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 , $V_C=-n{V}_i$

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.