Biasing a BJT: a simple case study

Look at circuit in figure 1: it's a simple circuit with a NPN BJT transistor and some resistors to bias transistor itself. The resistors are chosen to bring the transistor in active area. Remember that a transistor may operate in three areas: interdiction, active and saturation.

First and last areas are used to "transform" the BJT in a switch. Instead, the active area is used to make a transistor to operate as an amplifier. The question is: how much volts can be found on output pins when the transistor is in active area and no signal is present on input pins? Some calculations must be done.

First let calculate the base resistance, that is what an observer could "view" looking through the base; thank to Thevenin's theorem we can write:

R_b = R₁ || R₂ = (R₁* R₂) / (R₁ + R₂) = 99.01 ohm

Let consider the mesh equation from Vcc (12 Volts) to Gnd through the base - emitter of transistor:

V_cc - I_bR_b - V_be - I_eR₄ = 0

i.e. with numerical values

12 - 99.01 I_b - 0.7 - 4700 I_e = 0

The current flow into emitter is the sum of base's current and collector's current

11.3 - 99.01 I_b - 4700 (I_b + I_c) = 0

Assuming that the transistor's beta is 200 and that b=I_c / I_b , we have

11.3 - 99.01 * (99.01 / b) * I_c - 4700 ((1 / b) * I_c + I_c) = 0

i.e. with numerical value of beta

11.3 - 0.495 * I_c - 4.724 * 10³ * I_c = 0

and, solving for Ic, we have

I_c = 11.3 / 4.724*10³ = 2.392*10^-3

Finally, considering the mesh at the output side (V_cc - R₁ * I_c)

V_out = V_cc - 270 * 2.392*10^-3 = 11.35 V