Electronics Principles & Applications Eighth Edition

1. Electronics Principles & Applications Eighth Edition Charles A. Schuler Chapter 7 More About Small-Signal Amplifiers

2. INTRODUCTION • Amplifier Coupling • Voltage Gain • FET Amplifiers • Negative Feedback • Frequency Response • Positive Feedback

3. Capacitive coupling is convenient in cascade ac amplifiers. VCC These two points are at different dc voltages.

4. Direct coupling is required for dc gain. VCC

5. The Darlington is a popular dc arrangement. VCC

6. Transformer coupling offers the advantage of impedance matching. VCC 10:1 ZRATIO = TRATIO2 10 W P S = 102 = 100 ZCOLLECTOR = 100 x 10 W = 1000 W

7. Transformer coupling is used in 70.7 volt sound systems.

8. Transformer coupling can be used in bandpass amplifiers to achieve selectivity. VCC Gain fR

9. Amplifier coupling quiz Capacitive coupling is not useful for _________ amplifiers. dc Dc frequency response requires ________ coupling. direct Transformer coupling offers the advantage of _________ matching. impedance Tuned transformer coupling provides frequency _____________. selectivity A Darlington amplifier is an example of _________ coupling. direct

10. More about solving the practical circuit for its ac conditions: VCC = 12 V RB1 RL 22 kW = 2.2 kW C Zin = ? B E RB2 2.7 kW RE = 220 W

11. Zin is a combination of RB1, RB2, and rin of the transistor. VCC = 12 V Determine rin first: rin = b (RE + rE) RB1 RL 22 kW = 2.2 kW rin = 150 (220 W + 9.03 W) C rin= 34.4 kW B E RB2 2.7 kW RE = 220 W Note: rin = brE when RE is bypassed.

12. 1 Zin = 1 + + RB1 1 RB2 1 Zin = 1 1 1 1 + + 34.4 kW 2.7 kW 22 kW rin RB1, RB2, and rin act in parallel to load the input signal. VCC = 12 V RB1 RL 22 kW = 2.2 kW C B E RB2 2.7 kW RE = 220 W Zin = 2.25 kW

13. RL = 2.2 kW What happens when an amplifier is loaded? RL and the Load act in parallel. VCC = 12 V RP = 1.1 kW RB1 22 kW Load = 2.2 kW RB2 2.7 kW RE = 220 W

14. VCC RL + RE VCC = 12 V VCC RP + RE There are two saturation currents for a loaded amplifier. = 4.96 mA ISAT(DC) = ISAT(AC) = = 9.09 mA RL = 2.2 kW RB1 22 kW RP= 1.1 kW Load = 2.2 kW RB2 2.7 kW RE = 220 W

15. 100 mA 80 mA 60 mA 40 mA 20 mA 0 mA There are two load lines for a loaded amplifier. The DC load line connects VCC and ISAT(DC). 14 12 10 IC in mA 8 TEMPORARY AC 6 4 DC 2 6 16 2 4 10 12 0 14 18 8 VCE in Volts A temporary AC load line connects VCC and ISAT(AC).

16. 100 mA 80 mA 60 mA 40 mA 20 mA 0 mA The quiescent VCE is projected to the DC load line to establish the Q-point. The AC load line is drawn through the Q-point, parallel to the temporary AC load line. 14 12 10 IC in mA 8 AC 6 TEMP. AC 4 DC 2 6 16 2 4 10 12 0 14 18 8 VCE in Volts 5.3 V

17. 100 mA 80 mA 60 mA 40 mA 20 mA 0 mA The AC load line shows the limits for VCE and if the Q-point is properly located. 14 12 10 IC in mA 8 6 4 AC 2 6 16 2 4 10 12 0 14 18 8 VCE in Volts 5.3 V With loaded amplifiers, the Q-point is often closer to saturation.

18. VCC = 12 V RL = 2.2 kW RP AV = RE + rE 1.1 kW = 4.8 AV = 220 W + 9.03 W What about voltage gain for a loaded amplifier? RB1 22 kW RP= 1.1 kW Load = 2.2 kW RB2 2.7 kW RE = 220 W

19. When analyzing cascade amplifiers, remember: VCC 1st 2nd Zin of the 2nd stage loads the 1st stage.

20. Amplifier ac conditions quiz Emitter bypassing _________ an amplifier’s input impedance. decreases Loading at the output of an amplifier ________ its voltage gain. decreases A loaded amplifier has two load lines: dc and ___________. ac The clipping points of a loaded amplifier are set by its _______ load line. ac In a cascade amplifier, the Zin of a stage _______ the prior stage. loads

21. Common-source JFET amplifier. VDD = 20 V 20 V = 4 mA ISAT = 5 kW RL = 5 kW Drain Gate Input signal CC RG Source Phase-inverted output VGS = 1.5 V Fixed bias

22. N-channel JFET characteristic curves The Q-point is set by the fixed bias. 0 Load line 4 1 VP-P -0.5 3 -1.0 VGS in Volts ID in mA 2 -1.5 -2.0 1 -2.5 5 20 25 10 0 15 VDS in Volts 8 VP-P AV = 8

23. 1.6 mA Determining forward transfer admittance: 0 4 ID in mA -0.5 3 -1.0 VGS in Volts 2 -1.5 -2.0 1 -2.5 5 20 25 10 0 15 VDS in Volts DID Yfs = = 1.6 mS VDS DVGS

24. When the forward transfer admittance is known, the voltage gain can be determined using: VDD = 20 V RL = 5 kW AV = Yfs x RL D = 1.6 mS x 5 kW G = 8 CC RG S This agrees with the graphic solution. VGS = 1.5 V

25. Source bias eliminates the need for a separate VGS supply. VDD IS = ID RL D VGS = ID x RS G CC S RG RS This resistor also provides ac negative feedback which decreases the voltage gain.

26. JFET amplifier quiz In a common-source amplifier, the input signal goes to the _______. gate In a common-source amplifier, the input to output phase relationship is ____. 180o The voltage gain of a C-S amplifier is equal to Yfs x _________. load resistance Source bias is produced by current flow through the _______ resistor. source An unbypassed source resistor _______ the voltage gain of a C-S amp. decreases

27. Summing junction A Vout Vin AB +1 AVin Vin Vout A - AB AVin 1 = AB +1 = A simplified model = Vout AB +1 = Vout Vin AB +1 Vout A A negative feedback model A(Vin - BVout) Vin Vout A Vin - BVout A = open loop gain B Feedback BVout B = feedback ratio Vout = A(Vin - BVout) Vout = AVin - ABVout

28. The feedback ratio (B) for this circuit is easy to determine since the source and drain currents are the same. VDD RL = 5 kW D 800 W G B = = 0.16 5 kW CC S RG RS = 800 W

29. A Vout Vin AB +1 Use the simplified model: 8 A(WITH NEG. FEEDBACK) = = 3.51 (8)(0.16) + 1

30. CS The source bypass capacitor will eliminate the ac negative feedback and restore the voltage gain. VDD RL D G CC RG RS

31. DC reduces sensitivity to device parameters DC stabilizes operating point DC reduces sensitivity to temperature change AC reduces gain AC increases bandwidth AC reduces signal distortion and noise AC may change input and output impedances Amplifier Negative Feedback

32. Midband Amax Bandwidth The frequency response curve of an ac amplifier A -3dB 0.707 Amax f The gain is maximum in the midband. The bandwidth spans the -3 dB points which are called the break frequencies.

33. 6.8 kW 1 kW 10 mF 50 W 1k W 100 W 10 mF The emitter bypass capacitor in this amplifier has a significant effect on both gain and bandwidth.

34. BW1 BW2 Gain and bandwidthwithand without the emitter bypass 50 Gain in dB 0 100 MHz 10 Hz Frequency

35. Amplifier frequency response • The lower break frequency is partly determined by coupling capacitors. • It is also influenced by emitter bypass capacitors. • The upper break frequency is partly determined by transistor internal capacitance. • Both break frequencies can be influenced by negative feedback.

36. Positive Feedback • Is the opposite of negative feedback • Increases gain and reduces bandwidth • Can be used in some circuits to reduce the effects of noise • The next slide shows a circuit with a noise problem.

37. This circuit is supposed to convert the input signal to a rectangular output signal. It works, but the output waveform shows an extra pulse caused by noise. The trip points are equal.

38. R5 provides positive feedback back from the output amplifier to the input amplifier. This circuit has positive feedback and two trip points. The hysteresis is the difference between the trip points (UTP and LTP) and that makes this circuit less sensitive to noise. The output waveform is noise free. UTP LTP

39. REVIEW • Amplifier Coupling • Voltage Gain • FET Amplifier • Negative Feedback • Frequency Response • Positive Feedback