Questions & explanations
1. In a CE amplifier, V_CC = 12 V and load resistance R_C = 2 kΩ. For maximum symmetric output swing without clipping, the Q-point collector current I_C should be approximately:
- 6 mA
- 3 mA
- 4 mA
- 2 mA
Answer: 3 mA
For maximum symmetric swing, Q-point is at the centre of the DC load line. The load line equation is V_CE = V_CC - I_C R_C. At saturation, V_CE = 0, so I_C(sat) = V_CC / R_C = 12 V / 2 kΩ = 6 mA. Midpoint corresponds to I_C = I_C(sat)/2 = 3 mA. This gives equal room for swing towards saturation and cut-off.
2. Which device is best suited to convert sunlight into electrical energy in a solar panel?
- Zener diode
- Photodiode
- LED
- Solar cell
Answer: Solar cell
A solar cell is specifically designed to convert light energy into electrical energy. It operates in the photovoltaic mode, generating a voltage when illuminated. NCERT Class 12 Physics, Chapter 14, Section 14.8 describes the solar cell as a device that converts light into electricity.
3. A photodiode is operated in reverse bias. If the intensity of incident light is doubled, the photocurrent becomes:
- half
- unchanged
- four times
- double
Answer: double
In a photodiode, the photocurrent is directly proportional to the intensity of incident light, as long as the wavelength is suitable. Doubling the intensity doubles the number of photons, which doubles the number of electron-hole pairs generated, thus doubling the photocurrent.
4. In a common-emitter transistor NOT gate with V_CC = 5 V, R_C = 1 kΩ, and R_B = 10 kΩ, β = 100. For V_in = 5 V, what is V_out?
- 0.2 V
- 5 V
- 0.7 V
- 2.5 V
Answer: 0.2 V
When V_in = 5 V, base-emitter junction is forward-biased. Base current I_B = (5 - 0.7)/10k = 0.43 mA. With β = 100, I_C = 43 mA, but saturation occurs when I_C = V_CC/R_C = 5 mA. So transistor saturates, V_out = V_CE(sat) ≈ 0.2 V. Output is LOW, implementing NOT operation.
5. A regulated DC power supply uses a 12 V RMS transformer, bridge rectifier, capacitor filter, and a 6 V Zener regulator. What is the approximate output DC voltage?
- 12 V
- 6 V
- 17 V
- 24 V
Answer: 6 V
The bridge rectifier and capacitor filter produce a DC voltage close to the peak voltage (12√2 ≈ 17 V) minus two diode drops (≈ 1.4 V), giving about 15.6 V. However, the Zener regulator clamps the output to its breakdown voltage of 6 V, so the final output is 6 V.
6. Which of the following correctly matches the device with its bias condition and energy conversion?
- Photodiode: forward bias, light to electrical; LED: reverse bias, electrical to light; Solar cell: unbiased, light to electrical
- Photodiode: reverse bias, light to electrical; LED: forward bias, electrical to light; Solar cell: unbiased, light to electrical
- Photodiode: reverse bias, electrical to light; LED: forward bias, light to electrical; Solar cell: unbiased, light to electrical
- Photodiode: reverse bias, light to electrical; LED: forward bias, electrical to light; Solar cell: forward bias, light to electrical
Answer: Photodiode: reverse bias, light to electrical; LED: forward bias, electrical to light; Solar cell: unbiased, light to electrical
According to NCERT, a photodiode operates in reverse bias and converts light into electrical current. An LED operates in forward bias and converts electrical energy into light. A solar cell operates with no external bias and converts light into electrical energy.
7. In a common-emitter transistor amplifier, the output characteristic shows that collector current I_C is nearly independent of V_CE in the active region. This is primarily because:
- the base is thick and heavily doped, so recombination is high
- the emitter is lightly doped, so few carriers are injected
- the base is thin and lightly doped, so most carriers from emitter reach collector
- the collector is heavily doped, so it attracts all carriers
Answer: the base is thin and lightly doped, so most carriers from emitter reach collector
The near-constant I_C in the active region is due to the base being thin and lightly doped. This ensures that most charge carriers injected from the emitter cross to the collector without recombining, making I_C almost equal to I_E and independent of V_CE.
8. In a Zener voltage regulator, the input voltage is 15 V and the Zener diode has Vz = 6 V. If the series resistor is 300 Ω, what is the current through the Zener diode when the load is disconnected?
- 20 mA
- 50 mA
- 30 mA
- 40 mA
Answer: 30 mA
When the load is disconnected, the entire current from the series resistor flows through the Zener diode. The voltage across the series resistor is Vin - Vz = 15 V - 6 V = 9 V. Using Ohm's law, Iz = (Vin - Vz)/Rs = 9 V / 300 Ω = 0.03 A = 30 mA.
9. In a CE amplifier with V_CC = 12 V and R_C = 3 kΩ, the base current is fixed at 40 μA. The output characteristic for I_B = 40 μA gives I_C = 2 mA at V_CE = 6 V. What is the Q-point (I_C, V_CE)?
- (2 mA, 6 V)
- (4 mA, 0 V)
- (2 mA, 12 V)
- (4 mA, 6 V)
Answer: (2 mA, 6 V)
The Q-point is the intersection of the DC load line (V_CE = V_CC - I_C R_C) with the output curve for I_B = 40 μA. The given point (2 mA, 6 V) satisfies V_CE = 12 - 2*3 = 6 V, and lies on the I_B = 40 μA curve. So Q-point is (2 mA, 6 V).
10. In a common-emitter transistor amplifier, which terminal is common to both input and output circuits?
- Emitter
- Ground
- Collector
- Base
Answer: Emitter
In common-emitter configuration, the emitter is the terminal common to both input (base-emitter) and output (collector-emitter) circuits. The input is applied between base and emitter, and output is taken between collector and emitter.
11. In a CE amplifier with V_CC = 12 V, R_C = 2 kΩ, and β = 100, the Q-point is set at V_CEQ = 6 V. What is the voltage gain if the input resistance is 1 kΩ?
- 100
- 50
- 200
- 300
Answer: 200
First, find I_CQ: I_CQ = (V_CC - V_CEQ)/R_C = (12-6)/2000 = 3 mA. Then I_BQ = I_CQ/β = 30 μA. The voltage gain magnitude is A_v = β R_C / r_i = 100 × 2000 / 1000 = 200. The negative sign indicates phase reversal, but magnitude is 200.
12. In a bridge rectifier with a 12 V (rms) secondary voltage, what is the peak voltage across the load? (Assume ideal diodes.)
- 12 V
- 24 V
- 8.5 V
- 17 V
Answer: 17 V
For a bridge rectifier, the peak secondary voltage is V_m = √2 × V_rms = 1.414 × 12 V ≈ 17 V. Since two diodes conduct in series, but ideal diodes have zero drop, the load sees the full peak voltage. Thus the correct answer is 17 V.