Freewheeling Diode Sizing

[sturriff]

I'm posting this on behalf of a colleague. We're working on a high frequency (20kHz) PWM application and are looking at connecting a free wheeling diode to the circuit. We need some guidance on how to calculate the average forward current and maximum current that this diode might see. Also, if anyone has any tips on calculating the heat dissipation by the diode, that would be helpful as well. Thanks.

 

[ScottyUK]

Hi Sturriff-

Give us a clue as to the circuit configuration and switching strategy of the PWM switches. Makes it easier to visualise what is going on. Ratings etc aren't that important for what you are asking, if it is a secret...


Scotty

 

[jbartos]

Suggestion: Visit

http://www.fairchildsemi.com/an/AN/AN-9014.pdf

etc. for more info. The more accurate answer would need the more accurate original posting

 

[electricuwe]

The freewheeling diode will have to dissipate the following losses (Duty cycle of switching device):

Conduction losses:
(1-D)*Iload*Uf
Switching losses:
Qrr*Ud*fswitch

 

[Marke]

Hello sturriff

The freewheeling diode would generally be across an inductor or a switching element controlling an inductor and is there to provide a discharge path for the colapsing magnetic field after the driving source has been removed. (Switched OFF) The collapsing field will work to maintain the current flow at a constant level and so the current flowing through the diode will be equal to the current through the inductor before the switching element was opened.
There is loss due to this forward current, plus there is additional losses due to the stored charge in the diode. This is dependant on the diode design and at 20KHz can be significant. You need to look at the storage energy of the diode and ensure that the diode can cope with that loss at the frequency that you are operating it.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[avigodner]

I suppose that we are talking about H-bridge or 3-phase bridge topology.

High (low) side free-wheeling, or anti-parallel, diode contributes to turn-on switching loss of its low (high) side transistor counterpart via reverse recovery phenomenon - at turn-on transistor sees current pulse from the diode in addition to taking over motor current. Slow diode (slow recovery) causes more transistor turn-on loss.

Power MOSFETs have intrinsic slow body diode (this is not IGBT case). Therefore, there were suggestions to use two additional diodes to reduce switching loss - serial to neutralize intrinsic diode and parallel fast recovery one. This is probably not that practical as turn-on switching loss may be controlled to a certain extent by turn-on slope command set by MOSFET driver circuit.

 

[OperaHouse]

I read the post and what came to mind is you are driving a DC motor with PWM. If that is the case, you would be better off putting an inductor in series with the motor with the diode on the driver side.

 

[jbartos]

Eng-tip: Reference:
1. Muhammad H. Rashid "Power Electronics Handbook," Academic Press, 2001, page 19, Section: 2.6 Series and Parallel Connection of Power Diodes.