Question

In a CE transistor amplifier, the audio signal voltage across the collector resistance of 2kOmega is 2V. If the base resistance is 1kOmega and the current amplification of the transistor is 100, the input signal voltage is:

Answer 1

★ We know that:-

Collector resistance, $R _{C}$ = $2k\Omega= 2000\Omega$

Voltage of audio signal across the collector resistance, V = 2V

Current amplification factor of the transistor, $\beta = 100$

Base resistance, $R _{B}$ = $1k\Omega = 1000 \Omega$

Input Signal = $V_{i}$

Base current = $I _{B}$

We have amplification relation as,

$\frac{V}{V_{i}} = \beta \frac{R_{C}}{R_{B}}$

$\frac{2 \times 1000}{100 \times 2000} = 0.01V$

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

★ Below is the relation of base resistance:-

$R_{B} = \frac{V_{i}}{I_{B}}$

$\frac{0.01}{1000} = {10}^{ - 6} \times 10\mu A$

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

Answer 2

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Answer:

d) 10mV

Explanation:

Voltage = Rc = 2 k = 2 × 10³

(Given)Base resistance = Rb = 1 k

(Given)Vo=2VInput voltage (Vi) = ?

Thus, using the equation

-B = Ic/lb = 100VolcRc2V

Ic=2/Rc = 2/2 × 103 Q= 10-3 A =1mA

Calculating the base current

= lb = lc / B= 10-3/100= 10 × 10-6 A= 10 ΜΑ

Input resistance voltage

= Vi = Rb / lb= 1 x 103 x 10 x 10-6

= 10mV

Thus, the input signal voltage is 10mV,