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I followed a guide by FesZ Electronics on biasing a JFET amplifier and came up with the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab I then hooked up to a 0.01mA 150MHz signal source and simulated. enter image description here

But it seems that it does not amplify the given signal, at least I can not measure it. Did I do something wrong with my coupling and simulation? I know that the input and output impedance are not matched but does that create such a difference?

enter image description here

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    \$\begingroup\$ Show me your expected gain calculations given the drain tied to a voltage source. \$\endgroup\$ Jun 26, 2023 at 13:05
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    \$\begingroup\$ Just in case it isn't clear; your output is connected directly to the positive supply rail via a capacitor. \$\endgroup\$
    – Andy aka
    Jun 26, 2023 at 13:25
  • \$\begingroup\$ As drawn, this is a source follower, thus, the output is across R1. The gain is slightly less than unity as a source follower. \$\endgroup\$
    – qrk
    Jun 26, 2023 at 16:37
  • \$\begingroup\$ Ups yes thank you \$\endgroup\$
    – Phönix 64
    Jun 27, 2023 at 16:32

1 Answer 1

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You have the output capacitor C2 connected to the drain which is connected directly to the positive supply, so you're not going to measure any signal.

You can connect the capacitor to the source, but you won't get any voltage gain, what you'll have is a source follower which is good for matching a high impedance input to a low impedance output.

For voltage gain you'd want to use a common source amplifier, which will need some impedance in the drain circuit, either a resistor, RF choke or LC tank circuit, and the source resistor needs to be bypassed by a capacitor.

Also a 1 pF capacitor has a reactance of 1061\$\Omega\$ at 150 MHz, so that will be a problem, imagine a voltage divider with 1061\$\Omega\$ and 50\$\Omega\$, how much of the voltage are you going to see across that 50\$\Omega\$? For a simple capacitive coupled amplifier you need to choose the capacitor values to so that their impedance at the frequency of interest is 1/10th the load impedance or less. In an RF amp at 150 MHz you wouldn't normally do that though, you would use LC or transformer matching networks on the input and output. You would also feed the bias voltage through an RF choke so your input signal isn't attenuated by it.

Another option is to use a common gate amplifier, that's a whole subject unto itself so I won't go into specifics, but it's something you can do some research on.

I played around with this circuit in LTspice and got negative gain from it. After numerous modifications I came up with this version that shows around 20dB of gain at 150 MHz. enter image description here

You can see that it's quite different from the original. It could probably be improved on further if I had more time to put into it. I just used some impedance matching calculators to get an idea of what value capacitors and inductors to use and then tweaked them in simulation.

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  • \$\begingroup\$ Thank you very very much for the example and explanation. It helped a lot and I now got a design that works for me :) \$\endgroup\$
    – Phönix 64
    Jun 27, 2023 at 16:32

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