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Topic: Burned Mosfet and popped flyback diode.. help! Big DC motor PWM (Read 4654 times) previous topic - next topic


The motor the OP has is more like a starter motor so  a scrap surplus starter would be a good UUT Test Load for a prototype driver.
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Sorry R.. missed your question.....

since L di/dt = V,   ( standard equation for an inductor.  For a capacitor  C dv/dt = I   ....  textbook stuff)

so di/dt = V/L =  12v/0.011H  = 1091 A/s  .

If you add resistance the current will approach the final dc current such that

I(t)  =  I ( final ) (1- e^-t/(L/R)     .  L/R  is the time constant , in this case a few tens of ms.


A starter motor can easily pull 500A at switch on.

Not easy to drive . still  recommend ntc thermistors and 1kHz pwm with schottky flywheel and suitable snubbers



I'll send the resultsof simulations in my next post


Please show your work.

1091A/S  ?

please show how you obtained this result.
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Hi R..

you'lll note in my original sums I said that I presumed the motor was a short.  I also assumed the inductor had zero dc resistance.  The 1091 A/s then applies.

Even with added resistances , the inital slope will be this.

I enclose PSPICE circuit and resulting waveform, with included resistance. You'll note the risetime is far too slow to use a 1kHz PWM control.




I don't have PSpice so I found this online which only works if you enter the inductance and try increasing values for current until the result = 12.001 V and the value for current that obtains that result is the 1091A you cited.

In all honesty, math was never my strong suit. I learned basic electronics math including algebra but deriviatives is something I studied at DeVry but never mastered and frankly , don't remember.
I did find this, which helps, but frankly I am having a hard time grasping how the math can explain the typical ringing waveform you see across an inductor. I can generate that waveform and observe it on my scope,
(which I have done and taken photos of : see attached) but as yet , I don't understand how the math explains or describes the ringing seen in the photo.

One of the things that I am trying to understand (again) is:
You can see according to this formula that the voltage is directly proportional to the derivative of the current. Since the derivative of a constant is equal to 0, if the current is a direct current (DC), the current across the inductor will be equal to 0. So if the current is a DC current, the current flowing through the capacitor will always be 0. This, again, is because the derivative of a constant is always equal to 0. A constant does not change. So if a user simply enters in a current such as 10A or 20A or 30A, the current will be 0, for all these values. This shows that no voltage can be across an inductor if it is connected to a DC power source. There is only voltage across an inducttor when it is connected to an AC source.
It seems to me the above statement does not take into account the case of turn on and turn off.

Since the OP's motor is a DC motor , Reactance does not apply, and Cos(t) does not apply.

In the attached schematic, how should the diode or diodes be connected ?

Sorry about the price tag but this is an old photo taken before I removed the price tag. 8)

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Hi R

To get the ringing you see there must be some capacitance - do you have a circuit for the set-up for this waveform  ?

regards allan.


It was a diode test circuit where the diode is across a 100 uH inductor that is switched by a mosfet.
The point of the test was to compare the effects of different diodes in eliminating the back EMF.
This photo was supposed to represent the "CONTROL" where there is NO diode. The diodes being tested were schottky diodes. I had some caps across the diode originally but removed them. I honestly don't know if they were still in the circuit when this photo was taken.
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Hi R..

     well you asked for it.  There are loads of textboks which will do better than me.

Another way of looking at the characteristics of inductors and capacitors is this.

the impedance of an inductor is  Z(L) = jwL               L is the inductance in henrys

and a capacitor                       Z(C)  =  -1/jwC        C is the capacitance in farads

w is the rate of change in radians/second . There are 2*pi radians in a circle, so w = 2*pi*Freq/Hz

j is the square root of -1. An imaginary number. But useful.  Treat it as a vector length 1 at right angles to the 'real' ( ie ordinary numbers) axis. Think of an x-y graph , but now  horizontal is real  and  vertical imaginary.  +j is above the horizontal , -j below.

Inductors are above the horizontal, capacitors below ( because of the - )

Let's think of a resonance - that where the overall impedance of an inductor and capacitor adds to nothing. ie he inductor length above = the capacitor length below.

So we can say that jwL = -1/jwC.  Multiply thru by jwC

so  j^2 * w^2 * L * C = -1.

But j^2 = -1     by definition.

so w^2 * L * C = 1

or w = 1 / ( sqrt (L*C)

but since w =  2* pi * freq/ Hz

the frequency of resonance / Hz is  1/(2*pi*sqrt(L*C)) 

been a long time.....


Allan ..

I won't bore you any more if you don't want......



the frequency of resonance / Hz is  1/(2*pi*sqrt(L*C))   
so basically, it's the resonance that causes the ringing and the amplitude is diminishing because the energy is being "depleted/dissipated " ?
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Yes - the ringing you see is at the resonant freqency. The decay is because of resistance in the circuit -
whether fitted , or in the L-C components. Real components always have some resistance.



ps I generally use s for seconds, as S is for siemens - ie reciprocal ohms. Confusing, eh?



another example - just for fun.

suppose I have a resistor of 10 ohms in series with
an inductor of impedance 10 ohms at a particular frequency.

I now apply a voltage of 1v at this frequency - what current flows?

1/(10+10)    amps?

 No, because the total effective impedance has to take into account of the fact that the inductive impedance is at right angles to the resistive.

Hence the answer is that the actual impedance is sqrt(10^2 + 10^2) or 10 * sqrt(2) - it's just finding
the resultant of 2 equal vectors at right angles.

Hence the actual current will be 1/10* sqrt(2)  amps and it will it will lead the applied voltage by

arctan( 10/10)  ie 45 degrees

regards Allan.

mess about with a signal generator and a 2-channel scope and you'll see it's true.



Yes. I said I was rusty. I didn't say I forgot everything I learned LOL.

A pure resistance has no phase shift while an inductive reactance causes the current to lag voltage by 90 degrees.

BTW, I the 11 mH was based on the assumptions of 50 ms for the motor rampup time and 0.22 for the sum of the inductor and motor series resistances. The inductor value would be different if the windup time was different or the series resistance was different. I'm inclined to believe the method used was correct even the assummed values and resultant inductance are not, mainly because I got the formula from a Phd electrical engineer.

current LAGS voltage in an inductor and LEADS volyage in a capacitor .
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so if I apply 12v across your 11mH inductor ( taking the motor as a short for now)
current will rise at 1091/amps second . In 1mS ( the pwm period) it will only get to 1.09 amps.

Hence the inductor is far too large by a factor of at least 30 if we want to deal with 30 amp max.
with 1kHz pwm.

No, this is a complete non-sequitor.

What you should have said is that the inductor limits the rise of current to 30A to at least 30ms,
irrespective of the PWM frequency.

If a fast current control loop is wanted for motor control (ie for a servomotor) extra series inductance
is bad news, unless its upstream of a big capacitor.

Any way a 30A 11mH choke is probably a monster, perhaps as big as the motor itself!  That's enough to say
its not the solution to any reasonable problem!
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]


Hi MarkT...

      if you read through the thread you'll find it isn't my idea...  the quote is from a post where I pointed out it couldn't work

If you want to do  it an outline design is aooended.  But Iwouldn't do it like that- even 220uH at that current is a juicy device!   I've ommitted snubber stuff which would certainly be required..

My feeling is that big NTC thermistors could be used to limit inrush current.

take no notice of device types - it's just what PSPICE gives me.



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