# A question about Back EMF

1. Feb 12, 2009

### woodfich

So here is the circuit I came up with to represent back emf in a DC motor with rotational inertia.

When current is applied to the motor it spins. When it rotates, the motor acts as a generator and creates a voltage (back emf) as shown (sorry for the incorrect notation, I don't know the correct notation for a generator, so I just put the two bars of the battery on each end of the resistor that represents the motor). Therefore, as soon as the engine starts rotating, the reading on the voltmeter increases while the reading on the current decreases.

My question is, does the power dissipated by the motor increase or decrease or stay constant (since P=VI and V increases while I decreases)?
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An additional question. Is there any way to measure this back EMF?

If something is unclear please tell me so I can edit.

edit: I carried out a series of experiments (using the circuit arrangement shown in the diagram, where the M/G block represents a DC motor that is also acting as a generator as it rotates) that demonstrate that the power calculated from the reading of the ammeter and voltmeter is greater when the motor is rotating than when it is still when applying the same voltage to the circuit.

This is how I am explaining it to myself. The rotational inertia (which is considerable in the motor I am using) inputs electrical power into the system, therefore increasing the total electrical power measured. Is my argument sound? I am still confused because I thought back EMF would reduce the power output instead of increasing it. Am I making a mistake somewhere?

Last edited: Feb 12, 2009
2. Feb 12, 2009

### pallidin

Last edited by a moderator: May 4, 2017
3. Feb 13, 2009

### uart

Normally in that circuit the opposite situation would happen. If the supply voltage (battery in this case) is fairly stiff (as in doesn't drop much with increasing load) then the power is approximately proportional to current. So when the motor is stalled and there's no back EMF the current and power should be at maximum. Note that this power will be entirely lost as heat because there is no conversion to mechanical power taking place at stall.

The situation you describe (increasing power with increasing speed) is possible however in the case where the battery is a fairly weak source (battery voltage falls away markedly with increasing current demand). This of course means that the battery has a high internal resistance.

Specifically, if the internal resistance of the battery is larger than the internal resistance of the motor then you are likely to observe the power actually increasing with increasing motor speed. Otherwise if the internal resistance of the battery is lower then you should observe maximum input power at stall.

4. Feb 13, 2009

### woodfich

Thank you for your answer. The last part most probably applies to my experiment. I am NOT using a battery as a current source (the internal resistance of the cell I am using is considerable). Could you or someone please explain why the quoted happens? (I am still confused about the physics of why it happens)

5. Feb 13, 2009

### woodfich

Anyone? Please, I've been thinking about this all day. It still doesn't make sense to me since the induced emf would produce power in the other direction of the one supplied to the motor. Wouldn't the induced power "cancel out" the original power, thereby decreasing the total power?

6. Feb 13, 2009

### Gnosis

It’s true that the back emf is the voltage generated by the rotating armature as it cuts through the magnetic fields of the stationary permanent magnets however, the polarity of the back emf voltage is the same polarity as the applied voltage that’s causing the armature's rotation. Often, the term “opposing voltage” tends to mislead those studying DC motors and they presume the "polarity" of the back emf voltage is reversed compared to the applied voltage, but this isn't so.

Since the generated back emf voltage has the same polarity as the applied voltage (and this can easily be verified), it effectively reduces the difference in applied voltage across the armature windings thereby causing a reduction in the amount of current drawn through the armature windings.

Picture two batteries that are about to be connected in parallel (positive to positive, negative to negative). One is a fully charged 12 volt battery while the other is a fully discharged 12 volt battery therefore, it has 0 volts across its terminals. If we connect these two batteries in parallel, the fully charged 12 volt battery is given good cause to deliver as much current as the discharged battery can pull because there’s an initial difference of 12 volts.

Now picture the same scenario, but instead of a fully discharged 12 volt battery, this time it’s only partially discharged and producing just 8 volts. This time when we connect the fully charged 12 volt battery in parallel with the 8 volt battery, there’s a lesser difference in the two potentials (just 4 volts), so the 12 volt battery is given less cause to deliver as much current.

Now picture the same scenario with two fully charged 12 volt batteries. When we connect the two batteries in parallel, there’s no difference in potential whatsoever therefore neither battery is given reason to allow current flow. The tow voltage sources possess the same polarities, just differences in potentials, which causes differences in current flow. This best describes what’s happening with the generated back emf voltage verses the applied voltage.

If you do not believe the back emf has the same polarity as the applied voltage, simply apply a DC voltage to the motor’s terminals taking note of the polarity applied to each terminal as well as noting the direction of the armature’s rotation. Next, disconnect the applied voltage, then mechanically rotate the armature shaft in the same direction with a voltmeter attached per the same polarity as the previously applied voltage had been, and you’ll see that the generated back emf produces the same polarity as the applied voltage!

Many are under the false impression that the back emf has the opposite polarity of the applied voltage simply because the term “opposing voltage” is used. The two 12 volt batteries in parallel with each other are the same polarity, but current flow is "opposed" because they allow no voltage difference to exist between the two battery voltages. The same thing happens between applied voltage and the generated back emf voltage in a DC motor.

The faster the armature rotates, the greater will be the voltage potential of the generated back emf voltage, which reduces the difference between the back emf and applied voltages. This is turn causes less current to be drawn through the armature’s windings. This is why DC motors draw just a fraction of the current at their no-load RPM compared to the rather heavy current they draw at start-up or when the armature has been stalled.

7. Feb 14, 2009

### uart

Well imagine for a moment that you were driving the motor with a current source instead of a voltage source. Now a current source by definition will keep the current constant regardless of how much back EMF it encounters. So since P = V I it should be clear that if we keep the current constant then the power drawn by the motor will increase with increasing terminal voltage (which in turn will increase with increasing back EMF).

So just to summarize the situation. When voltage-source driven the power intake at the motor terminals should decrease monotonically with increasing motor speed. When current-source driven the motor input power should increase monotonically with increasing motor speed.

In the situation of a "real source" we are somewhere in between an ideal voltage source and an ideal current source. Typically if the source impedance is low compared to the load impedance then you'd best describe it as a non-ideal voltage source (that is, it's closer to a voltage source than a current source). On the other hand if the source impedance is large compared to the load impedance then it would be better to think of it as a non ideal current source. So qualitatively (non rigorously) you might guess that for R1 < R2 it should behave more like the voltage driven scenario and that for R1 > R2 it should behave more like the current driven scenario, at least at low speed. (BTW, here I'm denoting the source internal resistance as R1 and the motor internal resistance as R2, ok).

If you like you can make it a rigorous conclusion by analysing the circuit with R1 and R2 included and the back EMF modelled as a voltage source of magnitude E = k w. Where "k" is a constant and "w" is the rotational speed of the motor. If you follow through with this you will find the equation for the motor input power is :

$$P = \frac{V^2 + (\alpha - 1)V k \omega - \alpha k^2 \omega^2}{R_2 ( 1 + \alpha)^2}$$

Where "V" is the open circuit source voltage and alpha is the ratio of source resistance to load resistance, that is $\alpha = R1 / R2$.

Now you can solve the above equation and show that P is a maximum at :

$$\omega = \frac{(\alpha - 1) V}{2 \alpha k}$$

This last equation tells us that if alpha < 1 (that is, R1 < R2) then the max power point occurs at a negative rotational speed (you'd have to manually turn the motor backwards against the motor torque to achieve this) so the motor input power decreases monotonically with increasing positive speed. If however alpha > 1 (that is, R1 > R2) then the motor input power starts out rising with increasing motor speed, up to the speed ("w") given by this last equation, after which it decreases with increasing speed.

Last edited: Feb 14, 2009
8. Feb 15, 2009

### woodfich

Thanks for the responses.

I still have a question. The explanation implies that the voltage difference across the motor should decreases once it starts rotating. However, when I measure the voltage with the setup I depicted above, I get a higher voltage difference reading when the motor is rotating.

Should I subtract the induced back EMF from the voltage dropped by the motor when it is still? I am thinking this as two batteries connected in parallel (the first battery being the original voltage source and the second one being the back emf of the motor).

V (motor stalled) + V (back emf)= V(read)
V (motor stalled) - V (back emf)= V(read)- 2 V(back emf)

Or, does the voltage reading itself represent the voltage drop at the motor?