**Answer 1.**

**Answer 2. **

If RC >> T, then the circuit acts as an integrator.

If RC << T, Here the time constant is very small, so the capacitor will be charged very fast and the output voltage almost follows the input voltage.

**Answer 3.**Given a DAC circuit so the value 1010 corresponds to the 10 in decimal and it is 4-bit. Therefore the least significant bit will be equaled to Vred/16 = 8/16 = 0.5V. Hence the value 1010 is 0.5*10 = 5V. One more method by using a superposition theorem also it can be solved.

**Answer 4.**Vthe = 400mV. The maximum current that can flow through the resistor R=1kOhm is 1mA. In order to find the entire current which flows through the resistor R, the MOSFET M1 should be turned OFF otherwise M1 will consume some amount of current. So to make M1 OFF

Vgs1 < Vth,

Vg – Vs = Vth,

Vg – Vth = Vs,

2-0.4=Vs. therefore Vs = 1.6V is the minimum value. Vs should be greater than 1.6V.

**Answer 5.** We know the transfer characteristics of the inverter.

The inverter gives the output voltage VDD/2 when the input voltage is equaled to VDD/2.

**Answer 6.** PMOS transistor does not allow the strong “0” to pass and the NMOS transistor does not allow the strong “1” to pass through it. Hence at the buffer output, the output logic will be degraded by the threshold voltage Vth as shown in the figure.

**Answer 7.** The circuit with the BJTs will have a higher output impedance compared to the circuit with MOSFETs since the gm of BJT is higher than that of MOSFET.

**Answer 8.**In the above current mirror circuit, the current through both the transistors M1 and M2 will be the same, so they will have equal currents. Assuming both transistors are in the saturation region, therefore Id will be 0.5µCox(W/L)(Vgs-Vth)^2.

Id1 = Id2

0.5µCox(4)(5-Vx-1)^2 = 0.5µCox(1)(Vx-1)^2

2(4-Vx) = Vx-1

Vx = 3V

**Answer 9.**Because of the positive feedback dominates, the output voltage will be +Vsat.

**Answer 10.** Here the absolute value depends on the MOSFET technology.

**Answer11.** Option 2 the offset current is known. Since Vout = (IB1 – IB2)*1MOhm.

**Answer 12.** From the circuit

(2V1-V2)/R = I

I * R = 2V1 – V2

and also

Vout = V1 + Vr

Vout = V1 + 2V1 – V2

Vout = 3V1 – V2.

**Answer 13.**From the circuit

Vout = 4.7k * 1.1mA + 2.2V

Vout = 7.37V

**Answer 14.**

**Answer 15.** The voltage across the capacitor can be given as

Vc = (1/C)int(i dt)

Vc = (1/30µF)int_0to30s(Vi/440k dt)

5.4 = (1/30µF) * (Vi/440k) * 30

Vi = 2.376V

**Answer 17.** Overall gain = Vout/Vin = 80000

**Answer 18.** UTP = 1.45, LTP = 0.45

**Answer 19.**

**Answer 20.** The time period of the square wave is 0.27µs

**Answer 21.** Option C is the correct answer.

**Answer 22.** Let us assume both the diodes are ON.

Vout = 0V

I = 0.5mA

**Answer 23.**

**Answer 24.** Assuming Vsat = 0V for BJT.

**Answer 25.** Assuming the parasitic capacitance at the output. The fall time will be more compared to the rise time.

**Answers 26.** Option D SAR ADC is the correct answer.

**Answer 27. ** Vout = (1C/5C)*5 = 1V.

**Answer 28.** Here the charge capacitor can be replaced by an uncharged capacitor in series with the voltage source.

Applying capacitance division formula,

(C1/(C1+C2))*V = Vc2

V3c = (1C/4C)*8 = 2V

V1c = (3C/4C)*8 = 6V

Vc = 8 – V1c

Vc = 8 – 6V

Vc = 2V

V3c = Vc = 2V

Vc –> voltage across 1C capacitor in the actual circuit

Vc1–> voltage across 1C capacitor in the equivalent circuit.