This is a fundamental question which I am struggling to answer after familiarizing myself with the MOSFET and analyzing different circuits containing MOSFETS.

In terms of GDS channels, what enables a MOSFET to be used as an amplifier?

I know that

  1. Transconductance relates the output current to input voltage.
  2. Voltage gain of a MOSFET is directly proportional to the transconductance and to the value of the drain resistor.
  3. Gradually increasing the positive gate-source voltage VGS, the field effect begins to enhance the channel regions conductivity and there becomes a point where the channel starts to to conduct.
  4. We can control how the MOSFET operates by “enhancing” its conductive channel between the source and drain regions.

However, I am unable to form a logical analysis as to what really goes on in the Gate, Drain and Source channels to enable a MOSFET to be used as an amplifier.

  • 1
    \$\begingroup\$ Your point 3 makes little sense for an amplifier. In an amplifier the \$V_{GS}\$ will be varied such that the output current will vary around a certain average value. So the point where there is no channel is irrelevant for an amplifier. That is only relevant when using a MOSFET as a switch. \$\endgroup\$ Feb 6, 2018 at 15:27
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    \$\begingroup\$ Also realize that the MOSFET by itself cannot work as an amplifier. It needs to be in a circuit and biased in a certain way for amplification to happen. So understanding how the MOSFET works as an amplifier also means understanding how it works in a circuit. \$\endgroup\$ Feb 6, 2018 at 15:30
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    \$\begingroup\$ It also needs a power supply - it doesn't magically amplify a weak signal to a strong signal - the strong signal is ultimately sourced from the power supply and all the MOSFET does is control how energy is taken from that power supply and delivered to a load. \$\endgroup\$
    – Andy aka
    Feb 6, 2018 at 15:32
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    \$\begingroup\$ @Bimpelrekkie, thanks for the clarification. I do realize the MOSFET needs to be biased in a certain way and that we can control how the MOSFET operates by “enhancing” its conductive channel between the source and drain regions, but I do not yet understand if this is the main reason for signal amplification. \$\endgroup\$
    – rrz0
    Feb 6, 2018 at 15:33
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    \$\begingroup\$ "I do not yet understand if this is the main reason for signal amplification" Yes, the gate voltage modulates the channel, giving you transconductance. The transconductance times the drain impedance gives you a voltage. that results in a change in output voltage for a change in input voltage. If the amplifier is designed correctly the output delta will be bigger than the input delta and you will have voltage gain. \$\endgroup\$
    – John D
    Feb 6, 2018 at 15:38

1 Answer 1


Consider this random picture of a MOSFET characteristic I took off the internet: -

enter image description here

This is the bare bones characteristic of a MOSFET used in a very simple circuit like this: -

enter image description here

You set a gate-source voltage (\$V_{GS}\$) and plot what the drain current is for various values of \$V_{DS}\$.

Now consider what happens if you put a resistor in series with the drain and used a fixed 40 volt power supply feeding the resistor and drain. If the MOSFET is fully off there will be no current through that drain resistor and you get point A (below).

If the drain resistance was 10 ohms you would get 20 volts across it when 2 amps passed. This allows you to draw a load line on the first picture: -

enter image description here

So, for this particular set-up with a 10 ohm drain resistor (see load line in red) and a \$V_{GS}\$ of 5 volts, the \$V_{DS}\$ would be about 23 volts and, for a \$V_{GS}\$ of 6 volts, \$V_{DS}\$ would be about 13 volts.

Can you see that if you had an input signal that was a sinewave going between 5 volts (bottom of sine) and 6 volts (top of sine), the output would be also a sinewave changing between a trough of 13 volts and a peak of 23 volts.

That is a signal voltage gain of 10.

Ignoring DC offsets and just concentrating on the output signal, it has an RMS voltage of 3.536 volts and an RMS current of 0.7071 i.e. a power output of 2.5 watts. It's not an amazing performance but you have generated an output signal power of 2.5 watts by varying the input voltage at the gate by 1 volt p-p.

The input signal power needed to do this is a few tens of microwatts. You have made a massive power gain and this is the important thing for such things as audio power amplifiers.

  • \$\begingroup\$ Would you like to explain @Rrz0 why you have unaccepted this answer. Maybe you want clarification about something? \$\endgroup\$
    – Andy aka
    Feb 9, 2018 at 17:57

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