Electronics Fundamentals/Semi Conductor Devices/Transistor/Electronic DC Amplifier

Configuration

Assume that the transistor is ideal.

V_{B} = V_{i} \frac{R_{2}}{R_{2} + R_{1}}

V_{E} = V_{B} - 0.7 = V_{B}

I_{E} = \frac{V_{E}}{R_{4}} = \frac{V_{B}}{R_{4}}

I_{C} = I_{E}

V_{C} = I_{C} R_{4}

V_{C} = V_{i} (1 - \frac{R_{3}}{R_{4}} \frac{R_{2}}{R_{2} + R_{1}})

When $R_{2} > > R_{1}$ or R₁ = 0. The equation above is reduced to

V_{C} = V_{i} (1 - \frac{R_{3}}{R_{4}})

R₃ = 0 ,

V_{C} = V_{i}

R₃ = R₄ ,

V_{C} = 0

R₃ = 2R₄ ,

V_{C} = - V_{i}

R₃ = (n+1)₄ ,

V_{C} = - n V_{i}

When $R_{3} = R_{4}$ . The equation above is reduced to

V_{C} = V_{i} (1 - \frac{R_{2}}{R_{2} + R_{1}})

When R₁ = 0 or R₂ = 0,

V_{C} = V_{i}

. Transistor operates like a closed switch

When R₂ >> R₁

V_{C} = 0

. Transistor operates like an opened switch

With the right biased voltage at the base, the transistor can be operated like a mechanical switch.