Designing a Diode-RC AM Demodulator: Component Values for 1MHz fc and 10kHz fm

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SUMMARY

This discussion focuses on designing a diode-RC AM demodulator with a carrier frequency (fc) of 1 MHz and a modulation frequency (fm) of 10 kHz. Key component values suggested include a capacitor (C) of 0.01 µF and a series resistor (R) of 500 ohms to optimize signal retention. The series resistor acts as a voltage divider, allowing for effective rejection of the 1 MHz component while preserving the 10 kHz signal. The circuit's performance is enhanced by ensuring the resistor across the capacitor discharges within the modulation period to avoid peak voltage detection.

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Hi everyone,

I'm working on a diode-rc demodulator (AM), but for my lab the design shows a resistor in series before the diode and the resistor and cap in parallel. I am supposed to figure out the component values, but I'm not sure what to do with the first resistor in series (a current limiter?). Here's what I have so far:

fc = 1 MHz
fm = 10kHz
C = 0.1 uF
R = 500ohms

Thanks!
 
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Looks pretty good. You have two time constants between successive 10 kHz peaks. I would also try a shorter time constant (~25 usec) depending on the depth of the am modulation. The first (series) resistor determines the charging time constant of the capacitor. This should not be too short.
Bob S
 
To fit this in with what you have been studying, maybe you could show the relevant formulas.

I think the 0.1 uF is a bit too large. Try 0.01 uF with a series 500 ohm resistor. I chose that to have 3 times as much reactance as the resistance of the series resistor at 10 KHz. This way I could retain 75% of the incoming signal as output.

The series resistor produces a voltage divider with the capacitor, so that you can reject most of the 1 MHz component while retaining as much as possible of the 10000 KHz component.

A 0.01 uF cap has a reactance of 16 ohms at 1 MHz and 1592 ohms at 10 KHz so it should give reasonable rejection of the 1 MHz component while retaining most of the 10 KHz component.

This is a series RC circuit, so you have to treat it like in AC theory, with complex numbers, but you can see that the 1 MHz component would be greatly reduced compared with the 10 KHz one. It works out that 95 % of the 10 KHz signal is retained while only less than 1 % of the 1 MHz signal is retained.

The resistor across the capacitor has to discharge the capacitor in the period of the 10 KHz modulation component. Otherwise the circuit would be working as a peak voltage detector.
So, if R * C = 1 / 10000 then R = ? (don't forget the R is in Mohms if C is in uF)
 
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