Multi-functional Packaged Antennas for Next
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Transcript Multi-functional Packaged Antennas for Next
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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