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Diode Connection Mode MOSFET (discuss.tchncs.de)
submitted 1 year ago* (last edited 1 year ago) by Lazyhotodogu@discuss.tchncs.de to c/askelectronics@discuss.tchncs.de

Hello! Someone suggested this "If the MOSFET is put into diode connection mode (connect the gate to the drain), you can force a small current (I recommend 1 uA–10 uA; 1 mA would probably be OK; 100 mA would result in significant power dissipation and could influence the results) and record Vds (which is equal to Vgs in this configuration).” There are just some questions i had in mind. *Is it possible to connect the gate and drain on a breadboard to put the MOSFET into diode connection mode? *How to force a small current in the drain. Is voltage divider will be enough? *Is it still possible to apply voltage source after putting it into a diode connection? Also, what is the expected measurement at the Vds terminal. If there will be no voltage source, will it be in the range of 0.1 to 1 V increase? *Why should the MOSFET be connected into a diode?

*Can a MOSFET handle low dosage for example, Cs-137 (666 keV)

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[-] opensesame11@discuss.tchncs.de 0 points 1 year ago

Yes you can absolutely breadboard it. Forcing a current is as easy as following ohm's law. Make sure there is a certain voltage across a resistor and ensure that only a negligible amount of that current is leaked elsewhere. A difference amplifier is a good way to ensure this, as long as you pay attention to the amplifier input currents.

If current regulation isn't super important, a highish voltage (say 24+ V) and a large resistor will also work because the variation of threshold voltage will be so small that the voltage across the resistor will be relatively stable.

I think there is some confusion about the word diode here. The transistor is effectively an inverting amplifier, that is that the drain voltage is reduced if the gate voltage is increased. By tying them together, they reach a stable configuration where the gate is just high enough to make the drain low enough for them to be equal. In this configuration, there are two terminals, hence the di in diode. Like a traditional diode, it has a very nonlinear voltage-current relationship. If you apply 10V to it, theoretically the current would be thousands of amperes. Practically that won't happen but you will blow up the transistor.

I don't know enough about radiation and semiconductor physics to answer your other questions but if I were you I would just build it and test it. MOSFETs and resistors are cheap and if you do have a radiating source on hand it might be easier to try and fail than to hope someone here can tell you what your part will do when exposed to conditions outside of the manufacturer recommendations.

[-] Lazyhotodogu@discuss.tchncs.de 1 points 1 year ago

Helloo!!! I've sent a message to you

this post was submitted on 24 Jun 2023
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