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For a Ce-transistor Amplifier, the Audio Signal Voltage Across the Collector Resistance of 2 kΩ is 2 V. Suppose the Current Amplification Factor of the Transistor is 100, Find the Input Signal Voltage and Base Current, If the Base Resistance is 1 kΩ. - Physics

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For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current if the base resistance is 1 kΩ.

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Solution 1

Given:
Output voltage, Vo = 2 V, output resistance, Ro = 2 kΩ, base resistance, Ri = 1 kΩ
Current amplification factor, β = 100

Then input signal voltage is calculated as:

`V_0/V_i=R_0/R_ixxbeta=>2/V_i=2/1xx100`

`=>V_i=10mV`

Now, collector current is

`I_C=V_0/R_0=2/2=1mA`

Therefore, base current is

`I_B=I_C/beta=(1mA)/100=10muA`

Solution 2

Collector resistance, RC = 2 kΩ = 2000 Ω

Audio signal voltage across the collector resistance, V = 2 V

Current amplification factor of the transistor, β = 100

Base resistance, RB = 1 kΩ = 1000 Ω

Input signal voltage = Vi

Base current = IB

We have the amplification relation as:

Voltage amplification = `V/V_' = beta R_C/R_B`

`V_i = V R_B/(beta R_C)`

`= (2 xx 1000)/(100 xx 2000) = 0.01 V`

Therefore, the input signal voltage of the amplifier is 0.01 V.

Base resistance is given by the relation:

`R_B = V_i/I_B`

`= 0.01/1000 = 10 xx 10^(-6)mu^6 A`

Therefore, the base current of the amplifier is 10 μA.

Concept: Junction Transistor - Transistor as an Amplifier (Ce-configuration)
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APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits
Q 9 | Page 510
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