Below is a schematic of an amplifier. The inductor used is a 1 μH inductor, emitter resistor is 2.2 kΩ. The bias resistors are both 200 kΩ. The capacitors are both 1 μF. I am planning to design a matching network to match it to a 50 Ω load and test it with a signal generator. How should the output impedance be calculated? I am planning to use this circuit as a buffer for a hartley oscillator.

enter image description here

Here is what I was thinking for calculating the equivalent output impedance (L is connected to ground):

enter image description here

  • \$\begingroup\$ Its connected to a 50 ohm antenna. Not a hw question. \$\endgroup\$
    – xcaliber99
    Commented May 1 at 16:26
  • \$\begingroup\$ Why does it need to be matched? Is there a long feed line to the antenna? \$\endgroup\$
    – Andy aka
    Commented May 1 at 16:44
  • \$\begingroup\$ rf. output imepdance probably != 50 ohms. therefore, matching is needed. \$\endgroup\$
    – xcaliber99
    Commented May 1 at 17:15
  • \$\begingroup\$ No it isn't. Matching is needed to prevent reflections down feed-lines and sometimes we match for maximum power transfer. \$\endgroup\$
    – Andy aka
    Commented May 1 at 18:05
  • \$\begingroup\$ rf. wavelength is small. small wavelength leads to reflection. therefore, matching is needed. \$\endgroup\$
    – xcaliber99
    Commented May 1 at 18:19

1 Answer 1


Your transistor model claims that the transistor output side is an ideal current source. The only output side impedance is inductor L. If you for some still unknown reason absolutely want that the output impedance of the amp with JUST EXACTLY THIS transistor is 50 ohms, put a 50 ohm resistor in parallel with the inductor.

Use so high inductance that inductor's reactance is much higher than 50 ohms. Or insert a parallel capacitor with the inductor, select the capacitance so that together they are a resonant circuit at the operating frequency. The remaining output impedance is the inserted resistance in parallel with the resonant circuit.

I wouldn't waste RF power to the inserted resistor, but that's your way, selected by you when you decided the unreaalistic equivalent circuit of the transistor and started to want 50 ohm output impedance. I would worry how to make my load such that it doesn't reflect the signal back and how to make my amp to deliver what's expected to be got to the load. If the load reflects anyway, the amp should be made robust enough to tolerate the reflected wave. I would also worry how to either keep the amplification free of distortion or how to achieve a high energy efficiency and how the filter out the distortion frequencies. And all this with practical transistors.

  • \$\begingroup\$ Thank you for the suggestion. I will try this. \$\endgroup\$
    – xcaliber99
    Commented May 1 at 21:32
  • \$\begingroup\$ I added a capacitor in parallel with the inductor. The output was reduced by half. Why is this? \$\endgroup\$
    – xcaliber99
    Commented May 1 at 23:16
  • \$\begingroup\$ If you were running into a 50 Ω load then it might indicate that you've got a 2:1 potential divider and your source impedance is now 50 Ω - which is what you wanted! \$\endgroup\$
    – Transistor
    Commented May 1 at 23:26
  • \$\begingroup\$ @xcaliber99 maybe you only say "the amp should have output impedance =50 ohm" but actually you expect that the amp should work with 50 ohm load somehow well. You imagined that amp output impedance = 50 ohm implies "working somehow well" which you couldn't describe. Now you got what you asked and wonder why the output voltage dropped 50%. Quite a typical result of making a classic XY-question (=asking something good looking, but obscure in terms of solving the actual problem) \$\endgroup\$ Commented May 2 at 0:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.