Question 22.8: What is the approximate efficiency in Example 22-7?

The load voltage is 9.2 V, and the load current is 230 mA. The output power is:

P_{out} = (9.2 V)(230 mA) = 2.12 W

The input voltage is 15 V, and the input current is approximately 230 mA, the value of the load current. Therefore, the input power is:

P_{in} = (15 V)(230 mA) = 3.45 W

The efficiency is:

Efficiency =I_{C} =\frac{2.12 \ W}{3.45 \ W }\times 100\%=61.4\%

We can also use Eq. (22-13) to calculate the efficiency of a series regulator:

Efficiency =\frac{V_{out}}{V_{in}}\times 100\%=\frac{9.2 \ V}{15 \ V}\times 100\% =61.3

This is much better than 25.8 percent, the efficiency of the shunt regulator in Example 22-3. Typically, a series regulator has an efficiency about twice that of a shunt regulator