Question 22.9: What is the approximate output voltage in Fig. 22-15? Why is...

With the equations of Fig. 22-9:

V_{out} = V_Z − V_{BE}                         (22-9)

V_{out} =\frac{2.7 \ kΩ + 2.2 \ kΩ}{2.7 \ k\Omega } (5.6 \ V) = 10.2 \ V

The load current is:

I_L =\frac{10.2 \ V}{4 \ \Omega } = 2.55 \ A

If an ordinary transistor with a current gain of 100 were used for the pass transistor, the required base current would be:

I_B =\frac{2.55 \ A}{100 } = 2.55 \ mA

This is too much output current for a typical op amp. If a Darlington transistor is used, the base current of the pass transistor is reduced to a much lower value. For instance, a Darlington transistor with a current gain of 1000 would require a base current of only 2.55 mA