**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 V_{CC} = 12V; I_{C} = 4.5 mA; V_{E} = 1.2V; V_{CE} = 6V; h_{FE} = 100.

Given V_{E} = 1.2V. Therefore R_{E} = V_{E} / I_{E} » V_{E} / I_{C} =266.67W; **R**_{E} **=270Ω**

Writing KVL for the Collector loop we get, V_{CC} = I_{C}R_{C} + V_{CE} + V_{E}

R_{C} = (V_{CC} – V_{CE} – V_{E}) / I_{C} = (12-6-1.2)V/4mA=1.06KW; **R**_{C }**= 1 K****W**

h_{FE} R_{E} = 10R_{2}

Assume **R**_{2}**=2.7KΩ,**

V_{B} = (V_{cc} x R_{2} ) / (R_{1} + R_{2})

Hence R_{1} = 14.14 KW ; **R _{1} = 15 K**

**W**

Use

**C**

_{C1}

**= 0.47**

**m**

**F**

Use

**C**

_{C2}

**= 0.47**

**m**

**F**

Use

**C**

_{E}

**=47**

**m**

**F**

**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 | V_{RC} |
V_{CE} |
V_{E} |
I_{CQ} |
V_{BE} |

Assumed | 4.8V | 6V | 1.2V | 4.5mA | 0.6V |

Practical |

- Select sine wave input and set the input signal frequency ≥10f
_{1}(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,**A****M****= V**_{o}**(p-p) / V**_{in}**(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 (
*f*_{1}). - 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 (
*f*_{2}). - Thus you have pre-determined
*f*_{1 }and*f*_{2}. Find the amplifier band width, BW =*f*_{2}*–**f*_{1} - 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
*f*_{1}and 5 readings above*f*_{1}, at least 5 readings in the mid band, at least 5 readings below*f*_{2}and 5 readings above*f*_{2}*.* - Plot graphs of AV versus Frequency,
*f*and /or M, dB versus Frequency,*f*on a. 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.*semi log graph paper*

**Observation: **

Use the tabular column separately for each circuit

V_{in} (P-P) = …….. V

A_{V}= V_{O}_{(P-P)}/V_{in}_{(P-P)}

M = 20log (A_{V}) dB

Frequenc y in Hz | 100 | 200 | 300 | 350 | 400 | 500 | 600 | 700 | 800 |

V_{O}_{(P-P)} in volts |
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A_{v} |
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M dB(A_{v} in dB) |

Frequenc y in Hz | 1k | 2k | 3k | 5k | 8k | 10k | 20k | 30k | 50k |

V_{O}_{(P-P)} in volts |
|||||||||

A_{v} |
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M dB(A_{v} in dB) |

Frequenc y in Hz | 100k | 200k | 300k | 400k | 500k | 600k | 700k | 800k | 900k |

V_{O}_{(P-P)} in volts |
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A_{v} |
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M dB(A_{v} in dB) |

**Expected graph:**