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.

#### Solution 1

Given:

Output voltage, *V** _{o}* = 2 V, output resistance,

*R*

*= 2 kΩ, base resistance,*

_{o}*R*

*= 1 kΩ*

_{i}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, *R*_{C} = 2 kΩ = 2000 Ω

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

Current amplification factor of the transistor, *β* = 100

Base resistance, *R*_{B} = 1 kΩ = 1000 Ω

Input signal voltage = *V*_{i}

Base current = *I*_{B}

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.