DC Permanent Magnet Motor Impedence PM

I don't think this is necessarily a hard and fast rule. It probably depends a lot on the air gap.

Anyway, you need an impedance bridge to measure it. Something like the General Radio 1650A.

Technically, impedance (in this context) is inductance plus resistance of the windings in the motor.

Jon

Inductive reactance plus resistance, right?

Does that mean that a given motor will exhibit somewhat higher impedance at high speed than at low speed?

--Winston

Inductive reactance plus resistance, right?

Right, motors always behave as an inductance, *unless* they are wound-rotor synchronous machines, in which case they can behave as inductances, pure resistances, or capactances, as a function of the machine's excitation.

The impedance of a PM motor will indeed be higher than that of a wound field motor as the wound field motor is using power to generate the field that the magnet does with the PM motor. As to the actual impedance, that is dependant upon the power being used to turn the motor. The output power does make a big difference in the actual impedance of the motor as the turning of the rotor generates a back-EMF that reduces the current flow through the motor.

-- Bob May Losing weight is easy! If you ever want to lose weight, eat and drink less. Works every time it is tried!

Yes. The reactance of a given inductance is proportional to frequency. For a DC motor, reactance will be zero at zero RPM, and increases with increasing RPM. That's because the commutator and brushes turn the DC into AC to make the motor work, and the frequency of the manufactured AC will be proportional to speed.

Gary

From: "Wild Bill" snipped-for-privacy@usachoice.net

How would the impedence of a PM motor be determined?

Do a Google search on "characterize" and "motor". You'll come up with some useful hits. Here's a brief quote from one that gives a simple technique if you have a signal generator.

"1) Measure the resistance of the armature.

Measure the static DC resistance of the armature and that will tell you what the max DC current (stall current) the motor will consume at a given voltage using the formula I = V/R.

2) Measure the inductance of the armature using a signal generator and a known resistance, like 100 ohms.

Set the signal generator to generate a sinusoidal signal and connect the resistor and motor in series with the output of the signal generator. (keep the motor from turning). Now adjust the frequency on the signal generator until the RMS voltage measured between the motor terminals is exactly half the measured voltage between the signal generator terminals. You've matched the impedence of the motor to be R ohms at the frequency on the generators. Use L = (2 * pi * F)/ R. "

Richard Coke

Not really. When the motor is spinning, there is also the induced "back" EMF, ie. voltage generated by the windings spinning in a magnetic field. That's why as the motor speeds up, the current drops.

Jon

Not quite true, they act as inductor and resistors, even at no load, and at high load, a lot more like a resistor. jk

As a former motor design engineer I am reading this thread -- and tryng to understand why there is so much interest in DC motor impedance.

If you are working on a high performance servo system, I guess that could be a consideration, but a Statotrol Jr. is a pretty simple drive with just a couple taps for stability adjustment if I recall correctly. Certainly not a high performance drive. Same could be said about the treadmill motor. Probably a good combinatiion for many applications, but as long as the motor and drive will run in a stable mode in the application who cares what the motor impedance is?

Measuring motor impedance with an impedance bridge will give unsaturated impedance, but the real impedance during motor operation is (magnetically) saturated impedance.

Bigger motors (and higher speed motors) generally have lower armature circuit inductance. When powered from a phase controlled SCR drive, at some size (maybe a few hundred HP at 10,000 RPM or 1250 HP at 1150 RPM with a 6 pulse supply) the armature circuit inductance starts to get low enough that current ripple begins to become a problem. Big DC motors do not like current ripple and at some level commutation problems will start to develop. The solution is to actually add a series reactor external to the motor.

Mill

Well I hate to stick my neck out, but I think the bit below is wrong. If you think about what impedence is, it is the AC resistance. So if you measure the AC voltage applied to the motor and measure the AC current and measure the phase angle. But if you just want to compare two motors just apply the same voltage and measure the currents. The one with the lower current has the higher impedence.

Now Jim Pentagrid will probably be able to point out what is wrong with this. And I will pay attention.

Dan

I think this is exactly correct. Measure applied voltage, and the current and phase angle that results, that gives you the exact impedance. This result will of course vary with the rotor condition - free running, loaded, locked, etc.

Jim

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Nah, it's right as far as it goes. It's an old trick for determining the inductance of a coil. Here it's being applied to a motor winding. As the frequency of the test signal increases the impedance of the coil increases. When the impedance matches the resistance of the known series resistor the voltage drop across the motor will be equal to half the applied signal voltage. Having determined the impedance at a known frequency, the inductance can be determined. Not the most accurate method in the world of testing but close enough for the purposes of the OP's examination. This is a _static_ test and only gives you the inductance of the coil. Why he wants the impedance of a small DC motor is another question.

Richard Coke

No. Impedance is the vector sum of resistance and reactance.

Z = R + jX

Resistance is the same whether you're dealing with AC or DC. The voltage and current are in phase, and their product can be expressed as power, representing the rate at which energy is dissipated or otherwise consumed by the resistance. (The resistance need not be dissipative, the energy consumption may be due to transformation of the energy into some other useful form of energy, such as shaft rotation.)

Reactance is a function proportional to frequency and the inductance (or capacitance) of the circuit. At zero frequency reactance is zero, and hence steady state DC circuits are purely resistive.

As frequency is increased, a given inductance (or capacitance) will show reactance of increasing magnitude (but opposite signs for inductive or capacitive reactance). Reactance is always non-dissipative, no net energy is consumed by a reactance, and no power is expressed. Reactance follows Ohm's Law, ie X = E/I, E = X * I, I = E/X, etc. But I^2 * X = 0, E^2/X = 0, and I * E = 0 in a purely reactive circuit.

The vector sum of the resistance and reactance is then the impedance. In AC circuits this follows Ohm's Law in the same way as a resistance, ie Z = E/I, E = Z*I, I = E/Z, etc. But I^2 * Z is not necessarily equal to P, nor is E^2/Z, nor is I * E.

Note that impedance can be purely resistive if its reactive component is zero, or purely reactive if its resistive component is zero, or anything in between if it has both non-zero resistive and reactive components. Ohm's Law can't tell you which. You have to observe the net energy flux to know.

Gary

Discussion of the voltag-current relationship of a PM motor requires you consider back EMF. Think resistor, inductor and battery in series.

Ted

What I was trying to convey is that impedance is the AC equivalent of resistance in DC circuits. Dan

I know, you said that, it's wrong. *Resistance* is the AC equivalent of resistance, in an AC or DC circuit. Reactance is unique to circuits with time varying currents. Impedance is the vector sum of the two.

Z = R + jX

Gary

For the case of pure resistance then

Z=R

and for the case of pure non-dissipative load

Z= jX

I think that's what he was getting at.

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Thanks for all the input on this topic. The GE Statotrol Jr drive has 4 different jumper settings for low impedance/lowest current, to high impedance/highest current.

The charts show examples of GE DC motors with wound fields to use the lower

3 jumper settings, and that for GE PM motors to use the 2 highest settings.

I don't have a wound field motor for comparison, so I wanted some confirmation that PM motors were in fact higher impedance.

the replies are much appreciated WB ..................

I think I discovered that the PM setting recomended in the manual isn't the optimal setting, probably due to the particular motor design, as you suggest.

I managed to get some run time on the motor today (yippee, chips) and changed the jumper setting in the drive to check the results for speed regulation and motor temperature. The motor was heating up (but not overheating) fairly quickly at the highest impedance setting on the drive. This appeared to be abnormal, since I was only drilling with a small drill, and the hole had previously been pilot drilled. At the next lower setting, there didn't appear to be any heating at all after numerous continuous cutting runs.

I didn't bother to 'scope the armature waveform since it's been so long since I've read DC drive troubleshooting manuals that I wouldn't know what I was seeing anyway.

I'm content with very good speed regulation and adequate cooling from the small muffin fan I'm using.

Thanks again for all the replies WB ...............