0
\$\begingroup\$

enter image description here

I am trying to see whether I can turn on 28 V voltage rail from a 5 V input signal using an N-channel MOSFET (NMOS). I am using IRF530N for now which has a threshold voltage of 4 V. I just modeled it using microcap and it seems as long as RL is big enough it'll work as long as I connect the return/gnd of the signal to the source of the IRF530N. Seems you are driving the source of the NMOS with two different supplies and I don't think it's recommended to use this way. I know PMOS is used for high-side switching and the V source is connected to power supply for that case but just wanted to know whether this is a possible configuration. Any feedback on this circuit?

\$\endgroup\$
3
  • \$\begingroup\$ Yes, this solution is e.g. used in a slow switching context with photovoltaic optocouplers. \$\endgroup\$
    – Jens
    Commented Apr 9 at 13:49
  • \$\begingroup\$ Ground is just a reference. Any node in your circuit can be ground, although it's a useful convenience when chosen wisely. Your FET isn't connected to ground, so it sees only what is applied to its terminals. This design is fine, since the gate voltage is reliably related to the source voltage. \$\endgroup\$ Commented Apr 9 at 14:32
  • 1
    \$\begingroup\$ "I am using IRF530N for now which has a threshold voltage of 4 V" - Which means it's unsuited for 5V gate drive. You need 10V Vgs to get the rated Rdson. \$\endgroup\$
    – marcelm
    Commented Apr 9 at 17:48

1 Answer 1

1
\$\begingroup\$

Seems you are driving the source of the NMOS with two different supplies and I don't think it's recommended to use this way.

What you did is basically "translation" or "elevation": you have an independent 5V voltage source, and its reference (GND/return or whatever) is connected to the MOSFET source pin (Practically, the 1R gate-stopper resistor is still in series with the gate so I assumed the stopper resistor is connected between the gate and the positive of the 5V source.).

In practice, if the 5V source is isolated by default then it's totally fine to drive the MOSFET as shown in your circuit because you are free to connect this isolated source's reference to anywhere you want. But if the 5V source is normally GND-referenced (i.e. shares the same GND with 28V source) by default then connecting the 5V source's reference the MOSFET's source will short the load (10R resistor in your circuit).

If you want to drive an NMOS as a high-side switch, several methods exist:

  • Bootstrapping
  • Using gate-drive transformers
  • Using dedicated gate driver ICs
  • Using boosters (e.g. charge-pump).

The first two are useful when there's non-zero-frequency switching (e.g. PWM). If you want to drive the load indefinitely (e.g. no periodic on-offs or no pre-set on-time for the load) then these methods may not work or may not be practical.

The third one is usually a bit more advanced application of the first one.

The last one is basically generation of a higher voltage from a lower input voltage but this requires constant independent switching i.e. you need a separate oscillator to drive the pump and you can use the output of the pump which is going to be a DC and can share the same ground as 28V source (There are also ICs available for this job). Example: You can get 1.5 times the input voltage with a charge pump (~40V), and you can switch that voltage to the MOSFET's gate directly.


Buck converters usually use NMOS as the main switch which works as a high-side switch (the difference is the load is a more complex circuit there). To drive it, bootstrapping is one of the most common methods today. Bootstrapping is useful because it's basically a switching application.

\$\endgroup\$
1
  • \$\begingroup\$ I need to process this information. Thank you very much \$\endgroup\$
    – nmr
    Commented Apr 9 at 16:52

Not the answer you're looking for? Browse other questions tagged or ask your own question.