Low Voltage burns out electric motor - How ?

Factual Questions
1

I’ve heard the P = I*E explanation (P is constant, E decreases, so I must increase (in violation of Ohm’s Law?); and I’ve also heard that it’s garbage. (I.e., if that were true, and that’s all there was to it, a light bulb would burn brighter if you lowered the voltage …)

So what’s really going on?

I’ve read stuff about starter coils not disengaging in some cases, and back-EMF somehow playing a role in others, but … what’s the … STRAIGHT DOPE ?

Is there an electrical genius in the house?

Or at least a good motor destroyer? :slight_smile:

2

From what I remember, it’s because the motor never gets started right.

V = I/r

When the voltage is low, the motor can not turn over at all (doesn’t start), or will run more slowly than it should. If there’s no actual movement, then there’s no load on the coils of the armature. If movement is slow, then there’s very little load on the coils of the armature. Voltage doesn’t change, but resistance goes down, so current spikes. High current = high heat, and the motor burns out.

I don’t have a firm grasp of the electromagnetic principles that say why no/slow movement of the motor assembly results in low resistance, sorry. I know when we were in Jamaica, our fridge was plugged into a box that would cut out if the house voltage was low. Figured it was better for the fridge to be unpowered for awhile rather than having to keep replacing the electric motor.

3

Oh, good motor overview, with pretty pictures:

4

This looks like it has some nice info to, but I have to head to bed:

http://howthingswork.virginia.edu/supplements/electric_motors.pdf
(Warning - PDF)

5

As usual, it’s not the voltage that burns out the motor. It’s the current. As an electric motor speeds up, it creates an increasing electromagnetic field that acts counter to the torque created by the applied voltage, until the back emf causes net torque on the motor to be zero.

The easiest way to understand it is to imagine that an electric motor will attempt to accelerate as quickly as possible towards an equilibrium rpm. At 0 rpm, an electric motor will draw the most current, thus generating the most heat, but its current draw will decrease as rpm increases until it reaches the equilibrium rpm. If you apply a load to the motor, it will slow down, current will increase and equilibrium rpm will be reached again (or close to it). This is how the motors in cassette or CD players always rotate at the same speed given different loads (weight of CDs, start of cassette play vs end, etc.).

IMO, low voltage should not burn out an electric motor unless it is stalled or constantly speeding up/slowing down. If the motor can reach an equilibrium rpm, then only enough current will be drawn to overcome mechanical losses. Since the equilibirum rpm will be lower, an electric motor would draw less current with a lower voltage. Of course, if the motor stays at 0 rpm it will draw max current and could easily burn out as it struggles to overcome starting friction.

6

Low voltage can burn out a motor if it doesn’t start or if it turns too slowly to provide enough cooling airflow through the motor.

When a motor runs the armature is a coil of wire turning in a magnetic field. Therefore there is a voltage generated in the armature that opposes the applied voltage. This is the back emf.

In the case of a commutator motor, dc, with the field coil in parallel with the armature coil, when it runs with no load the speed increases until the applied voltage minus the back emf allows enough current to flow to overcome friction and windage.

As the load is increased the armature rotation slows which reduces the back emf and allows more current to flow so as to provide power to drive the load. The power required is a function of the load and not of the motor. So if the applied voltage is low, the armature has to turn even slower so as to allow more current to flow so that the product of voltage times current equals the power needed to drive the load and overcome losses. The heat produced in the armature is proportional to the square of the current times the armature resistance. It’s easy to see, then, that the heat generated rises quite rapidly as the motor slows down because of the low applied voltage.

The same thing is true of AC induction motors.

7

This is probably the case I was talking about. The device we plugged our fridge into was supposed to protect the motor in the case of a brownout. If the voltage is low enough and the load high enough such that the RPMs never reach high enough, wouldn’t the current remain at a fairly high level for long periods of time? Seems like that could burn out a motor in a pretty short period of time.

Great answers, though.

8

Just to comment on this specific part of the OP:

The part where you are going wrong is assuming that P is a constant. If you lower the voltage on a light bulb, the P definately drops. It is not a constant. Ohm’s Law is not violated, and in fact you can use Ohm’s Law to figure this out. If you take Ohm’s Law (E=IR), rearrange it to I=E/R, and put it in P=IE, you get P=E(squared)/R. R is constant, so you can see that if you change E that P is going to change also.

You might be getting transformers mixed up in this. Transformers don’t absorb or generate energy (ok, technically they absorb some energy and convert it into heat, but let’s ignore that for now), so in a transformer the power going into the transformer has to equal the power going out of the transformer. This is the case where you were probably taught P is a constant.

9

Thanks for all your responses. I may understand this yet.

OK, here’s my understanding of motors, EMF, back EMF, and current. Please tell me if I’m understanding this right:

(1) A jammed motor burning out - explanation:

I = E / R; R, being just the wire, is negligible; the motor is jammed, so there’s no motion of the armature in the magnetic field, so backEMF = 0.

This results in high I, high current.

Heat generation is high; heat dissipation is low, due to no cooling from the relative wind of spinning; insulation on the wires melts, etc. Result is a burned out motor.
(2) A motor with no load (no physical load):

EMF is constant. Initially, when it’s turned on, I is high (similar to the situation in (1) above); but then the motor spins up easily since there’s no load.

Now we have a back EMF.
The back EMF is proportional to the rotational speed of the armature (conductor moving in a magnetic field generates the back EMF) [This is the key point, isn’t it?]

The motor spins faster and faster until EMF = backEMF. (Or back EMF almost equals EMF, since friction also acts against the EMF.) At this point, Net EMF is small, and current is low.
(3) A motor with a load (a physical load, something to turn):

Suppose we take our motor spinning with no load, and add a load.

That slows down the armature, which reduces the back EMF.
Less back EMF means more Net EMF, so current increases.
Increased current means more torque, which gets the load spinning.
So the armature gets back up to speed, which increases back EMF.
And we’re back to the Net-EMF is small, current is low, and now the load is spinning.
(4) Effect of voltage being "too low."

Voltage is low. Current is lower than if voltage was normal. This means torque is low. Now there are two possible results.

(a) Case 1: the motor still has enough torque to get up to a speed where EMF = BackEMF
It runs slower than it would at full voltage.
It may take longer to get up to a steady speed.
It does not burn out, because once it gets up to speed, current is low, so heat is low.
(b) Case 2: the motor does not have enough torque to get up to speed; BackEMF < EMF
It runs slower or not at all
High current situation. Even though the EMF is lower, the back EMF is even less.
Less cooling from slower than normal speed. Result: overheat.
(5) I did some reading. There are also some motors that have a separate starter winding to help with the initial start up. This winding shuts off after the motor gets up to 70-80% normal speed. If the motor never gets up to speed, it never shuts off, and the wires aren’t designed to carry current all the time.

[(6) The whole P=E*I thing is what other people told me: P is constant, E goes down, so I goes up. I felt it was wrong, didn’t make sense, except for transformers (which I will leave for another day :slight_smile: ). But I didn’t know what was right, what was actually going on in electric motors.]
Final questions:

(A) Is my understanding correct?
(B) Is the key point the Back EMF being proportional to rotational speed?
(C) So then normal rotational speeds, and when exactly does EMF = back EMF, are basically design issues – ?

10

In the interest of fighting ignorance: A cassette player motor, at least the cheaper ones, have an internal governor that rely on flyweights to open and close the motor circuit and a flywheel to minimize the variations in speed. The more expensive one will have a closed loop servo system that senses the speed of the motor and constantly fine tune it.
A CD player does not revolve the platter at a constant speed. It uses a closed loop servo system to spin the platter so that the tracks pass over the laser read head at a constant linear velocity. A CD reads from the inside to the outside. It’s spinning at maximum velocity at the start of the disk and minimum by the time it reaches the end.

To address your other point, an overloaded electric motor will burn up even if it’s not stalled if the current through the windings exceeds the current rating for the motor.

11

The resistance of a lightbulb filament increases as it heats up. I just measured the room temperature resistance of a 90 watt bulb and I measured 11 ohms.
If you assumed that the resistance stayed constant, than by placing the bulb in a 120 volt circuit you’d be dissipating approximately 1300 watts. However, because the resistance of the filament is non-linear, when its reaches its operating temperature, its resistance has increased to approximately 130 ohms.

12

Many really good motors, such as those in my evaporative air cooler, have a thermal switch that disconnects the input power when the temperature exceeds some safe value. The burn-up is usually a failure of the interwinding insulation from overheating. This leads to internal short circuits and smoke rising.

13

According to pages 36-37 of this document,

14

:confused: How the heck did that get into a manual that seems on the surface to be sensible? [I didn’t read through it, so it may be a spoof in which case consider me whooshed.]

I remember an occasion some years ago, when I was a kid and my parents were driving us on a family vacation. We saw a sign saying that the road was narrowing from two lanes down to one. My mother announced that this was a good thing, since it would mean that we would go faster, just like when a river narrowed. :dubious:

In similar vein: I have here in my hand a lightbulb that is labeled “60 Watts”. I’ve just measured the voltage across it, and it’s “zero volts”. Therefore, by my calculation, the current through it right now must be infinite! I think I’m going to put it down before it 'splodes. :dubious:

15

There’s some confusion here.

A d.c. motor, like a computer fan motor, or a universal motor like a mains powered electric drill motor, can run at reduced speed on a reduced voltage, without burning out.

An induction motor, like an air-conditioner or refrigerator motor, will not run properly on a reduced voltage.

Induction motors like to run close to their “synchronous” speed. At 50 Hz, that’s 3,000 rpm, or at 60 Hz, 3,600 rpm (assuming a simple two-pole design). An induction motor will draw a small no-load current, plus a current that’s proportional to its “slip”.

Slip is defined as:

slip = (synchronous speed - actual speed) / synchronous speed.

For example, a 60 Hz motor running at 3,300 rpm has a slip of

(3,600 - 3,300) / 3,600 = 0.083, or 8.3%.

Now, ignoring the small no-load current, the motor has an impedance of Z[sub]0[/sub]/(1-slip), where Z[sub]0[/sub] is a constant for a particular motor. This is particularly important. It means that you can’t assume an I = V / Z relationship, where I will decrease with decreasing V, because Z isn’t a constant.

So what happens when we reduce the voltage? It depends on the speed/torque characteristic of the fan or pump or whatever the motor is driving, but as a first order approximation, this is what happens:

Reduce voltage by 1%
Slip increases by 2%, reducing impedance by 2%
Current increases by 1%

So effectively, the motor really does behave as a constant power device.

Large, expensive motors will have undervoltage protection to trip them off if the voltage drops below about 85% of normal. That’s because while the current increases roughly linearly with decreasing voltage, the I[sup]2[/sup]R losses in the windings, and therefore the heating in the motor, increase quadratically.

16

Cheers to Desmostylus for bringing up induction motors. I was (and am) speaking strictly about simple dc motors.

Regarding cassette player motors: Aside from a servo-controlled system, variations in speed are handled by the dc motor itself. Whatever governing system is in place is there to prevent the tape from breaking and the motor from stalling, not to regulate the speed of the motor as tensions increase/decrease within tolerance.

As for CD players… :smack: Even though newer CD-ROM drives tend to be CAV, they are probably servo-controlled as well.

Another good point. However, is this possible for a dc motor which is merely in a low voltage situation, and not stalled or somehow experiencing angular acceleration? It is possible that the lowered rpm can result in increased heat/increased resistance but wouldn’t this increase the voltage drop across the windings without increasing current?

17

Sort of. As has noted when the motor slows down, both ac or a dc motors, the current increases to try to maintain constant power output in order to drive the load.

However, the “constant power” business is really irrelevant. What counts is the maximum temperature that the insulation and other materials in the motor can withstand and the fact that the temperature is a function of the square of the armature current. If the input voltage falls low enough that the motor can’t provide enough torque to continue to turn the output power falls to zero. Nevertheless it will still burn up if the heat generated by the square of the armature current multiplied by the armature resistance raises the temperature high enough. Burn-up is quite likely if the motor stalls or turns much too slowly since many motors have a fan attached to the rotor to provide cooling.

18

How about low voltage and heating elements? I bought a heated dog dish in January so the dog’s water wouldn’t freeze outside. By the end of February, it was no longer working, so I took it back, and the clerk flipped it over and showed me the sticker that said “Do not use with extension cords.”, which of course is what I had done. Why would using a 12’ heavy duty extension cord cause the heating element to fail?

19

No, you’re on the wrong track altogether.

There’s a fundamental difference between the way a dc or a universal motor operates, and the way an induction motor operates. This difference is the key to the OP’s question.

You’ve noted that if a dc motor slows down, current will increase. The same is true for an ac motor. But the amount of the increase is different, and that’s why you can use a varying voltage to control the speed of a dc motor or a universal motor, but not an induction motor. You have to use a variable frequency controller with an induction motor.

With a dc motor, if you hold the voltage constant, but reduce the speed by increasing the load, the current will increase. That’s the situation you’re describing as “constant power”. That’s not the correct usage of the term.

With a dc motor, if you reduce the voltage, speed will decrease and current will also decrease. With an induction motor, if you reduce the voltage, speed will decrease but current will increase. (It’s not really a constant power relationship, more a constant VA relationship).

20

I did not write about universal dc motors but rather a parallel connected dc motor.

I’m going to have to go over this a little because it has been a long time. I’m not so sure that you are right about about a parallel connected dc motor. At no load if you reduce the voltage in such a motor you also reduce the field strength which results in less back emf and the armature current increases temporarily until the speed increases to bring it back down to that current sufficient to overcome losses. However you also reduce the torque so that under load the motor will slow down and the current increase. It does seem to me that if you reduce the voltage in such a motor the current could increase enough to damage it.

DC motor speed control is either with armature current control or field current control but the armature and field are not in parallel but are separately excited. In that case you can either vary the field current to change the stationary magnetic field or the armature current so as to vary the rotating magnetic field. The the better of the two is armature current control but it requires handling larger currents in the control curcuit than is the case with field control. The best method to use is the amplidyne system if you want armature current control. In that case you drive a generator with a prime mover and the generator supplies armature current to the speed-controlled motor. You vary the much smaller field current on the generator to control the armature current. That way you have the best of both worlds, excellent speed control of the motor and small current in the control circuits.

With the development of high power semiconductor switches this might have changed and good speed control of large dc motors might be done using such switches these days. In fact I believe such switches allow good speed control of ac inductions motors in modern diesel-electric locomotives.