In the circuit shown, assume that the BJT in the circuit has very high \( \beta \) and \( V_{BE} = 0.7 \) V, and the Zener diode has \( V_Z = 4.7 \) V. The current \( I \) through the LED is ________ mA (rounded off to two decimal places).
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When a Zener diode is used to bias a BJT base, the base voltage is clamped at the Zener voltage. For high $\beta$, the collector current is approximately equal to the emitter current, which can be determined from the emitter voltage and emitter resistance.
Step 1: Determine the base voltage (\(V_B\)).
The Zener diode is connected between the base and ground and is in breakdown, so \(V_B = V_Z = 4.7 \, \text{V}\).
Step 2: Determine the emitter voltage (\(V_E\)).
The base-emitter junction is forward-biased, so \(V_E = V_B - V_{BE} = 4.7 \, \text{V} - 0.7 \, \text{V} = 4.0 \, \text{V}\).
Step 3: Determine the emitter current (\(I_E\)).
\(I_E = \frac{V_E}{R_E} = \frac{4.0 \, \text{V}}{1 \, \text{k}\Omega} = 4.0 \, \text{mA}\).
Step 4: Determine the collector current (\(I_C\)).
For a BJT with very high \( \beta \), \(I_C \approx I_E = 4.0 \, \text{mA}\). The current through the LED is \(I = I_C\).
Step 5: Round off to two decimal places.
The current \(I\) through the LED is 4.00 mA.
