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Part 2: Calculation of volt\textasciicircum{\textasciicircum}DORIN>tput 5 15V The single-stage BJT amplifier in Figure 2 is identical to the amplifier in Figure 1, except for the addition of the emitter bypass capacitor. Note that the dc analysis for the two amplifiers is identical (so no need to recalculate). However, the addition of emitter bypass capacitor increases the gain to > 100. Assume that the input voltage to the amplifier is a 60 mV (peak), 5 kHz sinusoidal signal, measure the output voltage and voltage gain for circuit Figure 2. Analyze the circuit and then complete Table 1 section relevant to Figure 2. R R Vo Signal Generator \nu 10 \mu F 10 \mu F 50 \Omega \nu RE R3 Load 100 \mu F Figure 2: BJT CE Amplifier with emitter bypass capacin HAND CALCULATIONS (Gain, $V_o$, and $R_{in}$) for Figure 2 Upon completion of calculations, do Multisim simulations and record the results in Table 1. R 500 $R_{IB}$ $\Gamma_{\pi} = 16.244 \Omega$ 3RL V $R_{OZ}$ $A_{CE} = \frac{g_m}{1 + g_m R_E} \left( \frac{R_{A} \parallel R_{IB}}{R_I + R_{A} \parallel R_{IB}} \right) = \frac{-10(16m)(18.03k)}{1 + 10(16m)(0)} = \frac{103.125 \text{k} \parallel 18.244}{50 - 103.125 \text{k} \parallel 16.244}$ $=-182.5$ $R_{in} = 150k \parallel 3 \text{k} \Omega \parallel 16.24 \text{k} \Omega$ $= 14.03k \Omega$ prification)}

          Part 2: Calculation of volt\textasciicircum{\textasciicircum}DORIN>tput
5
15V
The single-stage BJT amplifier in Figure 2 is identical to the
amplifier in Figure 1, except for the addition of the emitter
bypass capacitor. Note that the dc analysis for the two amplifiers
is identical (so no need to recalculate). However, the addition of
emitter bypass capacitor increases the gain to > 100. Assume
that the input voltage to the amplifier is a 60 mV (peak), 5 kHz
sinusoidal signal, measure the output voltage and voltage gain for
circuit Figure 2. Analyze the circuit and then complete Table 1
section relevant to Figure 2.
R
R
Vo
Signal Generator
\nu 10 \mu
F
10 \mu
F
50 \Omega
\nu
RE
R3 Load
100 \mu
F
Figure 2: BJT CE Amplifier with emitter bypass capacin
HAND CALCULATIONS (Gain, $V_o$, and $R_{in}$) for Figure 2
Upon completion of calculations, do Multisim simulations and record the results in Table 1.
R
500
$R_{IB}$
$\Gamma_{\pi} = 16.244 \Omega$
3RL
V
$R_{OZ}$
$A_{CE} = \frac{g_m}{1 + g_m R_E} \left( \frac{R_{A} \parallel R_{IB}}{R_I + R_{A} \parallel R_{IB}} \right) = \frac{-10(16m)(18.03k)}{1 + 10(16m)(0)} = \frac{103.125 \text{k} \parallel 18.244}{50 - 103.125 \text{k} \parallel 16.244}$
$=-182.5$
$R_{in} = 150k \parallel 3 \text{k} \Omega \parallel 16.24 \text{k} \Omega$
$= 14.03k \Omega$
prification)}

Part 2: Calculation of voltˆˆDORIN>tput
5
15V
The single-stage BJT amplifier in Figure 2 is identical to the
amplifier in Figure 1, except for the addition of the emitter
bypass capacitor. Note that the dc analysis for the two amplifiers
is identical (so no need to recalculate). However, the addition of
emitter bypass capacitor increases the gain to > 100. Assume
that the input voltage to the amplifier is a 60 mV (peak), 5 kHz
sinusoidal signal, measure the output voltage and voltage gain for
circuit Figure 2. Analyze the circuit and then complete Table 1
section relevant to Figure 2.
R
R
Vo
Signal Generator
ν10 μF
10 μF
50 ΩνRE
R3 Load
100 μF
Figure 2: BJT CE Amplifier with emitter bypass capacin
HAND CALCULATIONS (Gain, Vo, and Rin) for Figure 2
Upon completion of calculations, do Multisim simulations and record the results in Table 1.
R
500
RIB
Γπ = 16.244 Ω
3RL
V
ROZ
ACE = (gm)/(1 + gm RE)( (RA∥ RIB)/(RI + RA∥ RIB)) = (-10(16m)(18.03k))/(1 + 10(16m)(0)) = (103.125 k∥ 18.244)/(50 - 103.125 k∥ 16.244)
=-182.5
Rin = 150k ∥ 3 kΩ∥ 16.24 kΩ
= 14.03k Ω
prification)

Added by Jaime J.

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