1
$\begingroup$

40 MHz crystal Oscillator for VHF Fm transmitter

I am working on a project designing a 40 MHz radio transmitter, but I need some clarifications. I need to know how to calculate its output impedance.

I know how to calculate common collector's output impedance, but this circuit has 2 capacitors (C2,C3=100pF) in series which are connected to the emitter.

  • What will be the output impedance, and how will I calculate it? I need the specific formula for it if I change those capacitors somehow.
  • I also want to know how to calculate its output voltage, because I will connect this circuit with a buffer to amplify its power.
$\endgroup$
  • 1
    $\begingroup$ The output of the oscillator must drive a substantially higher impedance to avoid disturbing the oscillator. If the load impedance is not sufficiently large, the load may prevent the oscillator from starting and the load may affect the oscillator frequency. Therefore, the relatively low output impedance of the oscillator circuit is not important to know. $\endgroup$ – Brian K1LI Nov 19 at 13:08
  • 1
    $\begingroup$ @BrianK1LI I think you should post what you put in the comment as an answer, and then delete the comment. $\endgroup$ – rclocher3 Nov 19 at 16:24
2
$\begingroup$

The output impedance is not that important: the (lowest) allowed load impedance is important. And the assumption is right: from the output impedance it is possible to calculate the effect of load on the output voltage. But be aware of frequency deviation as a function of load variation, and also eventual stop of oscillation with serious load.

This is not a linear circuit that can be described only with its output impedance. Insert an impedance buffer when using the signal. And indeed: the effect of the buffer rises the same question, but at a different level.

Your question is the output impedance. It is an emitter follower with a resonant circuit. Put it in a simulator. I guess it will be a few hundred Ohm's or lower. PA0FSB

| improve this answer | |
$\endgroup$
  • $\begingroup$ The actual output voltage depends highly on the crystal activity, or better series loss resistance at resonance. In the past I did simulations with equivalent circuits in Simetrix (also the free version is ok) and I remember output voltages in the range 1 to 3 volts p-p. And 1 dB attenuation of the amplitude when loading with 1000 Ohm. And finally a feeling: the 100 pF is (for 40 MHz) on the larg side: try also 40 pF. $\endgroup$ – F. Sessink Nov 18 at 11:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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