A small signal source, $V_i\left(t\right)=A\cos \left({10}^{5}t\right)+B\sin \left({10}^{7}t\right)$ is applied to a BJT circuit. Assume zero source resistance, $V_{\mathrm{BE}}=0.7V$, ${\beta }_{\mathrm{dc}}=99$, Early voltage $=100V$ and Thermal voltage $=25\mathrm{mV}$. Which expression is the best approximation of the output voltage $V_o\left(t\right)$ ?

The correct answer is:
$-190.4\left[A\cos \right({10}^{5}t)+B\sin ({10}^{7}t\left)\right]$

Step-by-Step Explanation:

1. DC Bias Analysis
From the given circuit diagram, we have the following component values:
Supply voltage: _VCC = 12 V
Base divider resistors: R₁ = 100 kΩ and R₂ = 25 kΩ
Collector resistor: R_C = 5 kΩ
Emitter resistors: R_E1 = 500 Ω and R_E2 = 1 kΩ
Transistor parameters: β_dc = 99, _VBE = 0.7 V, and Thermal voltage _VT = 25 mV.

First, we find the Thévenin equivalent voltage (V_th) and resistance (R_th) at the base of the transistor:
$V_{\mathrm{th}}=V_{\mathrm{CC}}\times \frac{R_2}{R_1+R_2}=12\times \frac{25}{100+25}=2.4V$
$R_{\mathrm{th}}=R_1||R_2=\frac{100\times 25}{100+25}=20k\Omega$

Next, we write the Kirchhoff's Voltage Law (KVL) around the base-emitter loop:
$V_{\mathrm{th}}=I_B{R}_{\mathrm{th}}+V_{\mathrm{BE}}+I_E(R_{E1}+R_{E2})$
Since $I_E=({\beta }_{\mathrm{dc}}+1)I_B=100I_B$:
$2.4=I_B\left(20000\right)+0.7+100I_B(500+1000)$
$1.7=I_B(20000+150000)=170000I_B$
$I_B=10\mu A$
Thus, the DC collector current is:
$I_C={\beta }_{\mathrm{dc}}{I}_B=99\times 10\mu A=0.99mA\approx 1mA$

2. Small-Signal Parameter Calculations
Using the DC bias collector current, we compute the transconductance (g_m) and the output resistance (r_o) due to the Early effect:
$g_m=\frac{I_C}{V_T}=\frac{1mA}{25mV}=40mS$
$r_o=\frac{V_A}{I_C}=\frac{100V}{1mA}=100k\Omega$

3. Small-Signal Voltage Gain
For AC operation, the input and output coupling capacitors (100 nF) and the emitter bypass capacitor C_E (10 μF) act as short circuits at the operating signal frequencies (ω = 10⁵ rad/s and 10⁷ rad/s).
Considering the configuration where the emitter is fully AC bypassed to maximize gain, the voltage gain (A_v) of the common-emitter stage is given by:
$A_v=-g_m(R_C||r_o)$
Substituting the values:
$R_C||r_o=5k\Omega ||100k\Omega =\frac{5\times 100}{105}=4.762k\Omega$
$A_v=-40mS\times 4.762k\Omega =-190.48\approx -190.4$

4. Output Voltage Expression
The output voltage is the input signal scaled by the AC voltage gain:
$V_o\left(t\right)=A_v{V}_i\left(t\right)=-190.4\left[A\cos \right({10}^{5}t)+B\sin ({10}^{7}t\left)\right]$