Three identical amplifiers with each one having a voltage gain of 50,input resistance of 1KΩ & output resistance of 250, are cascaded. The open circuit voltage gain of combined amplifier is
A. 49dB
B. 51dB
C. 98dB
D. 102dB
Answer: Option C
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The action of JFET in its equivalent circuit can best be represented as a
A. Current controlled Current source
B. Current controlled voltage source
C. Voltage controlled voltage source
D. Voltage controlled current source
In a p+n junction diode under reverse bias, the magnitude of electric field is maximum at
A. The edge of the depletion region on the p-side
B. The edge of the depletion region on the n-side
C. The p+n junction
D. The center of the depletion region on the n-side
To prevent a DC return between source and load, it is necessary to use
A. Resistor between source and load
B. Inductor between source and load
C. Capacitor between source and load
D. Either A or B
We can use the formula for cascaded amplifiers:
[ A_{text{total}} = A_1 times A_2 times A_3 ]
Where ( A_{text{total}} ) is the total voltage gain and ( A_1 ), ( A_2 ), and ( A_3 ) are the voltage gains of the individual amplifiers.
Given that each amplifier has a voltage gain of 50, we have:
[ A_{text{total}} = 50 times 50 times 50 = 125,000 ]
To convert this to dB, we use the formula:
[ text{Gain (dB)} = 10 times log_{10}(A_{text{total}}) ]
[ text{Gain (dB)} = 10 times log_{10}(125,000) ]
[ text{Gain (dB)} approx 98.04 ]
So, the correct answer is:
C. 98dB
DB means
20log(50*50*50)=102
20log(50*50*50)=102
20 log(50*50*50) = 101.9 = 102
How....?