Q. No. 1 - 10 Carry One Mark Each
1. The ripple factor for a power supply is given by
(A) \(\frac {P_{dc}}{P_{ac}}\) (B) \(\sqrt {(\frac {I_{rms}}{I_{dc}})^2 - 1}\) (C) \(\sqrt {(\frac {I_{dc}}{I_{rms}})^2 - 1}\) (D) \(\frac {I_{dc}}{I_{rms}}\)
Correct Answer : B
(A) \(\frac {P_{dc}}{P_{ac}}\) (B) \(\sqrt {(\frac {I_{rms}}{I_{dc}})^2 - 1}\) (C) \(\sqrt {(\frac {I_{dc}}{I_{rms}})^2 - 1}\) (D) \(\frac {I_{dc}}{I_{rms}}\)
Correct Answer : B
2. Biasing is done to establish
(A) A fixed level of current (B) A fixed level of voltage
(C) A fixed Q port on the characteristics curve (D)All of these
Correct Answer : D
(A) A fixed level of current (B) A fixed level of voltage
(C) A fixed Q port on the characteristics curve (D)All of these
Correct Answer : D
Solve-2. The term biasing is used for the application of dc voltages to establish a fixed level of current and voltage. For transistor amplifiers the resulting dc current and voltage establish an operating point on the characteristics that define the region that will be employed for the amplification of the applied signal.
3. Consider the following statements:
FET when compared to BJT has
1. High input impedance
2. Current flow due to majority carriers
3. Current flow due to minority carriers
4. Low input impedance
Which of the statements is/are incorrect?
(A) 1 and 3 (B)3 and 4 (C) 1 and 2 (D)2 and 4
Correct Answer : C
FET when compared to BJT has
1. High input impedance
2. Current flow due to majority carriers
3. Current flow due to minority carriers
4. Low input impedance
Which of the statements is/are incorrect?
(A) 1 and 3 (B)3 and 4 (C) 1 and 2 (D)2 and 4
Correct Answer : C
Solve-3. In Field effect transistors, the input impedance is much higher and current flow takes place due to majority carriers.
4. Choose the correct output waveform for the circuit shown below.
Correct Answer : A
Correct Answer : A
Solve-4. Since Zener diodes act as voltage regulators, so the correct alternative is (A). This circuit basically acts as a 40 Volt peak to peak ac regulator.
5. The 'h' parameter equivalent circuit of a junction transistor is valid for
(A) High frequency, large signal operation
(B) High frequency, small signal operation
(C) Low frequency, small signal operation
(D)Low frequency, large signal operation
Correct Answer : C
(A) High frequency, large signal operation
(B) High frequency, small signal operation
(C) Low frequency, small signal operation
(D)Low frequency, large signal operation
Correct Answer : C
Solve-5. The 'h' parameter equivalent circuit of a junction transistor is valid for low frequency, small signal operation only. Hence alternative 'C' is the correct answer.
6. A transistor is said to be in a quiescent state when
(A) No signal is applied to the input
(B) It is unbiased
(C) No current is flowing
(D) Emitter junction bias is equal to collector junction bias
Correct Answer : A
(A) No signal is applied to the input
(B) It is unbiased
(C) No current is flowing
(D) Emitter junction bias is equal to collector junction bias
Correct Answer : A
Solve-6. A transistor is said to be in a quiescent state when no signal is applied to the input.
7.7. Consider the following statements regarding damper circuit
1. Add or subtracts a dc voltage from a wave form
2. Does not change the shape of the waveform
3. Amplifies the waveform
Which of these statements are correct?
(A) 1 and 2 (B)2 and 3 (C)1 and 3 (D)1, 2 arid 3
Correct Answer : A
1. Add or subtracts a dc voltage from a wave form
2. Does not change the shape of the waveform
3. Amplifies the waveform
Which of these statements are correct?
(A) 1 and 2 (B)2 and 3 (C)1 and 3 (D)1, 2 arid 3
Correct Answer : A
8. The output waveform for the circuit given below is
Correct Answer : B
Correct Answer : B
Solve-8. During the positive half cycle of the sinusoidal input voltage, the diode \(D_2\) is OFF and \(D_1\) is also OFF until \(V_i\) is greater than 6 Volt.
When \(V_i>6\) Volt, the diode \(D_1\) conducts. So then at the output the upper portion of 6 Volt will be clipped off.
Similarly during negative half cycle the diode \(D_2\) is ON when \(V_i\) < —4V
So in a similar way the portion below —4 Volt will be clipped off and the output voltage waveform will be similar as presented in the alternative
When \(V_i>6\) Volt, the diode \(D_1\) conducts. So then at the output the upper portion of 6 Volt will be clipped off.
Similarly during negative half cycle the diode \(D_2\) is ON when \(V_i\) < —4V
So in a similar way the portion below —4 Volt will be clipped off and the output voltage waveform will be similar as presented in the alternative
9.If diodes \(D_1\) and \(D_2\) are practical Si diodes with cut in voltage = 0.7 Volt and \(V_i = V_2 = 10\) volt then \(V_o\) will be ___ (V).
Correct Answer : From: 8.055 To: 7.895
Correct Answer : From: 8.055 To: 7.895
Solve-9. Diode \(D_1\) is OFF as it is reverse biased and diode \(D_2\) is ON. So the circuit appears as,
Apply KCL at mode A
\(\frac {10 - 0.7 - V_o}{2} + \frac {4 - V_o}{6}\) = 0 \(\implies \frac {9.3 - V_o}{2} = \frac {V_o - 4}{6}\)
\(\implies 27.9 - 3V_o = V_o - 4\) \(\implies 31.9 = 4V_o\)
\(\implies V_o = \frac {31.9}{4}\) = 7.975 volt
\(V_o = 7.975\) volt
Apply KCL at mode A
\(\frac {10 - 0.7 - V_o}{2} + \frac {4 - V_o}{6}\) = 0 \(\implies \frac {9.3 - V_o}{2} = \frac {V_o - 4}{6}\)
\(\implies 27.9 - 3V_o = V_o - 4\) \(\implies 31.9 = 4V_o\)
\(\implies V_o = \frac {31.9}{4}\) = 7.975 volt
\(V_o = 7.975\) volt
10. Consider the following statements
1. CMOS circuit is similar to a class B amplifier because one MOSFET conducts, while the other is OFF.
2. CMOS inverter consists of a p channel MOSFET and an n channel MOSFET.
3. The average dynamic power consumption in a CMOS device is greater than the quiescent power consumption.
4. CMOS circuits are often used In battery powered applications.
Which of these statements are correct?
(A) 1 (B) 1 arid 3 (C) 1 and 2 (D)All of these
Correct Answer : D
1. CMOS circuit is similar to a class B amplifier because one MOSFET conducts, while the other is OFF.
2. CMOS inverter consists of a p channel MOSFET and an n channel MOSFET.
3. The average dynamic power consumption in a CMOS device is greater than the quiescent power consumption.
4. CMOS circuits are often used In battery powered applications.
Which of these statements are correct?
(A) 1 (B) 1 arid 3 (C) 1 and 2 (D)All of these
Correct Answer : D
Solve-10. All of the four statements are correct regarding the characteristics of a complementary metal oxide semiconductor field effect transistor (CMOSFET).
Q. No. 11 - 20 Carry Two Mark Each
11. Two junction diodes (identical) are connected as shown in figure. Find voltage across diode \(D_1\) and \(D_2\). Given, \(I_s = 0.2μA\), η = 1, \(V_T = 25\)mV, supply voltage 15V.
(A)14.9527V, 0.0173V (B) 0.03V, 0.0173V
(C) 0.0173V, 14.9527V (D) 14.9527V, 0.03V
Correct Answer : C
(A)14.9527V, 0.0173V (B) 0.03V, 0.0173V
(C) 0.0173V, 14.9527V (D) 14.9527V, 0.03V
Correct Answer : C
Solve-11. When two diodes are reverse connected the current in the series circuit is the reverse saturation current \(I_o\).
\(\implies I_o\) = \(I_o(e^{\frac {V_1}{ηV_T}} - 1)\) \(\implies e^{\frac {V_1}{ηV_T}}\) = 2 \(\therefore V_1 = ηV_Tln2\) = \(1 \times 0.025 \times 0.693\) = 0.0173; \(\therefore V_{D1}\) = 0.0173V
Voltage across resistor = \(150 \times 10^3 \times 0.2 \times 10^{-6}\) = 0.03V
\(\therefore\) Voltage across Diode \(D_2\) = 15 - (0.03 + 0.0173) = 14.9527
\(\implies I_o\) = \(I_o(e^{\frac {V_1}{ηV_T}} - 1)\) \(\implies e^{\frac {V_1}{ηV_T}}\) = 2 \(\therefore V_1 = ηV_Tln2\) = \(1 \times 0.025 \times 0.693\) = 0.0173; \(\therefore V_{D1}\) = 0.0173V
Voltage across resistor = \(150 \times 10^3 \times 0.2 \times 10^{-6}\) = 0.03V
\(\therefore\) Voltage across Diode \(D_2\) = 15 - (0.03 + 0.0173) = 14.9527
12. In the circuit shown in figure, diode have cut in voltage of 0.7V. The diodes in ON stage are
(A) Only \(D_1\) (B) Only \(D_2\) (C) Both \(D_1\) and \(D_2\) (D) None
Correct Answer : C
(A) Only \(D_1\) (B) Only \(D_2\) (C) Both \(D_1\) and \(D_2\) (D) None
Correct Answer : C
Solve-12.Applying nodal equation at a, we get
\(\frac {V_a - 5.4 + 0.7}{12}\) + \(\frac {V_a + 0.7 - 5}{6}\) + \(\frac {V_a}{8}\) = 0
\(\therefore V_a = \frac {\frac {5.4 - 0.7}{12} + \frac {5 - 0.7}{6}}{\frac {1}{12} + \frac {1}{6} +\frac {1}{18}}\) = 3.63V
\(\therefore I_{D_1} = \frac {5.4 - 0.7 - 3.63}{12}\) = 89mA > 0, So \(D_1\) is ON
\(I_{D_2} = \frac {5.07 - 3.63}{6}\) = 112mA > 0, So \(D_2\) is ON
\(\frac {V_a - 5.4 + 0.7}{12}\) + \(\frac {V_a + 0.7 - 5}{6}\) + \(\frac {V_a}{8}\) = 0
\(\therefore V_a = \frac {\frac {5.4 - 0.7}{12} + \frac {5 - 0.7}{6}}{\frac {1}{12} + \frac {1}{6} +\frac {1}{18}}\) = 3.63V
\(\therefore I_{D_1} = \frac {5.4 - 0.7 - 3.63}{12}\) = 89mA > 0, So \(D_1\) is ON
\(I_{D_2} = \frac {5.07 - 3.63}{6}\) = 112mA > 0, So \(D_2\) is ON
13. What is the Q point of the circuit shown below?
(A) 12mA, 11V (B) 12mA, 12V (C) 0mA, 11V (D) 12mA, -3V
Correct Answer : A
(A) 12mA, 11V (B) 12mA, 12V (C) 0mA, 11V (D) 12mA, -3V
Correct Answer : A
Solve-13. As gate source both are earthed, so
\(V_{GS} = 0\); \(I_D = I_{DSS} = 12\)mA; \(V_{GS} = V_{DD} - I_DR_D\) = \(20 - 12 \times 10^{-3} \times 0.75 \times 10^3\) = 11V
\(V_{GS} = 0\); \(I_D = I_{DSS} = 12\)mA; \(V_{GS} = V_{DD} - I_DR_D\) = \(20 - 12 \times 10^{-3} \times 0.75 \times 10^3\) = 11V
14. The circuit using a BJT with β =50 and \(V_{BE}\) = 0.7V is shown in the figure. What is the value of base current \(I_B\) and collector voltage \(V_L\)?
(A) 40μA, 14V (B) 25mA, 18V (C) 38.5mA, 18V (D) 38.5μA, 10.4V
Correct Answer : D
(A) 40μA, 14V (B) 25mA, 18V (C) 38.5mA, 18V (D) 38.5μA, 10.4V
Correct Answer : D
Solve-14. From E-B loop
\(2 \times 10^3 \times 51 \times I_B\) + 0.7 + \(400 \times 10^3 \times I_B\) - 20 = 0
or \(I_B = \frac {19.3}{502 \times 10^3}\) = 38.5μA; \(V_C\) = 20 - \(5 \times 10^3 \times 50 \times 38.5 \times 10^{-6}\) = 10.4V
\(2 \times 10^3 \times 51 \times I_B\) + 0.7 + \(400 \times 10^3 \times I_B\) - 20 = 0
or \(I_B = \frac {19.3}{502 \times 10^3}\) = 38.5μA; \(V_C\) = 20 - \(5 \times 10^3 \times 50 \times 38.5 \times 10^{-6}\) = 10.4V
15. For the circuit shown in figure, each diode has \(V_T\) = 0.7V. The output voltage \(V_o\) is ___ (V)
Correct Answer : From: 4.343 To: 4.257
Correct Answer : From: 4.343 To: 4.257
Solve-15. Diodes \(D_2\) and \(D_3\) are ON. As \(D_3\) ON and so \(D_1\) is OFF. Hence \(V_o\) = 5 -0.7 = 4.3V
16. For the transistor in figure, value of voltage \(V_{EC}\) is ____ (V). (β = 50)
Correct Answer : From: 6.135 To: 6.013
Correct Answer : From: 6.135 To: 6.013
Solve-16. \(I_E\) = 1mA; β =50 \(\therefore I_c = \frac {β}{β + 1}I_E\) = \(\frac {50}{51}\).1mA = 0.98mA
\(V_c = I_cR_c -9\) = \(0.98 \times 4.7 - 9\) = -4.394V
\(I_B = \frac {I_E}{β +1}\) = \(\frac {1mA}{51}\) = 19.6μA, \(V_{EB} = V_E - V_B\)
\(\therefore V_E = V_{EB} + V_B\) = 0.7 + \(I_B.R_B\) = 0.7 + \(50 \times 19.6 \times 10^{-3}\) = 1.68V
\(\therefore V_{EC} = V_E - V_C\) = 1.68 - (-4.394) = 6.074V
\(V_c = I_cR_c -9\) = \(0.98 \times 4.7 - 9\) = -4.394V
\(I_B = \frac {I_E}{β +1}\) = \(\frac {1mA}{51}\) = 19.6μA, \(V_{EB} = V_E - V_B\)
\(\therefore V_E = V_{EB} + V_B\) = 0.7 + \(I_B.R_B\) = 0.7 + \(50 \times 19.6 \times 10^{-3}\) = 1.68V
\(\therefore V_{EC} = V_E - V_C\) = 1.68 - (-4.394) = 6.074V
Common Data Questions: 17 and 18
\(R_g\) = 1kohm
17.For the BJT amplifier configuration \(h_{ie}\) = 1kΩ, \(h_{re} = 2.5 \times 10^{-4}\), \(h_{fe}\) = 150 and \(\frac {1}{h_{oe}}\) = 40kΩ. Determine current gain \(A_I\).
(A) -100 (B) 100 (C) -150 (D) 150
Correct Answer : A
Solve-17. 
Here current gain
\(A_I = -\frac {h_{fe}}{1 + h_{oe}R_L}\) = -\(\frac {150}{1 + \frac {20}{40}}\) = -100
Here current gain
\(A_I = -\frac {h_{fe}}{1 + h_{oe}R_L}\) = -\(\frac {150}{1 + \frac {20}{40}}\) = -100
18. Determine the input and output resistance \(R_i\) and \(R_o\) respectively.
(A) 500Ω, 160kΩ (B) 600Ω, 170kΩ (C) 500Ω, 2000Ω (D) 1000Ω, 2000kΩ
Correct Answer : A
Solve-18. \(R_i = H_{ie} + A_Ih_{re}R_L\) = 1000 - \(100 \times 2.5 \times 10^{-4} \times 20 \times 10^3\) = 500Ω
Output resistance;
\(R_o = \frac {R_g + h_{ie}}{R_g + h_{ie}h_{oe} - h_{fe}h_{re}}\) = \(\frac {1000 + 1000}{\frac {1}{40} + \frac {1}{40} - 150 \times 2.5 \times 10^{-4}}\) = 160kΩ
Output resistance;
\(R_o = \frac {R_g + h_{ie}}{R_g + h_{ie}h_{oe} - h_{fe}h_{re}}\) = \(\frac {1000 + 1000}{\frac {1}{40} + \frac {1}{40} - 150 \times 2.5 \times 10^{-4}}\) = 160kΩ
A current amplifier has an input resistance of l5kΩ and output resistance of 20KΩ and a current gain of 1000. It is fed by a current source having a source resistance of lOkΩ and its output is connected to a 10Ω load resistance.
19. The voltage gain will be
(A) 661.33 (B) 666.33 (C) None of these (D) 670.33
Correct Answer : B
20.The power gain will be
(A) \(661.33 \times 10^3\) (B) None of these (C) \(670.33 \times 10^3\) (D) \(666.33 \times 10^3\)
Correct Answer : D
19. The voltage gain will be
(A) 661.33 (B) 666.33 (C) None of these (D) 670.33
Correct Answer : B
Solve-19. \(V_i = I_i R_i = 15I_i\)
and \(V_o = \frac {βI_i R_o R_L}{(R_o + R_L)}\) = \(\frac {1000 \times 20 \times 10^3 \times 10}{(20 \times 10^3 + 10)}I_i\)
= \(\frac {2 \times 10^8}{20010}\) ≃ \(9995I_i\)
\(\therefore A_v = \frac {V_o}{V_i}\) = \(frac {9995}{15}\) = 666.33
and \(V_o = \frac {βI_i R_o R_L}{(R_o + R_L)}\) = \(\frac {1000 \times 20 \times 10^3 \times 10}{(20 \times 10^3 + 10)}I_i\)
= \(\frac {2 \times 10^8}{20010}\) ≃ \(9995I_i\)
\(\therefore A_v = \frac {V_o}{V_i}\) = \(frac {9995}{15}\) = 666.33
20.The power gain will be
(A) \(661.33 \times 10^3\) (B) None of these (C) \(670.33 \times 10^3\) (D) \(666.33 \times 10^3\)
Correct Answer : D
Solve-20. Power gain = \(A_p = A_v \times A_I\) = \(666.33 \times 1000\) = \(666.33 \times 10^3\)