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Example
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Example contd….
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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.
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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
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Analysis of four-resistor bias circuit contd.
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Example of a Four-resistor bias circuit
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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
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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
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