Understanding Amplifier Frequency Response in Analog Circuits

1. Teaching Materials of Analog Circuits Chap 5 Weiwei Xia College of Physics Science and Technology Yangzhou University Yangzhou, 225002 Email: wwxia@yzu.edu.cn

2. Chapter 4 Amplifier Frequency Response

3. § §4.0 Preview Thus far in our linear amplifier analyses, we have assumed that coupling capacitors and bypass capacitors act as short circuits to the signal vol- tages and open circuits to dc voltages. However, capacitors do not change instantaneously from a short circuit to an open circuit as the frequency approaches zero. We have also assumed that tran- sistors are ideal in that output signals respond instantaneously to input signals. However, there

4. § §4.0 Preview are internal capacitances in both the bipolar tr- ansistor and field-effect transistor that affect the frequency response. The major goal of this chapter is to determine the frequency response of amplifier circuits due to circuit capacitors and transistor capacitance. Initially, we derive transfer functions, using the complex frequency s, of several passive circuits as a basic review of frequency response. We introduce Bode plots of the transfer function magnitude and phase, and a time constant technique

5. § §4.0 Preview for determining the corner, or 3dB, frequencies of the circuit response. The goal of this analysis is to help the reader become comfortable with basic fre- quency response analysis and sketching Bode plots. When there is more one capacitor in a circuit, computer simulation becomes an attractive analysis tool for determining the frequency response.

6. § §4.1 The Concept of Frequency Response 一、 一、Amplifier gain versus frequency All amplifier gain factors are functions of signal frequency. These gain factors include voltage, current, transconductance, and transresistance. Up to this point, We have assumed that the signal frequency is high enough that coupling and the bypass capacitors can be treated as short circuits and, at the same time, we have assumed that the signal frequency is low enough that parasitic, load,

7. § §4.1 The Concept of Frequency Response and transistor capacitances can be treated as open circuits. In this chapter, we consider the amplifier response over the entire frequency range. In general, an amplifier gain factor versus frequency will resemble that shown in figure 4.1. Both the gain factor and frequency are plotted on logarithmic scales ( the gain factor in terms of decibels). Three frequency ranges, low, midband, and high, are indicated.

8. § §4.1 The Concept of Frequency Response High-frequency range Low-frequency range │A(jf)│dB │Am│dB 3dB 3dB Midband fL fH O f(Hz) (log scale) figure 4.1 As we will shown, the gain at f=fLand at f=fH is 3dB less than the maximum midband gain. The bandwidth of the amplifier ( in Hz ) is defined as

9. § §4.1 The Concept of Frequency Response BW=fH－ fL. For an audio amplifier, for example, signal frequencies in the range of 20Hz ＜ f ＜20kHz need to be amplified equally so as to reproduced the sound as accurately as possible. Therefore, in the design of a good amplifier, the frequency fL must be designed to be less than 20Hz and fH must be designed to be greater than 20kHz. 二、 二、Equivalent Circuits

10. § §4.1 The Concept of Frequency Response 二、 二、Equivalent Circuits Each capacitor in a circuit is important to only one end of the frequency spectrum. For this reason, we can develop specific equivalent circuits that apply to the low-frequency range, to midband, and to high freqnency range. 1. Midband Range The equivalent circuits used for calculations in the midband range are the same as those co-

11. § §4.1 The Concept of Frequency Response nsidered up to this point in the text. As already mentioned. The couping and bypass capacitors in this region are treated as short circuits. The stray and transistor capacitances are treated as open circuits. In this frequency range there are no ca- pacitances in the equivalent circuit. These circuits are referred to as midband equivalent circuits. 2. Low-Frequency Range

12. § §4.1 The Concept of Frequency Response In this frequency range, we use a low-frequency equivalent circuit. In this range, coupling and by- pass capacitors must be included in the equivalent circuit and in the amplification factor equations. The stray and transistor capacitances are treated as open circuits. The mathematical expressions obtained for the amplification factor in this frequency range must approach the midband results as f approaches the midband frequency range , since in this limit the capacitors approach short-circuit conditions.

13. § §4.1 The Concept of Frequency Response 3. High-Frequency Range In the high-frequency range, we use a high- frequency equivalent circuit. In this region, coupling and bypass capacitors are treated as short circuits. The transistor and any parasitic or load capacitances must be taken into account. In this equivalent cir- cuit. The mathematical expressions obtained for the amplification factor in this frequency range

14. § §4.1 The Concept of Frequency Response must approach the midband results as f approaches the midband frequency range , since in this limit the capacitors approach open-circuit conditions. 三、 三、Frequency Response Analysis Using the three equivalent circuits just con- sidered rather than a complete circuit is an app- roximation technique that produces useful hand- analysis results while avoiding complex transfer functions. This technique is valid if there is a

15. § §4.1 The Concept of Frequency Response large separation between fLand fH, that is fH＞＞fL. This condition is satisfied in many electronic circuits that we will consider. Computer simulations, such as Pspice, can take into account all capacitances and can produce fr- equency response curves that are more accurate than the hand-analysis results. However, the computer results do not provide any physical insight into a particular result and hence do not provide any

16. § §4.1 The Concept of Frequency Response suggestions as to design changes that can be made to improve a particular frequency response. A hand Analysis can provide insight into the “ways and wherefores” of a particular response. In the next section, we introduce two simple circuits to begin our frequency analysis study. We initially derive the mathematical expressions relating output voltage to input voltage ( transfer function ) as a function of signal frequency. From these

17. § §4.1 The Concept of Frequency Response functions, we can develop the response curves. The two frequency response curves give the magnitude of the transfer function versus frequency and the phase of the transfer function versus frequency. The phase response relates the phase of the output signal to the phase of the input signal. We will then develop a technique by which we can easily sketch the frequency response curves

18. § §4.1 The Concept of Frequency Response without resorting to a full analysis of the transfer function. This simplified approach will lead to a general understanding of the frequency response of electronic circuits. We will rely on a computer simulation to provide more detailed calculations when required.

19. § §4.2 s-Domain Analysis 一、Transfer function and “poles” “zeros” In general, a transfer function in the s-domain can be expressed in the form: Y(s) (s－z1) (s－z2) … (s－zm) A(s)= =H0 (4－1) X(s) (s－p1) (s－p2) … (s－pn) Where H0is a constant, z1 , z2 , … zm are the transfer function “zeros”, and p1 , p2 , … pn are the transfer function “poles”, when the complex frequency is equal to a zero, s=zi, the transfer function is zero;

20. § §4.2 s-Domain Analysis when the complex frequency is equal to a pole, s=pi, the transfer function diverges and becomes infinite. The transfer function can be evaluated for physical frequencies by replacing s with jω. In general, the resulting transfer function A (jω ) is a complex function, that is, its magnitude and phase are both functions of frequency. These topics are usually discussed in a basic circuit analysis course.

21. § §4.2 s-Domain Analysis For a simple transfer function of the form: K A(s) = (4－2) s +ω0 We can rearrange the terms and write the function as: 1 A(s) = K1 (4－3) 1+ sτ1 Where τ1 is a time constant. Other transfer functions may be written as sτ2 A(s) = K2 (4－4) 1+ sτ2

22. § §4.2 s-Domain Analysis Whereτ2 is also a time constant. In most cases, we will write the transfer functions in terms of the time constant. To introduce the frequency response ana- lysis of transistor circuits, we will examine some simple circuits. 二、 二、Bode Plots A simplified technique for obtaining approximate plots of the magnitude and phase of a transfer fun- ction, given the poles and zeros or the equivalent time constants, was developed by H.Bode, and the resulting diagrams are called Bode plots.

23. § §4.2 s-Domain Analysis Example 4.1: For the circuit in Figure 4.2. Sketch the Bode plot of the transfer function. + Vi － + Vo － R1 · · C1 figure 4.2 1/sC1 R1+1/sC1 1 Vo(s)= Vi(s) = Vi(s) Solution: 1+sR1C1 Vo(s) 1 1 AVH(s)= = = Vi(s) 1+sR1C1 1+sτ

24. § §4.2 s-Domain Analysis Wherτ=R1C1is time constant. 1 1 1+ j2πf R1C1 · 1 fH= 2πR1C1 AVH= = 1+ j(f/fH ) 1 1+(f/fH)2 AVH=  =－arctg(f/fH) H   f＜＜fH → AVH≈1 or 0dB; ≈0o H f = fH → AVH=1/ 2 or －3dB; =－45o H  f＞＞fH→ AVH≈fH/f or －20lg(f/fH)/dB; ≈－90o H

25. § §4.2 s-Domain Analysis 20AVH/dB 3dB －20dB/ /decade －20 －40 f/Hz 0.01fH 0.1fH 10fH 100fH fH  H 0o f/Hz －45o －45o/decade －90o figure 4.3

26. § §4.2 s-Domain Analysis Example 4.2: For the circuit in Figure 4.4. Sketch the Bode plot of the transfer function. C2 Solution: + Vi － + Vo － · · R2 Vo(s) R2 Vi(s) R2+1/sC21+sτ s +1/τ sτ s AVL(s)= = = = figure 4.4 Whereτ=R2C2 is time constant. 1 2πR2C2 1 · fL= AVL= 1－j(fL/f) 1 1+(fL/f)2  =arctg(fL/f) AVL= L

27. § §4.2 s-Domain Analysis 20AVL/dB 0 3dB －20 20dB/ decade －40 f/Hz fL 0.01fL 0.1fL 10fL 100fL  L 90o 45o －45o/decade f/Hz 0o figure 4.5

28. § §4.3 High-Frequency Response of Transistor Circuits §4.3.1 The Frequency Properties of BJT 一、 一、The High Frequency Equivalent Circuits of BJT rb′c Ib b′ · § · · · · Ib Ic Ic Cb′c Cb′c b b′ c b c + · + · · rbb′ rbb′ Cb′e Cb′e rb′e Vb′e rb′e rce Vb′e e － e － figure 4.6 rbe=rbb′+rb′e  iC β0 vB′E IE (4－ －5) gm= = ≈ VT IE rb′ ′e VT  rb′e = ( 1+β0) VCE

29. § §4.3.1 The Frequency Properties of BJT 二、 二、The Frequency Parameter of BJT · · · β /dB Ib Ic Cb′c β0 b b′ c 0.7β0 + · · rbb′ Cb′e rb′e Vb′e － f 0 e fβ fT figure 4.8 · · figure 4.7 Ic Ib · · · β = = · Ic= (gm－jωCb′c)Vb′e Vce=0 1 1 ∥ jωCb′e · · gm－jωCb′c 1/rb′e+jω(Cb′e+Cb′c) Vb′e=Ibrb′e∥ jωCb′c

30. § §4.3.1 The Frequency Properties of BJT gmrb′e β0 · β ≈ = 1+jωrb′e(Cb′e+Cb′c) 1+jωrb′e(Cb′e+Cb′c) 1 β0 · fβ= (4－6) β≈ 1+(f/fβ)2 2πrb′e(Cb′e+Cb′c) · f＞＞fβ β ≈β0fβ/f (4－7) fT≈β0 fβ β(fT)=1 gm gm 2π(Cb′e+Cb′c) 2πCb′e fT= ≈ fT= (100～1000)MHz

31. § §4.3.2 High-Frequency Response of Common-Emitter Circuits 一、Miller,s Theorem 一、 A(s)=V2(s)/V1(s) Y(s) + + + + network network V2 (s) － V1 (s) － V2 (s) － V1 (s) － figure 4.9 1 1 1 (4－8) Y2 (s)= =Y(s)［1－ Z2 (s) ］ Y1 (s)= =Y(s)［1－A(s)］ Z1 (s) A(s) Y(s)［V1(s) －V2(s)］=Y(s)V1(s)［1－A(s)］= Y1 (s)V1(s) Y(s)［V2(s) －V1(s)］=Y(s)V2(s)［1－1/A(s)］= Y2 (s)V2(s)

32. § §4.3.2 High-Frequency Response of Common-Emitter Circuits 二、Miller,s Approximation Cb′c rbb′ · · 二、 C=Cbe′+ CM1 CM1 =(1+gmRC)Cb′c =(1－AV ′)Cb′c b′ c b + · + Rs · Cb′e Vb′e Vo RC rb′e + · Vs R=(Rs+rbb′)∥rb′e · － e － － · Vs ′= Vsrb′e/(Rs+rbb′+rb′e) rbb′ b b′ c R b′ c · + · + · · + · Rs + · + · rb′e RC Vo Vo RC + Vb′e Vs ′ · Vb′e － C Vs C － － － － － e e figure 4.9

33. § §4.3.2 High-Frequency Response of Common-Emitter Circuits 三、 High-Frequency Response and fH 三、 1 1+ jωRC · · · Vb′e= Vs′ R c b′ + · · + · · + · Vo=－gmVb′eRC · · Vo RC Vs ′ Vb′e － C － － Vs′= Vsrb′e/(Rs+rbb′+rb′e) e · · 1 AVM · = (4－9) Rs+rbb′+rb′e1+ jωRC 1+ j(f/fH ) Vo Vs rb′e · AVH= =－gmRC · Where · AVM=－gmRC rb′e · 1 fH= (4－11) 2πRC (4－10) Rs+rbb′+rb′e

34. § §4.3.2 High-Frequency Response of Common-Emitter Circuits 四、 四、 Gain-Bandwidth Product 1 · 2π RC rb′e · │AVM · fH│=gmRC (4－12) Rs+rbb′+rb′e R=(Rs+rbb′)∥rb′e C=Cb′e+CM1=Cb′e+(1+gmRC)Cb′c gmRC · │AVM · fH│= (4－13) 2π( Rs+rbb′) Cb′e+(1+gmRC)Cb′c 五、 五、Conclusion · · │AVM│↑ 1. │AVM · fH│=constance RC↑ → fH↓

35. § §4.3.2 High-Frequency Response of Common-Emitter Circuits 1 fH= 2πRC R=(Rs+rbb′)∥rb′e C=Cb′e+ CM1 = Cb′e+ (1+gmRC)Cb′c rbb′↓ Cb′e↓ Cb′c↓ 2. fH↑ 3. Rs↓→ fH↑ 4. fH ↑→ fT 5. emitter amplifier. The Miller effect reduced the bandwidth of common-

36. § §4.3.3 High-Frequency Response of Common-Collector Circuits VCC b b′ RB 1 rbb′ C1 Rs rb′e Cb′e · T C2 e Cb′c + + + Vo － · Rs · Vs RB 2 + vo － RL′ vi +vs － － RE RL c － RL′= RL∥RE · CM1= (1－AV )Cb′e b b′ CM1≈CM2 · rbb′ CM2= (1－1/AV )Cb′e Rs Cb′e · rb′e e + fH (CC) ＞fH (CE) · ·+ Vo － Vs RL′ fH (CC) → fT － figure 4. 10 c

37. § §4.3.4 High-Frequency Response of Common-Base Circuits · · Io e c T + － · + · Vb′e Rs Is rb′e Cb′e b′ Cb′c · RL′ vo RL′ R E Vo + Rs + vs － － rbb′ － b figure 4.11 1 rb′e · · · Is≈－Vb′e gm+ + jωCb′e Io Is α0 ≈ · 1+ jω Cb′e/ gm · · Io≈－ gmVb′e (4－14)

38. § §4.3.4 High-Frequency Response of Common-Base Circuits β0 gmrb′e α0= =1+ gmrb′e Where 1+ β0 · Io Is α0 gm ≈ · fH≈ = fT 2π Cb′e (4－15) 1+ jω Cb′e/ gm fH (CB) ＞ ＞ fH (CE) The common-base amplifier has a larger bandwidth because of a small Miller multiplication factor.

39. § §4.4 Low-Frequency Response of Transistor Circuits 一、 Low-Frequency Equivalent Circuit VCC 一、 · Ib C1 C2 RC · C2 βIb RB 1 + C1 + rbe + Rs + T · + RB RC RL Vo Rs +vs － + RL · vo RB 2 vi － + Vs－ RE CE CE RE － － · figure 4.12 Ib C2 ′ C1′ C2 C1 · · βIb + + Ib RC rbe Rs Rs · · RC rbe Vo RL RL Vo + － + · · · Vs－ Vs－ CE +βIbRC － －

40. § §4.4 Low-Frequency Response of Transistor Circuits 1 C1′ 1 1+ β CE 1 1 1 1 C1 CE = + C1 = + ≈ C2 C1′=(1+β)C1 + CE or C2′ CE C2 二、 二、 Low-Frequency Response and fL Vo βRL′ AVL= =－ Vs Rs+rbe1－j/ωC1′(Rs+rbe) 1－j/ωC2(RC+RL) · 1 1 · · · (4－15) · 1 2π(Rs+rbe) C1 · fL1= Vo Vs βRL ′ Rs+rbe · AVM= =－ · ω↑ → 1 2π(RC+RL) C2 fL2= · 1 AVL AVM = (4－16) · 1－j(fL1/f) 1－j(fL2/f) fL = fLmax(fL1，fL2)

41. § §4.5 Frequency Response of Multistage Amplifiers stage1 stage2 C + Vi － + Vo － · · · · AV1 AV2 figure 4.13 R AV AVM12 · 0.707AVM12 AV↑ AVM1 BW↓ 0.707AVM1 figure 4.14 fL1 fH1 fL fH O f (log scale)

42. § §4.5 Frequency Response of Multistage Amplifiers 一、 一、fH A A A (j )    vm1 vm2 vm n   A     vH 1 j  1 j  1 j  H1 H2 H n A ( )     , vm A A A A A 其中： vH vm vm1 vm2 vm n                         2 2 2                             1 1 1 H1 H2 H n A   ( ) vm 2 A vH H                         2 2 2                              1 1 1 2 2 H1 H1 H n

43. § §4.5 Frequency Response of Multistage Amplifiers 1   H (4-17) 1 1 1      2 2 2 H1 H2 H n                         2 2 2                         If ωH1=ωH2=…=ωHn,      1 1 1 2 H1 H1 H n 1 n (4-18)   H12   1 H 二、 二、fL A A A (j )    vm1   vm2   vm   n A vL 1 j  1 j  1 j  L1 L2 L n

44. § §4.5 Frequency Response of Multistage Amplifiers A ( )     , vm A A A A A 其中： vL vm vm1 vm2 vm n                         2 2 2                             1 1 1 L n L1 L2 (4-19)        2 2 2 L L1 L2 Ln If ωL1=ωL2=…=ωLn    L1 1 n (4-20) L  2 1 Conclusion:

45. § §4.5 Frequency Response of Multistage Amplifiers 1. fH < fHk (k=1,2,…n) , fL> fLk(k=1,2,…n); 2. BWk>BW (k=1,2,…n) ; BW=300Hz～ ～3.4kHz 1 4 3.4    1 130Hz   7.8kHz, 300 2 f f H L k k 1 4  2 1 fH = fHmin(fH1，fH2 ，…fHn) 3. fL = fLmax(fL1，fL2 ，…fLn)

46. Summary 1. In this chapter, we studied the frequency response of transistor circuits. We determined the effects due to circuit capacitors, includeing coupling, bypass, and also analyzed the expanded equivalent circuits of BJTs to determine the frequ- ency response of the transistors circuits.. 2. A time constant technique was developed so that Bode plots can be constructed with out the need of deriving complex transfer functions.The high and low corner frequen- cies or 3dB frequencies can be determined directly from the time constant .

47. Summary 3. Coupling and bypass capacitors affect the low-frequency characteristics of a circuit. In general, capacitance values in the microfarad range typically result in cutoff frequencies in the hertz or tens of hertz range. The transistor capacitors affects the high-frequency characteristics of a circuit. 4. An expanded hybrid-π model for the bipolar transistor were developed. The capacitances included in the model result in reduced transistor gain at high frequencies. The cutoff frequency is a figure of merit for the transistor and is defined as the frequency at which the magnitude of the current gain is unity.

48. Summary 5. The Miller effects is a multiplication of the base- collector capacitance due to feedback between the output and input of the transistor circuit. The band-width of the amplifier is reduced by this effect. 6. The common-emitter amplifier, in general, shows the greast effect of the Miller multiplication factor, so the bandwidth of this circuit is the smallest of the three basic types of amplifiers, The common-base amplifier has a larger bandwidth because of a smaller Miller multiplication factor. The cascode configuration, a combination of common-emitter and common - base stages, combines the advantanges of

49. Summary high gain and wide bandwidth. The collector-follower ampli- fier generally has the largest bandwidth of the three am- plifier configurations. Example 4.3: The transfer function of a amplifier as follows: 108(jf) · AV(jf) = (jf + 20) (jf + 106) (1) Sketch the Bode plot of the transfer function. · (2) Determine the values of AVM , fH , and fL. Solution: The transfer function can rewrite as follows:

50. Summary jf 10020 jf 20 · · (2) AVM =40dB fH =1M Hz fL = 20 Hz AV(jf) = (1 + ) · jf 106 (1 + ) AV(jf) /dB (1) jf 20 60 20lg 20lg100 40 20 20 f/Hz 10 100 1000 10k 100k 1M 10M 100M jf jf －20lg 1+106 －20lg 1+20 Figure 4.15