Operational Amplifiers Practice Questions

The given amplifier can be redrawn as

During positive hall cycle: $\therefore \mathrm{D}_{1} \rightarrow \mathrm{ON}$ and $\mathrm{D}_{2} \rightarrow \mathrm{OFF}$ Redraw the circuit : $\therefore V_0=-V_i$ $\quad (V_0)_{min}=-10V$ During negative half cycle : $\mathrm{V}^{+} \gt \mathrm{V}^{-}$ Zener diode $\rightarrow$ ON, $\mathrm{D}_{1} \rightarrow$ OFF and $\mathrm{D}_{2} \rightarrow \mathrm{ON}$ Redraw the circuit: $\therefore \quad (V_0)_{max}=5V$

Redraw the circuit: From circuit,

Redraw the circuit: From circuit,

Hence, $V_{out}=-V_{EE}$

Bode plot of negative feedback amplifier: Given amplifier is a voltage series feedback amplifier. Therefore, Output impedance is given by $Z_{out}=\frac{Z_o}{1+A\beta } =\frac{Z_o}{1+A } \;\;\;(\beta=1 \text{ for buffer})$ From Bode plot ; at low frequency, the open loop gain (A) is constant. when $\omega \uparrow, A \downarrow,Z_{out}\uparrow$ At $A=0, Z_{out}=Z_o\rightarrow constant$ Therefore, $z_{out}$ with frequency represented by

Discharging,

Given data: $R_{D}=4.7 K\Omega ,\; g_{m}=520 \mu A/V$ Voltage gain of CS amplifier $=-g_{m}R_{D}=-520\, \mu A/V\, \times \, 4.7\, k\Omega =-2.44$

As per GATE official answer key MTA (Marks to ALL)

$V_o=\frac{R_2}{R_1}(V_2-V_1)$ $\;\;=\frac{100k}{10k}(50\; mV-10\; mV)$ $\;\;=10(40\; mV)=400\;mV$

According to virtual ground,

$V_A=\frac{V_s}{2}$ $I_3=\frac{V_A}{R}=\frac{V_s}{2R}$ $V_B-V_A=I_3R$ $V_B=V_A+I_3R=\frac{V_s}{2}+\frac{V_s}{2}=V_s$ $I_2=\frac{V_B}{R}=\frac{V_s}{R}$ $I_1=I_2+I_3$ $\;\;=\frac{V_s}{R}+\frac{V_s}{2R}=\frac{V_s}{R}[1.5]$ $V_0-V_B=I_1R$ $\Rightarrow V_0=V_B+\frac{V_s}{R}[1.5]R$ $\;\;\;=V_s+1.5V_s=2.5V_s$

$V_{\in} \gt 0\;\; V_A=+V_{55},$ D on, $V_o=V_{\in}$ $V_{\in} \lt 0\;\; V_A=-V_{55},$ D off, $V_o=0$

$V_A=\left ( \frac{4}{5}V_1+\frac{1}{5}V_2 \right )$ $V_{out}=-9V_3+10V_A$ $\;\;=-9V_3+8V_1+2V_2$

$\frac{V_{out}}{V_{\in}}=-\left[\frac{\frac{ R_2\cdot \frac{1}{j\omega C}}{R_2+\frac{1}{j\omega C}}}{R_1} \right]$ $\frac{V_{out}}{V_{\in}}=-\left[ \frac{R_2}{R_1(R_2j\omega C+1)} \right]$ So the system is a low pass filter.

The circuit is Since circuit has negative feedback, so with help of virtual short $V_A=V_B$ So, KVL in the loop from A to B given, $I\left ( \frac{1}{sC} \right )-V_i+I\left ( \frac{1}{sC} \right )=0$ So, $I=\frac{V_i(sC)}{2}$ So, $V_A=-IR$ $V_A=-\frac{sCRV_i}{2}$ and $V_A=V_B$ So, $V_B=-\frac{sCRV_i}{2}$ Now, $\frac{V_B-V_o}{R}=I$ So, $V_o=V_B-IR$ $V_o=-\frac{V_i(sCR)}{2}-\frac{V_i(sCR)}{2}$ $\;\; =-V_i(sCR)$ So, $V_o=-j\omega RCV_i$ So, $V_o$ lag $V_i$ by $90^{\circ}$ or phase of $V_o$ w.r.t $V_i$ is $-90^{\circ}$ .

Currently no explanation available

The circuit is Now, the gain of op-amp is 1000 So, $V_o=1000(V_+-V_-)$ Since $V_+=0$ (grounded) $V_-=V_A$ The above circuit can be redrawn as So, $V_o=1000\;VA$ Now, KCL at node 'A' $\frac{V_A-V_i}{1000}=\left ( \frac{V_A-V_o}{1/sC} \right )$ $\frac{V_A-V_i}{1000}=\frac{V_A-(-1000V_A)}{1/sC}$ $\frac{V_A-V_i}{1000}=(1001V_A)sC$ $V_A-V_i=(1001)\times 1000V_AsC$ $V_A-V_i=s(1001000 \times 10^{-6}) V_A$ $V_A-s1.001V_A=V_i$ $V_A=\left[ \frac{V_i}{1-s(1.001)} \right]$ Since, $V_o=-1000V_A$ $\;\; =-\frac{1000}{1-s(1.001)} V_i$ Since, pole is at (1/1.001), so time constant is approax '1001'.

$V_i=2 \sin (2\pi \times 2000t)$ The transfer function of the system is $H(s)=\frac{1000}{(1000 \times 0.1 \times 10^{-6}s+1 )1000}$ $H(s)=\frac{-1}{(10^{-4}s+1)}$ The input is $2\sin (2\pi \times 2000t)$ $\omega =4000 \pi$ $V_0=2|H(j\omega )|_{\omega =4000\pi} \times \sin(4000\pi t-\phi )$ $Output=1.245 \sin (4000\pi t-51.46^{\circ})$ So, amplitude of output is 1.245.