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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.

15.51

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

\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