Change to Ukrainian interface versionChange to English interface versionChange to Russian interface versionHome pageClear last query resultsHelp page
Search for specific termsBrowse by subject categoryBrowse alphabetical list of titlesBrowse by organizationBrowse special topic issues

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
Open this folder and view contentsLesson 5 - Transistor Biasing
Open this folder and view contentsLesson 6 - Transistor Biasing II
View the documentSecond Evaluation
close this folderLesson 7 - Small Signal Amplifier
close this folderLesson Plan
View the documentSmall signal amplifier
View the documentCoupling capacitor
View the documentBypass capacitor
View the documentAmplifier analyzing method
View the documentWorksheet No. 7
View the documentExperiment No. 4
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
 

Amplifier analyzing method

From a given amplifier circuit first do -the dc analysis (recall lesson 6) and than do the ac analysis.


Fig. 7-2: CE amplifier circuit

DC equivalent circuit

For dc, all capacitors are acting like open switches; therefore we can draw the following dc equivalent circuit:


Fig. 7-3: DC equivalent circuit

Now the dc analysis can easily be done: (see Lesson 6)

VB = 1.8V
VE = 1.1V
IE = 1.1 mA
VC = 6.04V
VCE = 4.94V

AC equivalent circuit

For the ac all capacitors are shorted and the dc sources are reduced to zero:


Fig. 7-4: Ac equivalent, circuit

The top of the 10K and 3.6K resistors are grounded. The resistors 10K/2.2K and 3.6K/10K are in parallel so we can combine them:


Fig. 7-4: Simplified ac equivalent circuit

Now we got a really simple circuit for the ac analysis.

Voltage Gain

One of the most important characteristics for small signal amplifiers is the voltage gain (AV).

The lowercase letters are used to indicate ac values. The output voltage is given by:

Vout = ic * rc

The input voltage is given by:

Vin = ie * re

Substitute of these two expressions:

Because ic approximately equals ie:

AC emitter resistance (re)

The first step in calculating the voltage gain is to estimate the ac emitter resistance (re).

(formula derived by using calculus)

This relation applies to all transistors that means it is a universal formula.

Let's remember our example circuit (Fig. 7-4):

AC collector resistance

Due to the ac analyzing method we easily get the ac collector resistance (re). See Fig. 7-4:

rc = 2.65KΩ

So now we are ready to calculate the voltage gain:

HO: What will be the voltage gain for the following circuit?


Fig. 7-5: CE amplifier circuit

Solution:

DC analysis

VB = 3.6KΩ, VE = 2.9 KΩ, IE = 2.9 mA

VC = 9.5V, VCE = 6.6V

AC analysis

rc = 2.65KΩ



to previous section to next section

[Ukrainian]  [English]  [Russian]