A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 10 mV is added to the base-emitter voltage, the base current changes by 10µA and the collector current changes by 1.5 mA. The load resistance is 5 kΩ. The voltage gain of the transistor will be ________.

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Answered by

Vishal Baghel | Contributor-Level 10

3 months ago

I_b = 10 µA

I_C = 1.5 mA

R_L = 50 kW or (Rc)

Base – emitter voltage = 10 mv

$R_{B} = \frac{V_{B}}{I_{B}} = \frac{10 \times 1 0^{- 3}}{10 \times 1 0^{- 6}} = 1 0^{3} Ω$

$A_{v} = (\frac{Δ l_{C}}{Δ l_{B}}) \times (\frac{R_{C}}{R_{B}})$

= $\frac{1.5 \times 1 0^{- 3}}{10 \times 1 0^{- 6}} \times \frac{5 \times 1 0^{3}}{1 0^{3}}$

A_v = 750

Ib = 10 µAIC = 1.5 mARL = 50 kW or (Rc)Base – emitter voltage = 10 mv <math> <mrow> <msub> <mrow> <mi>R</mi> </mrow> <mrow> <mi>B</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mrow> <mi>V</mi> </mrow> <mrow> <mi>B</mi> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>I</mi> </mrow> <mrow> <mi>B</mi> </mrow> </msub> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mn>0</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <mn>1</mn> <mn>0</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>6</mn> </mrow> </msup> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mi>Ω</mi> </mrow> </math> <math> <mrow> <msub> <mrow> <mi>A</mi> </mrow> <mrow> <mi>v</mi> </mrow> </msub> <mo>=</mo> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mi>Δ</mi> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mi>C</mi> </mrow> </msub> </mrow> <mrow> <mi>Δ</mi> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mi>B</mi> </mrow> </msub> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mo>×</mo> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>R</mi> </mrow> <mrow> <mi>C</mi> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>R</mi> </mrow> <mrow> <mi>B</mi> </mrow> </msub> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math> = <math> <mrow> <mfrac> <mrow> <mn>1</mn> <mo>.</mo> <mn>5</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <mn>1</mn> <mn>0</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>6</mn> </mrow> </msup> </mrow> </mfrac> <mo>×</mo> <mfrac> <mrow> <mn>5</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math>  Av = 750

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A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 10 mV is added to the base-emitter voltage, the base current changes by 10µA and the collector current changes by 1.5 mA. The load resistance is 5 kΩ. The voltage gain of the transistor will be ________.

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