Operational Amplifiers Practice Questions

The circuit of op-amp '1' is a Schmitt trigger, therefore $V_{01}=\pm V_{sat}$ and the circuit of op-amp '2' is a non-inverting amplifier $\therefore \;\;\; \left (\frac{V_0}{V_{01}} \right )=\left ( 1+\frac{R}{R} \right )$ $\Rightarrow \;\;\;V_0=2V_{01}$ where, $V_{01}=\pm V_{sat}$ Therefore, the answer is $V_0=\pm V_{sat}$

This filter is considered as low-pass filter (LPF). The transfer function for LPF $\frac{V_o}{V_i}=\frac{\frac{1}{sC}}{R+\frac{1}{sC}}=\frac{1}{1+sCR}$ $\;\; =\frac{1}{1+j\omega (10)^3(10^{-6})}=\frac{1}{1+j\omega (10^{-3})}$ $\;\;=\frac{1000}{1000+j\omega }=\frac{1}{1+\frac{s}{1000}}$ $\omega _c$ (corner frequency)= $10^3$ rad/sec. tha gain at low frequencies, $A_V=\frac{1}{1+sCR}=\frac{1}{1}=1$

Op-amp '3' circuit is a differential amplifier so, $V_o=V_{01}-V_{02}...(i)$ Now, apply KCL at node '2' $\frac{V_2-V_1}{2R}+\frac{V_2-V_{02}}{R}=0...(ii)$ and apply KCL at node '1' $\frac{V_1-V_2}{2R}+\frac{V_1-V_{01}}{R}=0...(iii)$ From equation (i), (ii) and (iii), we get $V_0=2(V_1-V_2)$

CASE-I: $V_i \gt 0$ , the circuit will look like Hence $V_o= 0$ CASE-II: $V_i \lt 0$ , the circuit will look like $\frac{V_{01}}{V_i}=-\frac{R}{R} \;\;\;...(i)$ $V_{01}=-V_i\;\;\;...(ii)$ $and \;\; \frac{V_o}{V_{01}}=-\frac{R}{R}$ $V_o=-V_{01}\;\;\;...(iii)$ From equation (ii) and (iii), $V_o=V_i$

Currently no explanation available

Using the concept of vitual ground, V=0 $V_{out}=-0.7V$

Output of the first op-amp $V_{out1}=-2\left[ \frac{-1}{1} \right]+1\left[ 1+\frac{1}{1} \right]=2+1(2)=4V$ $V_{out}=4\left ( 1+\frac{1}{1} \right )=4(2)=8V$

$\frac{V_{\in}-0}{R-1+\frac{1}{sC}}=\frac{0-V_o}{R_2}$ $\frac{V_o}{V_i}=-\left ( \frac{R_2}{R_1+\frac{1}{sC}} \right )$ It is a high pass filter with $f_{3dB}=\frac{1}{R_1C}$ rad/sec.

It is band pass filter.

First section is differential amplifier having gain-1. Output is $V_o=\left (\frac{R_f+R_1}{R_g+R_2} \right )\frac{R_g}{R_1}V_2-\frac{R_f}{R_1}V_1$ $V_o=V_2-V_1=-V_i$ Second stage-schmitt trigger $V_t=\pm \frac{R_g}{R_f+R_g}V_o=\frac{V_o}{2}=\pm 6V$

$V_o=\left ( 1+\frac{2R}{R} \right )V$ $\;\;=(1+2)(2)=6V$

For emitter follower $A_V\cong 1; \; A_I$ is high $\Rightarrow A_P=A_V\cdot A_I$ is high $\Rightarrow P_{out} \gt P_{\in}$ For transformer, $P_{\in}=P_{out}$

When $V_i \lt V_1 \;(upto \;\; t_2); \;\; V_o=+ve$ When $V_i \gt V_1 \; (t_2\leq t\leq t_4); \;\; V_o=-ve$ $V_1=\frac{1}{1+2}V_o=\frac{1}{3}V_o$

$I_s=\frac{V_s-V_o}{R}$ $V_s=\frac{\frac{1}{j\omega C}}{R+\frac{1}{j\omega C}}V_o$ $I_s=-V_sj\omega C$ $\;\;=-j(10 \times 2 \pi \times 50 \times 10\times 10^{-6})$ $\;\;=-j10\pi...mA$ $\;\;=10\pi \; mA$ lagging by $90^{\circ}$

Voltage-series feedback arrangement or voltage-voltage feedback.

Currently no explanation available

$V_{\in}=\frac{(sCR_A)}{1+sCR_A}\times \frac{V_i}{2}$ $\therefore \;\;\; V_0=\frac{\frac{1}{sC}}{\frac{1}{sC}+\frac{R_A}{2}}V_{\in}=\frac{1}{\left ( 1+sR_A\frac{C}{2} \right )}V_{\in}$ $\therefore \;\;\; V_{0}=\frac{1}{\left ( 1+sR_A\frac{C}{2} \right )}\cdot \frac{sR_AC}{1+sR_AC}V_{\in}$ Which is similar to equation of a band pass filter.

$V_A=\frac{R_4}{R_4+R_3} \times V_i,$ $V_A=\frac{R_B}{R_B+R_B}\times V_i=\frac{V_i}{2}$ Let, $V_i$ only on onverting terminal $V_{01}=\frac{-R_f}{R_1}V_i$ Here, $R_f=R_2||\frac{1}{sC}=\frac{R_2}{1+sCR_2}$ Let, $V_i$ only only on non-inverting terminal $V_{02}=\left ( 1+\frac{R_f}{R_1} \right )\frac{V_i}{2}$ $V_0=V_{01}+V_{02}$ $\;\;=\left ( 1-\frac{R_f}{R_1} \right )\frac{V_i}{2}$ Putting the value of $R_f$ , we get $V_0=\left ( 1-\frac{R_2}{R_1(1+sCR_2)} \right )\frac{V_i}{2}$ Here, $(R_1=R_2=R_A)$ $\therefore \;\;\; V_0=\frac{1}{1+\frac{1}{sCR_A}}\times \left ( \frac{V_i}{2} \right )$ So, it is high pass filter.