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Example.

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Presentation on theme: "Example."— Presentation transcript:

1 Example

2 Example contd….

3 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.

4 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 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

5 Analysis of four-resistor bias circuit contd.

6 Example of a Four-resistor bias circuit

7 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

8 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 = mS rπ = βVT/ICQ = 150x26 mV/10 mA = 390 Ω


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