mosfet diode

I watched picture using mosfet like diode in this website and try it.
But it didn't work

https://www.onsemi.com/pub/Collateral/2N7000-D.PDF

I buy 2n7000 n mosfet transistor and try voltage 1.5v.
I short drain and gate. I add 1k register in circuit.
and get 0.11V. (When I inverse 1.5v battery inverse, I get 0.9V)

Why it isn't work?

Can you make a simple schematic (pen(cil) and paper will do fine!)? Schematic says more then a thousand words :slight_smile:

I fail to see how the design presented on that website would do anything other than burn out the MOSFET if you tried to pass much current through it. If you look at the next few posts in the thread, it's people telling the OP that it won't work.

In order to turn on the MOSFET, you need Vgs to be high enough (datasheet will tell you what this needs to be; it starts turning on an Vgs(th), but to be fully on, it should be at least the lowest voltage for which Rds(on) is specified 4.5v for a typical logic level MOSFET). But if the FET is on, then resistance between drain and source would be very low, so Vgs wouldn't be high enough to turn the FET on.

What would happen with that circuit is that the voltage across the FET would settle to a voltage where the FET is part-way on, and the high Rds would result in the MOSFET heating up and burning out.

A MOSFET can be used as an ideal diode, but that circuit will not achieve it - and you need active control driving the gate. You can't just wire it up as simply as you would wire a diode (think about it, if this worked, who would use a diode?)

With JFETS, this configuration was often used to create current limiting devices.

But those JFETs were depletion mode devices operating at pinch-off and the gate was connected to source, not drain, which is utterly different.

JFETs are nice, too. If I connect gate to source,
I get IDSS (see data sheet for range of current).
If I connect gate to drain, I get rDS(on), minimum
D to S resistance!
Herb