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?
Best Answer
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.
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.