UC3842 SMPS short circuit protection

saqib

New member
i designed 230V AC to 12v 5A SMPS based on UC3842 forward converter, which works fine but i want output short circuit or overcurrent protection through pin 3 but as my current sense resistor is on the primary side so primary side equivalent of 5A never reaches enough to give 1V to pin 3 and thus short circuit or overcurrent never works and heats up the mosfet (IRF740), i also tried a Schmitt trigger based on 2 pnp transistors which works fine on low input voltage (24V DC ) but it only hiccups the circuit instead on tripping. my only goal is to trip the supply on output short circuit or current >5A. How should i do that? will appreciate any help from you....
voltage divider of 1.8k and 8k to pin 3 is not included in the circuit.
it is my first ever question on any forum so forgive my shortcomings..
 

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Silvio

Active member
i designed 230V AC to 12v 5A SMPS based on UC3842 forward converter, which works fine but i want output short circuit or overcurrent protection through pin 3 but as my current sense resistor is on the primary side so primary side equivalent of 5A never reaches enough to give 1V to pin 3 and thus short circuit or overcurrent never works and heats up the mosfet (IRF740), i also tried a Schmitt trigger based on 2 pnp transistors which works fine on low input voltage (24V DC ) but it only hiccups the circuit instead on tripping. my only goal is to trip the supply on output short circuit or current >5A. How should i do that? will appreciate any help from you....
voltage divider of 1.8k and 8k to pin 3 is not included in the circuit.
it is my first ever question on any forum so forgive my shortcomings..
 

Silvio

Active member
Hi Saqib, I think you are missing something here. Let us consider your forward converter.
So the power to be absorbed by this forward converter is 12v @ 5A so the power is W=V x I = 60watts. Taking account of some losses we must add around 20% in the primary to compensate for the losses so the power at the primary side is 60w + 20% = 72W

Now considering what current is reflected on the primary .
The input voltage is 320v more or less so to absorb 72w at this voltage then we calculate 72w / 320v = 0.225A
So when an output of 60w is absorbed at the output then there is 0.225A passing from the mosfet in the primary.
It seems that a 0.2 ohm resistor is too small to sense this current adequately so if your current sense voltage at the wanted output current of 5A at the secondary then we can make a calculation for the required sense resistor

The trigger voltage is 1v so using ohms law V=IR then V/I = R so 1v / 0.225A = 4.44R (4.7R)
now calculating the wattage of this resistor W= I^2 x R = 0.225 x 0.225 x 4.7R = 0.23794 W so a half watt resistor is good.

Just a last note for you. It is advisable to make a preset to adjust the current setting instead of the divider resistors tied at the reference voltage. this way you can have some adjustment to the wanted current.

I hope this helps you understand better :)
Regards Silvio
 

saqib

New member
Thanks a lot for quick reply. :). i was just trying to avoid increasing sense resistance preoccupied by the idea of heat losses without going into I2R calculations first. OK i should try this 4.7R resistance now.
regarding the divider resistors, i didn't use it as i am not sure that a preset would really be able to adjust current in case of UC3842. would it?
 

saqib

New member
I have another concern that mosfet IRF740 is heating a lot even for a primary current of almost 0.21A (which i calculated from voltage drop of 48mV across sense resistor which is still 0.22R). although this mosfet is rated for 10A continuous current. as for now i am only testing this supply so not using any heat sink for the moment because primary current is too low. is it that i am driving mosfet directly from 3842 output? does this chip really provide 1A output current? or is it the lack of heat sink ? snubber problem? what am i missing this time? forgot to tell that mosfet doesn't heat up even a little when there is no load at output except an indication LED. then why only 0.21A current heats it up a lot?
 
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