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)

V_B = 1.8V
V_E = 1.1V
I_E = 1.1 mA
V_C = 6.04V
V_CE = 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 (A_V).

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

V_out = i_c * r_c

The input voltage is given by:

V_in = i_e * r_e

Substitute of these two expressions:

Because i_c approximately equals i_e:

AC emitter resistance (r_e)

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

(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:

r_c = 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

Solution:

DC analysis

V_B = 3.6KW, V_E = 2.9 KW, I_E = 2.9 mA

V_C = 9.5V, V_CE = 6.6V

AC analysis

r_c = 2.65KW