Question

An ac voltage V_i = 12 sin(100 πt)V is applied between points A and B in a network of two ideal diodes and three resistors as shown in figure. During the positive half-cycle of the input voltage V_i supplied to the network.

1. Identify which of the two diodes will conduct and why?
2. Redraw an equivalent circuit diagram to show the flow of current.
3. Calculate the output voltage drops V₀ across the three resistors when the input voltage attains its peak value.

Answer 1

a. During positive half-cycle:

A is positive w.r.t. B.

Diode D₂ becomes forward biased.

Diode D₁ is reverse biased.

D₂ conducts because it is forward biased.

b. Since, D₂ conducts → acts as short circuit

D₁ off → open circuit

The equivalent network becomes:

A is directly connected to R through D₂.

Resistive path: 1 kΩ (P-R), 2 kΩ and 3 kΩ

Current flows:

A → D₂ → R → resistor network → B

c. Peak input voltage:

V_peak = 12 V

Now resistors form a series-parallel network.

Between P and R:

1 kΩ

Lower branch from P to B:

2 kΩ

Lower branch from R to B:

3 kΩ

Since current enters through R (via D₂), voltage division occurs between 3 k and the parallel branch.

Equivalent resistance of P-B branch:

RPB = 2 kΩ

Total series seen from R:

R_total = 3 kΩ + (1 kΩ + 2 kΩ)

= 6 kΩ

Voltage division across 1 k (output across the middle resistor):

V₀ = `V_"peak" xx 1/(1 + 2 + 3)`

= `12 xx 1/6`

= 2 V