Amplifier Teaching Aid (DED Philippinen, 86 p.)
 Lesson 9 - Small Signal Amplifier III
 Lesson Plan
 (introduction...) Other configurations Frequency response of an amplifier The AC load line

### (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 Vmax.
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: VB = 1.8V, VE = 1.1V, IE = 1.1 mA, VC = 6.04V VCC = 10V, VCE = 4.94V

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

With load: rc = 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

VCE (cut) = VCC - VE = 8.9V

2. Calculate and draw the Q point

IC = 1.1 mA

VCE = 6.04V

3. Draw a -temporary ac load line

VCE (cut) = VCC

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