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I am trying to design an isolated LED driver with primary side sensing. I want my output voltage to be 48-56 V and the current to be 350 mA. The current and the voltage are controlled by current sense resistors and feedback resistors which have correct values according to my calculations.

The problem is that every time I start the circuit the LED panel lights up correctly but after some time the output rectifying diode (in my case ES2J) burns up. The ratings of the diode are up to the marks which is 1A forward current and 600V blocking voltage which should be more than enough for this application. I've tried a single diode and two diodes in parallel but the results are the same. What am I doing wrong here?Circuit Schematic

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    \$\begingroup\$ which diode? can you display the values of the resistors instead of the dimensions. No body can make out the design by looking at the present picture. \$\endgroup\$
    – User323693
    Jan 19, 2017 at 7:26
  • \$\begingroup\$ May bad Umar, the diodes I am talking about are D9 and D10 on the output side. The resisters marked as RU and RL are there for voltage feedback and the resistors RS are there to sense and set current. (The values are 240k for RU and about 17k for combined RL and 0.75R for combined current sense) The thing is that I've made this exactly as the reference design the IC manufacturer gave me but the diodes are still not working which shouldn't happen unless I have made a mistake. \$\endgroup\$
    – Rutwij M
    Jan 19, 2017 at 11:27
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    \$\begingroup\$ Parallel diodes may not work well. Silicon does not like to share current. A larger diode should work better, and maybe a faster one. \$\endgroup\$ Jun 28, 2018 at 17:27

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Your Diode is burning up due to speed because your volts and amps are OK .Paralleling them wont do much here proving that it is not a peak current issue .The Flyback is cheap but does have high peak currents .If you experimentally place 2 diodes in series and you find that things are a little cooler then you are on the right track .Getting a fast diode with better reverse recovery times will help .The better diode will also make the switching device run cooler .When using cheap normal Si fast diodes the lower voltage types are faster .This observation can by justified by semiconductor physics .Do your homework and find what voltage you need for the output diode .It wont be anywhere near 600V .They make 200V shottkies .Increasing the gate resistor will cool the diode and warm up the Fet and help EMC .

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  • \$\begingroup\$ Thank you Autistic for answering! I reckoned that I do not need such high voltage diodes but I've used this diodes because this circuit is based on a reference design which uses these diodes for the same circuit(this is for 18W and 350mA). I think they used it because it might be in their existing inventory. Since their design is working I was wondering why this isn't. My transformer has a slightly different turn ratio(2.54 instead of 2.44) but I have compensated it by adjusting the feedback resistor values. Do you think could it be because of slightly wrong resistor values for feedback? \$\endgroup\$
    – Rutwij M
    Jan 19, 2017 at 11:36
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Temperature rise \$ΔT[°K]\$ (or °C) above ambient depends peak current I and duty cycle, d diode thermal resistance \$R_{thA}\$ which includes junction to case and case to board and board to Ambient.

The diode Rs must rise as current drops in half so sharing 2 diodes in this case does not make a significant difference.

The spec states \$R_{thA}\$ = 50°K/W for pad size 50mm² for each terminal.

Thus \$ΔT[°K]=d(I^2Rs+(V_fI))*R_{thA}\$ This is what I believe is your temp rise.

The datasheet show Vf =0.65V @I=1mA, I computed Rs at 1A assuming a duty cycle of 1/3 with Ipk/avg=3

Either use diodes with lower Vf,Rs RthA or bigger pad sizes.or your cap size needs to be reduce the Ipk.

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    \$\begingroup\$ p.s. although the OP is perhaps long gone, others may have a similar issue. \$\endgroup\$ Jul 24, 2017 at 1:10

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