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
- Transconductance relates the output current to input voltage.
- Voltage gain of a MOSFET is directly proportional to the transconductance and to the value of the drain resistor.
- 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.
- 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.
Best Answer
Consider this random picture of a MOSFET characteristic I took off the internet: -
This is the bare bones characteristic of a MOSFET used in a very simple circuit like this: -
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: -
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.