Example

Example contd….

Example Thus the fixed bias circuit has the limitation that changes in  does not change the base current. Thus the transistor can switch from active region operation to saturation or cut-off very easily.

Analysis of four-resistor bias circuit The voltage divider using R1 and R2 is useful in maintaining a constant bias to the base irrespective of changes in . This can be done by choosing the resistors so that the base current is a small fraction of the total current flowing through them Also, since the base in not directly connected to the supply or ground an ac signal can be easily coupled through a coupling capacitor Very small resistance can lead to overheating. Normally resistors are chosen so that 10-20 times base current flows through them. The Thevenin equivalent resistance RB is a parallel combination of R1 and R2 and given as The Thevenin voltage VB is given as

Analysis of four-resistor bias circuit contd.

Example of a Four-resistor bias circuit

Small signal equivalent circuit: Signal notations The instantaneous values of current and voltage, iB(t) and vBE (t), respectively, are given as: Recall that for a diode the dynamic resistance rd is given as rd = nVT/IDQ. For BJT, the base-emitter behaves as the diode and the dynamic resistance is called rπ, where At room temperature: VT = 26 mV. Thus, for a typical β = 100, and typical IC = 1 mA, rπ = 2600 Ω

Small signal analysis Since ib (t) and Vbe (t) are ac quantities, they are related by the equation: From the relationship between collector and base current we also have: Using the relationships and The equivalent circuit can be drawn as in part (a) of the figure at the side Defining a quantity called transconductance gm as gm = β/rπ = ICQ/VT, we obtain The equivalent circuit can then be drawn as in part (b) of the figure at the side Example: A t room temperature a certain transistor has β = 150. Calculate gm and rπ if ICQ = 10 mA. Solution: We have gm = ICQ/VT = 10 mA/ 26 mV = 384.6 mS rπ = βVT/ICQ = 150x26 mV/10 mA = 390 Ω