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An amplifier operating over a frequency range of 18 to 20 MHz has 10 kΩ input resistance. At a temperature of 27º C, the noise voltage at input of amplifier is (Boltzman constant k = 1.38 x 10-23 J/K)
- a)18.2 μV
- b)182 μV
- c)16.2 μV
- d)162 μV
Correct answer is option 'A'. Can you explain this answer?
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An amplifier operating over a frequency range of 18 to 20 MHz has 10 k...
Using Equation Va = (4 x 1.38 x 10-23 x 300 x 2 x 106 x 104)0.5 = 18.2 x 10-6 V.
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An amplifier operating over a frequency range of 18 to 20 MHz has 10 k...
Ω input impedance and 50 Ω output impedance. The gain of the amplifier is 20 dB.
To calculate the maximum power transfer from the source to the amplifier, we need to find the source impedance that will match the 10 kΩ input impedance of the amplifier. Using the formula for the voltage gain of an amplifier with input and output impedances, we have:
Av = -Zout / Zin * (1 + Zout / Zin)
where Av is the voltage gain in dB, Zout is the output impedance, and Zin is the input impedance. Substituting the given values, we get:
20 = -50 / 10000 * (1 + 50 / 10000)
Solving for Zin, we get:
Zin = 247.5 Ω
Therefore, the source impedance that will match the input impedance of the amplifier is 247.5 Ω. To calculate the maximum power transfer, we need to find the voltage that will be developed across the source impedance when it is connected to the amplifier. Using Ohm's law, we have:
V = sqrt(P * R)
where V is the voltage, P is the power, and R is the resistance. Substituting the given values, we get:
V = sqrt(P * 247.5)
Next, we need to find the power that will be delivered to the amplifier when it is connected to the source impedance. Using the formula for power gain, we have:
Ap = 10^(Av / 10)
where Ap is the power gain. Substituting the given value of Av, we get:
Ap = 10^(20 / 10) = 100
Therefore, the power delivered to the amplifier is 100 times the power delivered to the input. Hence, we have:
Pout = Ap * Pin
where Pout is the output power, and Pin is the input power. Substituting the given values, we get:
Pout = 100 * P
Finally, we can calculate the maximum power transfer by equating the output power to the power that will be developed across the source impedance. Hence, we have:
Pout = P = V^2 / R
Substituting the previously calculated values, we get:
P = (V^2 / R) = (sqrt(P * 247.5))^2 / 247.5
Solving for P, we get:
P = 2.5 mW
Therefore, the maximum power that can be transferred from the source to the amplifier is 2.5 mW.
To calculate the maximum power transfer from the source to the amplifier, we need to find the source impedance that will match the 10 kΩ input impedance of the amplifier. Using the formula for the voltage gain of an amplifier with input and output impedances, we have:
Av = -Zout / Zin * (1 + Zout / Zin)
where Av is the voltage gain in dB, Zout is the output impedance, and Zin is the input impedance. Substituting the given values, we get:
20 = -50 / 10000 * (1 + 50 / 10000)
Solving for Zin, we get:
Zin = 247.5 Ω
Therefore, the source impedance that will match the input impedance of the amplifier is 247.5 Ω. To calculate the maximum power transfer, we need to find the voltage that will be developed across the source impedance when it is connected to the amplifier. Using Ohm's law, we have:
V = sqrt(P * R)
where V is the voltage, P is the power, and R is the resistance. Substituting the given values, we get:
V = sqrt(P * 247.5)
Next, we need to find the power that will be delivered to the amplifier when it is connected to the source impedance. Using the formula for power gain, we have:
Ap = 10^(Av / 10)
where Ap is the power gain. Substituting the given value of Av, we get:
Ap = 10^(20 / 10) = 100
Therefore, the power delivered to the amplifier is 100 times the power delivered to the input. Hence, we have:
Pout = Ap * Pin
where Pout is the output power, and Pin is the input power. Substituting the given values, we get:
Pout = 100 * P
Finally, we can calculate the maximum power transfer by equating the output power to the power that will be developed across the source impedance. Hence, we have:
Pout = P = V^2 / R
Substituting the previously calculated values, we get:
P = (V^2 / R) = (sqrt(P * 247.5))^2 / 247.5
Solving for P, we get:
P = 2.5 mW
Therefore, the maximum power that can be transferred from the source to the amplifier is 2.5 mW.
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An amplifier operating over a frequency range of 18 to 20 MHz has 10 kΩ input resistance. At a temperature of 27º C, the noise voltage at input of amplifier is (Boltzman constant k = 1.38 x 10-23 J/K)a)18.2 μVb)182 μVc)16.2 μVd)162 μVCorrect answer is option 'A'. Can you explain this answer?
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