In the circuit of figure, if the current amplification factor β = 130, what is the output voltage Vout ?
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This is a common-emitter amplifier circuit. The output voltage Vout is measured at the collector. Since the base is biased at 5 V and the emitter is grounded, the base-emitter junction is forward-biased. The base current IB is (5 V - 0.7 V) / 100 kΩ = 43 μA. The collector current IC is β × IB = 130 × 43 μA ≈ 5.59 mA. The voltage drop across the 1 kΩ collector resistor is ICRC = 5.59 V. Therefore, Vout = VCC - ICRC = 10 V - 5.59 V = 4.41 V, which is closest to 4.1 V.
Final Answer: 4.1 V