One of the common phrases I live by is it ain't what folks know that gets them in trouble it's what they know that ain't so ( courtesy of Mark Twain). I just ran into that big time with a MOSFET. I always thought MOSFETs had a linear region it was just very very hard to reach, apparently not. I verified this to myself with this experiment, which failed miserably. Oh well, back to the drawing board.

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basically using a 10 turn pot I was unable to hit anything other than a one or a zero on the output.

 

If your schematic reflects the device polarity and how it's connected, the body diode would always be conducting.

The ohmic region only covers a small range of drain-source voltages before the device goes into saturation:



This info was from an undergraduate paper at UTAR by Gabriel Gan Kok Wai in 2011.

 

Change R2 from 100k to 10k or 1k you will see the linear region change.

 

Try adding some negative feedback. Connect the top end of the pot to the drain instead of VDD (Vcc in your diagram).

 

I seem to always have problems with MOSFET polarities. I may give the 1K ohm a try.

 

I'm hoping I can find a resistive region similar to a JFet. I'm going for something between 10K ohm and 100K ohms.

 

Wendy, what is the part number of your Mosfet?
Is your Vcc positive or negative?
Your schematic shows a P-channel Mosfet that needs a negative supply.

 

I'm hoping I can find a resistive region similar to a JFet. I'm going for something between 10K ohm and 100K ohms.

What JFET are you using to get that much resistance?

 

Wendy, what is the part number of your Mosfet?
Is your Vcc positive or negative?
Your schematic shows a P-channel Mosfet that needs a negative supply.

IRF540 (I will double check tomorrow)

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corrected drawing.​

 

I'm trying very hard not to buy any Jfets. Which is why I'm trying to make do with MOSFETs.

 

A power MOSFET such as the 540 is actually many smaller FETs in parallel. I suggest you start with a more "normal" part, such as a 2N7000. IIR this has a minimum Rdson of around 5 ohms. With a 100 ohm drain resistor you should see a noticeable non-saturated region as you adjust Vgs.

ak

 

Again, I'm trying to use parts I have on hand.

i'm not sure what i am trying to do is practical or possible

 

Looking at the datasheet, you will not see a large dependence of current on Vgs until you are at 1 or more amps of drain current.

Bob

 

Here's the LTspice simulation of the Vgs versus drain current for a small N-MOSFET with a 10kΩ load.
As you can see the high transconductance of the MOSFET generates a large change in current for a small change in Vgs.
Your pot resolution is likely way too coarse to see this change, and thus just goes from 0V out to 12V with the minimum resolution change in gate voltage.

 

When I was checking out the use of MOSFETs for the first time I used this circuit and plotted this graph:

 

Normally it's far too easy to find the linear region. You know you've found it when you think it has enough gate voltage to switch it fully on, but it gets very hot and blows up.
Presumably, if you want the MOSFET in the linear region, then it is because you are making some sort of linear circuit such as an amplifier. When you apply feedback, the feedback will keep the MOSFET in its linear region.

 

Normally it's far too easy to find the linear region. You know you've found it when you think it has enough gate voltage to switch it fully on, but it gets very hot and blows up.

Ah, yes - the empirical method.

ak

 

The IRF540 is a power device. Its threshold Vgs is spec'd at 2V to 4V at a drain current of only 0.25mA. With a Vgs of 6V then its drain current is typically 70 Amps!
Such a low current of 0.25mA is when the Mosfet is almost turned off. With the very high drain load of 100k ohms then the drain current will be a little less than half at 0.12mA.

 

One of the common phrases I live by is it ain't what folks know that gets them in trouble it's what they know that ain't so ( courtesy of Mark Twain). I just ran into that big time with a MOSFET. I always thought MOSFETs had a linear region it was just very very hard to reach, apparently not. I verified this to myself with this experiment, which failed miserably. Oh well, back to the drawing board.

View attachment 255746​

basically using a 10 turn pot I was unable to hit anything other than a one or a zero on the output.

If you simply want to see, observe, measure the device's behavior over a range of gate voltages, then remove the diode. That may have some purpose in some design you have but it interferes with the goal of measuring the device's behavior, that circuit above is measuring the device's behavior when it has a forward biased diode across its D & S terminals, that's a different thing altogether.

 

When I was checking out the use of MOSFETs for the first time I used this circuit and plotted this graph:
View attachment 255778

View attachment 255779

I'd be interested to see that Dave with a diode added as Wendy has in her OP.