In a diode detector , output circuit consists of ` R = 1 M omega and C = 1 pF`. Calculate the carrier frequency it can detect.

Answer

To solve the problem of calculating the carrier frequency that a diode detector can detect, we will follow these steps:

### Step 1: Identify the values of R and C
Given:
- Resistance (R) = 1 MΩ = \(1 \times 10^6 \, \Omega\)
- Capacitance (C) = 1 pF = \(1 \times 10^{-12} \, F\)

### Step 2: Calculate the time constant (τ) of the RC circuit
The time constant (τ) is given by the product of resistance and capacitance:
\[ \tau = R \times C \]
Substituting the values:
\[ \tau = (1 \times 10^6 \, \Omega) \times (1 \times 10^{-12} \, F) = 1 \times 10^{-6} \, s \]

### Step 3: Calculate the cutoff frequency (f_c)
The cutoff frequency (f_c) can be calculated using the formula:
\[ f_c = \frac{1}{2\pi \tau} \]
Substituting the value of τ:
\[ f_c = \frac{1}{2\pi \times 1 \times 10^{-6}} \approx \frac{1}{6.2832 \times 10^{-6}} \approx 159.15 \, kHz \]

### Step 4: Determine the carrier frequency that can be detected
According to the information provided, the diode detector can detect frequencies greater than or equal to \( \frac{1}{RC} \):
\[ f_{detectable} \geq \frac{1}{RC} \]
Since we already calculated \( RC \) as \( 1 \times 10^{-6} \, s \), we can conclude that:
\[ f_{detectable} \geq 1 \, MHz \]

### Final Answer
The carrier frequency that can be detected by the diode detector is greater than or equal to \( 1 \, MHz \).

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