A small signal source, \(V_i(t)=A\cos(10^5t)+B\sin(10^7t)\) is applied to a BJT circuit as shown in the figure. Assume zero source resistance, \(V_{BE}=0.7\) V, \(\beta_{dc}=99\), Early voltage \(=100\) V and thermal voltage \(=25\) mV. Effect of internal parasitic capacitances of the BJT may be neglected. Which expression is the best approximation of the output voltage \(V_o(t)\)?
Show Hint
Check whether the coupling and bypass capacitors act as short circuits at both signal frequencies.
Step 1: Find the DC bias point.
The base divider is \(R_1=100\ k\Omega\) from \(V_{CC}\) to the base and \(R_2=25\ k\Omega\) from the base to ground. The Thevenin voltage and resistance seen by the base are
\[
V_{th}=V_{CC}\cdot\frac{R_2}{R_1+R_2}=12\times\frac{25}{125}=2.4\ \text{V}
\]
\[
R_{th}=R_1\parallel R_2=\frac{100\times25}{125}=20\ k\Omega
\]
The total DC emitter resistance is \(500+1000=1500\ \Omega\) (the bypass capacitor blocks DC, so both emitter resistors carry the bias current). Writing the base loop with \(\beta_{dc}=99\):
\[
I_B=\frac{V_{th}-V_{BE}}{R_{th}+(\beta_{dc}+1)(1500)}=\frac{2.4-0.7}{20000+100\times1500}=\frac{1.7}{170000}=10\ \mu A
\]
So \(I_C=\beta_{dc}I_B\approx0.99\ \text{mA}\approx1\ \text{mA}\) and \(I_E=(\beta_{dc}+1)I_B=1\ \text{mA}\).
Step 3: Check which part of the emitter resistor is bypassed at each frequency.
The bypass capacitor is \(10\ \mu F\) across the \(1\ k\Omega\) resistor only, the \(500\ \Omega\) resistor is never bypassed. Its reactance at the two signal frequencies is
\[
X_C(10^5)=\frac{1}{\omega C}=\frac{1}{10^5\times10\times10^{-6}}=1\ \Omega,\qquad X_C(10^7)=\frac{1}{10^7\times10\times10^{-6}}=0.01\ \Omega
\]
Both values are tiny next to \(1\ k\Omega\), so at both \(\omega=10^5\) and \(\omega=10^7\) the \(1\ k\Omega\) resistor is fully shorted by the capacitor. The unbypassed emitter resistance is therefore the same \(500\ \Omega\) for both frequency components of the input.
Step 4: Check the coupling capacitors.
Each \(100\ nF\) coupling capacitor has reactance
\[
X_C(10^5)=\frac{1}{10^5\times10^{-7}}=100\ \Omega,\qquad X_C(10^7)=\frac{1}{10^7\times10^{-7}}=1\ \Omega
\]
Both are small compared to the input resistance the amplifier presents (several \(k\Omega\)) and there is no load resistor on the output side, so neither coupling capacitor causes any noticeable attenuation or phase shift at either frequency.
Step 5: Compute the mid-band voltage gain.
With \(500\ \Omega\) of unbypassed emitter resistance, the common-emitter gain with emitter degeneration is
\[
A_v=-\frac{R_C\parallel r_o}{r_e+R_{E,ac}}
\]
\[
R_C\parallel r_o=\frac{5000\times100000}{105000}=4761.9\ \Omega
\]
\[
A_v=-\frac{4761.9}{25+500}=-\frac{4761.9}{525}\approx-9.1
\]
Step 6: Apply the same gain to both terms of \(V_i(t)\).
Because the effective emitter resistance and the coupling capacitor behaviour are the same at \(\omega=10^5\) and at \(\omega=10^7\), both the \(A\cos(10^5t)\) part and the \(B\sin(10^7t)\) part of the input see the identical gain of about \(-9.1\), with the same inverting sign.
Step 7: Analyze the options.
(A) \(-9.1[A\cos(10^5t)+B\sin(10^7t)]\): Matches the gain found for both terms with the correct inverting sign. Correct.
(B) \(9.1[A\cos(10^5t)-B\sin(10^7t)]\): Has the wrong overall sign and wrongly flips the sign of only one term. Incorrect.
(C) \(-190.4[\ldots]\): This magnitude would follow from forgetting the \(500\ \Omega\) emitter degeneration entirely and using only \(r_e\) in the denominator. Incorrect.
(D) \(190.4[\ldots]\): Same wrong magnitude as (C), with the wrong sign as well. Incorrect.
Step 8: Final conclusion.
\[
\boxed{V_o(t)\approx-9.1\left[A\cos(10^5t)+B\sin(10^7t)\right]}
\]
