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FET Killer

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so 1 ohm would be the R in the time constant calculation. so if your measurement finds that the inductance is 1mH for instance, then that time constant would be 0.001S. one tenth of that would be 0.0001S, so the PWM frequency would be 1/0.0001 or 10khz. lower frequencies will work, but your losses will be higher. some of those losses will be as heat, some of those losses will be in the form of audio emitted from the motor. just out of curiousity, what is the manufacturer and model/part number of this motor. the inductance may be in the data sheet for the motor. that would save a lot of time and effort, although i admit letting you learn how to measure it yourself would teach you a lot. i keep a bunch of caps and inductors set aside for measuring unknowns with a sine wave generator and oscope.
 
so 1 ohm would be the R in the time constant calculation. so if your measurement finds that the inductance is 1mH for instance, then that time constant would be 0.001S. one tenth of that would be 0.0001S, so the PWM frequency would be 1/0.0001 or 10khz. lower frequencies will work, but your losses will be higher. some of those losses will be as heat, some of those losses will be in the form of audio emitted from the motor. just out of curiousity, what is the manufacturer and model/part number of this motor. the inductance may be in the data sheet for the motor. that would save a lot of time and effort, although i admit letting you learn how to measure it yourself would teach you a lot. i keep a bunch of caps and inductors set aside for measuring unknowns with a sine wave generator and oscope.
Thank You Unclejed613,
The resistance lead to lead was .1, (1/10) ohm. I spent some time making screen shots and marking the up for people to look at and for me the crunch some numbers on and see if I can figure it out my self, but I may have to ask for some help.
The following link has 11 drawings, 9 screen shots and 2 schematic drawings of what I've been up to. Found another Cap, 50,000 uf. Measured 60,000 measured with my fluke 87 at 3000 uf / sec at the 400 ohm range for 20 sec. Here a link to the motor I have. **broken link removed** , again good tip, I see if I can get a data sheet on it.
Oh yea, when you look at the screen shots, the first ones are with 54,000 uf and the last ones are with 120,000 uf. Labeled on the attached drawing.
One more set of numbers/measurements, via fluke wall rat= 9.07v, 8 ohms 8.62v, motor 173.3 mv, caps 75.8 mv current 1082 ma
Thanks,
Kinarfi

PS. as a motor gets bigger and stronger, does it's inductance increase or decrease?

**broken link removed**
 
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The inductance will remain the same.

But the motor is not a pure inductor. You have a magnet inside and a core. So you are going to have a frequency response, like a band-pass filter.
 
The inductance will remain the same.

But the motor is not a pure inductor. You have a magnet inside and a core. So you are going to have a frequency response, like a band-pass filter.

I think you misunderstood my last question. Smaller motor vs larger motor, low current motor vs high current motor, power vs weak motor, which motor would have the more henrys?
It seems to me that the larger, more powerful motor would have more induction and less resistance? but now I'm wondering if resistance and induction would be proportional.
Kinarfi
 
i did a search, and i found a 3rd party controller for this motor. switching frequency is 20khz. that motor looks similar to a motor i've seen in 1970's vintage computer cassette tape drives as a direct drive capstan motor.

i would go with the 20khz switching frequency.
 
After dusting out the Mathematician cobwebs in my brain, using advise and suggestions from this forum, I think I have it.
CORRECTIONS MADE

Hayato, I used you suggestion to calculate L, R was determined by measuring the DC voltage drop across the leads while sending 1000 ma DC through it, 0.0791 ohm. THEN Freq = 60 Hz, and current phase across an external 8 ohm resistor, and calculating phase shift by manipulating the screen shots so 1 cycle was a multiple of 360 pixels and counting the pixel between wave forms. One wave form had to be flipped because of the way I hooked the scope leads.

Unclejed, what should I see when I power up the motor with the PWM on the scope? and would you mind checking my Freq calculations?

This sure got away from the FET killing problem, Which I DO think was answered along the way, Thanks again everyone,
Kinarfi
 

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I didn't understand how you calculated R.

Are you using just the leads resistance?

Motor.GIF

Remember.

Motor impedance is:
Zm = r' + Zind
Zind = jωL


Voltage at B is:

vB = Zm*Vp/(Zm+R1)

Phase at B:
Phase = arctan(Zm/R1).
 
To find phase shift, I manipulated a screen shot by stretching it until one cycle was 360 pixels, then counted the pixels from resistor trace to motor trace, 33 pixels or 33 degrees. Also, because the scope ground was between the motor and resistor, one trace needed to be inverted, so measurements were taken at 0 crossing. Using the term "ELI the ICE man", E leads I in an inductor, the motor trace was 33 pixel ahead of the resistor trace.
Using a dc power supply, I set the current to 1000 ma and measured the dc voltage of the motor, 79.1 mv and calculated .0791 ohm. Direct reading with ohm meter was not accurate, .1 ohm was as close as it could get.
kinarfi
 
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Ok.

Tan(33°) = Zm/R => 0.65 = Zm/8

Zm = 5.2 Ohms

Zind + 0.0791 = 5.2 => Zind = 5.121 Ohms.

|j*2*3.1415*60*L| = 5.121

L = 13.58 mH
 
Ok.

Tan(33°) = Zm/R => 0.65 = Zm/8

Zm = 5.2 Ohms

Zind + 0.0791 = 5.2 => Zind = 5.121 Ohms.

|j*2*3.1415*60*L| = 5.121

L = 13.58 mH

The 8 ohm was the external resistor, the phase shift would be the Zm created by the the pure L and pure R of the motor and the 8 ohm would have no effect. it was used to limit current and give a 0 phase trace.
Tan 33 = motor l/ motor r
 
There is no way to measure internal phase shift with your scope probes.

What you measure with probes is the Zm phase shift. And then you substract the internal resistance.

Without the 8 ohm resistor is impossible to measure the phase shift.
Why? Because the phase shift is on the current, so you need a external resistor to "sense" this current phase shift.
 
There is no way to measure internal phase shift with your scope probes.

What you measure with probes is the Zm phase shift. And then you substract the internal resistance.

Without the 8 ohm resistor is impossible to measure the phase shift.
Why? Because the phase shift is on the current, so you need a external resistor to "sense" this current phase shift.

When you measure the 8 ohms with the scope, you get the phase of the current, when you measure the motor, you get the phase shift of the impedance Z of the motor which is the vector sum of the internal R & the internal L combined to create the overall phase shift. I determined R and calculated L.
I believe my calculations are correct.
Kinarfi
 

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When you measure the 8 ohms with the scope, you get the phase of the current, when you measure the motor, you get the phase shift of the impedance Z of the motor which is the vector sum of the internal R & the internal L combined to create the overall phase shift. I determined R and calculated L.
I believe my calculations are correct.
Kinarfi

Your concept is correct. But the calculation is wrong.

You are only going to have the 33° shift due the 8 ohm resistor.

If there were just the internal resistance and the inductance, you would get something near 90°.

The term Arctan(Zm/R1) applies to the mesh only.
 
OOPs,:eek: I think I errored,:eek: what I said was L is XL (.0541)
The trace from the motor will be the vector addition of R (.0791 ohms)*I at 0 degrees and XL (.0514 ohms)*I at 90 degrees and the 8 ohm resistor has nothing to do with the motor trace. It is a separate trace used to show the phase of the current, only!
XL= 2*3.1415*60*L
.0541 = 2*3.1415*60*L
L = 136 uh

You wrote" |j*2*3.1415*60*L|" What is "j"
Kinarfi
 
136 uH is too low for a motor.

You wont be able to calculate the inductance w/o considering the 8 ohms.

The j is to indicate that the inductor impedance is imaginary.
 
136 uH is too low for a motor.

You wont be able to calculate the inductance w/o considering the 8 ohms.

The j is to indicate that the inductor impedance is imaginary.

This is a pretty large motor
**broken link removed**,
That's why I have asked if the henry value changes with motor sizes.
When running, it pulls 12 amps and stall current is over 100 amps, Low R, Low XL, Low henrys?

I must disagree with you about the 8 ohms, except the phase of the current and limiting the current, the 8 ohms has nothing to do with motor calculations.
I have always used "i" for imaginary numbers, that why I didn't recognize it, but then I can't speak Spanish either, so that probably the difference.
Thanks for your help, when I try to explain something to someone, I always find my errors or convince myself I'm correct, and I found my error, but believe I'm correct also, thanks,
Kinarfi
 

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I don't speak Spanish. And in Electrical Engineering, j is used for complex numbers.
 
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I have always used "i" for imaginary numbers, that why I didn't recognize it, but then I can't speak Spanish either, so that probably the difference.
In school I observed that mathematics professors used i and the engineering professors used j.

i and j are quite unlike, though both called 'imaginary'. i is sqrt(-1), whatever that is. j is the reactance, current, or voltage that is 90 degrees apart from the reference.
 
Some of my math were wrong.

I'll show you why you have to consider, let's imagine the following circuit, were you want to measure voltage and phase at point B:

Example.png

Ok.

R1 = 10 Ω
R2 = 0.1 Ω
L1 = 100mH
ZL = j*2*pi*60*100mH = j37.7 Ω

Z1 = R2 + ZL

As it is known, the voltage VB at B is given by the divider network, considering the Vsource = 1 Vp:

VB = Z1/(Z1+R1).

So we are going to have:
VB = (0.1 + j37.7)/(10.1 + j37.7).

Changing to phasors:
VB = (37.7 /_ 89.85º) / (39.03 /_ 75°)

VB = 0.966 /_ 14.85°

So we are going to have a 14.85° of phase at point B.

The simulation result is 14.846° for 60 Hz.

Phase.png

I'm going to redo my math to look for a possible error on your motor calculation.
 
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Ok.
I was wrong, you were right. Somehow I got confused. Sorry. By calculations it is, indeed, about 136 uH.
 
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