurate results from the two motor method without a tachometer and an adjustable power supply.

Asymmetric brush effects are fairly easy to test for - if, for both directions at the same voltage, you get near enough the same unloaded speed and current, the brushes must be symmetric.   As the brushes may cock slightly in their holders, if the motor has been running in one direction for a significant period (as a part of the brush lifespan), its likely to require running in in the other direction before frictional losses become equal, which may confuse the issue. However as there would be no point in deliberately building a motor with very small brush asymmetry, one would expect to see a gross difference for a motor constructed with deliberate asymmetry.

One thing to note:  Copper has a notable positive temperature coefficient of resistivity of approximately 0.4%/°C, so if you run the motor long enough to heat up significantly, you will get different results as the winding resistance increases.  It may be useful to measure the motor temperature, and directly measure the DC resistance between runs.

[fishandchips]

It may be useful to measure the motor temperature, and directly measure the DC resistance between runs.

I tried to measure the resistance directly by connecting the probes from the multi-meter to the motor (see the two areas pointed by the arrows in the photo). Unfortunately, even I have tried three different multi-meters, I got 0 Ohm in each case. So, I used the calculated resistance of 0.1 Ohm.

https://bbqbbq2bbq.smugmug.com/My-First-Gallery/i-fpFQz8H/A

[fishandchips]

About "Back EMF = Applied Voltage minus (Input Current times Measured Resistance)"

I conducted another set of 10 experiments. I set the applied voltage from a power supply to 11.1V and measured the voltage across the two terminals of the DC motor as well as the current in series with the motor. Meanwhile, I measured the turning speed using a tachometer.

The averaged voltage measured across the motor was 10.18V.
The averaged current measured in series with the motor was 1.45A
...
Is the Back EMF = 11.1V - (1.45A * 0.1Ohm) = 10.96V?

For your reference, if I put back the gearhead to the motor and repeated the experiment, I got the following data:

The averaged voltage measured across the motor was 10.2V (also less than the supplied voltage of 11.1V).
The averaged current measured in series with the motor was 1.2A.

The Back EMF = 11.1V-(1.2A*0.1 Ohm) = 10.98V.

Are these calculated Back EMF values correct? I need to enter a Back EMF value to simulate the motor in Simulink. Are the instruments I used to do the measurements and to supply the voltage good enough to provide valid results?

[Ian.M]

It may be useful to measure the motor temperature, and directly measure the DC resistance between runs.

I tried to measure the resistance directly by connecting the probes from the multi-meter to the motor (see the two areas pointed by the arrows in the photo). Unfortunately, even I have tried three different multi-meters, I got 0 Ohm in each case. So, I used the calculated resistance of 0.1 Ohm.

Most multimeters cant perform useful measurements down in the milliohms.  You need a good bench milliohmmeter that uses Kelvin connections for that sort of stuff.
Otherwise, its apply a forcing current and measure the voltage, as you did, but that has complications from self-heating unless you keep the duty cycle very low.

[fishandchips]

Thanks Ian. Do you mean to get an accurate resistance, I should get a better bench milliohmeter that uses Kelvin connections and measure the resistance directly? From the manufacturer, the "dynamic" resistance at a lower test voltage is also about 0.1 Ohm. In this case, can I just use 0.1 Ohm to save some money? The kind of meter you mentioned costs hundreds of dollars.

How about the measured voltage across the motor and the current I measured. Are those data valid?

If I use the measured voltage across the motor (10.18V) instead of the 11.1V I measured before I connected the motor to the power supply, the Back EMF (without gearhead installed) becomes:

10.18V - (1.45A * 0.1 Ohm) = 10V

With the gearhead installed and using the measured voltage across the motor (10.2V) instead of the 11.1V I measured before I connected the motor to the power supply, the Back EMF becomes:

10.2V - (1.2A* 0.1 Ohm) = 10.08V

If the 0.1 Ohm is acceptable, which calculated Back EMF value is the correct one to use?

[Ian.M]

It depends on how much accuracy you need.  If you only need four digits, its possible to home-brew one for an affordable price.  e.g. http://hackaday.com/2017/01/24/milliohm-meter-version-1-5/   Settle for three digits and you could cheapen it further e.g. set up a 100mA LM317 current source and a x10 differential amplifier using an ordinary singe supply OPAMP with a common mode input and output range that goes down to ground, and 0.1% resistors and use an ordinary DMM on mV for readout.

Without having measured the temperature, you could easily have had a 10% change in resistance between your first and last run in a series - that's only a 25°C wiinding temperature change.

[fishandchips]

It depends on how much accuracy you need.  If you only need four digits, its possible to home-brew one for an affordable price.  e.g. http://hackaday.com/2017/01/24/milliohm-meter-version-1-5/   Settle for three digits and you could cheapen it further e.g. set up a 100mA LM317 current source and a x10 differential amplifier using an ordinary singe supply OPAMP with a common mode input and output range that goes down to ground, and 0.1% resistors and use an ordinary DMM on mV for readout.

Without having measured the temperature, you could easily have had a 10% change in resistance between your first and last run in a series - that's only a 25°C wiinding temperature change.

Thanks. As I recall, the motor was not even warm. I paused between runs.

[rstofer]

About "Back EMF = Applied Voltage minus (Input Current times Measured Resistance)"

I conducted another set of 10 experiments. I set the applied voltage from a power supply to 11.1V and measured the voltage across the two terminals of the DC motor as well as the current in series with the motor. Meanwhile, I measured the turning speed using a tachometer.

The averaged voltage measured across the motor was 10.18V.
The averaged current measured in series with the motor was 1.45A
...
Is the Back EMF = 11.1V - (1.45A * 0.1Ohm) = 10.96V?

For your reference, if I put back the gearhead to the motor and repeated the experiment, I got the following data:

The averaged voltage measured across the motor was 10.2V (also less than the supplied voltage of 11.1V).
The averaged current measured in series with the motor was 1.2A.

The Back EMF = 11.1V-(1.2A*0.1 Ohm) = 10.98V.

Are these calculated Back EMF values correct? I need to enter a Back EMF value to simulate the motor in Simulink. Are the instruments I used to do the measurements and to supply the voltage good enough to provide valid results?

Again, you're making this harder than it is.  External to the motor, you have absolutely no way to directly measure Back EMF.  The reduced voltage you are supplying to the terminals is caused by factors external to the motor - poor PS regulation, excessive voltage drop in the cables, etc.  In no way does it have anything to do with Back EMF.  The only voltage number you care about is measured at the terminals and this is the applied voltage.  Measured at the terminals...

Running a second motor as a generator provides no useful information because there is no load and Back EMF is always a function of load.

You did a locked rotor experiment and that gave you a measure of the winding resistance.  Use that value.  It was derived from Applied Voltage divided by Input Current with the rotor locked.

All you need to do is measure the terminal voltage and input current then use the formula for Back EMF.  At no load the current will be less than the simple Applied Voltage divided by Measured Resistance would suggest.  This is because the windings are generating a voltage internal to the motor which is the Back EMF.

The degenerate case of zero Back EMF occurs at locked rotor where that voltage is 0.  This is where you can get a measure of the winding resistance by working through Ohm's Law for the terminal voltage and input current.

[rstofer]

Your value for Back EMF above seems mathematically correct.  What is the full load current of the motor?  Given your 0.1 Ohm value and 12V supplied, it would seem like the locked rotor current would be 120 Amps.  That's a big honking motor!  I don't recall ever seeing any nameplate data discussed in the thread.  I have been thinking about toy motors where it seems we're talking about golf cart motors.

That would explain why your unloaded run current is so high.  Maybe a snapshot of the motor so we can get an idea what you're working with?

If the locked rotor current IS 120 Amps then it's no wonder you are dropping voltage in the power leads.

[fishandchips]

Your value for Back EMF above seems mathematically correct.  What is the full load current of the motor?  Given your 0.1 Ohm value and 12V supplied, it would seem like the locked rotor current would be 120 Amps.  That's a big honking motor!  I don't recall ever seeing any nameplate data discussed in the thread.  I have been thinking about toy motors where it seems we're talking about golf cart motors.

That would explain why your unloaded run current is so high.  Maybe a snapshot of the motor so we can get an idea what you're working with?

If the locked rotor current IS 120 Amps then it's no wonder you are dropping voltage in the power leads.

Thanks rstofer. So, forget about the method of using two motors to get the Back EMF and also forget about using 11.1V in the calculation.

When I stalled the motor to calculate the resistance, I did 10 runs. At 4V (again I got this measured value when I measured it directly from the output of the power supply before it was connected to the motor), the averaged measured voltage was 0.55V and the averaged measured current was 5.83A. At 5V, the averaged measured voltage was 0.64V and the averaged measured current was 7.35A. At 6V, smoke started to come out and I stopped the experiment immediately.

Photo of the disassembled motor is as follows:
https://bbqbbq2bbq.smugmug.com/My-First-Gallery/i-cbVKK5j/A

[rstofer]

You have some massive voltage drop in your setup.  Only terminal voltage counts but you are losing a lot of voltage in the PS and cables.  Wimpy!  But that's good!  If you didn't have that voltage drop, the motor would have gone up in smoke a long time back.

You don't need very much voltage to measure the internal resistance and after accounting for voltage drop, you don't have much.

It seems to me that your measured voltages and currents are directly related.

0.55V/5.83A -> 0.094 Ohms
0.64V/7.35A -> 0.087 Ohms

So, the winding resistance (including the brushes) is about 0.09 Ohms.  Given that number, you can get your Back EMF versus load (or RPM to the extent that RPM is a function of load) using the formula above.  But, again, it is a function of load.  If you could lash up something like a Prony Brake you could get a nice graph of torque versus current and hence Back EMF.

http://www.instructables.com/id/Bench-top-dynomometer/

[fishandchips]

Thanks rstofer. So, the Back EMF will be 10.1V because

Without gearhead case: 10.18V - (1.45A * 0.09 Ohm) = 10.05V

With gearhead case: 10.2V - (1.2A* 0.09 Ohm) = 10.1V

How come having the gearhead installed does not affect the Back EMF much. I thought having gears means more friction.

In the case without the gearhead, with the averaged measured speed of 2358.55 rpm from the tachometer, I could calculate the Back EMF constant as:

10.1V/23583.55 rpm = 10.1V/247 rad/s = 0.04 V-s/rad

Since Kt = Ke, the Torque Constant is also 0.04 but in N.m/Amp?

[rstofer]

It doesn't make sense that the current goes down when the gearhead is installed.  Maybe it doesn't have to...

The voltmeter is 1.3%+3 counts and the ammeter is 1.2%+5 counts.  The readings aren't far enough apart to say the problem isn't just the accuracy moving around.  Separating 1/10 volt or 1/10 amp isn't something you can count on with these meters.

My HP3478A bench multimeter is 0.004% of reading + 5 counts in 5-1/2 digit mode for DC voltage and 1% plus 30 counts for current (not all that good).  Bought used from eBay.

The EEVblog Brymen is 0.3% + 2 digits for voltage and 0.7% plus 3 digits for current on a 3-5/6 digit meter.

The GW Instek GDM8251A bench meter is 0.012% plus 5 digits for voltage and 0.2% plus 5 digits with a 20,000 count display (5-1/2 digits).  Also bought used and calibrated from eBay for about $100.

So, the HP for voltage and the Instek for current.

Maybe a better way to measure current is with a shunt resistor.  These usually have a full scale voltage drop of 50 mV for industrial units and are measured in milliohms for current sense resistors:
http://www.mouser.com/Passive-Components/Resistors/_/N-5g9n?Keyword=shunt&FS=True

0.001 Ohms times 10A is 10 mV and good meters have excellent resolution at lower voltages.  Fifty milliohms is also available but it will drop 0.5V.  Either way, the resistor needs to be upstream of the voltmeter at the motor termnals.

[fishandchips]

Thanks. Do you think having a clamp meter such as the Klein Tools's CL600 sold in Home Depot would help to measure the current more accurately? I double checked the output of the power supply. The max amp from the ports at the front is 5A. Should be sufficient?

I used very similar kind of hooked wires to connect the motor to the multi-meters and power supply. Could such thin wires be the source of the problem? I suppose thin wires do not allow much current to pass through. As a result, the voltage measured across the motor might be lower than the 11.1V I got when measuring the voltage directly from the power supply.
https://picclick.com/5-pairs-Wire-Kit-Test-Hook-to-Hook-201502669119.html

[Zero999]

Thanks. Do you think having a clamp meter such as the Klein Tools's CL600 sold in Home Depot would help to measure the current more accurately?

It's completely unsuitable for what you're doing. It can only measure AC current. You need something capable for measuring DC current.

Clamp measures AC current

http://www.kleintools.com/catalog/clamp-meters/digital-clamp-meter-ac-auto-ranging-600a

[rstofer]

Thanks. Do you think having a clamp meter such as the Klein Tools's CL600 sold in Home Depot would help to measure the current more accurately? I double checked the output of the power supply. The max amp from the ports at the front is 5A. Should be sufficient?

I used very similar kind of hooked wires to connect the motor to the multi-meters and power supply. Could such thin wires be the source of the problem? I suppose thin wires do not allow much current to pass through. As a result, the voltage measured across the motor might be lower than the 11.1V I got when measuring the voltage directly from the power supply.
https://picclick.com/5-pairs-Wire-Kit-Test-Hook-to-Hook-201502669119.html

Measure the voltage at the PS and again at the motor.  Every little bit that is missing is being dropped in the cables.  Those cables are totally unsuitable for higher currents.  Just measure the voltage drop...  Zero is a good answer, maybe 0.5V could be tolerated but you have several volts lost.  Those cables have to be getting hot.

You need precision measurements, not more of the +- several percent variety.  If you use the shunt resistors, you get your meter into something it does fairly well, measuring DC voltage, not current.  And if you want the right answer, you'll probably need to buy a meter with a whole lot more digits and accuracy.