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I'm trying to build a simple class a amplifier using a mosfet. I have read few articles about how to do it, but I'm just confused now. I'm not sure how to select resistor values. The articles I've read show how to calculate this values, but the requirement is conduction parameter "k" which is not in documentation for transistor I have(IRFZ44N). I tried calculating this value from equation k = Id/(Vgs-Vth)^2 after measuring current and voltages for different gate voltage, but the "k" I get isn't constant.

I also tried selecting resistor values without calculations and with that I have the most success, however the output signal is not amplified, it is diminished.

I chose Rd so that Vd is Vcc/2. R1 and R2 set gate voltage to 4.8V, because gate threshold voltage is 3V and sawtooth amplitude is 500mV(I can set Vg lower, but it doesn't change much). I'm not sure about Rs. The 8.2 Ohm resistor is a "speaker".

Why is the signal not amplified? I realise that the answer might be complex, so I would be also grateful for some good study material books/tutorials etc..

enter image description here

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  • \$\begingroup\$ electronics.stackexchange.com/questions/321904/… \$\endgroup\$
    – G36
    Jun 3, 2020 at 16:21
  • \$\begingroup\$ @G36 This answer is more or less how I chose my resistor values. Still it doesn't answer all of my questions, for one I still don't know if conduction parameter is constant or not. And still it doesn't address the issue of diminished output wave, or I'm not seeing it. \$\endgroup\$
    – xstmpx
    Jun 3, 2020 at 16:30
  • \$\begingroup\$ electronics.stackexchange.com/questions/474723/… \$\endgroup\$
    – G36
    Jun 3, 2020 at 16:33
  • \$\begingroup\$ Kp is constant if MOSFET is a saturation region and we ignore the channel length modulation. \$\endgroup\$
    – G36
    Jun 3, 2020 at 16:44

1 Answer 1

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Why is the signal not amplified? I realise that the answer might be complex

Maybe not too complex...

Class A amplifiers have to have a standing bias current that is of the same order as the load current. See the note I've added on your picture: -

enter image description here

So the 560 ohm resistor forms a potential divider with your load (8.2 ohms) and the biggest peak voltage that can be produced is: -

$$\text{12 volts}\cdot\dfrac{8.2 \text{ ohms}}{8.2 \text{ ohms} + 560 \text{ ohms}} = \text{173 mV}$$

I'll leave that to sink in.

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  • \$\begingroup\$ Well that explains why the signal isn't amplified, but now if I want to have drain voltage set to 6V i need to keep that 560 ohm resistor, right? What do you suggest I should do to keep drain voltage at 6V and create bigger output voltage? \$\endgroup\$
    – xstmpx
    Jun 3, 2020 at 15:30
  • \$\begingroup\$ What if you used an 8.2 ohm resistor and adjusted the gate-source bias voltage to give 6 volts at the drain under quiescent conditions? This is the price of simplicity and class A - high quiescent current. \$\endgroup\$
    – Andy aka
    Jun 3, 2020 at 17:55

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