1
\$\begingroup\$

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:

demonstration

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:

enter image description here

Is this the correct method to use?.

\$\endgroup\$

1 Answer 1

3
\$\begingroup\$

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}\$.

\$\endgroup\$
5
  • \$\begingroup\$ 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) \$\endgroup\$
    – ir.imad
    Oct 22, 2015 at 1:07
  • \$\begingroup\$ 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 ?. \$\endgroup\$
    – Bilal
    Oct 22, 2015 at 13:21
  • \$\begingroup\$ 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. \$\endgroup\$
    – ir.imad
    Oct 24, 2015 at 21:45
  • \$\begingroup\$ 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 \$\endgroup\$
    – ir.imad
    Oct 24, 2015 at 21:56
  • \$\begingroup\$ @ir.imad For 800MHz you need a different time constant. Same circuit. \$\endgroup\$
    – user207421
    Jul 12, 2016 at 0:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.