I'm working on controlling the supply of an amplifier board with 5V 2.4A power in a manner which tries to eliminate noise from the speaker. I asked a few professors at my university and they said that an optocoupler would be the best way. Since I need a higher amperage, I also need a MOSFET or transistor or some sort.

The following circuit is the result of my research and testing in Livewire. The idea is that this will then be my first custom PCB which we'll etch in the lab and then incorporate into my refurbishing of an old vintage radio.

Circuit Diagram

Q1: IRL520 (or similar) MOSFET

Q2: 4N35 Optocoupler

CN1: Line to 5V 2.4A power supply

CN2: Cables to On/Off switch

CN3: Power line to amplifier, needs at least 5V 2A

CN4: Control lines to 3.3V microcontroller, (Rasp0W or Arduino)

R1: 1k Ohms to make sure the gate isn't triggered unless microcontroller is HIGH

R2: ~250 Ohms to make sure the optocoupler's LED (Vf~=1.5V) only gets 10mA off a 3.3V line.

When I simulate the circuit by:

  • adding a 5V source at CN1

  • shorting the terminals in CN2

  • adding a lamp/led at CN3

  • adding a 3.3V source at CN4

The circuit remains closed when the microcontroller is LOW, and when it is pulled high, the voltage to the gate of the MOSFET reads 3.3V (wire marked with purple). However, it doesn't indicate that the light bulb at CN3 lights up.

Simulated Circuit Diagram

My line of thinking is that Livewire doesn't know that I'm using a logic-level MOSFET, therefore thinks that I need at least 5V to trigger the gate.

Will this circuit be able to deliver 5V 2.4A to the amplifer at CN3? Are the parts adequate or are there better options out there? Thanks!


3 Answers 3


This will not work. You need a p-channel MOSFET (logic-level) to switch the high side with low drop. Your circuit will drop at Vgs(on) at some current that depends on the load. Also you should increase R3 considerably since your phototransistor is barely turning on. Try 10K or 20K (or, much better, work it out from the datasheet worst-case values since you are presumably an engineering student).


simulate this circuit – Schematic created using CircuitLab

You may be able to get rid of the optocoupler and just use an ordinary NPN BJT such as a 2N4401 since the control signal is now ground-referenced.

  • \$\begingroup\$ I am indeed an engineering student, however I'm in civil engineering and have never taken anything electric or electronic beyond the basic introduction in Physics 101. So based on your comment, I have redrawn the circuit. If I'm understanding correctly, when the opto is lit up from the microcontroller, the gate voltage is pulled to 0, because Vgs is -5V. However in the simulator I'm using there is no difference between the opto being lit or not, so I'm not sure if the circuit is bad or it's the simulator. new MOSFET - IRF9630 \$\endgroup\$
    – daniel_l
    Aug 25, 2017 at 1:55
  • \$\begingroup\$ CN2 is shorted. Other than that, looks good. \$\endgroup\$
    – AnalogKid
    Aug 25, 2017 at 2:16

You have the wrong coupler

In order to use a N-channel FET here, you need to be able to pull its gate above its source, which your 4N35's transistor output has no ability to do. What you need instead of the 4N35 is a device called a photovoltaic isolator that replaces the transistor output with a tiny solar cell that can actually generate the voltage needed when the LED shines on it. The resulting circuit is shown below in its simplest form.


simulate this circuit – Schematic created using CircuitLab

(Of course, the PMOS high side switch will work for a power switching application, but the NMOS approach is somewhat more efficient/versatile.)


"Will this circuit be able to deliver 5V 2.4A to the amplifer at CN3?"

No. A logic level MOSFET will not solve the basic design flaw. For the output you want, the source has to be at +5 V. But the maximum voltage available for the gate is +5 V, so the Vgs is 0 V. You are using the transistor as a follower, so the gate must be above the source. For a logic level FET, you want 10 V on the gate to turn it "on". That is 10 V above GND, which is 5 V above the source.

The solution is to change to a p-channel FET. With the source tied to the +5 V input and the drain tied to the output, pulling the gate to GND will pull the drain up to the source (minus the Rdson losses).


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