Question

An a.c. voltage of peak value 20 V is connected in series with a silicon diode and a load resistance of 500 \(\Omega\). The forward resistance of the diode is 10 \(\Omega\) and the resistive voltage is 0.7 V. Find the peak current through the diode and peak voltage across the load.

Hint:

For an a.c. circuit with a diode, subtract the resistive voltage across the diode from the peak supply voltage to find the voltage across the load. Then, apply Ohm's law to calculate the current and voltage across the load.

Answer 1

Given: - Peak voltage of the a.c. source, \( V_{\text{peak}} = 20 \, \text{V} \) - Load resistance, \( R_L = 500 \, \Omega \) - Forward resistance of the diode, \( R_f = 10 \, \Omega \) - Resistive voltage across the diode, \( V_d = 0.7 \, \text{V} \) Step 1: Calculate the total resistance in the circuit.
The total resistance in the circuit is the sum of the forward resistance of the diode and the load resistance:
\[ R_{\text{total}} = R_L + R_f = 500 \, \Omega + 10 \, \Omega = 510 \, \Omega \] Step 2: Calculate the peak current through the diode.
The peak current through the diode can be calculated using Ohm's law:
\[ I_{\text{peak}} = \frac{V_{\text{peak}} - V_d}{R_{\text{total}}} \] Substituting the given values:
\[ I_{\text{peak}} = \frac{20 \, \text{V} - 0.7 \, \text{V}}{510 \, \Omega} = \frac{19.3 \, \text{V}}{510 \, \Omega} \approx 0.0379 \, \text{A} \] Thus, the peak current through the diode is approximately \( 0.0379 \, \text{A} \).
Step 3: Calculate the peak voltage across the load.
The peak voltage across the load can be calculated using Ohm’s law:
\[ V_{\text{L peak}} = I_{\text{peak}} \times R_L \] Substituting the values:
\[ V_{\text{L peak}} = 0.0379 \, \text{A} \times 500 \, \Omega \approx 18.95 \, \text{V} \] Thus, the peak voltage across the load is approximately \( 18.95 \, \text{V} \).