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 $V_B = V_Z = 4.7 \, {V}$. Step 2: Determine the emitter voltage ($V_E$). The base-emitter junction is forward-biased, so $V_E = V_B - V_{BE} = 4.7 \, {V} - 0.7 \, {V} = 4.0 \, {V}$. Step 3: Determine the emitter current ($I_E$). $I_E = \frac{V_E}{R_E} = \frac{4.0 \, {V}}{1 \, k\Omega} = 4.0 \, {mA}$. Step 4: Determine the collector current ($I_C$). For a BJT with very high $\beta$, $I_C \approx I_E = 4.0 \, {mA}$. 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.
