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close this bookAmplifier Teaching Aid (GTZ, DED; 86 pages)
View the documentPreface
View the documentIntroduction
Open this folder and view contentsLesson 1 - Semiconductor Review
Open this folder and view contentsLesson 2 - Bipolar Transistor
Open this folder and view contentsBipolar Transistor II
View the documentFirst Evaluation
Open this folder and view contentsLesson 4 - Transistor Fundamentals
close this folderLesson 5 - Transistor Biasing
close this folderLesson Plan
View the documentTransistor biasing
View the documentWorksheet No. 5
Open this folder and view contentsLesson 6 - Transistor Biasing II
View the documentSecond Evaluation
Open this folder and view contentsLesson 7 - Small Signal Amplifier
Open this folder and view contentsLesson 8 - Small Signal Amplifier II
Open this folder and view contentsLesson 9 - Small Signal Amplifier III
Open this folder and view contentsLesson 10 - Large Signal Amplifier
View the documentThird Evaluation
 

Transistor biasing

Emitter Bias

The analysis of base biased circuits depends on the current gain which can vary in a wide range. In an amplifier we need circuits whose Q-points are immune to changes in current gain. The solution for this problem is the emitter biased circuit:


Fig. 5-1: Emitter biased circuit, β = 100

Find the Q-point:

Given : VBB = 5V, VBE = 0.7V, VCE = 15V

RE = 2.2KΩ, RC = 1KΩ

Calculation:

VE = VBB - VBE = 5V - 0.7V = 4.3V

IE = IC (close approximation)

VC = VCC - (RC * IC)


= 15V - (1KΩ - 1.95 mA)

= 13.1V



VCE = VC - VE = 13.1V - 4.3V = 8.8V

Q-point coordinates:

>

IC = 1.95 mA

VCE = 8.8V

An emitter biased circuit is immune to changes in current gain. Analysing summary:

1. get VE
2. calculate IE
3. find VC
4. VCE = VC - VE[SFARSIT]

At no time we need the current gain!

Tip for troubleshooter:

Don't measure direct VCE, because the common lead of the voltmeter is grounded, so you will short the emitter to ground.

[inceput284]1. Measure VC
2. Measure VE
3. Subtract VCE = VC - VE

HO: What is the collector voltage?


Fig. 5-2: Emitter biased CE connection

Solution:

VE = VBB - VBE = 2V - 0.7V = 1.3V

IE = IC (approx.)

VC = VCC - (RC * IC) = 10V - (910 Ω * 7.2 mA) = 3.4V

HO: What is the collector-emitter voltage?


Fig. 5-3: Emitter biased CE connection

Solution:

VE = VBB - VBE = 1.1V

IE = IC (approx.)

VC = VCC - (RC * IC) = 9.4V

VCE = VC - VE = 8.35V

Effect of Small Changes

For example, tolerances of resistors (+/- 10%) are small changes.

See Fig. 5-4 on the next page.


Fig. 5-4: Emitter biased CE connection

Before we can analyse the effects of small changes we have to find out which values are dependent or independent.

independent values: VBB, VCC, βdc, RE, RC
dependent values : VE , VC, IB, IC, IE

Suppose the independent values will increase one after another. What will be the effect on the dependent values:

increase

dependent

   

VE

IE

IB

IC

VC

VCE

 

VBB

U

U

U

U

D

D

 

VCC

N

N

N

N

U

U

independent

RE

N

D

D

D

U

U

 

RC

N

N

N

N

D

D

 

hfe

N

N

D

N

N

N

U = up
D = down
N = no change
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