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
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.
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