Looking for a good high voltage MOSFET for a PWM design

Thread Starter

Hamlet

Joined Jun 10, 2015
517
I'm running a 115v DC motor, 1/3hp. I keep blowin' fets, even tried a parallel config...

It will run an IRF640 all day, 4 or 5A, if I keep a small fan on the heat
sink. But I have to start it gently, going from 5% up to 70% or so.

My problem is starting the motor with a load. So, if I am running a load of 4A, switch
off for whatever reason, then switch it on without crawling upon it, thar she blows!

I've used 200v, 250v, 600v, and 900v mosfets, all with an rdson of 0.45 or less, with
amp ratings of up to 18amps (IRF640). Maybe I'm not using the right mosfet, or not
big enough? Maybe IGBT?

upload_2017-2-12_22-41-11.png

Any help in mosfet selection would be much appreciated, thank yous.

EDIT: I would like to add that I'm running the adj. for the freq. at about 15kH.
Any higher and I get too much mosfet heating, any lower, I get annoying harmonics...

EDIT: Forgot the snubber diode in the schematic. It's been there all along.
 
Last edited:

crutschow

Joined Mar 14, 2008
32,849
You might add an automatic ramp up whenever the motor is turned on, such as with a large resistor (say 100kΩ) in series from the pot to the (+) LM339 input along with a capacitor from the (+) input to ground.
The turn on time will be roughly the RC time-constant of those two added parts.
You then turn the motor on and off with a switch for the Vcc voltage to the pot.

It sounds like it's the large startup current that's killing the FET.

Reducing the value of R8 may reduce the MOSFET heating.
 

Thread Starter

Hamlet

Joined Jun 10, 2015
517
You might add an automatic ramp up whenever the motor is turned on, such as with a large resistor (say 100kΩ) in series from the pot to the (+) LM339 input along with a capacitor from the (+) input to ground.
The turn on time will be roughly the RC time-constant of those two added parts.
You then turn the motor on and off with a switch for the Vcc voltage to the pot.

It sounds like it's the large startup current that's killing the FET.

Reducing the value of R8 may reduce the MOSFET heating.
Auto-ramp-up? Sounds good! I want to use this circuit on/off, so once I determine my load, I can set and forget it.
I can see where R8 could be better at 1K, I will try it.

Startup surge does seem to be the trouble. I don't a IRFP250, but I might order some for robustness.
Any recommendations for where to start with the cap value?
I'll admit, I ain't great with the math. I'm guessing 1/2 second should be plenty...
 

crutschow

Joined Mar 14, 2008
32,849
The LM339 can only sink about 6mA so R8 should be at least 2kΩ with a 12V supply.

You could start with the series resistor to the LM339 (+) input of 100kΩ, with a capacitor to ground at the input of 10μF.
That should give a start-up of about 3/4 of a second in you circuit, and go from there.
The capacitor can be a 16V aluminum electrolytic. Be sure and observe the polarity (negative to ground) as they are a polarized type capacitor.
 

tindel

Joined Sep 16, 2012
933
I will start by saying that I'm no motor expert - but I do have a lot of experience with FETs used as switches and their gate drive circuitry. I have also used a motor a few times, but in low power situations.

First - when you're testing is your motor loaded? It might help to remove the load - just to discern if it's the motor current or the gate drive circuitry that's hurting you.

A fully loaded 1/4W hp 115V motor will draw somewhere on the order of 5.8A. At least according to this http://www.engineeringtoolbox.com/elctrical-motor-full-load-current-d_1499.html. (Remember I'm no motor expert). At those current you'll have significant power losses in the FET. So per the IRF640 datasheet you're worst case power dissipation in the motor is: P=I^2*R = 5.8^2*0.18=6.1W. That's a lot of power - without adding in switching losses. You're power may not be this high because your motor is probably not fully loaded.

The thermal resistance of the part from the internal junction of the silicon to the case of the part (thetajc) is important when you're talking about these kind of powers. It's only 1degC/W - so this is pretty good. I have no idea what you're case temperature is though. This is probably very important for you to measure either with a FLIR camera, laser temperature gun, or something else of the sorts.

The junction to ambient air thermal resistance is 62degC/W - this is poor, but not uncommon in these types of packages. But to make my point - with 6W of dissipation in free air - the junction will be 375degC above ambient - or 400degC at ambient temps. The maximum specified temp of this part is 150degC. You're part will blow up in open air with maximum load. And this just highlights the need for a heatsink - which you have, but it may not be enough of a heatsink.

So, I'd start by measuring the temps. Startup currents will be much higher than at steady state. So this may be why you have a hard time starting up.

The IRF640 is an old part these days - but it's super cheap. And that's why folks still use it. Here is a list of 19 transistors with the same package that have at least half the Rds,on so you'll be dissipating more like 3.0W. Reducing your steady state thermal issues in half. http://www.digikey.com/products/en/discrete-semiconductor-products/transistors-fets-mosfets-single/278?FV=9780001,97c0064,97c0065,97c0069,97c0072,97c0074,97c0075,97c0077,97c007d,97c0082,97c002f,97c0030,97c0031,97c0032,97c0033,97c0034,97c0035,97c0036,97c0037,97c0038,97c0039,97c003a,97c003b,97c003c,97c003d,97c003e,97c003f,97c0040,97c0041,97c0042,97c0043,97c0044,97c0045,97c0049,97c0058,97c0059,97c0060,142c002a,142c0051,142c0368,1f140000,ffe00116,9902745,990082d,9900a3b,9900dd5,9900191,99001bc,99011ee,99011f4,99001fa,9901494,99016b7,9901894,9901db3,9901eab,9901f1d,9901f94,990229e,99025fd,9902647&mnonly=0&newproducts=0&ColumnSort=1000011&page=1&stock=0&pbfree=0&rohs=0&quantity=&ptm=0&fid=0&pageSize=25

Also, this is only half the battle - switching losses could also be part of (or all of) the problem - I second crutschow's recommendation to reduce R8. I'd probably also experiment with swapping Q1 and Q2 and putting a resistor between the base of each. This would reverse the logic though - in which case, swapping the pins of the comparator inputs may be in order. The experiment may also point you in the direction of the right solution. It might also blow up the drive transistors due to shoot-thru - but it might be worth a quick experiment to see what happens.

I'd also add a bit of hysteresis to the comparator - if it doesn't already have any. This also could be your problem.

Good luck!
 
Last edited:

crutschow

Joined Mar 14, 2008
32,849
I'd probably also experiment with swapping Q1 and Q2 and putting R8 between the base of each.
Don't see how that can work.
You need one resistor for each base.
And I think that will be worse, since emitter followers have a lower output impedance to better drive the MOSFET gate capacitance than common emitter circuits do.
 

Thread Starter

Hamlet

Joined Jun 10, 2015
517
The LM339 can only sink about 6mA so R8 should be at least 2kΩ with a 12V supply.

You could start with the series resistor to the LM339 (+) input of 100kΩ, with a capacitor to ground at the input of 10μF.
That should give a start-up of about 3/4 of a second in you circuit, and go from there.
The capacitor can be a 16V aluminum electrolytic. Be sure and observe the polarity (negative to ground) as they are a polarized type capacitor.
Ah, Senior crutschow, the following are my corrections, yes or no?
upload_2017-2-13_10-33-0.png

Corrections in red. Where is my blunder?
 

ian field

Joined Oct 27, 2012
6,536
You might add an automatic ramp up whenever the motor is turned on, such as with a large resistor (say 100kΩ) in series from the pot to the (+) LM339 input along with a capacitor from the (+) input to ground.
The turn on time will be roughly the RC time-constant of those two added parts.
You then turn the motor on and off with a switch for the Vcc voltage to the pot.

It sounds like it's the large startup current that's killing the FET.

Reducing the value of R8 may reduce the MOSFET heating.
Some SMPSU chips have built in soft start function, its not impossible to use a SMPSU chip for motor control.

The tricky bit is finding one with low enough UVLO.
 

EM Fields

Joined Jun 8, 2016
583
Ah, Senior crutschow, the following are my corrections, yes or no?
View attachment 120463

Corrections in red. Where is my blunder?
It's recorded here:

U2a is a voltage comparator running open-loop, so no matter how gently you treat its inputs,
its output will snap and either turn U2a -7 hard ON or hard OFF, ruining the possibility of soft-starting.

Consequently, when abruptly turned ON, Q3 has to pass the motor's full locked-rotor current until the motor starts spinning and can generate some back EMF. If the motor is mechanically loaded before power is applied that'll delay or prevent spinup, aggravating the problem.
 

tindel

Joined Sep 16, 2012
933
Don't see how that can work.
You need one resistor for each base.
And I think that will be worse, since emitter followers have a lower output impedance to better drive the MOSFET gate capacitance than common emitter circuits do.
Yeah - I realised that shortly after my first post - and edited it about the time you replied. You're totally right it probably won't work. it's not driving the FET on (pulling the comparator low, and driving the high side pnp would work pretty well - it's discharging the capacitance that would be harder - driving a low-side npn. My thought was it was something the OP could try that might lead us into the right direction to a good solution. I have used common emitter circuits to drive similar circuits - although without the humongous startup currents.

Question - I'm not a motor expert... I know just enough to be dangerous. How can a soft start reduce the the startup current? You've got to put enough energy into the motor to surpass the moment of inertia somehow (moment of inertia might not be the right term - but the idea is to that the current has to be high enough to get the motor moving initially). I'd think that you have to pull the startup current no matter what. Pulsing it may allow your transistor to cool - but only if your transition losses are less than the resistive losses.
 

crutschow

Joined Mar 14, 2008
32,849
U2a is a voltage comparator running open-loop, so no matter how gently you treat its inputs,
its output will snap and either turn U2a -7 hard ON or hard OFF, ruining the possibility of soft-starting.
No.
You guys are missing the point.
That voltage controls the PWM duty-cycle, so the duty-cycle will start at 0% and then go slowly to 100% giving the ramp start for the motor.
The motor inductance averages the PWM pulses so the current corresponds to the PWM duty-cycle.
The motor current will slowly ramp until it's sufficient to start turning, much less current than if you abruptly applied full voltage.

The op already stated that a slow startup works, as stated in his first post:
It will run an IRF640 all day, 4 or 5A, if I keep a small fan on the heat
sink. But I have to start it gently, going from 5% up to 70% or so.
 

tindel

Joined Sep 16, 2012
933
No.
The motor inductance averages the PWM pulses so the current corresponds to the PWM duty-cycle.
But you still have to hit exceed the startup current required to start the motor spinning - right? How does additional pulses from the PWM improve efficiency if you still have to meet the startup current?
 

crutschow

Joined Mar 14, 2008
32,849
I know just enough to be dangerous. How can a soft start reduce the the startup current? You've got to put enough energy into the motor to surpass the moment of inertia somehow (moment of inertia might not be the right term - but the idea is to that the current has to be high enough to get the motor moving initially). I'd think that you have to pull the startup current no matter what. Pulsing it may allow your transistor to cool - but only if your transition losses are less than the resistive losses.
Yes, you've reached the danger zone. ;)
The motor needs a certain current to start which likely is a lot less than if you instantaneously applied the voltage to the motor when there's no back EMF from rotation (the motor looks like a transformer with a shorted secondary).
In that case the initial startup current is limited only by the static winding resistance and perhaps a small winding leakage inductance.
Once it starts to rotate, it generates a back-emf which reduces the current for a given applied voltage.

They often connect large power resistors in series when starting large industrial motors to limit the otherwise huge startup current.
The resistors are then removed (shorted across) when the motor is nearly up to speed.
 

Thread Starter

Hamlet

Joined Jun 10, 2015
517
Okay, I've made the changes. I've got the delay we're looking for, but, I'm still blowing mosfets.
With the first mosfet, I played with R5, testing the delay, with things going smoothly. Then I loaded
it until mosfet failure, which was 8A or so. Okay, good to know. I replace the mosfet.

With R5 turned low, I apply 120v ac to my two power supplys, one for the motor, and the 12v supply
for the device. I roll the pot on to the halfway mark, wait a few moments, and it spins up the motor. Fine.
Then, while leaving the R5 at the new setting, I switch everything off (yank the plug out of the wall),
pause for 30 seconds, then plug back in, expecting a short delay, then a mild ramp up to speed...
but instead, the mosfet blows, the motor runs away at full speed.

Mind, in this test, the motor was not loaded. At the half setting on R5 on an unloaded motor, it only
draws one amp, running about 3/4 speed.

upload_2017-2-13_21-1-2.jpeg

I have three IRF640 mosfets remaining. I feel a little like "Flight of the Phoenix" over here, down to my last starting
cartridges...
 
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