Problem set 42

1. The input signal for the equivalent circuit shown below can have a frequency between 10 Hz and 50 kHz, then the value of the coupling capacitor is:
   
1. $1\mu F$
2. $10 pF$
3. $1 pF$
4. $10\mu F$
For coupling capacitive reactance $X_C$ must be low. Since, $X_C=\frac{1}{2\pi fC}$, $X_C$ is inversely proportional to frequency $f$. Hence, value of $C$ is decided by the lowest frequency of operation. Now, in a circuit, for a coupling, the value of $C$ should be such that $X_C=\frac{\text{total resistance of circuit}}{10}$. $${\scriptstyle \text{total resistance of circuit}=R=12k\Omega||24k\Omega||3k\Omega||6k\Omega}$$ $$\frac{1}{R}=\frac{1}{12}+\frac{1}{24}+\frac{1}{3}+\frac{1}{6}=\frac{15}{24}$$ $$R=\frac{24}{15}$$ $$X_C=\frac{R}{10}=\frac{24}{150}k\Omega$$ $$\begin{align*} C&=\frac{1}{2\pi fX_C}\\ &=\frac{1}{2\times3.14\times10\times\frac{24}{150}\times10^3}\\ &=100\mu F \end{align*}$$ However, if we take ${\scriptstyle X_C=\text{total resistance of circuit}=R=\frac{24}{15}}$, we get, $$\begin{align*} C&=\frac{1}{2\pi fX_C}\\ &=\frac{1}{2\times3.14\times10\times\frac{24}{15}\times10^3}\\ &=10\mu F \end{align*}$$ Hence, answer is (D)
2. In a 3-input OP-AMP summing amplifier shown below, the output voltage $(v_0)$ is
   
1. -3 V
2. +3 V
3. +6 V
4. -9 V
Applying KCL law to inverting node $$\frac{v_1}{R_1}+\frac{v_2}{R_2}+\frac{v_3}{R_3}+\frac{v_0}{R_f}=0$$ $$\begin{align*} v_0&=-R_f\left(\frac{v_1}{R_1}+\frac{v_2}{R_2}+\frac{v_3}{R_3}\right)\\ &={\scriptstyle -10^3\left(\frac{2}{3\times10^3}+\frac{3}{3\times10^3}+\frac{4}{3\times10^3}\right)=-3}\\ \end{align*}$$ Hence, answer is (A)
3. In the circuit given below what is the approximate ac voltage across the output resistor:
   
1. 15 mV
2. 150 mV
3. 15 $\mu$V
4. 15 V
This is a single stage amplifier. $$v_0=A_vv_{in}$$ where, $A_v$ is ac voltage gain given by $A_v=$
4. The input impedance $(Z_{in(total)})$ of the common-emitter amplifier given below is:
   
1. $5k\Omega$
2. $4k\Omega$
3. $2k\Omega$
4. $20k\Omega$
As $\beta=200$, the value of the transistor base impedance $=\beta(r_e+R_E)=\beta r_e$, since $R_E=0$.
transistor base impedance $=\beta r_e=200\times25=5k\Omega$ $$Z_{in(total)}=20||\beta r_e$$ $$\begin{align*} \frac{1}{Z_{in(total)}}&=\frac{1}{20}+\frac{1}{\beta r_e}\\ &=\frac{1}{20}+\frac{1}{5}=\frac{1}{4} \end{align*}$$ $$Z_{in(total)}=4k\Omega$$ Hence, answer is (B)
5. The load voltage in a Zener circuit shown below with $V_z=15V$ is approximately
   
1. 15 V
2. 10 V
3. 14.3 V
4. 15.7 V
$$V_L=V_Z+V_{BE}=15-0.7=14.3 V$$ $V_{BE}$ negative because it has polarity opposite to that of $V_Z$.
Hence, answer is (C)