Chapter 13 Frequency Response Microelectronic Circuit Design Richard C. Jaeger Travis N. Blalock 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Chapter Goals • Review transfer function analysis and dominant-pole approximations of amplifier transfer functions. • Learn partition of ac circuits into low and high-frequency equivalents. • Learn short-circuit and open-circuit time constant methods to estimate upper and lower cutoff frequencies. • Develop bipolar and MOS small-signal models with device capacitances. • Study unity-gain bandwidth product limitations of BJTs and MOSFETs. • Develop expressions for upper cutoff frequency of inverting, non-inverting and follower configurations. • Explore gain-bandwidth product limitations of single and multiple transistor circuits. 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Chapter Goals (contd.) • Understand Miller effect and design of op amp frequency compensation. • Develop relationship between op amp unity-gain frequency and slew rate. • Understand use of tuned circuits to design high-Q bandpass amplifiers. • Understand concept of mixing and explore basic mixer circuits. • Study application of Gilbert multiplier as balanced modulator and mixer. 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Transfer Function Analysis F ( s) L s L s L ... s L Z1 Z 2 Zk s L s L ... s L P1 P2 Pk F ( s) H a a s a s2 ... amsm N ( s ) 2 Av (s) 0 1 D(s) b b s b s2 ... b sn n 0 1 2 A F ( s) F ( s) H mid L Amid is midband gain between upper and lower cutoff frequencies. 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock 1 s 1 s ...1 s H H H Z 1 Z2 Zl s s s 1 1 ... 1 H H H P1 P2 Pl F ( j ) 1 for H , H ,i =1…l H Zi Pi A ( s) A F ( s) L mid L F ( j ) 1 for L , L ,j =1…k L Zj Pj A ( s) A F ( s) H mid H Copyright © 2005 – The McGraw-Hill Companies srl Low-Frequency Response s F ( s) L s L 1 2 P2 2 2 2 2 Z 2 Z1 Z 2 1 Z1 P2 Pole P2 is called the dominant low-frequency pole (> all other poles) and zeros are at frequencies low enough to not affect L. If there is no dominant pole at low frequencies, poles and zeros interact to determine L. A ( s) A F ( s) A L mid L mid 1 2 s s Z 1 Z 2 s s P1 P2 A mid For s=j, at L,A( j L ) 2 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock L2 Z12 L2 Z 22 L2 P12 L2 P22 2 4 L L 2 2 2 2 P2 P1 P2 1 P1 2 4 L L Pole L > all other pole and zero frequencies 2 L P1 P2 2 2 2 2 2 Z1 Z2 In general, for n poles and n zeros, L 2 2 2 n Pn n Zn Copyright © 2005 – The McGraw-Hill Companies srl Transfer Function Analysis and Dominant Pole Approximation Example • Problem: Find midband gain, FL(s) and fL for s 1 100 A ( s) 2000 L 0.1s 1s 1000 s • Analysis: Rearranging the given transfer function to get it in standard form, A ( s) 200 ss 100 L s 10s 1000 Now, A ( s) A F ( s) L mid L s(s 100) F ( s ) and A 200 L (s 10)(s 1000) mid Zeros are at s=0 and s =-100. Poles are at s= -10, s=-1000 1 f 102 10002 2(02 1002 158Hz L 2p All pole and zero frequencies are low and separated by at least a decade. Dominant pole is at =1000 and fL =1000/2p= 159 Hz. For s frequencies > a few rad/s: A ( s) 200 L 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock s 1000 Copyright © 2005 – The McGraw-Hill Companies srl High-Frequency Response s F ( s) L 1 (s / H P3 ) P3 1 ( 2 / 2 ) 1 ( 2 / 2 ) 1 H H Z1 Z2 2 2 2 2 1 ( / ) 1 ( / 2 ) H H P1 P2 2 2 4 H Pole P3 is called the dominant high1 H H 2 2 2 2 1 frequency pole (< all other poles). Z1 Z2 Z1 Z 2 2 2 2 4 H H H 1 If there is no dominant pole at low 2 2 2 2 P1 P2 P1 P2 frequencies, poles and zeros interact Pole H < all other pole and zero frequencies to determine H. 1 A ( s) A F ( s) H mid H ) 1 ( s / ) 1 ( s / Z 1 Z2 A mid 1 ( s / ) 1 ( s / ) P1 P2 A mid For s=j, at H,A( j H ) 2 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock H 1 1 2 2 2 2 2 2 P1 P2 Z1 Z2 In general, H 1 1 1 2 2 2 n Pn n Zn Copyright © 2005 – The McGraw-Hill Companies srl Direct Determination of Low-Frequency Poles and Zeros: C-S Amplifier R Vo(s) Io(s) R gmVgs (s) 3 gm ( R R ) 3 D s s 1 D R R (1/ sC ) R 3 3 2 3 Vgs (s) C (R R ) 2 D 3 s C R 1 G Vg (s) V (s) s C ( R R ) 1 i 1 I G s (1/ C R ) 3 S Vgs (s) Vg - Vs Vg (s) 1 s C (1/ gm ) R 3 S V (s) Av(s) o A F (s) mid L V (s) i R G A gm ( R R ) 3 R mid D G RI 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Direct Determination of Low-Frequency Poles and Zeros: C-S Amplifier (contd.) 2 F (s) L s s (1/ C R ) 3 S 1 1 s s C ( R R ) C (1/ g ) 1 I G m 3 1 s R C2 ( RD S R ) 3 The three zero locations are: s = 0, 0, -1/(RS C3). 1 The three pole locations are: s , 1 1 , C ( R R ) C (1/ g ) R C ( R R ) 1 I G 2 D 3 m S 3 Each independent capacitor in the circuit contributes one pole and one zero. Series capacitors C1 and C2 contribute the two zeros at s=0 (dc), blocking propagation of dc signals through the amplifier. Third zero due to parallel combination of C3 and RS occurs at frequency where signal current propagation through MOSFET is blocked (output voltage is zero). 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Short-Circuit Time Constant Method to Determine L • Lower cutoff frequency for a network with n coupling and bypass capacitors is given by: n 1 L Midband gain and upper and lower cutoff frequencies that define bandwidth of amplifier are of more interest than complete transfer function. 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock i 1 RiS Ci where RiS is resistance at terminals of ith capacitor Ci with all other capacitors replaced by short circuits. Product RiS Ci is shortcircuit time constant associated with Ci. Copyright © 2005 – The McGraw-Hill Companies srl Estimate of L for C-E Amplifier Using SCTC method, for C1, R R ( R RCE ) R ( R rp ) I B in B 1S 2 For C2, R R ( R RCE ) R ( R ro ) 2S 3 C out 3 C R R 3 C For C3, rp R th CC R R Rout R E E 1 3S o rp ( R R ) I B R E o 1 3 1 L i 1 RiS Ci 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Estimate of L for C-S Amplifier Using SCTC method, For C1, R R ( R RCS ) R R I 1S G in S G For C2, R R ( R RCS ) R ( R ro ) D out D 2S 3 3 R R 3 D For C3, 1 R R RCG R 3S S out S g m 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Estimate of L for C-B Amplifier Using SCTC method, For C1, 1 R R ( R RCB) R ( R ) I E in I E g 1S m For C2, R R ( R RCB )R R 2S 3 C out 3 C 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Estimate of L for C-G Amplifier Using SCTC method, For C1, 1 R R ( R RCG) R ( R ) I I 1S S in S g m For C2, R R ( R RCG )R R D out 2S 3 3 D 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Estimate of L for C-C Amplifier Using SCTC method, For C1, R R ( R RCC ) I B in 1S R R rp o 1 R R E 3 I B For C2, R R (R E 2S 3 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock CC R ) R Rout 3 E rp R th o 1 Copyright © 2005 – The McGraw-Hill Companies srl Estimate of L for C-D Amplifier Using SCTC method, For C1, R R ( R RCD) R R I I 1S G in G For C2, 1 R R R RCD R R 2S 3 S out 3 S g m 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Frequency-dependent Hybrid-Pi Model for BJT Capacitance between base and emitter terminals is: Cp gmt Capacitance between base and collector terminals is: Cmo Cm 1 (V / ) CB jc Cmo is total collector-base junction capacitance at zero bias, Fjc is its built-in potential. 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock F tF is forward transit-time of the BJT. Cp appears in parallel with rp. As frequency increases, for a given input signal current, impedance of Cp reduces vbe and thus the current in the controlled source at transistor output. Copyright © 2005 – The McGraw-Hill Companies srl Unity-gain Frequency of BJT Ic (s) ( gm sCm )V (s) be The right-half plane transmission zero Z = + gm/Cm occurring at high frequency can be neglected. o o ( s) s(Cp Cm )rp 1 (s / ) 1 = 1/ rp(Cm + Cp ) is the beta-cutoff frequency (s) rp ( gm sCm )I (s) b s(C C )r 1 p m p sC m o 1 g m I c (s) ( s) I (s) s(Cp Cm )rp 1 b 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock o T s s o gm where o T (Cp Cm )rp Cp Cm and fT = T /2p is the unity gain bandwidth product. Above BJT has no current gain. Copyright © 2005 – The McGraw-Hill Companies srl Unity-gain Frequency of BJT (contd.) Current gain is o = gmrp at low frequencies and has single pole rolloff at frequencies > f, crossing through unity gain at T. Magnitude of current gain is 3 dB below its low-frequency value at f. Cp gm T 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Cm 40I C C m T Copyright © 2005 – The McGraw-Hill Companies srl High-frequency Model of MOSFET o gm o T (C C )r C C p m p p m I (s) ( gm sC )Vgs (s) GD d ( g m sC ) GD I (s) b s(C C ) GS GD I (s) ( s) d T I g (s) s 1 T s 1 (C GS 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock /C ) GD f T mnCox " V V L GS TN W (2/3)Cox "WL 3 mn VGS VTN 2 L2 Copyright © 2005 – The McGraw-Hill Companies srl Limitations of High-frequency Models • • • • Above 0.3 fT, behavior of simple pi-models begins to deviate significantly from the actual device. Also, T depends on operating current as shown and is not constant as assumed. For given BJT, a collector current ICM exists that yield maximum fTmax. For FET in saturation, CGS and CGD are independent of Q-point current, so gm I T D 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Effect of Base Resistance on Midband Amplifiers Base current enters the BJT through external base contact and traverses a high resistance region before entering active area. rx models voltage drop between base contact and active area of the BJT. 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock To account for base resistance rx is absorbed into equivalent pi model and can be used to transform expressions for C-E, C-C and C-B amplifiers. i gmv gm rp v g 'v rp rx be m be r o gm ' gm p rp rx rp rx rp ' rp rx o' o Copyright © 2005 – The McGraw-Hill Companies srl Direct High-Frequency Analysis: C-E Amplifier R R R 100kΩ 4.3kΩ L 3 C R R R 30kΩ 10kΩ B 1 2 The small-signal model can be simplified by using Norton source transformation. R B v v th i R R I B v th is R rx th 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock R R R I B th R R I B rpo rp ( R rx ) th Copyright © 2005 – The McGraw-Hill Companies srl Direct High-Frequency Analysis: C-E Amplifier (Pole Determination) From nodal equations for the circuit C Cp T in frequency domain, V (s) Is(s) 2 ( sC m - g m ) s2 Cp Cm C C C L p L s Cp g Cm g g m gp C gpo g gpo L L L L High-frequency response is given by 2 poles, one finite zero and one zero at infinity. Finite right-half plane zero, Z = + gm/Cm > T can easily be neglected. For a polynomial s2+sA1+A0 with roots a and b, a =A1 and b=A0/A1. 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Cm 1 g m R L R L rpo R L C L rpo A 1 0 P1 A r C 1 po T gm gm P2 C 1 (C / C ) C Cp C L p L m L Smallest root that gives first pole limits frequency response and determines H. Second pole is important in frequency compensation as it can degrade phase margin of feedback amplifiers. Copyright © 2005 – The McGraw-Hill Companies srl Direct High-Frequency Analysis: C-E Amplifier (Overall Transfer Function) ( sC - g ) V (s) m m th R rx g g 1 ( s / ) 1 ( s / ) th L po P1 P2 1 ( s / ) V (s) Z Vo(s) th ( gm R rpo ) L R rx g gpo 1 (s / ) 1 ( s / ) th L P1 P2 V (s) g R r m L po Vo(s) - th R rx 1 ( s / ) th P1 o R L A A Vo (s) mid mid A ( s) R rx rp vth V (s) 1 (s / ) th th P1 Vo(s) 1 P1 r C po T Dominant pole model at high frequencies for C-E amplifier is as shown. 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Direct High-Frequency Analysis: C-E Amplifier (Example) • • • Problem: Find midband gain, poles, zeros and fL. Given data: Q-point= ( 1.60 mA, 3.00V), fT =500 MHz, o =100, Cm =0.5 pF, rx =250W, CL 0 Analysis: gm =40IC =40(0.0016) =64 mS, rp = o/gm =1.56 kW. gm Cm 19.9pF f 1 1.56MHz P1 2pr C 2pf po T T g 1 1 1 R R R 100kΩ 4.3kΩ 4.12kΩ m 1 L 3 C P2 R C C g r g R m p m po m L R R R 7.5kΩ 1kΩ 882Ω L B I th gm P 2 f 20.4GHz f 603MHz rpo rp ( R rx ) 656W Z P 2 th 2pCm 2p R R o R L L 156pF L C C Cp Cm 1 gm R A 153 T L L r vth rpo R rx rp po th Cp Overall gain is reduced to -135 as vth =0.882vs. 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Gain-Bandwidth Product Limitations of C-E Amplifier • If Rth is reduced to zero in order to increase bandwidth, then rpo would not be zero but would be limited to approximately rx. GBW Av H • o R L R rx rp th 1 rpoC T If Rth = 0, rx <<rp so that rx = rpo and CT Cm ( gmRL ) GBW 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock 1 rxCm Copyright © 2005 – The McGraw-Hill Companies srl High-Frequency Analysis: C-S Amplifier R R R I G th R R R L D 3 R G v v th i R R I G C C T GS C 1 gm R GD L R L R th 1 P1 R C th T g 1 1 1 m 1 P2 R C C gm R gm R L GD GS L th g m Z C GD 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Miller Multiplication Vo(s) AV (s) Is(s) sC V (s) Vo (s) 1 1 I ( s) Y(s) s sC (1 A) V ( s) 1 For the C-E amplifier, Total input capacitance = C(1+A) because C Cp Cm (1 A) Cp Cm (1 gmR ) T L total voltage across C is vc = vi(1+A) due to inverting voltage gain of amplifier. 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Miller Integrator 1 A RC 1 A V Ao Av (s) o 1 V s o s 1 RC(1 A) where Assuming zero current in input terminal of amplifier, V V 1 in sC (V V ) o in R Vo AV in 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock o 1 RC(1 A) For frequencies >> o, assuming A>>1, A o 1 Av ( s) s sRC which is the transfer function of an integrator. Copyright © 2005 – The McGraw-Hill Companies srl Open-Circuit Time Constant Method to Determine H At high frequencies, impedances of coupling and bypass capacitors are small enough to be considered short circuits. Open-circuit time constants associated with impedances of device capacitances are considered instead. 1 m H RioCi i 1 where Rio is resistance at terminals of ith capacitor Ci with all other capacitors open-circuited. For a C-E amplifier, assuming CL =0 Rpo rpo 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock R Rmo rpo (1 gmR L ) L r ix po 1 1 H R C R C po p mo m rpoCT vx Copyright © 2005 – The McGraw-Hill Companies srl Gain-Bandwidth Trade-off Using Emitter Resistor R R o L A L mid R rx rp ( o 1) R R E E th for rp Rth rx and gmRE 1 gain decreases as emitter resistance increases and bandwidth of stage will correspondingly increase. To find bandwidth using OCTC method: R rx R E Req th ix 1 gm R E R rx R E Rpo rp Req th 1 gm R E vx 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Gain-Bandwidth Trade-off Using Emitter Resistor (contd.) Test source ix is first split into two equivalent sources and then superposition is used to find vx =(vb - vc). Assuming that o >>1 and R rx rp th ( o 1) R E gmRL RL v x Rmo ( R rx )1 R r th ix 1 gm R x E th H 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock 1 Cp ( R rx ) th 1 gm R E R g R E C 1 m L 1 R rx m 1 gm R E th R L R rx th Copyright © 2005 – The McGraw-Hill Companies srl Dominant Pole for C-B Amplifier R R R E I th R R R L C 3 R rx R rx vx th Rpo rp rp th ix 1 gm R 1 gm R th th Using split-source transformation Assuming that o >>1 and rx << rp v b vc gmRL Rmo rx 1 RL ix 1 gm R th H 1 R gm R C p 1 th C 1 L C R rx m rx 1 gm R m L 1 gm R th th Neglecting first term of order of 1/ T and since last term is dominant. 1 H C R m L 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Dominant Pole for C-G Amplifier R R R th 4 I R R R L D 3 R 1 th R GSo 1 gmR th Gth gm R R L GDo 1 1 H C C R GD L GS C R GD L G gm th 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Dominant Pole for C-C Amplifier R R R B I th R R R L E 3 R rx R R rx R vx L L th Rpo rp rp th ix 1 gm R 1 gm R L L Rmo ( R rx ) RCC ( R rx ) rp ( o 1)R in th th L ( R rx ) th 1 H Cp ( R rx R ) ( R rx )Cm L 1 g R th th m L A better estimate is obtained if we set RL =0 in expression for Rpo. H 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock 1 Cp ( R rx ) th 1 gm R L Cm 1 GBW (1) H C r mx Copyright © 2005 – The McGraw-Hill Companies srl Dominant Pole for C-D Amplifier R R R G I th R R R L S 3 Substituting rp as infinite and rx as zero in expression for emitter follower, H 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock 1 1 R C th C C R GS C R 1 gm R GS GD th th 1 g R GD L m L Copyright © 2005 – The McGraw-Hill Companies srl Frequency Response: Differential Amplifier CEE is total capacitance at emitter node of the differential pair. Differential mode half-circuit is similar to a C-E stage. Bandwidth is determined by the rpoCT product. As emitter is a virtual ground, CEE has no effect on differential-mode signals. For common-mode signals, at very low R frequencies, C Acc (0) 2R 1 EE Transmission zero due to CEE is 1 s Z C 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock R EE EE Copyright © 2005 – The McGraw-Hill Companies srl Frequency Response: Differential Amplifier (contd.) P rp rx 1 R 2R EE 1 g EEO o m 1 Cp rx 1 2 gm R EE 2R EE 1 rx gm R C C 1 m 1 2 gm R EE R C C EE rx 2gm As REE is usually designed to be large, Common-mode half-circuit is similar to a C-E stage with emitter resistor 2REE. OCTC for Cp and Cm is similar to the C-E stage. OCTC for CEE/2 is: 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock 1 1 P C C p EE C ( R r ) Cm ( RC rx ) m C x 2 gm Copyright © 2005 – The McGraw-Hill Companies srl Frequency Response: CommonCollector/ Common-Base Cascade REE is assumed to be large and neglected. r r 1 CC 1 Rout p1 x11 g o1 m1 r r 1 CB 2 Rin p2 x12 g o2 m2 Sum of the OCTC of Q1 is: C p1 r x1 1 1 g m1 g m2 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock C p1 C r C m1 x1 2 m1 Copyright © 2005 – The McGraw-Hill Companies srl Frequency Response: Common-Collector/ Common-Base Cascade (contd.) Sum of the OCTC of Q2 is: C p2 r x2 1 1 g m2 g m1 gm R C C 1 1 m 2 1 g m2 g m1 C R m2 C C r p 2 x2 2 C 1 m2 gm R R C C 2 r x2 Combining the OCTC for Q1 and Q2, and assuming that transistors are matched, H 1 rx Cp Cm 2 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock gm R R C C 2 rx Copyright © 2005 – The McGraw-Hill Companies srl Frequency Response: Cascode Amplifier OCTC of Q1 with load resistor 1/ gm2 : R C R C r C r C C 1 po1 p 1 mo1 m1 po1 T1 po1 p 1 m1 g As IC2 = IC1, gm2 = gm1, gain of first stage is unity. Assuming gm2 rpo1 >>1, C r C 2C po1 T1 po1 p 1 m1 r OCTC of Q1, a C-B stage for ro1 >> RL and mf>>1: C C R C p 2 (r R )C po2 p 2 mo2 m 2 g x2 L m 2 m2 R Assuming matched devices, 1 H r C 2C r R C m x L m po1 p 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock 1 m1 g g r m2 m2 po1 Copyright © 2005 – The McGraw-Hill Companies srl Frequency Response: MOS Current Mirror rpo 1 g R r L o2 m1 Cm C GD 2 Cp C C GS1 GS 2 For matched transistors, 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock P1 1 2C GS g m1 2C r GD2 o2 1 2C r GD2 o2 Copyright © 2005 – The McGraw-Hill Companies srl Frequency Response: Multistage Amplifier • Problem:Use open-circuit and short-circuit time constant methods to estimate upper and lower cutoff frequencies and bandwidth. • Approach: Coupling and bypass capacitors determine low-frequency response, device capacitances affect high-frequency response. At high frequencies, ac model for multi-stage amplifier is as shown. 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Frequency Response: Multistage Amplifier (Estimate of L) SCTC for each of the six independent coupling and bypass capacitors has to be determined. R R R R R R R ( R R ) 10kΩ 1MΩ I 1S G in1 1.01MΩ 1 1 R R 200W 66.7W 2S S1 g 0.01S m1 R R R R R 3S D1 O1 B2 in2 R r R r 2.69kΩ D1 o1 B2 p 2 R R R r 571W th2 B2 D1 o1 R r th2 p 2 19.4W R R 4S E 2 1 o2 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock 5S C 2 O2 B3 in3 R r R r ( 1)( R R ) o3 E3 L C 2 o2 B3 p 3 18.4kΩ R R R r 3.99kW th3 B3 C2 o2 R r th3 p 3 311W R R R L E 3 1 6S o3 n 1 3300rad/s L i 1 RiS Ci f L 530Hz L 2p Copyright © 2005 – The McGraw-Hill Companies srl Frequency Response: Multistage Amplifier (Estimate of ) H OCTC for each of the two capacitors associated with each transistor has to be determined. For M1, R R L1 r 478W I12 p 2 1 g R C R C C R th T1 th GS1 GD1 m1 L1 1.07 107s R r 570W I12 o1 th2 r r ( R r ) 610W po2 p 2 th2 x2 R R R L2 I 23 in3 r p 3 ( o3 1)( RE 3 I 23 3.54kΩ 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock 1.74 107s For Q3, R R r 3.99kW I 23 o2 (R r ) R C R C th3 x3 C ( R r )C p 3O p 3 m 3O m 3 1 g R p 3 th3 x3 m 3 m3 EE 1.51108s 1 m 3.38 106 rad/s H RioCi i 1 f H 538kHz H 2p th3 For Q2, R R L1 R th R R r C r C C 1 g R L2 po2 T 2 po2 p 2 m 2 m2 L2 r po2 R ) L Copyright © 2005 – The McGraw-Hill Companies srl Single-pole Op Amp Compensation • Frequency compensation forces overall amplifier to have a single-pole frequency response by connecting compensation capacitor around second gain stage of the Ao basic op amp. B A ( s) T v 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock s B s B Copyright © 2005 – The McGraw-Hill Companies srl Three-stage MOS Op Amp Analysis Input stage is modeled by its Norton equivalent- current source Gmvdm and output resistance Ro. Second stage has gain of gm5ro5= mf5 and follower output stage is a unity-gain buffer. Vo(s) = Vb(s) = - Av2Va(s) Gm Ro A Vo (s) - Av2Va (s) v2 Av (s) V ( s) V ( s) 1 sRoC (1 A ) C v2 dm dm Ao T B s s B B Gm A 1 v2 B R C (1 A ) T C (1 A ) o C C v2 v2 Gm For large Av2 T C C 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Transmission Zeros in FET Op Amps Incorporating the zero determined by gm5 in the analysis, 1 ( s / ) Z A ( s) ( g r ) m5 o5 1 ( s / ) vth P1 g g m 5 m5 Z C C Tg C GD5 m2 1 P1 R C o T r C C (C C )1 m o5 T GS 5 C GD5 f 5 R o This zero can’t be neglected due to low ratio of transconductances of M2 and M5. Zero can be canceled by addition of RZ =1/ gm5. Z 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock (1/ g ) R m5 Z C C Copyright © 2005 – The McGraw-Hill Companies srl Bipolar Amplifier Compensation • Bipolar op amp can be compensated in the same manner as a MOS amplifier • Transmission zero occurs at too high a frequency to affect the response due to higher transconductance of BJT that FET for given operating current. I m 5 C5 Z C TI C C2 g • 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Unity gain frequency is given by: g 40I 20I m 2 C 2 C1 T C C C C C C Copyright © 2005 – The McGraw-Hill Companies srl Slew rate of Op Amp • Slew-rate limiting is caused by limited current available to charge/discharge internal capacitors. For very large Av2, amplifier behaves dv (t ) likeI an C integrator: B C dvo (t ) C1 C dt C dt I dvo (t) T 1 SR C dt max C Gm / I1 amplifier, • For CMOS I T 1 SR Gm / I 1 T K n2 SR • For bipolar amplifier, 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock T T 20 Gm / I 1 Copyright © 2005 – The McGraw-Hill Companies srl Tuned Amplifiers • Amplifiers with narrow bandwidth are often required in RF applications to be able to select one signal from a large number of signals. • Frequencies of interest > unity gain frequency of op amps, so active RC filters can’t be used. • These amplifiers have high Q (fH and fL close together relative to center frequency) • These applications use resonant RLC circuits to form frequency selective tuned amplifiers. 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Single-Tuned Amplifiers • RLC network selects the frequency, parallel combination of RD, R3 and ro set the sC g VQ ( s)and bandwidth. m GD Av ( s) o V ( s) G s(C C ) (1/ sL) P GD i G go G G P D 3 • o plane Neglecting right-half s Q zero, Av (s) A mid 2 s s o o 2 Q o 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock 1 L(C C ) GD R Q o R (C C ) P P GD L o Copyright © 2005 – The McGraw-Hill Companies srl Single-Tuned Amplifiers (contd.) • At center frequency, s = jo, Av = Amid. A gmR gm(ro R R ) P mid D 3 o o 2 L 1 BW Q R (C C ) R P P GD 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Use of tapped Inductor- Auto Transformer C and r can often be small enough to GD o degrade characteristics of the tuned amplifier. Inductor can be made to work as an auto transformer to solve this problem. V ( s) Vo(s) nV1(s) n2 1 I2(s) Is(s) / n Is(s) Z s ( s) n2 Z p ( s) These results can be used to transform the resonant circuit and higher Q can be obtained and center frequency doesn’t shift significantly due to changes in CGD. Similar solution can be used if tuned circuit is placed at amplifier input instead of output 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Multiple Tuned Circuits • Tuned circuits can be placed at both input and output to tailor frequency response. • Radio-frequency choke(an open circuit at operating frequency) is used for biasing. • Synchronous tuning uses two circuits tuned to same center frequency for high Q. • BWn BW 21/ n 1 Stagger tuning uses1 two circuits tuned to slightly different center frequencies to realize broader band amplifiers. • Cascode stage is used to provide isolation between the two tuned circuits and eliminate feedback path between them due to Miller multiplication. 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Mixers: Conversion Gain • Amplifiers discussed so far have always been assumed to be linear and gain expressions involve input and output signals at same frequency. • Mixers are nonlinear devices, output signal frequency is different from input signal frequency. • A mixer’s conversion gain is the ratio of phasor representation of output signal to that of input signals, the fact that the two signals are at different frequencies is ignored. 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Single-Balanced Mixer • Eliminates one of the two input signals from the output. i I I sin t EE EE 1 1 4 v (t ) sin n t 2 2 n odd np I R sin n t EE C 2 4 Vo (t) I R IR 1 C cos(n )t 1 C cos(n )t n p n odd 2 1 2 1 2 2 • No signal energy appears at1 , but 2 appears in output spectrum, so circuit is single-balanced. • Up-conversion uses component (2-1) and down-conversion uses (2+1) component. 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Double-Balanced Mixer/ Modulator: Gilbert Multiplier • Double-balanced mixers don’t contain spectral components at either of the two input frequencies. Vm Vm i I sin mt i I sin mt C 2 BB 2R C1 BB 2R 1 1 R 4 Vo(t) Vm C cos(n )t cos(n )t c m c m R 1 n odd np • Modulator applications give double sideband suppressed carrier output signal. Amplitude-modulated signal can also be obtained if v Vm(1 M sin mt) 1 R 4 M M vo(t) Vm C sin n t cos( n ) t cos( n ) t c c m c m R 2 2 1 n odd np 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl