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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.

Figure 1
Figure 1 (Picture 1.png)

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.

Figure 2
Figure 2 (Picture 5.png)

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.

Figure 3
Figure 3 (graphics1.png)
Figure 4
Figure 4 (graphics2.png)
Figure 5
Figure 5 (graphics3.png)

Figure 6
Figure 6 (graphics4.png)

Figure 7
Figure 7 (graphics5.png)

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.

Figure 8
Figure 8 (Picture 3.png)

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:

Figure 9
Figure 9 (graphics6.png)
Figure 10
Figure 10 (graphics7.png)
Figure 11
Figure 11 (graphics8.png)
Figure 12
Figure 12 (graphics9.png)
Figure 13
Figure 13 (graphics10.png)

High Frequency INCREMENTAL CIRCUIT of the CC Amplifier is:

Figure 4 is given as supplementary of AE_lecture5_PartC

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

The

Figure 14
Figure 14 (graphics11.png)
Thevenin Equivalent Voltage at the input of the amplifier is:

RTh = RS ││RB

Figure 15
Figure 15 (graphics12.png)
Figure 16
Figure 16 (graphics13.png)

Figure 17
Figure 17 (Picture 7.png)
v

Figure 18
Figure 18 (Picture 9.png)

Figure 19
Figure 19 (graphics14.png)

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

Figure 20
Figure 20 (Picture 10.png)

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

Figure 21
Figure 21 (graphics15.png)
Figure 22
Figure 22 (graphics16.png)
Figure 23
Figure 23 (graphics17.png)

τµo=(0.5pF x 1.1kΩ)=0.54nsec

Circuit for finding R2o

Figure 24
Figure 24 (Picture 12.png)

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

Total current drawn is i0 = i1 +i2 ;

Figure 25
Figure 25 (graphics18.png)
Figure 26
Figure 26 (graphics19.png)
Figure 27
Figure 27 (graphics20.png)
Figure 28
Figure 28 (Picture 14.png)
Figure 29
Figure 29 (graphics21.png)

And βf0i1 gmvπ

Figure 30
Figure 30 (Picture 15.png)

Req =

Figure 31
Figure 31 (graphics22.png)
This expression is derived as follows(Refer to Figure 10):

Figure 32
Figure 32 (graphics23.png)
Figure 33
Figure 33 (graphics24.png)
Figure 34
Figure 34 (graphics25.png)
Figure 35
Figure 35 (graphics26.png)
Figure 36
Figure 36 (graphics27.png)
Figure 37
Figure 37 (graphics28.png)
Figure 38
Figure 38 (graphics29.png)
Figure 39
Figure 39 (graphics30.png)

Figure 40
Figure 40 (graphics31.png)

Figure 41
Figure 41 (graphics32.png)
Figure 42
Figure 42 (graphics33.png)
Figure 43
Figure 43 (graphics34.png)
Figure 44
Figure 44 (graphics35.png)

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

Table 1
Configuration B.W.
CE 1.56MHz
CE Degenerate 10.7MHz
CB 15.9MHz
CC 159MHz

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