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 Reverse recovery of a diode

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T O P I C    R E V I E W
wasssup1990 Posted - Sep 25 2010 : 11:43:55 PM
For anyone interested, I made a simple boost converter that boosts 14V up to 60V to power a 100W light bulb. The reason why I am posting this is to show what the reverse recovery time (trr) of a silicon diode looks like on a DSO. I used a FR304 (Fast Recovery Diode) found in a modern SMPS from a kitchen refrigerator.

Channel 1 (yellow trace) is the ripple voltage across the output capacitor. Channel 2 (blue trace) is the positive going edge of the gate voltage on a RFP50N06 MOSFET. You can see as the MOSFET tries to turn on the diode needs roughly 152nS to be able to block the change in polarity across it.

Download Attachment: fr304_trr.png
9.28 KB



deltaX is the time period between the two cursor positions. deltaX is shown to be 152nS which is almost spot on with what the data sheet says.


Download Attachment: FR304_DSheet.png
60.03 KB

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wasssup1990 Posted - Sep 26 2010 : 02:39:37 AM
If you are wondering how I made the inductor take a look at this collage. I removed the ferrite core from an unused T.V.'s HV flyback transformer. The winding itself had its original iron core removed - it was from a model train power supply.

Download Attachment: BoostSMPS_resize.jpg
198.45 KB

wasssup1990 Posted - Sep 26 2010 : 01:59:12 AM
FYI,
Boost converter SMPS experiment measurements:


V(in) = 24.1V
I(in) = 470mA
V(out) = 59.2V
I(out) = 184.4mA

Efficiency = 96.38%

T(ambient) = 21.1C
T(ferrite) = 24.2C
T(winding) = 26.3C
T(heat-sink MOSFET) = 23.3C
T(MOSFET) = 23.5C
T(output capacitor) = 21.8C
Tmax(load light bulb glass) = 128C aprox

The light bulb is a 100W 240V type operating at 24.67% rated voltage. I cannot produce 240V out of my boost converter due to the 60V reversed biased Zener diode from Drain to Source in the MOSFET.

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