Question 3.33: In the circuit of Fig. 3.45(a), Vin has a nominal value of 5......
In the circuit of Fig. 3.45(a), V_{in} has a nominal value of 5 V, R_{1} = 100 Ω , and D_{2} has a reverse breakdown of 2.7 V and a small-signal resistance of 5 Ω. Assuming V_{D,on} ≈ 0.8 V for D_{1}, determine the line and load regulation of the circuit.
We first determine the bias current of D_{1} and hence its small-signal resistance:
I_{D1}={\frac{V_{i n}-V_{D,o n}-V_{D2}}{R_{1}}} (3.101)
= 15 mA. (3.102)
Thus,
r_{D1}={\frac{V_{T}}{I_{D1}}} (3.103)
= 1.73 Ω. (3.104)
From the small-signal model of Fig. 3.44(b), we compute the line regulation as
{\frac{v_{o u t}}{v_{i n}}}={\frac{r_{D1}+r_{D2}}{r_{D1}+r_{D2}+R_{1}}} (3.105)
= 0.063. (3.106)
For load regulation, we assume the input is constant and study the effect of load current variations. Using the small-signal circuit shown in Fig. 3.45(c) (where v_{in} = 0 to represent a constant input), we have
\frac{v_{o u t}}{(r_{D1}+r_{D2})||R_{1}}=-i_{L}. (3.107)
That is,
\left|\frac{v_{o u t}}{i_{L}}\right|=(r_{D1}+r_{D2})||R_{1} (3.108)
= 6.31 Ω. (3.109)
This value indicates that a 1-mA change in the load current results in a 6.31-mV change in the output voltage.
