Question

In a diode AM-detector, the output circuit consists of $\mathrm{R}=1\mathrm{k}\Omega \mathrm{and}\mathrm{C}=10\mathrm{pF}.$ A carrier signal of $100\mathrm{kHz}$is to be detected. Is it good?

• A. Yes
• B. No
• C. Information is not sufficient
• D. None of the above

Answer 1

The correct option is B

No

Step 1: Given data:

In a diode AM-detector, the resistance of the output circuit is given as $\mathrm{R}=1\mathrm{k}\Omega$

Capacitance of the output circuit is given as $\mathrm{C}=10\mathrm{pF}=10\times {10}^{-12}\mathrm{F}$

The frequency of the carrier signal given to be detected ${\mathrm{V}}_{\mathrm{c}}=100\mathrm{kHz}$

Step 2: Formula applied:

The expression given for better demodulation can be given as:

$\left(\frac{1}{{\mathrm{V}}_{\mathrm{c}}}\right)<<\mathrm{RC}$

Step 3: Checking the condition for demodulation:

$\frac{1}{{\mathrm{V}}_{\mathrm{c}}}=\frac{1}{100\mathrm{KHz}}={10}^{-5}\mathrm{Hz}$

Also,

$\mathrm{RC}={10}^3\times {10}^{-11}={10}^{-8}$

From the given data, we can see that

$\left(\frac{1}{{\mathrm{V}}_{\mathrm{c}}}\right)>\mathrm{RC}$

Hence, the given condition is not good for demodulation

Thus, as $\left(\frac{1}{{\mathrm{V}}_{\mathrm{c}}}\right)<<\mathrm{RC}$ , does not holds good for the given condition, hence it is not good. So, option B is correct.