Fet for turning on high voltage.

[jmgelba]

Unfortunately I do need to switch the high side. I thought the whole point of a logic level fet was to be able to turn on higher voltages than those seen at the gate.

Using your example, if the nfet was connected to the pfet gate and ground, I'd still need more than 24v at the gate as the pull up is pulling the drain to 24v, right?

[HenrikOlsson]

Hi,

I thought the whole point of a logic level fet was to be able to turn on higher voltages than those seen at the gate.

Yes and no. The idea with a logic level MOSFET is to allow it to turn on and fully saturate with a Vgs (voltage between gate and source) of ~5V when a "standard" MOSFET requires ~10V.

Using your example, if the nfet was connected to the pfet gate and ground, I'd still need more than 24v at the gate as the pull up is pulling the drain to 24v, right?

No. The N-channel drain is pulled up to 24V but its source is permanently tied to GND so when the gate gets 5V from the PIC (or whatever) you get a Vgs (voltage between gate and source) of 5V and the N-channel MOSFET will turn on (check the datasheet of the MOSFET to make sure 5V is enough). When the N-channel MOSFET turns on it pulls the gate of P-channel to GND making the voltage between its gate and source -24V which will turn IT on.

You need to check the maximum allowed Vgs of the P-channel MOSFET, if it's less than 24V in this case you can't switch the gate directly to GND. You could use a simple voltage divider with a much lower total resistance than the pullup. Switch the bottom of the voltage divider to GND with the NPN or N-channel MOSFET.


If you put the N-channel on the high side then its drain will be at 24V permanently, the source will basically be pulled down to GND thru the sensors internal circuitry. So far everything looks fine. Then you apply 5V to the gate of the MOSFET and it begins to conduct. Now the voltage at the source starts to climb towards 24V but the voltage at the gate is still only 5V. For the N-channel to keep conducting the voltage at the gate will have to "rise with" the voltage at the source so it always stays Vgs (voltage between gate and source) volts above the voltage at the source.

There are discrete solutions and integrated circuits which does this for you. But a P-channel and a NPN is probably easier.

/Henrik.