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Vsig Rsig Rin RB1 RB2 If Vsig = 20 mV value of Vout. ww Vcc www peak' Rc Rout RE RL = The common-emitter discrete amplifier shown has the following values: RB1 500 ΚΩ, RB2 = 100 ΚΩ, Rc = 5 ΚΩ, RE 200 , Rsig = 1 ks, R₁ = 10 kſ and Vcc= 22 V. Assume VBE = 0.7 V and Bß = 90. Also assume that the capacitors are shorts for ac. + Vout = The bias circuit has been analyzed and Ic = 2.63 mA, Rin = 0.856 k and Rout = Rc. determine the peak

Vsig Rsig Rin RB1 RB2 If Vsig = 20 mV value of Vout. ww Vcc www peak' Rc Rout RE RL = The common-emitter discrete amplifier shown has the following values: RB1 500 ΚΩ, RB2 = 100 ΚΩ, Rc = 5 ΚΩ, RE 200 , Rsig = 1 ks, R₁ = 10 kſ and Vcc= 22 V. Assume VBE = 0.7 V and Bß = 90. Also assume that the capacitors are shorts for ac. + Vout = The bias circuit has been analyzed and Ic = 2.63 mA, Rin = 0.856 k and Rout = Rc. determine the peak

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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### Common-Emitter Discrete Amplifier Analysis The schematic diagram illustrates a common-emitter discrete amplifier. Below are the specified parameters for each component in the circuit: - \( R_{B1} = 500 \, k\Omega \) - \( R_{B2} = 100 \, k\Omega \) - \( R_C = 5 \, k\Omega \) - \( R_E = 200 \, \Omega \) - \( R_{sig} = 1 \, k\Omega \) - \( R_L = 10 \, k\Omega \) - \( V_{CC} = 22 \, V \) Assumptions: - Base-Emitter Voltage, \( V_{BE} = 0.7 \, V \) - Current Gain, \( \beta = 90 \) - Capacitors are considered short circuits in AC analysis. After analyzing the bias circuit, the following values are obtained: - Collector Current, \( I_C = 2.63 \, mA \) - Input Resistance, \( R_{in} = 0.856 \, k\Omega \) - Output Resistance, \( R_{out} = R_C \) Given: - Input Signal, \( v_{sig} = 20 \, mV_{peak} \) **Objective:** Determine the peak value of the output voltage, \( v_{out} \). In this circuit, the task involves calculating the voltage gain and using it to determine the peak \( v_{out} \) based on the provided \( v_{sig} \).
Transcribed Image Text:### Common-Emitter Discrete Amplifier Analysis The schematic diagram illustrates a common-emitter discrete amplifier. Below are the specified parameters for each component in the circuit: - \( R_{B1} = 500 \, k\Omega \) - \( R_{B2} = 100 \, k\Omega \) - \( R_C = 5 \, k\Omega \) - \( R_E = 200 \, \Omega \) - \( R_{sig} = 1 \, k\Omega \) - \( R_L = 10 \, k\Omega \) - \( V_{CC} = 22 \, V \) Assumptions: - Base-Emitter Voltage, \( V_{BE} = 0.7 \, V \) - Current Gain, \( \beta = 90 \) - Capacitors are considered short circuits in AC analysis. After analyzing the bias circuit, the following values are obtained: - Collector Current, \( I_C = 2.63 \, mA \) - Input Resistance, \( R_{in} = 0.856 \, k\Omega \) - Output Resistance, \( R_{out} = R_C \) Given: - Input Signal, \( v_{sig} = 20 \, mV_{peak} \) **Objective:** Determine the peak value of the output voltage, \( v_{out} \). In this circuit, the task involves calculating the voltage gain and using it to determine the peak \( v_{out} \) based on the provided \( v_{sig} \).
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