A common emitter amplifier has a voltage gainof 50, an input impedance of 100Ω and anoutput impedance of 200Ω. The power gain ofthe amplifier is [2010]a)500b)1000c)1250d)50Correct answer is option 'C'. Can you explain this answer?

Question

A common emitter amplifier has a voltage gainof 50, an input impedance of 100&Omega; and anoutput impedance of 200&Omega;. The power gain ofthe amplifier is [2010]a)500b)1000c)1250d)50Correct answer is option 'C'. Can you explain this answer?

Answer

Ohms, and an output impedance of 1 kilohm. The amplifier is biased to operate in the active region of the transistor.

To calculate the DC bias point of the amplifier, we need to use the following equations:

IB = (VCC - VBE) / RB

IC = β * IB

VCE = VCC - IC * RC

where VCC is the supply voltage, VBE is the base-emitter voltage of the transistor, RB is the base resistor, β is the transistor's current gain, IC is the collector current, and RC is the collector resistor.

Assuming VBE = 0.7V and β = 100, we can solve for the bias point:

IB = (10V - 0.7V) / RB = 9.3V / RB

IC = β * IB = 100 * 9.3V / RB = 930V / RB

VCE = 10V - IC * 1kΩ = 10V - 930V / RB * 1kΩ

Next, we can calculate the small-signal parameters of the amplifier. The voltage gain is given by:

Av = -gm * RC

where gm is the small-signal transconductance of the transistor. For a common emitter configuration, gm is given by:

gm = IC / VT

where VT is the thermal voltage, approximately 26 mV at room temperature.

Thus, gm = 930V / RB / 26mV = 35.77 / RB

Av = -35.77 * RC / RB

The input impedance is given by:

Zin = RB

The output impedance is given by:

Zout = RC

Finally, we can calculate the maximum output voltage swing before the transistor enters saturation. Since the maximum collector current is limited by the saturation current, we have:

Vout,max = VCC - IC,sat * RC

where IC,sat is the saturation current of the transistor, which is typically around 10% of the maximum collector current.

Assuming IC,max = 10 mA, we have:

IC,sat = 1 mA

Vout,max = 10V - 1 mA * 1 kΩ = 9V

Therefore, the maximum output voltage swing is 9V peak-to-peak.

Answer

Power gain = (voltage gain)^2 * Ro/Ri is the correct formula

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