SERIES AND SHUNT FEEDBACK AMPLIFIERS

AIM:

To design the series and shunt feedback amplifiers and plot the frequency responses and also calculate the input and output impedances.

APPARATUS REQUIRED:

Sl.No. / Apparatus / Specifications / Quantity
1. / Transistor / BC107BP / 2
2. / Resistors / 1.14K, 3K, 4K, 5K, 10K, 11.6K, 14K, 33.6K, 56K, / Each One
3. / Capacitors / 5F
10F / 3
2
4. / RPS / (0-30) V / 1
5. / Function Generator / - / 1
6. / CRO / - / 1
7. / Bread Board / - / 1
8. / Connecting Wires / - / Few

THEORY:

In current series feedback, a voltage is developed which is proportional to the output current. This is called current feedback even though it is a voltage that subtracts from the input voltage. Because of the series connection at the input and output, the input and output, resistances are increased. This type of amplifier is called transconductance amplifier. One of the most common methods of applying the current-series feedback is to place a resistor Re between the emitter lead of a common emitter amplifier and ground. As the common emitter amplifier has a high gain, this is most often used with series negative feedback so that it can afford to loss some gain.

Current-shunt feedback is a series-derived, shunt-fed feedback. The shunt connection at the input reduces the input resistance and the series connection at the output increases the output resistance. This is true current amplifier. As this type of feedback has the least desirable effects, this connection will not be considered at all for practical applications.

In the voltage series feedback, the input to the feedback network is in parallel with the output of the amplifier. A fraction of the output voltage through the feedback network is applied in series with the input voltage of the amplifier. The shunt connection at the output reduces the output resistance Ro. The series connection at the input increases the input resistance Ri. In this case, the amplifier is a true voltage amplifier.

A voltage shunt feedback is a shunt-derived, shunt-fed feedback connection. Here, a fraction of the output voltage is supplied in parallel with the in put voltage through the feedback network. The feedback signal If is proportional to the output voltage Vo. This type of amplifier is called a trans-resistance amplifier.

DESIGN:

Design for Current Series Feedback Amplifier:

Specifications:

VCC = (3 to 5) VCE, IR1 = (10 to 100) IB, IC = 1mA, β = 100, VBE = 0.7V

Apply KVL at the output loop, ------(1)

VCC = ICRC + VCE + IERE

Assume, VCE =5V and RE = 5KΩ

Therefore, RC = 15KΩ.

We know that, IB = IC / β

Assume that the current through R1 is, IR1 = 50IB

Therefore, IR1 = 0.5mA.

The current through R2 is, IR2 = IR1 - IB

Therefore, IR2 = 0.49mA.

Apply KVL at the input loop,

VCC = VR1 + VR2 = IR1R1 + IR2R2 ------(2)

But, VR2 = VBE + IERE = 5.7V

Therefore, R2 = 11.6KΩ

From equation (2), VR1 = IR1R1 = VCC - VR2

Therefore, R1 = 38. 6KΩ

XCE ≤ 0.1RE = 500Ω

CE ≥ 6.37µF

Choose, CE = 10µF

Choose, Cin = Cout = 5µF.

Design Current Shunt Feedback Amplifier:

Specifications:

VCC = (3 to 5) VCE, IR1 = (10 to 100) IB, IC = 1mA, β = 100, VBE = 0.7V

Apply KVL at the output loop, ------(1)

VCC = ICRC + VCE + IERE

Assume, VCE =5V and RE = 5KΩ

Therefore, RC = 15KΩ.

We know that, IB = IC / β

Assume that the current through R1 is, IR1 = 50IB

Therefore, IR1 = 0.5mA.

The current through R2 is, IR2 = IR1 - IB

Therefore, IR2 = 0.49mA.

Apply KVL at the input loop,

VCC = VR1 + VR2 = IR1R1 + IR2R2 ------(2)

But, VR2 = VBE + IERE = 5.7V

Therefore, R2 = 11.6KΩ

From equation (2), VR1 = IR1R1 = VCC - VR2

Therefore, R1 = 38. 6KΩ

XCE ≤ 0.1RE = 500Ω

CE ≥ 6.37µF

Choose, CE = 10µF

Choose, Cin = Cout = 5µF.

Feedback Resistance Rf = 5KΩ

Design for Voltage Series Feedback Amplifier:

Specifications:

VCC = 15V, VCE = 1V, IC = 1mA, β = 100, VBE = 0.7V

Apply KVL in the output loop, ------(1)

VCC = VCE + IERE

Therefore, RE = 14KΩ.

Voltage across R2 is, VR2 = VCC R2/(R1 + R2) = VBE + IERE ------(2)

Assume, R2 = 56 KΩ,

From equation (2), R1= 1.14KΩ.

Choose, Cin = Cout = 10µF.

Design for Voltage Shunt Feedback Amplifier:

Specifications:

VCC = 20V, VCE = 10V, IC = 1mA, β = 100, VBE = 0.7V, IB = 10µA

Apply KVL in the output loop,

VCC = ICRC + VCE

Therefore, RC = 10KΩ

Apply KVL in the input loop,

VCC = ICRC + IBRB + VBE

Therefore, RB = 930 KΩ

Choose, RB = 1MΩ.

CIRCUIT DIAGRAMS:

Circuit Diagram for Current Series Feedback Amplifier:

Circuit Diagram for Current Shunt Feedback Amplifier:

Circuit Diagram for Voltage Series Feedback Amplifier:

Circuit Diagram for Voltage Shunt Feedback Amplifier:

TABULATIONS:

Current Series Feedback Amplifier - Vin = 1mV (p-p)

Without Feedback

/

With Feedback

Frequency (Hertz) / Output Voltage (Vo) (Volts) / Gain in dB = 20log(Vo/Vin) / Output Voltage (Vo) (Volts) / Gain in dB = 20log(Vo/Vin)

Input Impedance

/

Without Feedback

/

With Feedback

Initial input value

Initial DRB value
Final input value
Final DRB value

Output Impedance

/

Without Feedback

/

With Feedback

Initial output value

Initial DRB value
Final output value
Final DRB value
Current Shunt Feedback Amplifier - Vin = 10mV (p-p)

Without Feedback

/

With Feedback

Frequency (Hertz) / Output Voltage (Vo) (Volts) / Gain in dB = 20log(Vo/Vin) / Output Voltage (Vo) (Volts) / Gain in dB = 20log(Vo/Vin)

Input Impedance

/

Without Feedback

/

With Feedback

Initial input value

Initial DRB value
Final input value
Final DRB value

Output Impedance

/

Without Feedback

/

With Feedback

Initial output value

Initial DRB value
Final output value
Final DRB value
Voltage Series Feedback Amplifier - Vin = 10mV (p-p)

Without Feedback

/

With Feedback

Frequency (Hertz) / Output Voltage (Vo) (Volts) / Gain in dB = 20log(Vo/Vin) / Output Voltage (Vo) (Volts) / Gain in dB = 20log(Vo/Vin)

Input Impedance

/

Without Feedback

/

With Feedback

Initial input value

Initial DRB value
Final input value
Final DRB value

Output Impedance

/

Without Feedback

/

With Feedback

Initial output value

Initial DRB value
Final output value
Final DRB value
Voltage Shunt Feedback Amplifier - Vin = 10mV (p-p)

Without Feedback

/

With Feedback

Frequency (Hertz) / Output Voltage (Vo) (Volts) / Gain in dB = 20log(Vo/Vin) / Output Voltage (Vo) (Volts) / Gain in dB = 20log(Vo/Vin)

Input Impedance

/

Without Feedback

/

With Feedback

Initial input value

Initial DRB value
Final input value
Final DRB value

Output Impedance

/

Without Feedback

/

With Feedback

Initial output value

Initial DRB value
Final output value
Final DRB value

PROCEDURE:

  1. Connect the common base amplifier as shown in the figure
  2. Switch on the power supply
  3. Set the input voltage Vin = 10mV p-p
  4. Apply the input to the circuit
  5. Vary the frequency in steps and observe the respective output
  6. Calculate the gain using the formula, Gain in dB = 20 x log (Vo/Vin)
  7. Plot the graph between frequency versus gain
  8. Calculate the input impedance and output impedance
  9. Repeat the above steps for other circuits

RESULT:

The series and shunt feedback amplifiers were designed and the frequency responses were plotted and the input and output impedances were calculated.