Question 9.11: Objective: Design a voltage reference source with an output ...

Consider the circuit shown in Figure 9.42. For this example, we need
\frac{V_{O}}{V_{Z}} = \left( 1 + \frac{R_{2}}{R_{1}} \right) = \frac{10.0}{5.6}
Therefore,
\frac{R_{2}}{R_{1}} = 0.786
We know that
I_{F} = \frac{V_{O}  −  V_{Z}}{R_{F}}
If we set I_{F} equal to the minimum bias current, we have
1  mA = \frac{10  −  5.6}{R_{F}}
which means that R_{F} = 4.4  k\Omega. If we choose R_{2} = 30  k\Omega, then R_{1} = 38.17  k\Omega

Resistors R_{3} and R_{4} can be determined from Figure 9.43. The maximum Zener current supplied by V_{S} , R_{3} and R_{4} should be no more than 0.2 mA. We set the current through D_{1} equal to 0.2 mA, for V_{S} = 10  V. We then have
V_{2}^\prime = V_{Z} + 0.7 = 5.6 + 0.7 = 6.3  V
Also,
I_{4} = \frac{V_{2}^\prime}{R_{4}} = \frac{6.3}{R_{4}}

and
I_{3} = \frac{V_{S}  −  V_{2}^\prime}{R_{3}} = \frac{10  −  6.3}{R_{3}} = \frac{3.7}{R_{3}}

If we set I_{4} = 0.2  mA, then
I_{3} = 0.4  mA \qquad R_{3} = 9.25  k\Omega \qquad R_{4} = 31.5  k \Omega
Comment: Voltage V_{S} is used as a start-up source. Once the Zener diode is biased in breakdown, the output will be maintained at 10.0 V, even if V_{S} is reduced to zero