Question

In the linear regulator circuit shown, the base to emitter voltage \(V_{BE}\) of the BJT is \(0.6\) V. The Zener diode clamps the base voltage to \(5.4\) V. Ignore the biasing current of the Zener diode and the BJT.

The input supply is \(10\) V and the load current is \(I_L=100\) mA. The maximum possible efficiency of the regulator circuit is % (round off to one decimal place).

Hint:

Find Vout as VZ minus VBE, then compare the output power to the input power, since both use the same load current IL.

Answer 1

Correct Answer: 48

Step 1: Picture the regulator.
A Zener diode fixes the base of a BJT at $5.4$ V. The emitter follows the base minus the $V_{BE}$ drop, so the output sits at a fixed voltage below the source.

Step 2: Get the output voltage a different way.
Since the emitter voltage equals $V_Z - V_{BE}$, we get $V_{out} = 5.4 - 0.6 = 4.8$ V. This is also the voltage dropped across the transistor measured from the top rail: the collector sits at $10$ V and the emitter at $4.8$ V, so the transistor itself drops $10 - 4.8 = 5.2$ V while carrying the load current.

Step 3: Use the voltage ratio shortcut for efficiency.
Because ignoring the bias currents means the same current $I_L$ flows both into the regulator and out to the load, the power ratio collapses to a voltage ratio:
\[ \eta = \frac{V_{out} I_L}{V_{in} I_L}\times100 = \frac{V_{out}}{V_{in}}\times100 \]

Step 4: Plug in the numbers.
\[ \eta = \frac{4.8}{10}\times100 = 48.0\% \]
This also matches checking it as one minus the fraction dropped across the transistor: $\frac{5.2}{10}=0.52$, so 52 percent is lost across the transistor and 48 percent reaches the load.
\[ \boxed{48.0} \]