Question 2.SP.13: The circuit of Fig. 2-31(a) is an ‘‘inexpensive’’ voltage re......

We determined in Problem 2.7 that each diode can be modeled as a battery, V_F = 0.5  \text{V}, and a resistor, R_F = 500  Ω, in series. Combining the diode strings between points a and b and between points b and c gives the circuit of Fig. 2-31(b), where
V_{F1} = 2V_F = 1  \text{V} \quad   V_{F2} = 4V_F = 2  \text{V} \quad   R_{F1} = 2R_F = 100  Ω \quad   R_{F2} = 4R_F = 200  Ω

By KVL,                    I_b = \frac{V_b  –  V_{F1}  –  V_{F2}}{R  +  R_{F1}  +  R_{F2}}

whence                    V_o = V_{F2} + I_bR_{F2} = \frac{(V_b  –  V_{F1}  –  V_{F2}) R_{F2}}{R  +  R_{F1}  +  R_{F2}}

For V_{b1} = 4  \text{V} and V_{b2} = 6  \text{V},
V_{o1} = 2 +  \frac{(4  –  1  –  2)(200)}{2000  +  100  +  200} = 2.09  \text{V} \quad V_{o2} = 2 +  \frac{(6  –  1  –  2)(200)}{2000  +  100  +  200} = 2.26  \text{V}

and (2.6) gives
\text{Reg} ≡ \frac{(\text{no-load}  V_{L0})  –  (\text{full-load}  V_{L0})}{\text{full-load}  V_{L0}}          (2.6)
\text{Reg} = \frac{V_{o2}  –  V_{o1}}{V_{o1}} (100 \%) = 8.1 \%