Replacing a BJT with a FET

ian field

Joined Oct 27, 2012
6,536
Yup. That's what I meant by "tricky." Most of the time it seems like 100Ω is adequate to squelch the RF oscillations.
The tricky bit is biasing it.

The same idea as collector derived base bias can be used, but as the gate doesn't draw current you have to add a second resistor to make a voltage divider.

The method I like is; put a current sensing resistor in the source lead, dimensioned to bias a B/E junction at the desired operating point current. The current sensing transistor has a collector resistor, which also feeds bias to the gate (preferably) via a bootstrap resistor.

Decoupling is best done at the collector of the current sensing transistor - which sort of makes the bootstrap resistor a required part.
 

Thread Starter

jaydnul

Joined Apr 2, 2015
175
For a BJT I can look at the data sheet and roughly estimate the gain I'm going to get using the listed beta range, but what about a FET? Say I want to approximately predict what my drain current will be when I have 5V on the gate, do I just use the transconductance like this: \(I_d=(5)g_m\) ?
 

OBW0549

Joined Mar 2, 2015
3,566
For a BJT I can look at the data sheet and roughly estimate the gain I'm going to get using the listed beta range, but what about a FET? Say I want to approximately predict what my drain current will be when I have 5V on the gate, do I just use the transconductance like this: \(I_d=(5)g_m\) ?
No, for that you pretty much have to go by the drain current vs. gate voltage curves shown on the MOSFET data sheet, so that you can take the gate threshold voltage into account. Once you have an operating point established, then you might be able to roughly estimate the gain from gm. If the manufacturer doesn't provide that information, you're out of luck.

Trying to build amplifiers from MOSFETs, resistors and caps the same way you would do with BJTs is a great way to drive yourself insane, because of the huge unit-to-unit variation in gm and Vgs(th); it's very difficult to get a reliably predictable operating point.

All of the times I've used a MOSFET in linear mode, it's been as the control element in a precision voltage-to-current converter with a high-gain opamp driving the MOSFET's gate. There, so long as the opamp can provide adequate gate drive voltage to get the drain current you want, the individual MOSFET characteristics don't matter much.
 

Thread Starter

jaydnul

Joined Apr 2, 2015
175
Ok, one more question. In the Id vs Vgs plot, it gives the value that Vds was at in the test. But wouldn't a changing Id cause the voltage drop to change across the resistance that's between the power source and the drain?
 

OBW0549

Joined Mar 2, 2015
3,566
Ok, one more question. In the Id vs Vgs plot, it gives the value that Vds was at in the test. But wouldn't a changing Id cause the voltage drop to change across the resistance that's between the power source and the drain?
Yes, it will, and that in turn will change the drain-to-source voltage of the MOSFET, and that will have some effect on the drain current (very similar to the Early Effect in BJTs), at least slightly.

Along with the Id vs. Vgs plot, you will usually find another chart, probably labeled "Output Characteristics" that gives a family of Id vs. Vds curves, one curve for each value of Vgs over some range. These output characteristic charts usually show that for small to moderate levels of drain current, and given sufficient Vds, drain current depends mostly on gate voltage.
 

Thread Starter

jaydnul

Joined Apr 2, 2015
175
So wouldn't \(V_{ds} < (V_{gs}-V_{th})\) when the FET is completely opened up since there is only a negligible voltage drop across the drain-source? If so, why doesn't the FET go into triode mode when \(V_{ds}\) dips below \(V_{gs}\)?
 

crutschow

Joined Mar 14, 2008
34,281
Interesting... the article says that Mr Lilienfeld "is credited with the first patents on the field-effect transistor (1925)", but the transistor itself was not developed until 1947!
It wasn't until the basics of semiconductor physics was understood by Shockley and company at Bell labs (and Lilenfeld's patent expired) that a practical working device was developed.
The development of the MOSFET was further delayed until the problem of charge trapping at the gate insulator-semiconductor interface was solved.
These traps would greatly affect the threshold voltage of the MOSFETs as they were operated.
I believe it was determined that this was due to oxygen traps in that interface and subsequent fabrication changes mostly eliminated those traps.
 

Thread Starter

jaydnul

Joined Apr 2, 2015
175

Let's say \(V_{th}=1V\). Starting at \(V_g=0V, V_d=15V\) and the FET is cutoff. Now we instantaneously increase \(V_g to 2V\). Then \(V_d=15 - I_dR_d\), but to calculate Id we need to know the mode of operation since we will have two different equations for the drain current. BUT to know the mode of operation we need to know the current through Rd to find Vd.

So essentially it looks like we need Id to calculate Vd but we need Vd to KNOW how to calculate Id (which equation). Where am I going wrong?
 
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