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

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.