DER 286 design issue

5 posts / 0 new

We have made some samples using DER 286 design

In this we modify output from 30V,1A to 43V,700mA.

also output diode D1 is replaced by 2 parallel MUR460 of ON-semiconductor

Now in this case we are facing problem of Diode MUR460 burnout.

Please suggest us the appropriate solution...

 

The peak reverse voltage of the output rectifier will be 200 V, and the average output rectifier current will be less than 2A if the same DER286 transformer used.

 

Two paralleled MUR460's, which is ultrafast 600V, 4A diode, should not be burnout in this application if there is no layout issue, like long distance from transformer, output rectifier, and output capacitor.

 

You might want to use LQA300, which is 10A 300V schottky diode, or STTH10R04D, which is 10A 400V fast diode.

 

The diode burnout is not 100%,its about 8-10%. In burnout condition only single diode burns and is short circuited and output becomes zero.so still circuit is complete,so why there is no output?? what is the use of this output diode? As our output is 43V,700mA, so why should we use 10A diode with 300/400V.

 

If one of two paralleled diode is shorted, there would be no output because there is no output rectifier in the circuit.

 

The required voltage and current rating of output rectifier for the 43V_700mA would be 200 V and 2 A, but the required voltage rate of the diode might be higher with high leakage inductance due to transformer and layout.

 

It is recommended to measure the voltage stress of the output rectifier in the circuit, and it is better to reduce the voltage stress with proper layout and transformer design.

 

The diode burnout is not 100%,its about 8-10%. In burnout condition only single diode burns and is short circuited and output becomes zero.so still circuit is complete,so why there is no output?? what is the use of this output diode?

 

As our output is 43V,700mA, so why should we use 10A diode with 300/400V.