Unexplained voltage rise at terminals of small PM motor

[37pw56gy]

I have a five dollar bet with a co-worker on the correct answer to this question, so please give it your best!

Here’s the scenario: A small (about 1 HP) 120 VDC permanent magnet motor drives a reversible mechanism that imposes a varying mechanical load. The 120 VDC source is a simple full-wave rectifier without any sort of filtering. The rectifier is fed from 60 Hz and only one motor is the only load served by this source. Cable between the motor and its controller may be quite long (up to 2500 feet), but wire-wire capacitance is nil and AWG is selected to keep I*I*R drop to an acceptable level. The mechanism has a torque limiter that prevents motor current in excess of about 15-20 amps.

The problem: Under normal load conditions, the motor RPM is somewhat less than its nameplate speed and terminal voltage is reduced proportionate to I*I*R drop. No problem so far, but this seems to be an important clue.

Under light load conditions, however, something strange (?) happens. The terminal voltage increases to a level substantially over the 120 VDC no-load source voltage. The voltage can be as high as 160 VDC at the motor terminals while the motor is freewheeling. The motor, its controller, wiring, etc., are generously over-designed and in no jeopardy of failure. What causes this phenomenon? How do I justify doing nothing about it to those who have nothing better to do than question my work and delight in watching me squirm for the one answer that so far has eluded me?

I do have a theory: Consider the classical synchronous AC motor (or sync condenser) having an over-excited rotating DC field. In this condition, leading VAR’s are returned to the line to counteract lagging VAR’s thrown-off by predominately inductive loads. The motor is, in essence, a big capacitor, and readily adjustable too. Terminal voltage at the motor would rise substantially if corresponding inductive loads were not present or if the motor is grossly over-excited in proportion to the VAR situation.

How does this relate to a small permanent magnet DC motor? Here’s the idea: Under light torque load, the permanent magnets represent the over-excited field of the sync motor. As with any DC motor, a back EMF is generated in the spinning armature. The back EMF follows the full-wave cyclic waveform. Because the source is full-wave rather than pure DC, this back EMF raises the terminal voltage during the declining portion of each half cycle. With no other load to absorb this energy, the terminal voltage rises substantially.

Interestingly, the voltage does not rise under heavy load conditions. Back EMF is reduced as RPM drops.

To test this theory, two experiments are proposed. First, substitute a filtered 120VDC having little or no ripple. Does the overvoltage condition disappear? Second, feed the motor through a series diode and place a large capacitor across the motor terminals to absorb back EMF. In both cases, the overall voltage/current values should return to the usual I*I*R relationship.

Anyone with suggestions that might provide a more rigorous explanation? Many of those in my particular electrical engineering specialty have no experience with sync motors

 

[stardelta]

I cant comment on the electronics and physics side but can suggest something else to consider. PM motors can do all sorts of strange things when the magnets are cracked or damaged, as you can imagine it seriously upsets the field. Could this be a possibility.

 

[37pw56gy]

Another interesting point. In this particular instance, the voltage rise is consistent among several different motors on differing circuits that are geographically separated by many miles or more. Nothing to pursue here.

Aside from one instance of a loose magnet jamming against te armature, I hadn't previously given much thought to changes in stator flux patterns caused by defective magnets. Food for thought, however. I would expect that the net result is diminished torque output and increased brush arcing.

Some time ago, we were advised by the motor OEM that operation at voltages in excess of the nameplate rating could degauss the PM's, resulting in permament reduction in torque output, burnt brushes, etc. As things are, we have never had a complaint regarding weakened output (brushes are another story, however). Many of these motors (particularly the 24VDC counterpart of the 120VDC machine) are routinely operated at voltages well above their nameplate at 36VDC or more without distress. BTW, this type of machine is not subject to NEC or OSHA requirements that forbid operation of electrcial equipment in excess of nameplate ratings.

 

[moturcu]

one way to look at the problem is to analyze at frequency domain rather than time domain. if you can have have the frequency and associated phase information of the motor voltage and currents for dominant components than you can come up with reactive power information for each component.

Before going that much further, you can check current wave form. if it is pure dc due to the motor inductance this suggest zero reactive power.

 

[SERVOCAM]

How about quick, simple, might make sense rant:

I think by what you typed, I understand the situation. (Thought I did) Maybe REGEN! Take any PM DC motor (Brush or Brushless)..if you apply power to it, it turns...it is being a motor. If you take the motor and spin its shaft, it is now a generator, generating an output voltage.

Regenerating means that speed and torque are in opposite directions (one negative and the other is positive). So when the motor is decelerating, we are regenerating power back to the line.

Kinda an example:
The motors will have a Voltage Constant rating, Ke. Ke is usually specified in Volts per RPM or Volts per 1000 RPM. So say your 120VDC motor is rated for 3000 rpm, we can estimate that the Ke is 40 VDC/kRPM. So it takes ~40 VDC to go 1000 RPM, but if you turn the shaft 1000 rpm, the motor will produce ~40 VDC.

Regen Power (Watts) is equal to the change in Potential Energy divided by time. There are many variables that you need to know to calculate the regen power: Motor Inertia, Motor Resistnace, Friction Torque, Motor Voltage Constant, Motor Torque Constant, Load Inertia, and Motor Speed before Decel.

No that is just for the motor. If you want to size for a regen/dump resistor you also need to take in consideration of the amplifier driving it.

Now I'm not to worried about regen power damaging the motor, more worried about the smoke'n the electronics (but there aren't really any). If the regen power is too high, a shunt circuit should be added. We do this by shunting/short circuiting the DC bus through power resistors. This is more common and can handle higher transient loads than dumping it back on to the line (line regeneration).

So this calculated kinetic energy (Regen Power) must be stored or dissipated. The power supply has some capacity to store some of this energy, and some energy will be dissipated in the mechanism friction, motor windings, and drive circuitry. The remaining energy will need to be dissipated the regen resistor.


If we look at the Heavy load vs. light load strange phenomenon, ...hmmmm, you know, I thinkng I just spent the last 10 mins. babbeling about something that is no help.

Why would the motor voltage be higher when the load is lighter??? Not an EE, so struggling if the electronic could cause any problem....Is the motor running at a higher speed when the load is lighter? Now Regen will pump up the bus voltage (but you don't have anything to pump up), but when the load is lighter, there will be less regen if it is decelerating the same or slower than the heavy load. Maybe I will run this accross the Ph.D's at work and see what they think.

Just makes you wanna and . Just wish I had an .



Cameron Anderson - Sales & Applications Engineer
Aerotech, Inc. -

http://www.aerotech.com

"Dedicated to the Science of Motion"

 

[Skogsgurra]

Have you all forgotten about basic electricity?

The DC PM motor accelerates when subjected to armature current. And if the motor is not loaded it will accelerate each time an armature current pulse is received - i.e. each half-cycle.

It will do so until it reaches a speed that producec a counter-EMF that is just below the rectified peak voltage. The 160 V mentioned seem to be just right.

That is all there is to it.

BTW, I*I*R is not a voltage drop. You should use I*R.

 

[moturcu]

skogsgurra
I do not agree with you. Motor inductance just filters out the current pulses that you mentioned.

In addition, (for 37pw56gy)
relation between peak sinus voltage and voltage rise can be checked by running system with slightly reduced and known AC side voltage.

 

[Skogsgurra]

moturucu

What you say is true if the motor is loaded, i.e. running with continous current (the Germans use the expression "nicht lueckender Strom&quot.

A DC motor that idles does not run with continous current, the current flows in short pulses and is driven by the difference between rectified mains voltage and counter-EMF (the Germans call it "lueckender Strom&quot.

The motor and its inertia actually act as a capacitor and the rectifier/motor arrangement works like a peak rectifier, which explains the sqrt(2) factor between idling voltage and loaded voltage.

A simple check with a clamp-on current transducer and an oscilloscope will show you how it works. As I said before; That's all there is to it.

 

[cbarn24050]

Hi, this is normal operation, no damage will occur as long as the rectifier diodes have a suitable voltage rating (above 350v).Scogsgurra's explaination covers is correct.

 

[GusD]

Hi 37pw56gy

Just like Stardelta ,I cannot tell you much about the science that causes small permanent magnet motors to simply misbehave for no reason.(I call it magnetic disturbance).
We use many Tachometers that have permanent magnets in the fields.When you strike these units with a hard blow,they don't like it at all and stop working or working erractically.Also,when working on these motors ,one has to be extremely carefull not to upset the magnectic balance of the fields.
I don't know what you call it,but it is fairly common.
If there is any money left out of the five,I need new Golf balls.

Good Luck

GusD

 

[37pw56gy]

Oh no... I feel that five bucks slipping away. GusD: better make other plans for acquiring that new set of balls.

skogsgurra has two good points: first, the bridge rectifier tends to create a peak-holding circuit, and second, I've carelessly used power loss (I*I*R) instead of voltage drop (I*R). To help salvage my argument, I'd be interested in more comments on how/why the spinning motor is analogous to a capacitor.

As for regeneration and flywheel effect propositions, I tend to discount these factors. Intra-pulse acceleration (newly invented term?) must be just about negligible (the armature alone weighs several pounds, and it is connected to a substantial 189:1 geartrain). The mechanical load driven by this motor is not insignificant, even when freewheeling. The load is definitely not backdriving the motor.

 

[Skogsgurra]

Dear 37pw56gy

Sorry about your five bucks, but I think that the knowledge gained will be worth more than that.

There is an application note from Unitrode. It is about modelling electric motors and contains the information you asked for. The address is:

http://www.powerdesigners.com/InfoWeb/design_center/Appnotes_Archive/u120.pdf

And the title is:

A SIMPLIFIED APPROACH TO DC MOTOR MODELING FOR
DYNAMIC STABILITY ANALYSIS.


A short snip:

Once we can represent the mechanical load by means
of electric elements, we can draw an equivalent circurt of
the motor and its mechanical load. The armature has a
finite resistance RA and an inductance LA, through which
the torque-generating current lA must flow. These components
are not negligible. and must be included. An inertially
loaded motor can be represented as in Fig. 1,
where the moment of inertia J is the sum of the load’s JL
and the rotor’s JM.

There are diagrams and formulae to help you understanding the analogy. It works, I know. I have used it for more than thirty years to explain inrush current and exponential acceleration in uncontrolled DC drives.

Yours,

Skogsgurra

 

[37pw56gy]

skogsgurra: An excellent refernce - thanks! The particular system involved is not nearly as sophisticated as a servo-controlled disc drive (i.e. it is controlled by simple contactors and limit switches). Anyone else with information or anecdotes related to a PM motor's capacitor-like behavior when the source is not steady DC?

Mention of the motor's current-torque transfer function has me thinking about something else that might help me redeem that $5... another thread, however.