How would the impedence of a PM motor be determined? The motor I'm using is
a 130VDC PM 7000+ RPM, treadmill type labeled as 2 1/2 HP.
The drive I'm using is set up for 90VDC output.
The manual for a GE Statotrol Jr drive states that wound field motors have
lower impedence than PM motors do.
Is it reasonable that the PM motor would have higher impedence?
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.
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.
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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.
From: "Wild Bill" email@example.com
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. "
Not really. When the motor is spinning, there is also the induced
ie. voltage generated by the windings spinning in a magnetic field. That's
why as the motor speeds up, the current drops.
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
Measuring motor impedance with an impedance bridge will give unsaturated
impedance, but the real impedance during motor operation is (magnetically)
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.
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.
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,
please reply to:
JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com
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.
No. Impedance is the vector sum of resistance and
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
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.
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
For the case of pure resistance then
and for the case of pure non-dissipative load
I think that's what he was getting at.
please reply to:
JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com
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
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
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
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
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