The digital logic shown in the figure satisfies the given state diagram when Q_{1} is
connected to input A of the XOR gate.

**A. ** Input A is connected to

**B. ** Input A is connected to Q2

**C. ** Input A is connected to and S is complemented

**D. ** Input A is connected to

**Answer : ****Option D**

**Explaination / Solution: **

No Explaination.

Suppose the XOR gate is replaced by an XNOR gate. Which one of the following
options preserves the state diagram ?

No Explaination.

Workspace

Report

A good current buffer has

**A. ** low input impedance and low output impedance

**B. ** low input impedance and high output impedance

**C. ** high input impedance and low output impedance

**D. ** high input impedance and high output impedance

**Answer : ****Option B**

**Explaination / Solution: **

No Explaination.

No Explaination.

Workspace

Report

In the ac equivalent circuit shown in the figure, if i_{in} is the input current and R_{f} is very larger, the type of feedback is

**A. ** voltage-voltage feedback

**B. ** voltage-current feedback

**C. ** current-voltage feedback

**D. ** current-current feedback

**Answer : ****Option B**

**Explaination / Solution: **

From the circuit, we observe that output is Vout (Voltage). Feedback is current through resistance Rf , which is added to input current iin . Thus, the configuration is voltage-current feedback.

From the circuit, we observe that output is Vout (Voltage). Feedback is current through resistance Rf , which is added to input current iin . Thus, the configuration is voltage-current feedback.

Workspace

Report

In the circuit shown, the op-amp has finite input impedance, infinite voltage
gain and zero input offset voltage. The output voltage V_{out} is

**A. **
**B. **
**C. **
**D. **
**Answer : ****Option C**

**Explaination / Solution: **

Given that the op-amp has infinite voltage gain, i.e.

-I_{2} (R_{1} + R_{2})

I_{2}R_{2}

I_{1}R_{2}

-I_{1} (R_{1} + R_{2})

Given that the op-amp has infinite voltage gain, i.e.

A_{OL} =

and zero input offset voltage

V_{IO} = 0

So, we redraw the op-amp circuit as

Hence, the current I1 is drawn through resistance R2. So, the output voltage is

V_{out} = I_{1}R_{2}

Workspace

Report

In the figure, assume that the forward voltage drops of the PN diode D_{1} and
Schottky diode D_{2} are 0 7. V and 0 3. V, respectively. If ON denotes conducting
state of the diode and OFF denotes non-conducting state of the diode, then in
the circuit,

**A. ** both D_{1} and D_{2} are ON

**B. ** D1 is ON and D_{2} is OFF

**C. ** both D1 and D_{2} are OFF

**D. ** D1 is OFF and D_{2} is ON

**Answer : ****Option D**

**Explaination / Solution: **

Alternatively, we can solve the problem by considering the current through two diodes. Here, the correct case is only considered.

Alternatively, we can solve the problem by considering the current through two diodes. Here, the correct case is only considered.

Case : Diode D_{1} is OFF, D_{2} is ON. For this case. The equivalent circuit is

From the circuit, we have

I1 = 0

I2 = 10 - 2.3/1.02

= 9.7/1.02

= 9.5 mA

Since, the current I2 is positive, So our assumption is correct.

Workspace

Report

If fixed positive charges are present in the gate oxide of an n-channel enhancement type MOSFET, it will lead to

**A. ** a decrease in the threshold voltage

**B. ** channel length modulation

**C. ** an increase in substrate leakage current

**D. ** an increase in accumulation capacitance

**Answer : ****Option A**

**Explaination / Solution: **

If fixed positive charges are present is the gate oxide of an n-channel enhancement type MOSFET, it will lead to a decrease in the threshold voltage.

If fixed positive charges are present is the gate oxide of an n-channel enhancement type MOSFET, it will lead to a decrease in the threshold voltage.

Workspace

Report

In the following circuit employing pass transistor logic, all NMOS transistors
are identical with a threshold voltage of 1 V . Ignoring the body-effect, the
output voltages at P, Q and R are,

**A. ** 4V, 3V, 2V

**B. ** 5V, 5V, 5V

**C. ** 4V, 4V, 4V

**D. ** 5V, 4V, 3V

**Answer : ****Option C**

**Explaination / Solution: **

No Explaination.

No Explaination.

Workspace

Report

The doping concentrations on the p-side and n-side of a silicon diode are 1 × 10^{16} cm^{-3 }and 1 × 10^{17} cm^{-3}, respectively. A forward bias of 0 3. V is
applied to the diode. At T = 300 K, the intrinsic carrier concentration of silicon n_{i} = 1.5 × 10^{10} cm^{-3} and (kT/q) = 26 mV. The electron concentration at the edge of
the depletion region on the p-side is

**A. ** 2.3 × 10^{9} cm^{-3}

**B. ** 1 × 1016 cm-3

**C. ** 1 × 1017 cm-3

**D. ** 2.25 × 106 cm-3

**Answer : ****Option A**

**Explaination / Solution: **

Given the doping concentration on p-side

Given the doping concentration on p-side

NA = 1 × 1016 cm-3

V = 0.3 V

Intrinsic carrier concentration,

So, the equilibrium electron concentration on the p-side is

Therefore, the electron at the edge of the depletion region on the p-side is obtained
as

Workspace

Report

For a parallel plate transmission line, let v be the speed of propagation and Z be the characteristic impedance. Neglecting fringe effects, a reduction of the spacing between the plates by a factor of two results in

**A. ** halving of v and no change in Z

**B. ** no changes in v and halving of Z

**C. ** no change in both v and Z

**D. ** halving of both v and Z

**Answer : ****Option B**

**Explaination / Solution: **

No Explaination.

No Explaination.

Workspace

Report

Consider the feedback system shown in the figure. The Nyquist plot of G (s) is
also shown. Which one of the following conclusions is correct ?

**A. ** G (s) is an all pass filter

**B. ** G (s) is a strictly proper transfer function

**C. ** G (s) is a stable and minimum phase transfer function

**D. ** The closed-loop system is unstable for sufficiently large and positive k

**Answer : ****Option D**

**Explaination / Solution: **

Given the feedback system and the Nyquist plot of G (s)is

Given the feedback system and the Nyquist plot of G (s)is

For the given system, we have the open loop transfer function as

G (s) = KG (s)

Considering the open loop system G (s) is stable, we have no open loop poles in
right half plane

P = 0

From Nyquist theorem, we know that

N = P - Z

Where N is the number of encirclements of (-1 + j0), P is number of open loop
poles in right half plane, Z is number of closed loop poles in right half plane. For
stability, we must have

Z = 0

N = 0, if closed loop system is stable

N ≠ 0, if closed loop system is unstable

observing the Nyquist plot, we conclude that the plot of KG(s) encircles (-1 + j0)

if K> 1

Hence, N ≠ 0 for sufficient large and positive value of K . Thus, the closedsystem
is unstable for sufficiently large and positive K .

Workspace

Report