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The correct answer is (b) BASE

Easy EXPLANATION: The collector feedback configuration is done by connecting a RESISTOR from the collector to the base VOLTAGE. This is done to stabilize the biasing voltage against thermal runaway.

Right answer is (a) Not much effect

Easy EXPLANATION: The increase in COLLECTOR resistance doesn’t have much IMPACT on the output voltage SINCE the BIAS voltage is kept stable by the feedback operation. After all, one of the important applications of feedback is increasing stability and the output voltage is kept stable by this configuration.

Right option is (a) β+1/{1 + β * R1/(R1 + β)}

Easy explanation: The stability factor is determined by calculating the change in IC with respect to the change in ICO. Hence, we can SIMPLY apply the method of K.V.L. and derive a RELATION between these two currents. After DIFFERENTIATION, we’ll get the stability factor as β+1/{1 + β * R1/(R1 + β)}.

Correct CHOICE is (b) R1 << R2

For explanation I would say: The feedback resistance should be MUCH smaller than the COLLECTOR resistance SINCE we NEED to reduce the sensitivity of the collector current to the current gain. Typically, R2 should be lower than R1 by a factor of β.

The correct answer is (b) 1 + β * R1/(R1 + β)

To EXPLAIN I would say: This can be simply OBSERVED from the stability FACTORS of both cases. The correct factor is 1 + β * R1/(R1 + β).

The correct option is (c) β/(RB + Rc * (1 + β))

To EXPLAIN: This is easily derived from the COLLECTOR feedback configuration by using the METHOD of K.V.L. deriving a relation between the collector current and the base emitter voltage. It should be noted that the stability against VBE INCREASES in comparison to the base bias comparison.

Right answer is (d) 140

Easiest explanation: The Base resistance should be substantially lower than the COLLECTOR resistance by a FACTOR of β. 110 is a good CHOICE but provided 140 is present as an option, a better choice is a factor of 140.

The correct OPTION is (a) To maintain a stable q-point irrespective of β

Explanation: If we change the DEVICE, β changes. But we want to keep the q-point stable so that CIRCUIT if represented as a black box, would provide the same characteristics and not be highly DEPENDENT on the transistor. Hence, we want to MAKE the circuit insensitive to β.

Right answer is (b) False

To explain I WOULD say: The stability FACTORS are independent of WHETHER the transistor is of npn or pnp type. It is only dependent on β and the impedance connected to the terminals of the transistor of a hyperbola.

Right answer is (a) lower cut-off FREQUENCY and upper cut-off frequency

To elaborate: The frequencies are CALLED lower cut-off frequency and upper cut-off frequency. At these frequencies power of the SIGNAL becomes half of its original value, 3dB LESS than the MAXIMUM and they are also called 3dB cut-off frequencies.

Right option is (a) Difference between upper cut-off FREQUENCY and LOWER cut-off frequency

Explanation: 3DB bandwidth of an amplifier is the difference between upper cut-off frequency and lower cut-off frequency. The unity gain bandwidth is the difference between frequencies where gain is 1.

Right option is (B) 70.7% of MAXIMUM gain

Explanation: 3dB cut-off frequency is the frequency at which the power becomes HALF of its maximum value. That is the voltage gain becomes 0.707 times maximum voltage gain.

The correct OPTION is (a) HALF of MAXIMUM VALUE

The best explanation: 3DB cut-off frequency is the frequency at which the power becomes half of its maximum value. That is the voltage gain becomes 0.707 times maximum voltage gain. The importance of 3dB points is that for a human ear, it will not notice the fall in power of the signal up to 50% of maximum power.

Right answer is (a) 70.7

Best explanation: 3dB cut-off frequency is the frequency at which the power becomes HALF of its maximum value. That is the VOLTAGE GAIN becomes 0.707 times maximum voltage gain. THEREFORE if voltage gain is 100 then gain at 3dB frequencies will be

100 x 0.707 = 70.7.

Correct option is (a) 20dB/decade

Easy explanation: At lower and HIGHER frequencies the GAIN is decreasing. This fall or DECREASE in gain is known as roll-off rate. It is COMMONLY specified in dB/octave or dB/decade. For a single order FILTER, the roll-off rate is 6dB/octave or 20dB/decade.

Correct choice is (b) 0.25

The EXPLANATION is: The RELATION between dB and power gain is

 dB = 10 log(power gain)

 That is power gain for -6DB is

 10(-6 /10) = 0.251.

The correct choice is (b) -20dB

For explanation: The initial POWER GAIN in dB = 10 LOG (output power)

= 10 log(100) = 20dB

The final power gain in dB = 10 log(output power)

= 10 log(1) = 0 dB

So change in power = final power – initial power

= 0-20= -20dB.

Correct CHOICE is (a) 12dB

Easy explanation: Gain of the system is = output VOLTAGE/input voltage

= 48/12 = 4

Gain in dB = 20 LOG(voltage gain)

= 20 log(4) = 12.04 dB.