In the figure, Vs is a square wave of levels ±10 V, frequency 100 Hz and duty cycle 50%. The diode is ideal. What is the average value of V_L?

Concept:

We are given a square wave input V_S of ±10 V, 100 Hz frequency, and 50% duty cycle. The diode is ideal.

We need to find the average voltage across V_L, which is the voltage across the load resistor R_L.

Step-by-step Analysis:

When V_S = +10 V (50% of the time):

• The diode is forward-biased.
• Current flows through the upper 100 Ω resistor, diode, and lower 100 Ω + R_L.
• It acts like a voltage divider between two 100 Ω resistors.
• So, \(V_L = \frac{10}{2} = +5~V\)

When V_S = -10 V (other 50% time):

• The diode is reverse-biased, so it acts as an open circuit.
• No current flows; hence, \(V_L = 0~V\)

Average value over full cycle:

\(V_{avg} = \frac{1}{2}(+5~V) + \frac{1}{2}(0~V) = +2.5~V\)