DC motor freewheeling diode dimensioning.

[Odin]

Hi all!

I have a DC motor that I'm trying to control with a PIC and a FET, doing pwm. The motor is a two coil design. 24 V, consuming about 4 A with no load. In "real life" it takes about 15 A at full load. (the markings on the motor says 24 V / 20 A)

I've connected the coils in paralell, from +24V to drain on my FET. A IRF540 logic level type. Then a resistor between gate and the PIC output pin. (PIC16F870 with hardware pwm)

My problem is that the freewheeling diode for the motor keeps blowing up, or at least gets very hot. I first tried a 1N4007, and switching freq at approx 20 KHz. The literally exploded! Then tried a bigger diode, some 3 A type. It breaks in a couple of seconds.

Next try was to lower the switching freq to 244 Hz. Then it works pretty OK, but the diode still gets warm. And that's at 50% dc, 2,5 A and no load. The FET seems to be doing fine. Only a tiny bit warm. But a cold running FET doesn't help much, when my diodes are glowing...

So, how do I solve this problem? Bigger diode? Different circuit? I'd prefereably keep the switching freq above audible, but that seems impossible here. What is actually happening here? Wheen the FET close, the motor generates a negative spike, that the diode "kills", right? Or does the motor become a generator, sending the same amount of current trough the diode that it gets from the FET?
I'd might consider a h-bridge later, to change direction easy. How are the values for the FETs calculated? Low Rds is fine, but doesn't help much if the diode inside "scrubs off" 0,5 V when the FET isn't conducting. Please enlighten me!

Odin

 

[ljcox]

Odin,
It would help us if you posted a circuit. Either a photo or one drawn in a package such as PowerPoint and converted to a .gif.

DC motors do become a generator when the supply is switched off.

But the polarity will be the same as when the power is on so the diode should still be reverse biassed. So I don't know why it is hot. A circuit would help.

I don't understand your question

"How are the values for the FETs calculated? Low Rds is fine, but doesn't help much if the diode inside "scrubs off" 0,5 V when the FET isn't conducting"

The internal diode should be reverse biassed.

Len

 

[Styx]

THe freewheel diodrs have to be rated the same as the FETs

I have a DC motor that I'm trying to control with a PIC and a FET, doing pwm. The motor is a two coil design. 24 V, consuming about 4 A with no load. In "real life" it takes about 15 A at full load. (the markings on the motor says 24 V / 20 A)

...

My problem is that the freewheeling diode for the motor keeps blowing up, or at least gets very hot. I first tried a 1N4007, and switching freq at approx 20 KHz. The literally exploded! Then tried a bigger diode, some 3 A type. It breaks in a couple of seconds.

So basically you are putting in 3A diode in a situation where they will experience 4A !!!! I am not surpised they blew. You need 15A rated diode.
I mean your FET is ok (33A max rating) so you are you rating the diode so low

Think about what the diodes have to do. They have to allow the mochine winding current to continue to flow while the FET's are off- the widings are inductors. Thus they will be subjected to the full machine winding current

 

[Odin]

Odin,
It would help us if you posted a circuit. Either a photo or one drawn in a package such as PowerPoint and converted to a .gif.

Well, here is my circuit, drawn in Paint! (I really have to find a program to draw circuits...)

DC motors do become a generator when the supply is switched off.

But the polarity will be the same as when the power is on so the diode should still be reverse biassed.

I see, then there is not the generated current that causes the heat, but back emf.

I don't understand your question

"How are the values for the FETs calculated? Low Rds is fine, but doesn't help much if the diode inside "scrubs off" 0,5 V when the FET isn't conducting"

What I meant is that in a h-bridge, the FETs are not making so much heat (voltage loss) when they are conducting. Rds multiplied with the current (U=Rds*I) With a good FET we are talking maybe 0,1 V or less. But the diode junction of the FET loose 0,5 V or so. And my freewheeling diode looses 0,7 V. Guess I have to find a bigger diode, and mount it on a heatsink.

 

[ljcox]

Is the motor an armature type? If so, the the quote below is only true when the motor is starting. When it is running, the EMF generated by the armature is just a bit less than the applied voltage so the current is proportinal to the difference in these voltages divided by the resistance.

Think about what the diodes have to do. They have to allow the mochine winding current to continue to flow while the FET's are off- the widings are inductors. Thus they will be subjected to the full machine winding current

 

[spuffock]

diode doesnt turn off fast enough. There will be huge spikes of current in a 1n4001 when the fet switches on, before the diode turns off. Use a big schottky.

 

[Styx]

Is the motor an armature type? If so, the the quote below is only true when the motor is starting. When it is running, the EMF generated by the armature is just a bit less than the applied voltage so the current is proportinal to the difference in these voltages divided by the resistance.

Think about what the diodes have to do. They have to allow the mochine winding current to continue to flow while the FET's are off- the widings are inductors. Thus they will be subjected to the full machine winding current

Err no.
Look at his sketch in his last reply.
When the FET is turned ON, current will build up in the inductive load (ie the machine). Now he says it is rated at

4A no load
15A loaded
20A max

So lets say he switched the FET on and had 15A flowing through the FET. He switches the FET off. Since the load is an inductor the current MUST carry on flowing, the only route available is via the diode (that is why it is there). Thus the diode MUST take all of the load current - in this example 15A, since a 3A diode was used - it will fail.

Yes slow switching of the diode will also have an effect but it is still looking like a severly under-rated diode

 

[ljcox]

[/quote]

Err no.
Look at his sketch in his last reply.
When the FET is turned ON, current will build up in the inductive load (ie the machine). Now he says it is rated at

4A no load
15A loaded
20A max

So lets say he switched the FET on and had 15A flowing through the FET. He switches the FET off. Since the load is an inductor the current MUST carry on flowing, the only route available is via the diode (that is why it is there). Thus the diode MUST take all of the load current - in this example 15A, since a 3A diode was used - it will fail.

Yes slow switching of the diode will also have an effect but it is still looking like a severly under-rated diode[/quote]

You must be right as there must be an inductive spike, but I don't know why.

I am familiar with switching inductance and the fact that the current goes through the diode when the FET switches off. But motors generate an EMF when the FET turns off.

I once a designed a pulse width motor controller that used the EMF of the motor as the feedback signal. I used a diode to protect the switching transistor and had no problem. The EMF is in the same direction as the applied voltage. Any book on motor theory will show this.

(':?:')