I'm trying to come up with a way to do overcurrent protection for the following circuit: My first thought here was to monitor the increase in voltage drop over one of the mosfets (M1) and via a zener (D5) (that sets the cutoff level) activate a transistor that drains the gate. The zener is a 1.8V one and combined with the 1 voltage drop of D4 i thought it would cutoff at 2.8V which would be 3-4A for a irf740. But it's not cutting off anything.. any thoughts on how this function could be implemented into this circuit?

 

i thought it would cutoff at 2.8V which would be 3-4A for a irf740

That circuit looks familiar.

How did you arrive at that value?
It has a ON-resistance of 0.55Ω so 2.8V would give 5.1A.

One problem is that you have to monitor for both positive and negative voltages across the MOSFETs since they carry AC current.

Also D4 is backwards.

And there should be a resistor in series with Q3's base to limit the current.

What's the purpose of R6?

 

You might want to check out this 10¢ voltage monitor chip because it is supposed to be accurate and "decisive" (an engineering term?).

RHOM BU4818G-TR

Note the link to the datasheet on the left side of the page.

 

@crutschow oh how I was hoping to see Wally reply to this! Thanks again for your excellent replies to my other questions, simply priceless.

I arrived at it through eyeing the diagram below. But I was apparently very bad at following straight lines

I couldn't get pspice to simulate without R6 and got the suggestion to connect "unconnected" pins to gnd via a high resistance. That got the simulation running fine (not the overcurrent protection part obviously)

Both positive and negative voltages, true.. hmm, can you see a solution to that?