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

Configuration

Analysis

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 , $V_{C}=-nV_{i}$

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.