How to make high side switch using N channel mosfet? Supply (drain) voltage is 24V and so its source voltage is also goes to 24V when mosfet is fully turn on. I'm not using any PWM to gate of the mosfet. How to turn on the mosfet continuously?

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    \$\begingroup\$ What is your required load current? \$\endgroup\$
    – Jens
    Commented Dec 6, 2023 at 11:25
  • \$\begingroup\$ Maximum load current will be 10A. \$\endgroup\$ Commented Dec 6, 2023 at 11:38
  • \$\begingroup\$ There are integrated charge pump circuits for this use case or do you need a solution with discrete components? \$\endgroup\$
    – Jens
    Commented Dec 6, 2023 at 11:45
  • \$\begingroup\$ thanks for reply, i don't have any charge pump circuit. yes need a solution with discrete components. \$\endgroup\$ Commented Dec 6, 2023 at 11:58
  • \$\begingroup\$ The 9V battery can ensure a supplying. For on/off signal transfer use an optocoupler. You can use a small transformer or small isolated smps instead battery. \$\endgroup\$ Commented Dec 6, 2023 at 12:29

3 Answers 3


A simple solution is to use photovoltaic driver, like VOM1271, from Vishay. You need to supply 10 mA to the diode, from 24 V or any output, and typical open circuit voltage is 8.4 V, but only 15 uA. In case your MOSFET needs more gate voltage, you can use 2 in serial connection. The switching on will be slow, but for low speed application is Ok. enter image description here


To fully enhance ("turn on") an n-channel power MOSFET, the gate must be more positive than the source. Many devices are specified for their minimum Rdson with Vgs = 10 V, and a max allowed Vgs of 20 V. Since you are not driving the FET with a PWM signal and you don't say anything about a maximum turn-on time, there are several ways to do this.

One is to use a gate-driver IC. This has a charge pump in it, a form of switching power supply that is a low-current boost converter. The output can be either added to the supply voltage, so in a 24 V circuit the gate drive signal would be 34 V (-ish), or isolated. In that case, the driver IC could be running on 12 V, but the output can be floating on 400 V. This works well in things like motor drives and high-power switching power supplies.

Depending on the application and the rest of the circuit, you can grow your own charge pump driver from scratch

Another approach is to use an optocoupler that has a small, isolated, floating voltage output. The drive signal is relatively slow, especially with large FETs. The input capacitance of a large FET can be something like 10 nF. Driven by a current source of 15 uA, the turn-on time would be around 6 or 7 milliseconds. That would be a real problem if the FET were being driven with a 1 kHz square wave,, but for a static switch application it should be fine. A possible issue is the power level required. While the static current into a MOSFET gate is almost 0, The current to drive the LED in this type of device is often 10mA to 20 mA. If the control signal is coming from something like a CMOS gate that is rated for only 3 or 4 mA, you will have to add a driver transistor.


With a small transformer in flyback mode you can provide an isolated gate voltage. The important aspect is here to provide fast rising and falling gate signals to keep the transition time between on and off short.

The shown circuit has rise and fall times below 40 us, which is a bit faster than the typical photovoltaic couplers. This may be helpful with 10 A load current.

The combination of V5, R1, M3 and V4 just simulates a fast PWM, enabled and disabled with 1 kHz, coming from a uC. So we can see the switching transitions in the simulator.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ OP says I'm not using any PWM to gate of the mosfet. How to turn on the mosfet continuously? Also M1 will probably die because of the spikes. Better to tie one end of the pri to ground and drive from a totem-pole as the pri will never be floating. \$\endgroup\$ Commented Dec 6, 2023 at 14:29
  • \$\begingroup\$ @RohatKılıç If the 100 kHz PWM is present the FET is continuously on. \$\endgroup\$
    – Jens
    Commented Dec 6, 2023 at 14:33
  • \$\begingroup\$ I know how the circuit works. What I'm saying is that your circuit requires a pulse generator yet the OP doesn't have any, as far as I understood, otherwise he/she could have gone for a charge pump or something similar already. \$\endgroup\$ Commented Dec 6, 2023 at 14:53
  • \$\begingroup\$ The pulse generator for this can be as simple as a two-transistor oscillator or a TLC555, so I think this is a valid answer. Rohat has a point that you may want a snubber to protect M1 though. \$\endgroup\$
    – Hearth
    Commented Dec 6, 2023 at 14:57
  • \$\begingroup\$ Well, I understood the "not using PWM" as a hint, that no fast load switching is required, not that no PWM would be available at all. And yes, a snubber network on the primary side should be there. \$\endgroup\$
    – Jens
    Commented Dec 6, 2023 at 15:02

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