Isolated flyback current sense

isolated current sensing

This concerns a secondary current regulated flyback....

on first page of this App Note please see the circuit diagram at the bottom of page:-

http://www.powerint.com/sites/default/files/PDFFiles/di168.pdf

-there is a current sense resistor on the secondary side of 0.16 Ohms (total)

-the secondary current is supposed to be 1.67A maximum.

1...So how would such a small value resistor ever get Q2 to turn ON?

2. Also, if Q2 did turn ON.....Q3 would just turn it off again, and i cannot see what is the point of that ?

Any reader can answer the points 1 and 2 ?

...I would be most grateful.

The current sense resistor is not connected directly between the base and emitter of Q2. Q2 is biased through R15 and R13. Without Q3, Q2 would just tun on, turning on harder Q4, which would turn it on even harder, and so on. The circuit would behave like a thyristor. The role of Q3 is to make sure that Q2 does not turn on without any current flowing, by keeping its base at a value just below 0.6V. So Q2 is off, but its base voltage is not zero, its positive.

But as the current increases, at 1.67A the voltage drop across the current sense resistor begins to add to the voltage maintained by Q3 and so Q2 turns on. Since its base voltage was already positive, it only takes 0.25V to turn it on. This is what makes the circuit a low-drop current sense.

Personally I probably would not have built it that way. I mean, if you are going to use that many parts, you can build a differential amplifier and do a better job. But we are here to analyze the circuit, not criticize the designer.

thankyou so much for responding so helpfully,

here is the circuit again:-



...i read your interesting answer.....so i take it you are saying that Q3 , Q2 and Q4 never in fact fully turn off ?

That's right, they conduct, but they are not saturated.