Can a low voltage MOSFET switch higher voltage?

[uoficowboy]

Hi - can anybody tell me if it is possible to use a lower voltage MOSFET to switch the low side of a high impedance load that is connected to high voltage?

For example, let's say you have a 1M load. The high side of it is connected to 100V, while the low side is connected to the drain of 10V N-channel MOSFET. The source of the FET is grounded.

If it were a low impedance load, I'd expect the MOSFET to be destroyed. But what about this case where it is high impedance? The maximum power the 100V supply can deliver to the MOSFET is 10mW - so as long as the package can handle that power dissipation, will it survive?

If not, are there any other ways to use a low voltage MOSFET to switch the low side of a higher voltage load?

Thanks so much!!!

 

[keith1200rs]

You might not destroy your MOSFET, but you load will never switch off. The transistor will break down at some voltage above 10V, so you load will still have nearly 90V across it.

Keith

 

[umesh49]

I think leakage current off the FET should matter to decide about breakdown. (100V - Ileak*Road) should always be less than break down votlage of FET.

Umesh.

 

[divsec]

respected SIR.
how to implement analog Linear Feed Back Shift Register(LFSR).i have already done with the digital part.
how to generate analog pest waveforms using LFSR.?
i have attached the block diagram of analog Test Pattern Generator.please help me to implement the design using VHDL

 

[uoficowboy]

You might not destroy your MOSFET, but you load will never switch off. The transistor will break down at some voltage above 10V, so you load will still have nearly 90V across it.

Keith

Now that I'm thinking about it - you are certainly correct! But it's still very useful to know that the MOSFET wouldn't be destroyed!

 

[keith1200rs]

Power might kill it, if not current. 90V at 10mA is 0.9W. Not a lot, but it could be too much if the transistor wasn't expecting it e.g. a SOT23.

Keith.

 

[erikl]

... it's still very useful to know that the MOSFET wouldn't be destroyed!

You still have to be careful to protect the gate: when you switch off the MOSFET, its drain voltage ramps up from - say - 1V to 10V (or more, depending on the drain-bulk breakthrough voltage). Via the drain-gate capacitance, this ramp is fed back to the gate. You should delimit the surge to the max. allowed gate-source voltage, e.g. by using a low impedance source to control the gate, or by a delimiting diode to the (low voltage) VDD.

Added after 3 minutes:

90V at 10mA is 0.9W.
Keith.

Hey Keith: 100V via 1MΩ !

 

[keith1200rs]

90V at 10mA is 0.9W.
Keith.

Hey Keith: 100V via 1MΩ !

I wasn't working it out for 1MΩ, but my calculation is wrong anyway - the 90V would be in the load not the transistor!

Keith.