The lower 3dB frequency of the amplifier circuit shown below is
The lower 3 dB frequency ω_{L }
The efficiency of a class B amplifier with a supply voltage of 24V and peak output voltage V_{o} = 22V is _______%
For a class B amplifier with V_{CC} = 24 V and V_{O} = 22 V efficiency η (in %) is given by
In class AB amplifiers, the current flows through the active device for
For class A the angle of excursion is 360^{0}
For class B the angle of excursion is 180^{0}
For class AB the angle of excursion is >180^{0} and < 360^{0}
For class C the angle of excursion is less than 180^{0}
A class – A transformer coupled, the Transistor power amplifier is Required to deliver a power output of 10 watts, the maximum power rutting of the transistor should not be less than –
P_{iDC} = P_{oac} + P_{Tr}
P_{ooc} = 10 W
η = 50% for classA transformer coupled
In the following transistor circuit,V_{BE }= 0.7 V, r_{π} = 25 mV / I_{E}, and β and all the capacitances are very large
The midband voltage gain of the amplifier is approximately
Apply ac analysis to calculate gain
∴ Option (4) is correct.
A Transformer coupled class A amplifier has a loudspeaker of 8 Ω connected or across its secondary. The Q point of collector current is 140 mA. The Turns ratio of the transformer is 3:1. If the power delivered to the load is 0.48 W. Then the rms voltage at the primary and efficiency of the transformer is
Given
P_{AC}(secondary) = 0.48 W
I_{CQ} = 140 mA
R_{L} = 8 Ω
P_{AC(sec)} = P_{AC(primary)}
P_{AC(primary)} = 0.48 W
V_{1(rms)} = 1.96 × 3 = 5.88 V
output DC power
P_{DC} = V_{CC} I_{CQ}
= 10(140 × 10^{3})
= 1.4 W
The low frequency cutoff frequency due to source capacitance (0.47 uF) is ________ Hz
Take V_{T} = 26 mV. ( assume 0.47 uF capacitance at input is source capacitance )
The cutoff frequency due to source capacitance
Where R_{S} = source resistance
R_{i} = Input impedance
Where R_{i} = R_{1} // R_{2}// βr_{e
}
In DC bias values
Taking KVL in baseemitter path
V_{B}  V_{BE}  I_{E} (1.2 Ω) = 0
I_{E} = 0.917 mA
∴ R_{i} = R_{1} // R_{2} // βv_{e}
= 10 kΩ // 68 kΩ // (120 × 28.36) Ω
≅ 2.44 k Ω
= 103. 87 Hz
For a class A transformer coupled amplifier calculate the maximum power delivered to the load when its R_{L} = 10 Ω, n = 5 (turns ratio) and V_{CC} = 16 V
For class A transformer coupled amplifier
η = 50%
The amplifier is shown below as a voltage gain of 2.5, and input resistance of 10 kΩ and a lower 3 dB cutoff frequency of 20 Hz. Which on the following statements is TRUE when the emitter resistance R_{E} is doubled?
The Voltage Gain of the common emitter amplifier is equal to the ratio of the change in the input voltage to the change in the amplifiers output voltage. Then ΔV_{L} is V_{out }and ΔV_{B} s Vin. But voltage gain is also equal to the ratio of the signal resistance in the Collector to the signal resistance in the Emitter and is given as:
Voltage gain
∴ If R_{E} increase then voltage gain will decrease and –ve feedback increase
In the circuit shown in figure find the voltage drop (in V) in 1 kΩ Resistor. Assume β = 100
Assume that Q_{2} in an. It fallows that current will flow from ground through the 1kΩ Resistor into the emitter of Q_{2}. Thus the base of Q_{2} will be at –Ve voltage and base current will be flowing out of the through the 10 kΩ resistor and into +5V supply. This is impossible, since if the base is negative. Current in the 10 kΩ resistor will have to flow into the base. Thus we conclude that our original assumption that Q_{2} is ON is incorrect. It follows that Q_{2} will be off and Q_{1} will be ON.
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