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).
Show Hint
Find Vout as VZ minus VBE, then compare the output power to the input power, since both use the same load current IL.
Step 1: Identify the regulator topology.
This is a series pass voltage regulator built with a Zener diode and a BJT. The 10 V source feeds a 5 kΩ resistor, which biases the Zener diode and holds the base of the transistor at a fixed voltage of 5.4 V. The transistor emitter is the regulated output, and the load draws current \(I_L=100\) mA from there.
Step 2: Find the output voltage.
The BJT works as an emitter follower here. The base sits at the Zener voltage, and the emitter is one \(V_{BE}\) drop below the base.
\[
V_{out}=V_Z-V_{BE}
\]
Putting in the given values,
\[
V_{out}=5.4-0.6=4.8\text{ V}
\]
Step 3: Work out the input power.
We are told to ignore the small current that biases the Zener diode and the base current of the transistor, so all the input current is really just the current the transistor pulls to supply the load. This means the current drawn from the 10 V source is simply \(I_L\).
\[
P_{in}=V_{in}\times I_L=10\times0.1=1\text{ W}
\]
Step 4: Work out the output power.
\[
P_{out}=V_{out}\times I_L=4.8\times0.1=0.48\text{ W}
\]
Step 5: Compute the efficiency.
\[
\eta=\frac{P_{out}}{P_{in}}\times100=\frac{0.48}{1}\times100=48.0\%
\]
Final Answer:
The maximum possible efficiency of the regulator circuit is 48.0%.
\[ \boxed{48.0} \]
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