EECE 311 Lecture 2, Spring 2012

EECE 311: Electronic Circuits Part 2 AMPLIFIER FREQUENCY RESPONSE 2  High-Frequency Small-Signal Model EECE 311 – High-Frequency Models 3 MOSFET High-Frequency Model  The Basic N-channel MOSFET Structure  When B is shorted to S or body effect is negligible (the usual case) EECE 311 – High-Frequency Models 4  At low frequency, caps are open circuits  High-Frequency Small-Signal Model EECE 311 – High-Frequency Models 5 BJT High-Frequency Model  B’ is the internal base  rx is base spreading resistance B-B’ (small, negligible at low frequency)  Vp = Vb’e EECE 311 – High-Frequency Models 6 DEFINITION: Unity-Gain Frequency Used to find capacitance values 2 ( ) m T gs gd g f C C     MOSFET  BJT 2 ( ) m T g f C C      Frequency Response  At low frequency  capacitors are open-circuits (and inductors are short circuits)  amplifier characteristics are not affected by frequency (no frequency-dependent elements)  Capacitor impedance  Changes with frequency: , Z(jw) = -j /wC (infinite at DC)  As w increases, we need to study the “Frequency Response”  Input is  Output is since the system is linear  How do A (magnitude) and f (phase) vary with frequency w? EECE 311 – Frequency Response 7 1 ZC ( ) s sC  sin(  )t sin( ) A t  Example – How does one capacitor affect amplifier gain? EECE 311 – Frequency Response 8  Given: Blue block is the amplifier  Rs = 20 KW  Ri = 100 KW  Ci = 60 pF  m = 144 V/V  Ro = 200 W  RL = 1 KW  Find:  Vo/Vs  Plot  magnitude response  phase response  vo(t) when vs(t) = 0.1 sin10nt V for n = 2, 5, 6, 8 Solution EECE 311 – Frequency Response 9 i i s i s Z V V Z R    Express Vi as a function of Vs 1 1 1 1 1 ( ) i i i s s s i s s i s i R Z V V V V Z R R Y R sC        1 1 1 1 ( // ) s i i i R R s s i i s i s i V R V sR C R R sC R R         Express Vo as a function of Vi o L L o i L o i L o V R R V V R R V R R        Solution EECE 311 – Frequency Response 10 1 1 ( // ) o i L s L o i s i s i V R R V R R R R sC R R       0 ( ) 1 / o s V K T s V s      This is of the form: K 1/w0  Low-pass single-time-constant (LP-STC)  K is the low-frequency gain  w0 is the 3-dB frequency 1000 100K 144 100 1000 200 100K 20K i L L o i s R R K R R R R           6 0 12 3 1 1 10 rad/sec ( // ) 60 10 (100//20) 10 i i s C R R         Review: Low-Pass Single Time Constant  Gain (or Transfer) function:  Low-frequency gain = K  Upper 3-dB frequency = w0  -20 dB/decade drop at high frequency (w >> w0)  What is T at w0? 11 0 ( ) 1 / o s V K T s V s     Solution (continued) EECE 311 – Frequency Response 12  Bode Plot - Magnitude Solution EECE 311 – Frequency Response 13  Bode Plot - Phase Solution EECE 311 – Frequency Response 14  Output voltage vo(t) when  vs(t) = 0.1 sin102t V  vs(t) = 0.1 sin105t V  vs(t) = 0.1 sin106t V  vs(t) = 0.1 sin108t V  For w = 102 rad/sec: vo(t) = 10 sin(102t) V  For w = 105 rad/sec: vo(t) = 9.95 sin(105t – 5.7o) V  For w = 106 rad/sec: vo(t) = 7.07 sin(106t – 45o) V  For w = 108 rad/sec: vo(t) = 0.1 sin(108t – 89.4o) V 6 6 6 2 6 6 2 100 100 ( ) arctan( /10 ) 1 /10 1 ( /10 ) 100 ( ) 0.1 sin( arctan( /10 )) 1 ( /10 ) o T j j v t t                  15 Amplifier High-Frequency Response  We will not consider low-frequency effects In IC amplifiers, neither bypass nor coupling capacitors are used because of their large size: Amplifiers are directly coupled.  Bode Plot 16 0 1 2 1 2 1 2 1 2 ( ) ( ) ( ) 1 lim 1 1 ... 1 ( ) 1 1 ... 1 , ... are the zero frequencies and can be positive, negative or infinite. , . M H H s Z Z Zn H P P Pn Z Z Zn P P A s A F s F s s s s F s s s s                                                     .. Pn are the pole frequencies and are positive and real.   The High-Frequency Gain Function 17  The Dominant Pole Approximation H 2 2 2 2 1 2 1 2 1 1 1 1 1 ... 2 ... P P Z Z                         Better Approximation If one pole is obviously smaller than the other zeroes and poles. wp1 is 2 octaves down (less by a factor of 4) 18 2 2 1 2 2 2 1 2 2 2 2 2 2 2 1 2 1 1 ( ) 2 1 . 1 ... 1 2 1 . 1 ... let 1 1 ... 1 2 1 ... 1 H H H H Z Z H H P P H P P Pn Z Zn F j x x x x x x                                                                                 The Exact Solution: Solve |A|=|AM|/√2 ωH = √x 19  Example: Find H  5 4 4 1 10 ( ) 1 1 10 4 10 H s F s s s              5 1 2 4 1 4 2 10 rad/sec 10 rad/sec 4 10 rad/sec Z Z P P           Given:  Solution:  Dominant Pole? 4 10 rad/sec H   Better Approximation: H 2 2 2 2 1 2 1 2 1 1 1 1 1 2 9800 rad/sec P P Z Z                        Exact Solution: 2 2 2 1 2 1 1 1 2 1 9573 rad/sec P P Z H x x x x                          20  Example: Magnitude Bode Plot 5 4 4 1 10 ( ) 1 1 10 4 10 H s F s s s               Given: Pole: slope 20 dB/decade Zero: slope 20 dB/decade     Bode Plot Construction 21 5 4 4 1 10 ( ) 1 1 10 4 10 H s F s s s              P P Z Bode Plot Construction 22 5 4 4 1 10 ( ) 1 1 10 4 10 H s F s s s              What is the slope? Bode Plot Construction 23 5 4 4 1 10 ( ) 1 1 10 4 10 H s F s s s              What is the slope? What is this value (dB)? Bode Plot Construction 24 5 4 4 1 10 ( ) 1 1 10 4 10 H s F s s s              What is the slope? What is this value (dB)? 25 High-Frequency Response of Common Source and Common Emitter Amplifiers vsig Rsig VDD vI Q1 vO Q2 VBIAS Rsig vsig VDD Q1 vO Q2 VBIAS Cdb2 Cdb1 Cgd2 Cgs2 Cgs1 Cgd1 26 At signal frequencies Rsig vsig Q1 vO Q2 Cdb2 Cdb1 Cgd2 Cgs2 Cgs1 Cgd1 27 High-Frequency Response of Common Source and Common Emitter Amplifiers  High-Frequency Small-Signal Equivalent Circuits CS: CE:  RL represents the output resistance of the active load and any load resistance  CL represents the total capacitance between drain (or collector) and ground. It also includes the input capacitance of a succeeding amplifier stage, and parasitic/load capacitances.  We will analyze the CS circuit only since the analysis of the CE amplifier is exactly the same after relabeling 28 to , to , to gs gd gs C C C C V V      ' ' // sig sig sig x sig x sig r V V r R r R r R r          and replacing and by: sig sig V R Thevenin equivalent for circuit composed of Vsig, Rsig, rx, and rp 29  Common Source / Common-Emitter High-Frequency Small-Signal Equivalent Circuits Rsig vsig Q1 vO Q2 Cdb2 Cdb1 Cgd2 Cgs2 Cgs1 Cgd1 30  Exact Analysis in s-domain: YC = sC     ' ' ' KCL at drain: KVL at input: 0 KCL at gate: o gd gs o m gs L o L sig sig i gs i gs gs gd gs o V sC V V g V sC V R V R I V I sC V sC V V                     ' ' ' ' ' 2 ' ' ' 2 1 2 1 2 1 1 1 1 1 1 1 1 1 1 gd m L m o sig gs gd m L sig L gd L L gd gs L gd sig L M M Z Z o m Z sig gd P P P C g R s g V V s C C g R R C C R s C C C C C R R s s A A V g V b s b s C s s s                                                                             2 2 1 2 1 2 1 ' ' ' ' ' ' 2 ' ' 1 1 1 1 1 1 1 1 P P P P P P gs gd m L sig L gd L gs gd m L sig L gd L P L gd gs L gd sig L s s s C C g R R C C R C C g R R C C R C C C C C R R                                                  1 2 negligible ( 0) if P P     31 • Low frequency gain is –gmR’L • Zero frequency gm/Cgd is very high • What is the 3-dB frequency? 32 Miller’s Theorem  When Z is a capacitor C  C1 = C (1 – K)  C2 = C (1 – 1/K) 33  Example 2: Find / and . o sig H V V   Given:  Solution using Miller 1 pF Z  K  100 100 o i V V     1 1 1 9 1 1 1 1 1 1 10 k 101 p 100 100 1 (1010 n) 1 10 157.6 kHz 1010 2 i sig sig sig i sig o m o i sig H H H H V sC V sC R R sC V V s V A V V s V s f                       1 2 101pF 1.01pF C C   34  Find using Miller’s Theorem H 35  Miller’s Approximation in CS/CE   ' 1 in gs gd m L C C C g R    ' At midband, o m L gs V K g R V   The input circuit and the output circuit are low-pass RC circuits What is the pole frequency for a simple RC circuit? +CL 36  Miller’s Approximation in CS/CE   ' ' ' 1 1 1 H sig in sig gs gd m L R C R C C g R          ' 1 o m m m L sig H V A A g R s V     Input circuit pole at 1/(R’sig Cin) Output circuit pole at 1/(R’L Cout) Usually input circuit determines high frequency response +CL 37  Miller’s Approximation in CS/CE Input circuit pole at win = 1/(R’sig Cin) Output circuit pole at wout = 1/(R’L Cout) 2 2 1 1 1 in out H      If input pole is not much smaller than output pole +CL 38  Open-Circuit Time Constants Instead of finding A(s) to get wH, we’d like to determine wH directly from the circuit 1 H i io i C R   is capacitor " " in the circuit. is the resistance seen by capacitor with all other capacitors open-circuited and the signal sources set to zero. open-circuit time constant i io i i io i C i R C R C   1 1 2 2 H i io i i i f C R        39  Find using OCTC H  ' ' gs : find gs sig gs gs gs gs sig C R R C R C R      40   ' 1 1 ' ' 1 ' ' 1 : find ' ' ' ' ' ' gd gd gs x sig gs x o x m gs m gs L L x gd sig L m sig L x gd gd gd gd sig L m sig L C R V I R V V V I g V g V R R V R R R g R R I C R C R R g R R                  Frequently-used resistance calculation 41 RX RY gmVX VX + - RXY (1 ) where (1 ) where XY X Y m X Y XY Y X Y Y m Y XY X Y X X m X R R R g R R R R R g R R R R g R                42  Find using OCTC (cont’d) H  L : find CL C R 9 1 10 rad/sec 80 1000 1080 H    43 Open-Circuit Time Constants (cont’d) ' ' ' ' ' ' gs gs sig gd gd sig L m sig L L CL L C R R C R R R g R R C R R         gs gs gs gd gd gd cl L CL C R C R C R       1 H gs gd cl         Numberical Example: Find H .  1 pF 1pF gs gd C C   80 k neglect 1000 k gs L gd R C R      Given:  Solution: 80 ns 1000 ns gs gs gs gd gd gd C R C R       OCTC allows us to determine the circuit component(s) that limit the value of wH   ' ' ' ' ' ' L gs gd C gs sig gd sig L m sig L L L C R C R R g R R C R             44  Open-Circuit Time Constants (cont’d)   ' ' ' 1 ' ' ' H gs sig gd sig L m sig L L L C R C R R g R R C R       ' 1 o m m m L sig H V A A g R s V     45  Example 1: CMOS Common-Source Amp ' 2 ' 2 Given 3V 200 A/V 65 A/V 10 for all MOSFETs 0.6V 20V 10V 100 A DD n p W L tn tp An Ap REF V k k V V V V I              Find fH using:  Miller’s approximation  Open-circuit time constants (OCTC)  Transfer function Given 50 k 30 fF 10 fF 50 fF sig gs gd L R C C C      ' 1 ' 1 Miller's approximation input pole frequency: (1 ) 30 43.2 10 462 fF 1 6.89 MHz 2 ' output pole frequency: 1 1 1 10 1 50 60.24 fF 42.2 1 39 2 ' in gs m L gd in in sig out gd L m L out out L C C g R C f C R C C C g R f C R                                 .6 MHz input pole is dominant: 6.89 MHz Hf  46  Example 1: CMOS Common-Source Amp 1 ' 1 2 Given: 50 k ' here 30 fF 10 fF 50 fF Previously found: 0.633 mA/V // 66.67 K sig sig gs gd L m L o o R R C C C g R r r          47  Example 1: CMOS Common-Source Amp 1 ' 1 2 Given: 50 k 30 fF 10 fF 50 fF Previously found: 0.633 mA/V // 66.67 K sig gs gd L m L o o R C C C g R r r         ' ' 1 ' OCTC ' 50k ' (1 ) 50 43.2 66.67 2226.67k 66.67k 30 50 10 2226.67 50 66.67 1500 22266.7 3333.3 1 5.87 MHz 2 L L gs sig gd sig m L L C L H gs gs gd gd L C H H R R R R g R R R R C R C R C R f                            48  Example 1: CMOS Common-Source Amp 1 ' 1 2 Given: 50 k 30 fF 10 fF 50 fF Previously found: 0.633 mA/V // 66.67 K sig gs gd L m L o o R C C C g R r r        1 2 1 1 Exact values (from transfer function) 10.1 GHz, 5.96 MHz, and 386 MHz is a dominant pole 5.96 MHz Error in Miller's approximation = +15.6% Error in OCTC approximation = 1.51% z p p p H p f f f f very f f        Example 2: CMOS Common-Source Cascade  What is the overall low-frequency gain if the per-stage gain is –42?  What is the 3-dB frequency if the per-stage 3-dB fH is 6 MHz? Solution 50 Example 3: Common-Emitter Amplifier  Find the 3-dB frequency fH using Miller’s approximation  Assume that the input circuit determines the value of fH 51 vi Q3 IREF Q2 Q1 vo VCC Rsig vsig Solution 52