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## Q. 15.16

Objective: Determine the bias current, temperature-compensated reference voltage, and required resistor $R_{12}$ in a particular LM78LXX voltage regulator.

Consider the voltage regulator circuit in Figure 15.51. Assume Zener diode voltages of $V_{Z} = 6.3 V$ and transistor parameters of $V_{B E} (npn) = V_{E B} (pnp) = 0.6 V$. Design $R_{12}$ such that $V_{O} = 8 V$.

## Verified Solution

The bias current, neglecting base currents, is found as
$I_{C3} = I_{C5} = \frac{V_{Z} − 3V_{B E} (npn)}{R_{3} + R_{2} + R_{1}} = \frac{6.3 − 3(0.6)}{0.576 + 3.4 + 3.9} = 0.571 mA$
The temperature-compensated portion of the reference voltage, which is the input to the base of $Q_{7}$, is
$V_{B7} = I_{C3} R_{1} + 2V_{B E} (npn) = (0.571)(3.9) + 2(0.6) = 3.43 V$
From the voltage divider network, we have

$\left(\frac{R_{13}}{R_{12} + R_{13}} \right) V_{O} = V_{B8} = V_{B7}$
or

$\left(\frac{2.23}{R_{12} + 2.23}\right) (8) = 3.43$

which yields
$R_{12} = 2.97 k\Omega$
Comment: The voltage divider of $R_{12}$ and $R_{13}$ is internal to the IC. This means the output voltage of a voltage regulator is fixed