Understanding part of a crystal radio schematic

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Circuits123

Joined Dec 7, 2012
64
I'm trying to understand a basic crystal radio schematic I found in a book. I redrew part of the schematic and inserted it below. To the left, I omitted the antenna and tuning circuit. The diode below is labeled as the detector. The description in the book says the capacitor (ignore the value) "shorts the RF part of the signal to ground, leaving only audio-frequency energy at the output." My question is why does the capacitor do that? My understanding is that both RF and AF are a/c so why would one get shorted to ground and not the other?

Screenshot - 11192016 - 06:09:06 PM.png
 

MrChips

Joined Oct 2, 2009
19,914
You are omitting one part of the equation.

The output has impedance. In a crystal radio, the output is usually a high-impedance headset. Ideally the headset would have an impedance of 10k-20kΩ. The RC combination forms a low-pass filter with a certain frequency response. You want the filter to pass frequencies below 5kHz and reject the RF which is normally greater than 500kHz. Hence there is a lot of tolerance here.

10μF is way too large for this application. C should be in the order of 1nF.

In reality you don't need the capacitor because you cannot hear the signal above 20kHz and there will always be some capacitance in the circuit.

In fact, you don't even need the tuning coil and capacitor on the front end. The simplest crystal radio you can build consists of a single Ge diode between the antenna and ground with the high-impedance headset placed across the diode.

The trouble is there is no way of selecting the radio station. Works great if there is only one AM/MW radio station in your neck of the woods.
 

Ramussons

Joined May 3, 2013
798
I'm trying to understand a basic crystal radio schematic I found in a book. I redrew part of the schematic and inserted it below. To the left, I omitted the antenna and tuning circuit. The diode below is labeled as the detector. The description in the book says the capacitor (ignore the value) "shorts the RF part of the signal to ground, leaving only audio-frequency energy at the output." My question is why does the capacitor do that? My understanding is that both RF and AF are a/c so why would one get shorted to ground and not the other?

View attachment 115617
The "resistance" of a capacitor is inversely proportional to the frequency (Xc = 1/(2*pi*F*C) - which means that the "resistance" at Audio Frequency ( KHz) will 1000 times the "resistance" at RF ( MHz) because RF is 1000 times AF.
 
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