# introduction of common emitter amplifier

## Aim:

Design and set up the BJT common emitter amplifier using voltage divider bias with and without feedback and determine the gain- bandwidth product from its frequency response.

## Components and equipments required:

Transistor – SL100, Resistors – 470 Ω, 1KΩ, 10KΩ – 2nos, and 33KΩ, Capacitors 100µf, 0.22µf and 0.47µf, Power Supply, 10Hz –3MHz Signal generator, CRO, Connecting wires and Bread board/Spring board with spring terminals.

## Design:

Transistor: SL100
Let VCC = 12V; IC = 4.5 mA; VE = 1.2V; VCE = 6V; hFE = 100.
Given VE = 1.2V. Therefore RE = VE / IE » VE / IC =266.67W; RE =270Ω
Writing KVL for the Collector loop we get, VCC = ICRC + VCE + VE
RC = (VCC – VCE – VE) / IC = (12-6-1.2)V/4mA=1.06KW; RC = 1 KW
hFE RE = 10R2
Assume R2=2.7KΩ,
VB = (Vcc x R2 ) / (R1 + R2)
Hence R1 = 14.14 KW ; R1 = 15 KW
Use CC1 = 0.47mF
Use CC2 = 0.47mF
Use CE =47mF

## Circuit Diagram of amplifier without feedback: ## Circuit Diagram of amplifier with feedback. (introduce a resistor in the emitter circuit): ## Procedure:

Follow the same procedure for both circuits

• After making the connections, switch on the D.C. power supply and check the D.C. conditions without any input signal and record in table below:
 Parameter VRC VCE VE ICQ VBE Assumed 4.8V 6V 1.2V 4.5mA 0.6V Practical
• Select sine wave input and set the input signal frequency ≥10f1 (Say = 10 KHz. This will be a convenient ‘Mid – frequency’).
• Observe the input wave form and output wave form on a dual channel CRO.
• Adjust the input amplitude such that the output waveform is just undistorted (or in the verge of becoming distorted). Measure the amplitude of the Input Signal now. This amplitude is the Maximum Signal Handling Capacity of your amplifier.
• Decrease the input voltage to a convenient value such that the output is undistorted. Say 20mV. Measure the corresponding o/p voltage. Calculate mid-band gain, AM = Vo (p-p) / Vin (p-p).
• Keeping the input voltage constant, go on reducing the frequency until the output voltage reduces to 0.707 times its value at 10 KHz. The frequency at which this happens gives you the Lower Cut-off frequency (f1).
• Keeping the input voltage constant, go on increasing the frequency until the output voltage decreases to 0.707 times its value at 10 KHz. The frequency at which this happens gives you the Upper Cut-off frequency (f2).
• Thus you have pre-determined f1 and f2. Find the amplifier band width, BW = f2 f1
• Determine Gain Bandwidth product (GBW product) which is a Figure of Merit of your amplifier as GBW = AM x BW.
• Now repeat the experiment by recording values of output voltage versus frequency keeping the input voltage at a constant value convenient to you. You should take at least 5 readings below f1 and 5 readings above f1, at least 5 readings in the mid band, at least 5 readings below f2 and 5 readings above f2.
• Plot graphs of AV versus Frequency, f and /or M, dB versus Frequency, f on a semi log graph paper. From the graph determine: Mid –band – gain, Lower and Upper Cut-off frequencies and Band width. Compute the GBW product and verify with answer obtained earlier.

## Observation:

Use the tabular column separately for each circuit

Vin (P-P) = …….. V

AV= VO(P-P)/Vin(P-P)

M = 20log (AV) dB

 Frequenc y in Hz 100 200 300 350 400 500 600 700 800 VO(P-P) in volts Av M dB(Av in dB)
 Frequenc y in Hz 1k 2k 3k 5k 8k 10k 20k 30k 50k VO(P-P) in volts Av M dB(Av in dB)
 Frequenc y in Hz 100k 200k 300k 400k 500k 600k 700k 800k 900k VO(P-P) in volts Av M dB(Av in dB)

## Expected graph: ## Result:

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