MOSFET doesn't put out desired voltage

Question

I have an IRFB7430 which I want to use to switch 12V. There will be minimal load on the MOSFET, since it is only used to switch a signal. I know it is not its intended use, but it's what I have lying around.

The signal on the Gate is only 5V, which should leave the MOSFET with a higher-than-intended R_DSon, but since I only want to switch a voltage, not a current, I thought it might not be a problem.

However, the output is only around 3.6V, and even if I switch with a 12V signal (which should get the R_DSon to single digit mOhm), I only get 10.5V on the Drain.

I have a 26kOhm pull down resistor on the Gate, and a current limiting resistor of 220 Ohm. Removing either of these did not bring significant changes. +12V is connected to Drain and I am measuring between ground and Source.

I guess there is something in the design of the MOSFET which I don't understand and it is its regular behavior. What am I missing?

Answer 1

Are you by any chance trying to use it as a high-side switch?

I believe so - in which case you're actually using it as a source-follower instead of a switch. This is confirmed by your measurements, where Vs = Vg-1.5V approx. It's not switching, but operating in linear mode and dissipating a lot of power, supplying a fairly accurate Vg-1.5V to the load.

You have 3 options:
(1) generate a gate voltage ABOVE the supply (18 to 20V) to switch ON, and 0V to switch off.
(2) use it on the low side of the load with 12V gate voltage. This is the simplest - a conventional low side switch - and offers the best performance.
(3) replace it with a PMOS FET as a high side switch. Then 12V turns it off and 0V turns it on. But it might be difficult to find a PMOS with such a low ON resistance.

Answer 2

+12V is connected to Drain and I am measuring between ground and Source

To turn on a regular N channel MOSFET, the gate voltage has to be higher than the source voltage.

So, if you put 5 volts on the gate, the MOSFET will turn on when the source is at 0 volts for sure. It will still usually turn on when the source is at 1 or 2 volts but is becoming weaker in that the on-resistance is beginning to rise. With 3 or 4 volts on the source, the MOSFET is hardly turned on at all.

This is what I recommend for a 0 volt tied load: -

See also this SE.EE Q & A

Answer 3

If all you want to do is amplify the voltage for a signal from 5 V to 12 V rather than switch power to something on and off then you can do this:

^{simulate this circuit – Schematic created using CircuitLab}

This has a couple of big drawbacks over Andy aka's answer.

1. It will invert the sense of the signal, a low input will give a high output.
2. With a high output you have a series resistor in line with the signal, depending on what you need to drive this could be a big issue. The lower R1 the less of an issue this is.
3. When a low output you are dumping power through R1/M1. This is both a waste of power and depending on how 'on' M1 is could cause a lot of power loss and heat in M1. The larger R1 is the less of an issue this is.
4. When switching from a low to high output there will be a ramp up time based on R1 and the output capacitance. This may or may not be an issue depending on the load and value of R1.

You can avoid the inverting issue by putting two of these stages in a row. In that situation R1 for the first stage can be nice and large (say 10k) since there is no significant load.

The resistor in the final stage and how practical this approach is depends a lot on the load being driven. Calculate a maximum for R1 based on the expected current draw and maximum acceptable voltage drop at the load and then check if that maximum is too low in terms of acceptable efficiency and M1 power dissipation.

If you look at Andy aka's answer you will notice that it is using this exact circuit to generate the 12V signal needed to drive the gate of the p-channel part.