Chapter 6. BJT Amplifiers

Transcription

1 Basic Electronic Devices and Circuits EE 111 Electrical Engineering Majmaah University 2 nd Semester 1432/1433 H Chapter 6 BJT Amplifiers 1

2 Introduction The things you learned about biasing a transistor in Chapter 5 are now applied in this chapter where BJT circuits are used as small-signal amplifiers. The term small-signal refers to the use of signals that take up a relatively small percentage of an amplifier s operational range. Additionally, you will learn how to reduce an amplifier to an equivalent dc and ac circuit for easier analysis. 2

3 AC Quantities AC quantities are indicated with a lowercase italic subscript; V rms: 1/ avg: 2/π rms values are assumed unless otherwise stated. The figure shows an example of a specific waveform for the collector-emitter voltage. Notice the DC component is V CE and the ac component is V ce. rms avg V CE 0 0 v ce V ce V ce V ce V ce t Resistance is also identified with a lowercase subscript when analyzed from an ac standpoint. 3

4 Linear Amplifier A linear amplifier produces a replica of the input signal at the output. +V CC I c V b I CQ V BQ R 1 R C V ce C 2 R s C 1 I b V CEQ I BQ V s R 2 R E R L For the amplifier shown, notice that the voltage waveform is inverted between the input and output but has the same shape. 4

5 AC Load Line V CC /(R C R L +R E ) I C I b I BQ I c Operation of the linear amplifier can be illustrated using an ac load line. The ac load line is different than the dc load line because a capacitor looks open to dc but effectively acts as a short to ac. Thus the collector resistor R C appears to be in parallel with the load resistor R L. I CQ Q 0 V CE V ce V CEQ ac ground ac short; Z C = 1/( j ω C ) 5

6 Solution: Projections on the graph of Figure 6 4 show the collector current I c varying from 6 ma to 4 ma for a peak-to-peak value of 6 4 = 2 ma and the collector-to-emitter voltage V ce varying from 1 V to 2 V for a peak-to-peak value of 2 1 = 1 V. 6

7 Transistor AC Model very large very small 7

8 8

9 Transistor AC Model The five resistance parameters (r-parameters) can be used for detailed analysis of a BJT circuit. For most analysis work, the simplified r-parameters give good results. The simplified r-parameters are shown in relation to the transistor model. An important r-parameter is r e '. It appears as a small ac resistance between the base and emitter. ' r e 25 mv = I assuming an abrupt junction between the n and p regions, and an ambient temperature of 20 C E 9 B C E r e β ac I b B I b C r e E β ac I b

10 Quiz Q. The equation for finding the ac emitter resistance of a BJT is a. b. c. d. ' r e r ' e r ' e r ' e 25 mv = I B E B 25 mv = I E 0.7 V = I 0.7 V = I 10

11 I E = 25 mv / 8 Ω = ma 11

12 Comparison of β ac to β DC Curve of I C versus I B is nonlinear! slope slope 12

13 h (hybrid) parameters 13

14 Relationships of h Parameters and r Parameters forward current gain, common-base forward current gain, common-emitter voltage feedback ratio, common-emitter output admittance (conductance), common-emitter input impedance (resistance), common-emitter 14

15 The Common-Emitter Amplifier In the common-emitter (CE) amplifier, the input signal is applied to the base and the inverted output is taken from the collector. The emitter is common to ac input & output signals. V CC R 1 R C C 3 V out C 1 V in R L R 2 R E C 2 ac short; Z C = 1/( j ω C ) [bypass capacitor] 15

16 AC + DC (in amplifier) = V out AC (V p =5V) DC = 8 V (Q-point)

17 The Common-Emitter Amplifier

18 DC Analysis 18

19 The Common-Emitter Amplifier

20 AC Analysis 20

21 Signal (AC) Voltage at the Base 21

22 Input Resistance at the Base 22

23 Output Resistance The output resistance is the resistance seen looking back into the output terminal with V in =0. (r e ' is much smaller than r c ' ) (r c ' is much larger than R C ) 23

24 Why find Input Resistance? Why find Output Resistance? Supply Amplifier Amplifier Load Voltage division. We prefer high R in(tot) Voltage division. We prefer low R out 24

25 The Common-Emitter Amplifier This figure is mentioned in the next slide

26 (actually 3.58 ma on slide 18, but use 3.8 ma) < 10 mv! There is significant attenuation (reduction) of the source (supply) voltage due to the voltage division between the source resistance (R s ) and the amplifier s input resistance (R in(tot) ). 26

27 The Common-Emitter Amplifier

28 Voltage Gain This is the voltage gain from base to collector. To get the overall gain of the amplifier from the supply voltage to collector, the attenuation of the input circuit must be included. 28

29 Attenuation Attenuation is the reduction in signal voltage as it passes through a circuit and corresponds to a gain of less than 1. gain = 1 / attenuation gain = output / input attenuation = 1 / gain = input / output Example: If the signal amplitude is reduced by half, gain = 0.5 or attenuation = 1 / 0.5 = 2 Example: A source (supply) produces a 10 mv input signal, and the source resistance combined with the load resistance results in a 2 mv output signal. gain = output / input = 2 / 10 = 0.2 attenuation = input / output = 10 / 2 = 5 29

30 Overall Voltage Gain Voltage gain from base to collector = A v = V c / V b Attenuation from source (supply) to base = V s / V b The overall voltage gain of the amplifier, is the voltage gain from base to collector, V c / V b, times the reciprocal of the attenuation, V b / V s. (reciprocal of voltage division) 1 <

31 Effect of the Emitter Bypass Capacitor on Voltage Gain The emitter bypass capacitor provides an effective short to the ac signal around the emitter resistor R E, thus keeping the emitter at ac ground. With the bypass capacitor, the gain of a given amplifier is maximum and equal to R C /r e '. The value of the bypass capacitor (C) must be large enough so that its reactance (X C = 1/ (jωc)) over the frequency range of the amplifier is very small (ideally 0 Ω) compared to R E. A good rule-of-thumb is that the capacitive reactance, X C = 1/ (jωc), of the bypass capacitor should be at least 10 times smaller than R E at the minimum frequency for which the amplifier must operate. 10 X C R E 31

32 32

33 X C = 1/ (jωc) C = 1/ (ω X C ) 33

34 Voltage Gain Without the Bypass Capacitor To see how the bypass capacitor affects ac voltage gain, let s remove it from the circuit and compare voltage gains. Without the bypass capacitor, the emitter is no longer at ac ground. Instead, R E is seen by the ac signal between the emitter and ground and effectively adds to r e ' in the voltage gain formula. V b = I e (r e ' + R E ) I e (r e ' + R E ) The effect of R E is to decrease the ac voltage gain. < 34 X

35 X 35

36 Effect of a Load on the Voltage Gain ac ground small capital 36

37 Effect of a Load on the Voltage Gain A load is the amount of current drawn from the output of an amplifier or other circuit through a load resistance. When a resistor, R L, is connected to the output through the coupling capacitor C 3, it creates a load on the circuit. The collector resistance R c at the signal frequency is effectively R C R L, because the upper end of R C is effectively at ac ground. The total ac collector resistance is R c = R C R L Replacing R C with R c in the voltage gain expression gives: small letter R c < R C because of R L, so the voltage gain is reduced. However, if R L >> R C then R c R C and the load has very little effect on the gain. 37

38 38

39 Current Gain The current gain from base to collector is I c / I b = β ac However, the overall current gain of the common-emitter amplifier is I s is the total signal input current produced by the source, part of which (I b ) is base current and part of which (I bias ) goes through the bias circuit R 1 R 2 as shown in the figure. I s = I b + I bias The source sees a total resistance of R s + R in(tot). The total current produced by the source is Power Gain The overall power gain is the product of the overall voltage gain A v ' (= V c / V s ) and the overall current gain A i (= I c / I s ). A p = A v ' A i = V c I c / V s I s 39

40 The Common-Emitter Amplifier What is r e ' for the CE amplifier? Assume stiff voltage-divider bias. ' 25 mv r e = (we need I E ) I E V CC 27 kω +15 V VB = 15 V = 4.26 V 68 kω+ 27 kω V E = 4.26 V 0.7 V = 3.56 V I E ' r e V R E = = = E 3.56 V 2.2 kω 1.62 ma = 25 mv 25 mv I = 1.62 ma = 15.4 Ω E C µ F R 1 68 kω R 2 27 kω R C 3.9 kω R E 2.2 kω C 3 10 µ F C µ F R L 3.9 kω 40

41 The Common-Emitter Amplifier Notice that the ac resistance of the collector circuit is R C R L. What is the gain of the amplifier? ac ground small capital V CC +15 V = = = = 3.9 Ω 3.9 Ω 15.4Ω = 127 C µ F R 1 68 kω R 2 27 kω R C 3.9 kω R E 2.2 kω C 3 10 µ F C µ F ac short R L 3.9 kω 41

42 The Common-Collector Amplifier The common-collector amplifier (emitter-follower) has a voltage gain of approximately 1, but can have high input resistance and current gain. The input is applied to the base and output taken from the emitter. +V CC V in I in C 1 R 1 C 2 V out R 2 R E R L No phase inversion. 42

43 Voltage Gain where R e = R E R L If there is no load (R L ), then R e = R E. Notice that the gain is always < 1. If R e >> r e ', then a good approximation is: Since the output voltage is at the emitter, it is in phase with the base voltage, so there is no inversion from input to output. Because there is no inversion and because the voltage gain is approximately 1, the output voltage closely follows the input voltage in both phase and amplitude; thus the term emitter-follower. 43

44 Input Resistance Because of the high input resistance, it can be used as a buffer to minimize loading effects when a circuit is driving a low-resistance load. In a common-collector circuit, the emitter resistor R E is never bypassed because the output is taken across R e = R E R L. 44

45 Output Resistance With the load removed, the output resistance, looking into the emitter of the emitter-follower, is approximated as follows: R s is the resistance of the input source (ac supply). The derivation of the equation is relatively involved and several assumptions have been made. The output resistance is very low, making the emitter-follower useful for driving low-resistance loads. Amplifier Load Voltage division. We prefer low R out

46 Current Gain Power Gain 46

47 Power Gain The power gain is the ratio of the power dissipated in the load divided by the power delivered to the input resistance. A p = P L / P in This is approximately equal to the current gain (because A v 1). So, A p A i. You can also write power gain as a ratio of resistances: A p P V 2 L L L = = = 2 Pin Vin R R in( tot) A 2 v R in( tot ) R L V in C 1 R 1 V CC C 2 V out R 1 = RL R in( tot) in( tot) R L R 2 R E R L 47

48 Quiz Q. In a CC amplifier, the power gain is approximately a. one b. equal to the voltage gain c. equal to the current gain d. none of the above 48

49 The Common-Collector Amplifier Calculate the power gain to the load for the CC amplifier using a ratio of resistances. Assume A v = 1 and β ac = 200. Use r e ' = 2 Ω. R in(tot) = R 1 R 2 β ac (r e ' + R E R L ) = 39 kω 220 kω 200(2 Ω Ω) = 24.9 kω R L = 1.0 kω Rin ( tot) 24.9 kω Ap = = = R 1.0 kω 24.9 L V in R 1 C 1 39 kω 0.22 µ F R kω V CC +15 V C 2 V out R 3.3 µ F E R L 1.0 kω 1.0 kω 49

50 The Common-Collector Amplifier The input voltage-divider in the previous example is not rock-solid but the overall power gain is good. A rock solid stiff voltage-divider is not always the best design. Can you spot the problem illustrated here? V B = 10 / ( ) * 10 = 5 V V E = V B 0.7 = = 4.3 V I E = V E / R E = 4.3 / 4.3k = 1 ma r e ' = 25 mv / I E = 25 Ω R in(tot) = R 1 R 2 β ac (r e ' + R E R L ) V CC +10 V = 10 kω 10 kω 200(25 Ω kω) = 4.96 kω R L = 10 kω Rin ( tot) 4.96 kω Ap = = = RL 10 kω The problem is the power gain is less than 1! V in R 1 C 1 10 kω R 2 10 kω β = 200 R E C kω V out R L 10 kω 50

51 Quiz Q. A CC amplifier with a power gain less than 1 is a. a buffer b. an inverting amplifier c. unstable d. an example of poor design 51

52 Remember: 52

53 < 1 53

54 54

55 The Common-Base Amplifier The common-base (CB) amplifier is used in applications where a low input impedance is acceptable. It does not invert the signal, which is an advantage for higher frequencies. What is the purpose of C 2? 2 +V CC C 2 forces the base to be at ac ground. R 1 R C C 2 C 3 V out C 1 R L V in R 2 R E 56

56 57

57 Voltage Gain Notice that the gain expression is the same as for the common-emitter amplifier. However, there is no phase inversion from emitter to collector.

58 Input Resistance The resistance, looking in at the emitter, is = r e ' R E collector Output Resistance Looking into the collector, the ac collector resistance, r c ', appears in parallel with R C. But r c ' >> R C, so 59

59 Current Gain current gain = output current / input current A i = I c / I e I c I e A i 1 Power Gain A p = A v A i A v 60

60 61

61 62

62 Quiz Q. The amplifier shown is a a. differential amplifier b. CE amplifier +V CC c. CC amplifier d. CB amplifier R 1 R C C 2 C 3 C 1 V out R L V in R 2 R E 63

63 Quiz Q. An advantage to this amplifier is that it a. has high current gain b. has high input resistance +V CC c. is non-inverting d. all of the above R 1 R C C 2 C 3 C 1 V out R L V in R 2 R E 64

64 Comparison The current gains and the input and output resistances are the approximate maximum achievable values, with the bias resistors (R 1 & R 2 ) neglected. 65

65 Multistage Amplifiers To improve amplifier performance, stages are often cascaded where the output of one drives another. This is an example of a twostage direct-coupled amplifier, V CC +12 V in which the input and output signals are capacitively coupled. V S 100 mv pp 1.0 khz C µf V in R 1 10 kω R kω R C 1.0 kω Q 1 2N3904 R E1 100 Ω R E3 330 Ω C 3 Q 10 µf 2 2N3906 V out R L 330 Ω R E2 330 Ω C 2 47 µf 66

66 Selected Key Terms r-parameter Commonemitter ac ground Input resistance One of a set of BJT characteristic parameters that include α ac, β ac, r e ', r b ', and r c '. A BJT configuration in which the emitter is the common terminal to an ac signal. A point in a circuit that appears as a ground to ac signals only. The resistance seen by an ac source connected to the amplifier input. 67

67 Selected Key Terms Output resistance Commoncollector The ac resistance looking in at the amplifier output. A BJT configuration in which the emitter is the common terminal to an ac signal. 68