A silicon diode has a threshold voltage of 0.7 V. If an input voltage given by `2 sin(pi t)` is supplied to a half wave rectifier circuit using this diode, the rectified output has a peak value of

Options

A. 2 V
B. 1.4 V
C. 1.3 V
D. 0.7 V

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- The threshold voltage of the silicon diode (Vs) is given as 0.7 V.
- The input voltage (Vi) is given by the equation \( Vi = 2 \sin(\pi t) \).

### Step 2: Determine the peak value of the input voltage
The input voltage can be expressed in the standard form \( V = V_0 \sin(\omega t) \), where \( V_0 \) is the peak voltage. From the equation \( Vi = 2 \sin(\pi t) \), we can see that:
- The peak value \( V_0 = 2 \) V.

### Step 3: Calculate the rectified output voltage
In a half-wave rectifier circuit using a silicon diode, the output voltage is only present when the input voltage exceeds the threshold voltage. Therefore, the peak output voltage (Vout) can be calculated as:
- \( Vout = V_0 - Vs \)
- Substituting the values: \( Vout = 2 V - 0.7 V = 1.3 V \).

### Step 4: Conclusion
The peak value of the rectified output voltage is \( 1.3 V \).

### Final Answer
The rectified output has a peak value of **1.3 V**.

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