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Course by: Bijay_Kumar Sharma. E-mail the author

# AE_Lecture5_PartC_High Frequency Analysis of CB & CC Amplifier

Module by: Bijay_Kumar Sharma. E-mail the author

Summary: In AE-Lecture5_Part C we find the analytic relationship of the upper cutoff frequency of CB and CC amplifier.

High Frequency Response Of CB Amplifier.

CB Amplifier is almost an uni-lateral circuit. It has negligible reverse transmission. Hence it is ideal for RF applications where there is always a danger of parasitic oscillations.

In Figure 1, we have a CB BJT Amplifier with two battery biasing. This circuit has to be analyzed for its upper cut off frequency.

In Figure 2, we have given the high frequency model of CB BJT using the T-Model of BJT.

Following is the open circuit time constant method for arriving at the upper cut-off frequency of the amplifier.

HIGH FREQUENCY MODEL OF CC AMPLIFIER

CC Amplifier is also known as Emitter Follower. It is also known as Buffer. Buffer isolates the output system from input system.

It is used for driving transmission lines. It is a voltage controlled voltage source. Its output is a constant current source hence it is suitable for driving variable load.

In Figure 3, we have CC Amplifier with self bias. We will analyze the upper cut-off frequency of the amplifier.

The Trnsistor is operating at Q point (1.5mA,5V). Following Hybrid-π are given:

High Frequency INCREMENTAL CIRCUIT of the CC Amplifier is:

Figure 4 is given as supplementary of AE_lecture5_PartC

ibis the useful transistor current which takes part in transistor action and is the component of ib , the base current , which flows through rπ .

The

Thevenin Equivalent Voltage at the input of the amplifier is:

RTh = RS ││RB

v

What is τµo ( open-circuit time constant associated with Cµ ) and τπo ( open-circuit time constant associated with Cπ )?

τµo=(Cµ)(R10)

τπo=(Cπ)(R20)

Circuit for finding R1o

To find the equivalent resistance seen by Cµ, we apply a voltage source at base and ground.

Total current drawn from vo is i1+i2.

Therefore vo/(i1+i2)= R1o

τµo=(0.5pF x 1.1kΩ)=0.54nsec

Circuit for finding R2o

A voltage source is applied at the node pair where Cπ was connected.

Total current drawn is i0 = i1 +i2 ;

And βf0i1 gmvπ

Req =

This expression is derived as follows(Refer to Figure 10):

CC has unity gain hence largest 159 MHz largest B.W.

 Configuration B.W. CE 1.56MHz CE Degenerate 10.7MHz CB 15.9MHz CC 159MHz

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