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Exp 4 Current Series

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48 views7 pages

Exp 4 Current Series

Uploaded by

bvvvsathish098
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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EX.NO.

4 CURRENT SERIES FEEDBACK AMPLIFIER DATE:

AIM
To design and construct the current series feedback amplifier and observe the
frequency response of current series feedback amplifier with and without feedback for
the given specifications.

COMPONENTS/EQUIPMENTS REQUIRED

SI.NO COMPONENTS / EQUIPMENTS TYPE / RANGE QUANTITY


1. Transistor BC107/BC 548 1
2. Resistors
3. Capacitor
4. Function Generator (0-3) MHZ 1
5. RPS (0-30) V 1
6. CRO (0-20) MHZ 1
7. Connecting wires --- As Required
8 Bread board --- 1

FORMULA USED
𝑉𝑜
𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝐺𝑎𝑖𝑛 (𝐴𝑣) = 𝑛𝑜 𝑢𝑛𝑖𝑡
𝑉𝑖
𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝐺𝑎𝑖𝑛 (𝐴𝑣) 𝑖𝑛 𝑑𝐵 = 20 × 𝑙𝑜𝑔(𝐴𝑣)
𝐵𝑎𝑛𝑑𝑤𝑖𝑑𝑡ℎ = (𝑓𝑐1 − 𝑓𝑐2) 𝐻𝑧
Where,
𝑉𝑜 − 𝑜𝑢𝑡𝑝𝑢𝑡 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑖𝑛 𝑚𝑉
𝑉𝑖 − 𝑖𝑛𝑝𝑢𝑡 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑖𝑛 𝑚𝑉
𝑓𝑐1 − 𝑙𝑜𝑤𝑒𝑟 𝑐𝑢𝑡 − 𝑜𝑓𝑓 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
𝑓𝑐2 − 𝑢𝑝𝑝𝑒𝑟 𝑐𝑢𝑡 − 𝑜𝑓𝑓 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦

DESIGN SPECIFICATIONS

Given Values are

VCC = V, IB= A, β = , IC =

Assume VCE = VCC/2 V, VE = 10% of VCC


DESIGN EQUATIONS
PIN DETAIL AND SYMBOL

BC107 - NPN Transistor BC548 - NPN Transistor

CIRCUIT DIAGRAM - CURRENT SERIES FEEDBACK AMPLIFIER

WITHOUT FEEDBACK

WITH FEEDBACK
THEORY
Current Series Feedback Amplifier Circuit is also known as series-derived series fed
feedback amplifier. In such a feedback circuit, a part of the output current is made to
develop a voltage proportional to the output current and supplied back in series with
the input. Since feedback network is in series with the amplifier on the output end as
well as on the input end, both input and output impedance are increased with negative
feedback.

The current feedback can be obtained by removing the bypass capacitor across the
emitter resistor RE, as shown in CE amplifier circuit diagram. The emitter resistance
RE provides the negative current feedback. The current IE flowing through resistor
RE produces a voltage drop IERE across emitter resistance RE. This voltage is fed
back to the input and opposes the input signal as it is in opposition. This negative
voltage feedback is proportional to collector current because IE ≈ Ic. Thus negative
Current Series Feedback Amplifier is provided.
TABULATION - WITHOUT FEEDBACK
Vin= v

Frequency Output Voltage Gain (Av) Gain


(vo) = Vo/Vi = 20log(Vo/Vi)
(Hz) (Volts) (no unit) (dB)

TABULATION - WITH FEEDBACK RESISTOR (RL)

Frequency Output Voltage Gain


(vo) Gain (Av) = Vo/Vi
= 20log(Vo/Vi)
(Hz) (Volts) (no unit) (dB)

MODEL GRAPH

Note: Bandwidth representing A is for without feedback.


Bandwidth representing B is for with feedback.
PROCEDURE
1. Connections are made as per the circuit diagram.
2. Connect the function generator with sine wave of 30 mVp-p for 1 KHZ as input at
the input terminals of the circuit.
3. Connect the Cathode Ray Oscilloscope (CRO) to the output terminals.
4. Vary the frequency using frequency knob in function generator in steps and note
down the corresponding output voltage from CRO.
5. Remove the bypass capacitor CE and repeat the above step.
6. The readings are tabulated. Calculate the gain
7. Plot the frequency response on the semi log graph. From the frequency response
identify the 3 dB gain less from the maximum gain. Find the lower and higher cut
off frequencies, mid band gain and calculate the bandwidth using the given
formula.

BANDWIDTH CALCULATION

BW = fC2 (upper cut off frequency) – fC1 (lower cut off frequency)

INFERENCES
The feedback reduces the gain and increases the bandwidth.

RESULT
Thus, the current series feedback amplifier circuit was designed. The frequency
response was observed and the parameters are tabulated.

Parameter
Current Series Lower Cut-off Higher Cut-off
Feedback Amplifier MIdband Gain Bandwidth
frequency frequency
(in dB) (Hz)
(Hz) (Hz)

Without Feedback

With Feedback

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