# How to calculate R and C for AM Demodulation envelope detection?

How can I calculate the value of $R$ and $C$ for an AM envelope detection circuit.

Knowing that the carrier wave is $20\mathrm{KHz}$, $1\mathrm{Vpp}$ sine wave, and the message signal is a $32\mathrm{Hz}$, $2\mathrm{Vpp}$ triangular wave with a $3\mathrm{V}$ DC offset: I used the low pass fillter equation ($f_c=\frac{1}{2\pi R C}$) to calculate $R_1$ and $C_1$ and it was fine ~100 duplicate. The result was this: Is this the correct method to use?.

the cutoff frequency of RC filter is:

$$f_0 = \frac{1}{2\pi R C}$$

you make :

1. $f_0 << f_c$ (carrier frequency)
2. $f_0 >> f_d$ (maximum data frequency)

The optimal value:

$$f_0 = \sqrt{\left(f_c \times f_d\right)}$$

Example:

$f_c = 20\mathrm{kHz}$, $f_d = 32\mathrm{Hz}$, then:

$$f_0 = \sqrt{\left(20000 \times 32\right)} = 800 \mathrm{Hz}$$

Which can be achieved with for example $C = 39\mathrm{nF}$ and $R=5.1\mathrm{k\Omega}$.

• I suppose the data signal is sine wave, if it is square wave you must calculate for 5 fd or more ==> f0 = sqrt(fc * 5 * fd) Oct 22, 2015 at 1:07
• ok thank you , so we get f0 = 800Mhz , so what equation did you use to calculate r=5.1k and c=39nf the Rc filter equation or the T=r*c (time constant) and T=1/f ?. Oct 22, 2015 at 13:21
• Time const. T = R * C ; cutoff frequency f0 = 1/ (2 PI R C) where [PI=3.141]. you talk above about 20 kHz and 32 Hz. Oct 24, 2015 at 21:45
• Time const. T = R * C ; cutoff frequency f0 = 1/ (2 PI R C) where [PI=3.141]. you talk above about 20 kHz and 32 Hz. for 800 MHz you need more complicated circuits . sorry, i am not a specialist in RF applications Oct 24, 2015 at 21:56
• @ir.imad For 800MHz you need a different time constant. Same circuit. Jul 12, 2016 at 0:15