(introduction...)

Title: Small Signal Amplifier III

Objectives:

- Know the characteristics of CE, CB, and CC configurations
- Understand why the output voltage depends on the frequency
- Able to construct the ac load line

Figure

Other configurations

Up to now we discussed only the common emitter configuration, which is widely used. But for some circuit conditions the common base or the common collector configuration may be a better choice.

As we had already seen, the input/output impedance of an amplifier is a very important characteristic, because the internal impedance of signal sources vary widely:

Ex:

Antenna --- > approx. 50 W
Microphone --- > approx. 100000 W

To choose the best configuration let's have a look at its characteristics.

See Handout No. 2 (let the students complete)

Common base CB

- High voltage gain
- No current gain
- Low input impedance
- High output impedance
- No phase inversion

Common collector CC

- No voltage gain
- High current gain
- High input impedance
- Low output impedance
- No phase inversion

Common emitter CE

- High voltage gain
- High current gain
- Medium input impedance
- Medium output impedance
- Phase inversion

Frequency response of an amplifier

Fig. 9-1: Amplifier output voltage in terms of frequency

Fig. 9-1 shows the typical frequency response of an amplifier. At low frequencies the output voltage decreases because of coupling and bypass capacitors. At high frequencies, the output voltage decreases because of transistor and stray wiring capacitance.

Critical frequencies:

Where the output voltage is 0.707 of V_max.
Two critical frequencies -> f₁, f₂

Midband:

Is the band of frequencies between 10 * f₁ and 0.1 * f₂.
The midband is where an amplifier is supposed to be operated.

Ex: Find the midband of an amplifier with f₁ = 5 Hz and f₂ = 100 KHz.

10 * f₁ = 10 * 5 Hz = 50 Hz -- > lower end

0.1 * f₂ = 0.1 * 100 KHz = 10 KHz -- > upper end

Midband: 50 Hz - 10 KHz

The AC load line

In previous lessons we used the dc load line to analyze biasing circuits. But an amplifier sees two loads, a dc load and a ac load. Now we will use the ac load line to understand the large signal operations.

Fig. 9-2: CE amplifier

DC values: V_B = 1.8V, V_E = 1.1V, I_E = 1.1 mA, V_C = 6.04V V_CC = 10V, V_CE = 4.94V

Without load: DC an ac load line are the same.

With load: r_c = RC//RL

AC Load Line Construction

The following process shows you an easy method to get the ac load line:

1. Draw the dc load line

V_{CE (cut)} = V_CC - V_E = 8.9V

2. Calculate and draw the Q point

I_C = 1.1 mA

V_CE = 6.04V

3. Draw a -temporary ac load line

V_{CE (cut)} = V_CC

4. Construct the ac load line

- parallel to the temporary ac load line
- passing the Q point

Fig. 9-3: Construction of an ac load line

Handout No. 2

Figure

Worksheet No. 9

No. 1 An amplifier has -this critical frequencies:

f₁ = 2 Hz, f₂ = 200 KHz

What is the midband?

No. 2 See Worksheet No. 8, Problem No. 3.

What kind of transistor connection would you choose for the first stage of the amplifier? Explain!

No. 3 Construct the ac load line for the following circuit.

Figure