|Amplifier Teaching Aid (DED Philippinen, 86 p.)|
|Lesson 7 - Small Signal Amplifier|
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.
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 = 2.65KW
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
VB = 3.6KW, VE = 2.9 KW, IE = 2.9 mA
VC = 9.5V, VCE = 6.6V
rc = 2.65KW