I have made the mosfet circuit as I showed in a previous post.

I have made the mosfet circuit as I showed in a previous post. It ain’t working…
Some help please from some of you smart electronic engineers.

The issue is that the load is permanently switched on, no matter what…

I have an optoisolator 4N25 that takes the pid signal from my Ramps controller, and switches the gate on the SQM120N04 Mosfet.

The switching side is 24V, with a separate dedicated power supply to that of the ramps.

I have a zener diode 10V to drive the gate on the mosfet.

What am I doing wrong?

+Peter van der Walt​, @raykholo ​, @Ishaan_Gov@Stephen_Beeson ​, anyone else?

I mixed to the load Terminals, didn’t I??

The terminal on the photo markes as PS - should be swapped with the load… Right?

First try removing the resistor that pulls the mosfet gate high (the one that goes from the opto to the gate). See if it keeps turned on. With “removing” I mean pulling it out of the circuit, leaving an open circuit there, just leaving the transistor emitter unconnected. That’s just to test if that makes the mosfet turn off.

From the diagram you display, the zener diode is not doing anything. Its inverted. Lined side always is on the ground side.

Also I am not sure about what exactly you are trying to do with it. Gates and switches need a diode like that to drain the gate to get it to shut off. I don’t think that it is wired correctly​ for that.

@Alex_Hayden lol no thats just correct. zener with kathode to ground would short the gate to 0.5-0.7 volts… its just that voltage devide of 1/2 which is just not needed. just remove the upper resistor. also use smaller resistor like 1k. that bipolar transistor in the optocoupler may just not let pass so few current.

@VolksTrieb No. It’s not only a voltage divider: if you remove the upper resistor, you’ll have too much current flowing through the transistor and the zener diode. You also need the lower resistor to pull the gate low to turn it off.

@VolksTrieb ​ I haven’t made a switched circuit in some time. And had to seek help when I did it too. I didn’t know how to get it working. My uncle had to show me. My bad. Glad you could help.

@Gato1 what is the zenerdiode for anyways?

Got 24V?

Oh yeah overread that. Well just put a voltageregulator of 12-15V into the circuits supplyrail… Its easier and also would just work fine. A zenerdiode isnt that fast so remove that crap :wink: youlld also just need a pulldownresistor which is a bit smaller for fast switching…

@VolksTrieb The MOSFET gate tolerates 20v, the divider should be enough to keep it below that voltage. So it’s there only for protection.

@Gato1 yeah i know. Ive just overread the 24V supply. Well a voltageregulator at the supply would just work fine. No need for silly zeners in mid-fast-high-current switching applications… You would only do it for protection. Not for creating a voltage for switching.

@VolksTrieb you don’t need too fast switching, it’s just a heating bed, and the controller boards usually regulate temperature by turning them on/off at a few hertz.

@Gato1 yeah… But it would still create unneccesary heat on the mosfet. Just dont do it that way. Its not the fine way.

For a standard fet thatll be 0.5ms switching time for full gatevoltage. Naww man thats no the way to do it. Its not like impossible. But the thing is he wants to do it himself because its fun and he wants to understand things. So dont just tell him it will work or its fine. Tell him the alternatives…

@VolksTrieb Exactly, it’s there for protection only. The voltage divider halves the voltage, so the gate should be getting 12v (enough to turn it fully on). The zener isn’t absolutely necessary, but is there “just in case”. Using a voltage regulator just to get an adequate voltage for the gate is an overkill.

@Gato1 no its not. Its creating a fixed voltage. Thats common practice. Thats why you got supply rails for different things in electronics. Its not overkill, its more stable and precise

The only way it would be on all the time is the opto-isolator is on all the time, or the mosfet is bad or wired incorrectly. Measure the gate voltage with the opto on then off.

@VolksTrieb According to the specs:

Total gate charge @10v = 200nC
Which means its capacitance is = 200nC/10V = 20nF

The fall time with an impedance of 4.7KΩ is: 20nF x 4.7KΩ = 94µs, or around 0.1ms

The rise time is twice as fast, because the impedance is halved.