I am currently learning how to design/build a crystal radio receiver, transmitter and have a question for the 180 degrees phase shift for the carrier amplifier/oscillator system.

I commonly see this network setup: C1 to ground in series with the crystal and then another C2 to ground. The other networks have been left out for simplicity. I am not seeing how this works. If the RF signal is AC why doesn't the AC signal just go straight to ground? It's the exact same setup you see when placing an external crystal on a microcontroller. With the crystal in between two (usually 10pF) capacitors that are grounded. How does this work?

  • 1
    \$\begingroup\$ It's a Pierce oscillator. Look it up. \$\endgroup\$ Mar 22, 2013 at 4:07
  • \$\begingroup\$ Hey Leon thanks for the answer. I am (recently) familiar with the Pierce oscillator. What I do not understand is how the phase shift occurs without the AC signal going straight to ground. \$\endgroup\$
    – HelloWorld
    Mar 22, 2013 at 4:32
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    \$\begingroup\$ Please add a circuit digram to your question. \$\endgroup\$
    – jippie
    Mar 22, 2013 at 7:28
  • \$\begingroup\$ Capacitors are not simply shorts to AC. They have a frequency dependent impedance. If you do AC analysis of a circuit by replacing caps with shorts and it makes no sense then the assumption is wrong for at least some of the caps. \$\endgroup\$
    – Kaz
    May 22, 2013 at 2:31

2 Answers 2


C1 to ground in series with the crystal and then another C2 to ground.

It's not obvious exactly what your circuit looks like, but to clarify my answer, I'll show you what imagine it looks like based on your description:


simulate this circuit – Schematic created using CircuitLab

Incidentally, this circuit, with C1 in series with the crystal, is not what I think of as an especially common oscillator circuit.

If the RF signal is AC why doesnt the AC signal just go straight to ground?

The very simple description of a capacitor is "at low frequencies it's an open circuit and at high frequencies it's a short circuit." And this is a reasonable model for a lot of cases. But saying you have an AC signal doesn't mean you have a low or a high frequency, it just means the frequency is not exactly 0 Hz. And what's happening here is that the frequency you're operating at is neither a "low" frequency or a "high" frequency, it's somewhere in between.

Between the two extremes you need to look at the impedance model of the capacitor:

\$Z = \dfrac{1}{j2\pi{}fC}\$

This also tells you what is meant by "low" and "high" frequencies. When the frequency is low enough that Z is so large it doesn't affect your circuit differently than an open circuit would, that's a "low" frequency. When the frequency is high enough that it doesn't affect your circuit differently than a short circuit would, that's a "high" frequency.

  • \$\begingroup\$ It is a pierce oscillator but your response makes the most sense. Thank you. \$\endgroup\$
    – HelloWorld
    Jun 16, 2013 at 22:53

It's a Pierce oscillator. From the Wikipedia article: "The crystal in combination with C1 and C2 forms a pi network band-pass filter, which provides a 180 degree phase shift".

  • \$\begingroup\$ Here's the link: en.wikipedia.org/wiki/Pierce_oscillator "To understand the operation, note that at the frequency of oscillation, the crystal appears inductive. Thus, the crystal can be considered a large, high Q inductor. The combination of the 180 degree phase shift (i.e. inverting gain) from the pi network, and the negative gain from the inverter, results in a positive loop gain (positive feedback)." \$\endgroup\$
    – rdtsc
    May 25, 2015 at 19:16

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