Download 4. Replace the BJT with one of its small

JRE SCHOOL OF Engineering
Solution of CLASS TEST-1 EXAMINATIONS MARCH 15
Subject Name
ELECTRONIC
CIRCUITS
Roll No. of Student
Date
13 MARCH 2015
For ECE & AEI branches only
Subject Code
NEC 402
Max Marks
Max Duration
Time
30
1 Hr
9:20 AM-10:20 AM
SECTION – A
(3 marks * 5 Questions = 15 Marks)
Attempt any five:
1) Write the steps through which small signal equivalent circuit model can be used in
the analysis of transistor amplifier.
Soln. 1. Determine the dc operating point of the BJT and in particular the dc collector current I C.
2. Calculate the values of the small-signal model parameters:
3. Eliminate the dc sources by replacing each dc voltage source with a short circuit and each dc
current source with an open circuit.
4. Replace the BJT with one of its small-signal equivalent circuit models. Although any one of the
models can be used, one might be more convenient than the others for the particular circuit being analyzed.
5. Analyze the resulting circuit to determine the required quantities (e.g., voltage gain, input
resistance).
2) Draw the high frequency hybrid-π model of BJT.
Soln.
3) Discuss the various internal capacitances for BJT.
Soln.Various internal capacitances of BJT are as follows:
1) The Base-Charging or Diffusion Capacitance Cd
2) The Base–Emitter Junction Capacitance Cje
3)The Collector–Base Junction Capacitance C:
4) Derive the expression given below for the high frequency hybrid- π model of BJT:
Soln.
5) Draw the small signal models of BJT.
Soln.
6) Analyze the circuit as shown in Fig.1(Fig.2a) to
determine the voltages at all nodes and the
currents through all branches. Assume that the
transistor  is specified to be at least 50.
Soln.
Note that the circled numbers indicate the order of the analysis
steps.
1) Assuming active-mode operation, we have
2) Our assumption is wrong.In fact, the transistor has to be in
the saturation mode. Assuming this to be the case, we have
SECTION – B
(5 marks * 1 Questions = 5 Marks)
1) Explain the operation of a BJT as a switch?
Soln.To operate the BJT as a switch, we utilize the cutoff and the
saturation modes of operation. To illustrate, consider the commonemitter circuit shown in Fig (a-b), as the input vin is varied.
1) CUT-OFF REGION ( SWITCH IS OPEN/OFF MODE )
Here the operating conditions of the transistor are zero input base
current ( IB ), zero output collector current ( IC ) and maximum
collector voltage ( VCE ) which results in a large depletion layer and
no current flowing through the device. Therefore the transistor is
switched “Fully-OFF”.
Fig (a) Switch is OPEN
2) SATURATION REGION ( SWITCH IS CLOSED )
Here the transistor will be biased so that the maximum amount of base
current is applied, resulting in maximum collector current resulting in the
minimum collector emitter voltage drop which results in the depletion layer
being as small as possible and maximum current flowing through the
transistor. Therefore the transistor is switched “Fully-ON”.
Fig (b) Switch is CLOSED
OR
2) Analyze the transistor amplifier shown in Fig. 2(Fig.
3a) to determine its voltage gain. Assume =100.
Soln.1 . Determine the quiescent operating pointThe dc base current
will be
Fig 3(a) Circuit 3(b) dc analysis
The transistor will be operating in the active mode. The dc analysis is illustrated in Fig.3 (b).
2. Determine the small-signal model parameters:
3. Draw Equalent circuit:
4. To carry out the small-signal analysis it is equally convenient to employ either of the two hybrid-π
equivalent circuit models as shown in Fig.3(c).
5. Analysis of the equivalent circuit in Fig.3(c) proceeds as follows:
The output voltage vo is given by
Thus the voltage gain will be
3(c) small-signal model
where the minus sign indicates a phase reversal.
SECTION – C
(10 marks * 1 Questions = 10 marks)
1) A transistor amplifier is fed with a signal source having an open-circuit voltage vsig of
10 mV and an internal resistance Rsig of 100 kΩ. The voltage vi at the amplifier input
and the output voltage vo are measured both without and with a load resistance
RL=10 kΩ connected to the amplifier output. The measured results are as follows:
Find all the amplifier parameters.
Soln. First, we use the data obtained for RL = ∞ to determine
and
Now since
Therefore, Ri = 100 kΩ
Next, we use the data obtained when RL=10 kΩ is connected to the amplifier output to determine
And
The values of Av and Avo can be used to determine Ro as follows:
Ro=1.43 kΩ
Similarly, we use the values of Gv and Gvo to determine Rout from
Rout=2.86 kΩ
The value of Rin can be determined from
Thus,
Rin=400 kΩ
The short-circuit transconductance Gm can be found as follows:
and the current gain Ai can be determined as follows:
Finally, we determine the short-circuit current gain Ais as follows. From Equivalent Circuit , the
short-circuit output current is
However, to determine vi we need to know the value of Rin obtained with RL=0. Toward this
end, note that from Equivalent Circuit , the output short-circuit current can be found as
Now, equating the two expressions for iosc and substituting for Gvo by
and for vi from
Results in
Now, we use
To obtain
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