The circuit of Fig. 2-32 adds a dc level (a bias voltage) to a signal whose average value is zero. If v_S is a 10-V square wave of period T, R_L = R_1 = 10 Ω, and the diode is ideal, find the average value of v_L.
For v_L > 0, D is forward-biased and v_L = v_S = 10 \text{V}. For v_L < 0, D is reverse-biased and
v_L = \frac{R_L}{R_L + R_1}v_s = \frac{10}{10 + 10}(-10) = -5 \text{V}Thus, V_{L0} = \frac{10(T/2) + (-5) (T/2)}{T} = 2.5 \text{V}
For some symmetrical input signals, this type of circuit could destroy the symmetry of the input.