## Chapter 15

## 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.

## Step-by-Step

## 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

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