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EE332 LAB 2 REPORT

Single-Stage BJT Amplifiers

September 12, 2013GROUP 5 10ECE

Trng Hong Lnh L Quang Ha V Quang Tuyn

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PROCEDURE 1 - BIASING UP AN NPN STAGE

R1= 10 R2= 55.56 RC= 5 RE= 1

Measurement results:

Part 1: RCis not inserted

1. Unconnected collector

VB= 0.7087 V

VE=0.1 V

2. Connected Collector

VB= 1.5277 V

VE= 0.9116 V

Part 2: RCis inserted

VB= 1.5275 V VE= 0.9083 V VC= 5.5320 VVTH = 1.525 V

IB=.

.=179.7 uA

From your measured values of VE, VB, and VC, calculate the current flowingthrough each resistor, and the emitter, base, and collector currents for the

transistor. Verify that the terminal currents for the transistor sum to zero

(Kirchoffs Current Law). Calculate the value of for the transistor in this bias

state.

For a forward-active npn BJT, order the emitter, base, and collector terminals

in increasing voltage. For a forward-active pnp BJT, order the emitter, base,

and collector terminals in increasing voltage.

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9/12/2013 EE332 Lab 2 Report 2

IR1 = .+.k

= 152.7 uA

IR2=0

.+0k = 26.99 uA

IC=(0.0)

000= 1.3292

VCE=10 ICRC- IERE

IE= 1.2697 mV =

7.4

Mode : Forward activeBJT type NPN PNP

Applied voltage VE< VB< VC VC< VB< VE

PROCEDURE 2 - COMMON-EMITTER AMPLIFIER

For a forward-active npn BJT, order the emitter, base, and collector terminals

in increasing voltage.

For a forward-active pnp BJT, order the emitter, base, and collector terminals

in increasing voltage.

Starting from the biased-up npn BJT of procedure 1, add two capacitors C1 and C2

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Measurement results:

The gain of this amplifier: =

=

.

.

= 3.87

At Vin= 1.65 Vclipping occurs in negativepolarity.At Vin= 1.95 Vclipping occurs in positivepolarity.

At f = 1.13 MHz, when frequency increases, the amplitude decreases.Question & Answer:

The common-emitter amplifier is inverting amplifier as weve seen in the

waveform plot because as base-emitter current increases, so does collector-

emitter current, which pulls the collector towards the emitter, i.e. down, making it

an inverting amplifier.

To increase the upper clipping voltage we can increase the VCC or decrease the

VBE to increase the Q-Point.

To decrease the lower clipping voltage we can increase the VCC or decrease the

VBE to increase the Q-Point.

(a) Is the common-emier amplier inverng or non-inverng?

(b) Suggest a redesign of the amplier to increase the upper clipping voltage

(c ) Suggest a redesign of the amplier to decrease the lower clipping voltage.

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The Voltage gain:

=

.

.0= 4.296

The capacitor C1 function as a coupling capacitor.

PROCEDURE 3 - COMMON-EMITTER AMPLIFIER WITH BYPASSED EMITTER RESISTOR

Measurement and result:

Vin amp = 60mV

Gain = 6.16/0.048 = 128

(d) When the input is capacitor coupled through C1, what is the DC voltage gain of this

amplier?

(e) What funcon does the capacitor C1 serve?

Adjust the amplitude of the signal generator so that the output sinewave is as large as

possible, but not yet clipping on either polarity peak.

Calculate the voltage gain of the amplier by dividing the amplitude of the output

sinewave by the amplitude of the input sinewave and record the result in your lab

notebook

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At Vin = 67mVclipping occurs in negative polarity

At Vin = 107mVclipping occurs in positive polarity

This is an inverting amplifier because as base-emitter current increases, so does

collector-emitter current, which pulls the collector towards the emitter, i.e. down,

making it an inverting amplifier.

Increase the amplitude of the input sinewave from the signal generator unl the

output waveform just begins to clip at either the negave or posive peak.

(a)

Is this amplier inverng or non-inverng?

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Without the emitter resistor bypass With the emitter resistor bypassAt Vin = 1.65Vclipping occurs innegative polarity

At Vin = 67mVclipping occurs innegative polarity.

As we can easily see that the clipping point of common-emitter amplifier with the

emitter resistor bypass is much lower than (about 24.6 times smaller) that of

common-emitter amplifier without the emitter resistor bypass.

The presence of the bypass capacitor did affect the clipping levels.

XC =

=

= = 1

f = 000

= 15.915

Above this frequency, capacitor C2 forms an effective bypass for RE.

At 15.92Hz, it causes the capacitance of C2 to equal to the resistance of RE.

At 900 KHz, it causes the amplitude of the output sinewave fall to about 70

percent of its initial value.

Therefore, frequency range is from 15.92Hz to 900 KHz.

(b) How do the clipping points compare to those of the common-emier amplierwithout the emier resistor bypass? Does the presence of the bypass capacitor aect

the clipping levels?

(c) Calculate the frequency at which the impedance of the bypass capacitor is equalto the resistance of RE.

(d) What is the at frequency response range for this amplier?

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Bypassed common emitter amplifier Unbypassed common emitter amplifierAt f = 900 kHz, when frequency

increases, the amplitude decreases 70%.

At f = 1.13 MHz, when frequency

increases, the amplitude decreases by

70%

Obviously, the cut-off frequency (-3dB bandwidth) of bypassed CE amplifier is

lower than that of unbypassed CE amplifier. When frequency gets higher,

capacitance (XC =

) becomes smaller and approaches 0 so it will short emitter to

ground at high frequency. Bypass capacitor lowers the cut-off frequency ofcommon-emitter amplifier.

PROCEDURE 4 - COMMON-COLLECTOR AMPLIFIER (EMITTER-FOLLOWER)

Measurement and result:

Gain:

(e) How does the 3 dB bandwidth of the bypassed common-emier ampliercompare to the unbypassed case? Does the presence of the bypass capacitor have any

eect on the high frequency characteriscs?

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At Vin = 2Vclipping occurs in negative polarity

At f = 9MHz, when frequency increases, the amplitude decreases 70%.

The common-collector amplifier is non-inverting. Actually, this is an emitter

follower with gain is less than 1.

(a) Is the common-collector amplier inverng or non-inverng?

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The common collector amplifier also called the emitter follower is used for current gain.A typical example is the audio output stage of a surround sound or stereo receiver or

amplifier. The output voltage gain might be a little lower than 1 but the current gain

allows the amplifier to provide power gain for the speakers (load). It can also be used for

impedance matching as its input impedance is very high and its output impedance is

lowPROCEDURE 5 - COMMON-BASE AMPLIFIER

Measurement result:

a. Gain

(b) Since the voltage gain of this amplier is actually less than unity, what is theusefulness of this amplier?

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=.

0.0= 86

b. Clipping

At Vin= 367 mVclipping occurs in negative polarity

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At Vin = 420 mVclipping occurs in positive polarity

At f = 1.25 MHz, when frequency increases, the amplitude decreases 70% (From 3.8V to1.8V)

Question & Answer:

a. Is the common-base amplier inverng or non-inverng?

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Based on waveform displayed, common-base amplifier is non-inverting. Mathematically speaking

Voltage gain A= gmRC> 0

If we dont use bypass capacitor the voltage gain dramatically decreases

From CE amplifier with input at B node and output at C node, we got the high

voltage gain . In the case we reverse the input and output (B output and C

input), we will receive reverse gain

which is very small.

Therefore base terminal cannot be the amplifier output.

Again, CE amplifier with input at C node and output at E node, we got the high

voltage gain A. In the case we reverse the input and output (C output and E

input), we will receive reverse gain 1/A which is very small.

Therefore base terminal cannot be the amplifier output.

b. Speculate on the eect of not bypassing the base resistors R1 and R2. If youdont have any ideas, go ahead and try this in the lab if you have me.

c. From the results of this lab experiment and those of experiment 1, explainwhy the base terminal of a BJT is never useful as an amplier output.

d. From the results of this lab experiment and those of experiment 1, explainwhy the collector terminal of a BJT is never useful as an amplier input. As we can

clearly see that, the voltage gain of common collector amplier