LED Driver 76V/650mA - Diode shorted

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Hello sir,

     I need your help to solve this problem.For your reference ,Design is enclosed. 

In that design,D6 SB1A is shorted,while switchon the drilling machine and it is running in square wave inverter.



I can't tell where D6 is on the schematic because it's very small and it becomes unreadable if I zoom in. Please check first if the diode exceeds its voltage/current rating or is very close to it in normal operation. You can send a larger size schematic if you still need help.


PDF File enclosed for your reference.

One more problem is raised, it is shorted in mains Voltage also.kindly check whether I had done any mistake in this design.

Please check if the bridge rectifiers are shorted. I think there was a surge event which likely exceeded the voltage ratings of the bridge rectifiers as well as D6. We normally use 1kV rating for the bridge rectifier and 200V rating on D6.

The MOV, while rated at 325V will clamp above 700V. It won't protect the bridge from surge event. Please implement these changes and I suggest to do surge testing to verify the unit will pass.


 we are using 1KV rating bridge rectifier only,and it is normal. Only D6 is shorted.even this is happening in normal 230VAC also.Cant able to source 200V diode in our local source,Shall i try with two 100 V diode ( SB1A0 2 nos in series ) in series? 

You can try that or use a 400V ultrafast diode if available. Please re-verify that the voltage and current stress on D6 is not exceeded.


I had tried with 2Diode ( SB1A0 ) in series combination,till now i didnt face any problems.Also trying to get 400V Diode and will test it.

We had face one more problem,While working in Square wave inverter output load current is abnormally taken.So,Watts increases to double the time.The following is the result, that I had taken in our lab.

 Kindly review our design to working in both EB and Inverter.



Input AC Voltage (V)

Input  AC Current (mA)

Input AC Power (W)

Output Load Voltage (V)

Output Load Current (mA)







Square wave Inverter







LNK-PH is optimized to be used with sinusoidal input. It can still work with square wave but since the input characteristics is different, then the output current will also change so you need to adjust the IFB to get the desired load current.

 Thus, you can only have one type of input supply in order to meet one output current.

You can make it work if you can implement a circuit to detect if the input supply is operating with square wave or sine wave and control IFB accordingly. (I don't know what EB stands for, i assume it's sine wave)

Finally,Either the Driver  is used for EB or Squarewave.Then,we are using the series combination of SB1A0 Diode,till now we didnt face any problem.We have one more problem that the Driver OV Cutoff is 250V AC , But we need 285 V AC OV cutoff.Please suggest the changes in circuit.

Please adjust the values of the Vpin resistors. You can utilize the fine tuning section of the PIXLS designer for LinkSwitch-PH to optimize the values or use the formula below. Your RV1 comprise of R7, R8, and R9. RV1 is R10 on the schematic you provided. This will theoritically trigger line OV at about 257VAC. You can use the formula below to adjust the resistors.For instance, changing RV1 to 3.6Mohms will already increase the OV threshold to 300VAC. Please adjust the resistor combination to get good line regulation if necessary.


 ((112+3/RV2) * RV1 + 3.7)/1.414