The Miller Approximations The “exact” analysis is not particularly helpful for gaining insight into the frequency response ... consider the effect of Cµ on the input only It Cµ + Vt + gmVt R'out = ro roc RL Vout − −neglect the feedforward current Iµ in comparison with gm Vπ ... a goodapproximationIt = (Vt - Vo) / ZµVo = - gmVt RL / (RL + Rout’) = AvCµ Vtwhere AvCµ is the low frequency voltage gain across CµIt = Vt (1- Av) / ZµZeff = Vt / It = Zµ / (1 - Av)Zeff = -j--ω---1--C----µ- 1-----–-----1A----v--C----µ = ----------------------1----------------------- = -j--ω----1-C----M--- jω(Cµ(1 – AvCµ))CM = (1 – AvCµ)Cµ is the Miller capacitor EE 105 Spring 1997 Lecture 23

Generalized Miller Approximations An impedance Z connected across an ampliﬁer with voltage gain AvZ can be replaced by an impedance to ground ... multiplied by (1-AvZ)Z+ Avo + + Avo +Vi Vo Vi Zeff Vo−− − − Zeff = Z /(1 − Avo)s Common-emitter and common-source: AvZ = large and negative for Cµ or Cgd --> capacitance at the input is magnifieds Common-collector and common-drain: AvZ ≈ 1 --> capacitance at the input due to Cπ or Cgs is greatly reduced EE 105 Spring 1997 Lecture 23

Voltage Gain vs. Frequency for CE Ampliﬁer Using the Miller Approximations The Miller capacitance is lumped together with Cπ, which results in a single- pole low-pass RC filter at the input RSVs + + Cπ CM gmVπ + − R'out Vout rπ Vπ − − CM = Cµ(1 + gm R'out)Transfer function has one pole and no zero after Miller approximation: ω3–1dB = (rπ RS)(Cπ + CM) ω3–1dB = (rπ RS)[Cπ + (1 + gmro roc RL)Cµ] ω–31dB ≈ ω1–1 from the exact analysis (ﬁnal term Rout′Cµ is missing) EE 105 Spring 1997 Lecture 23

Multistage Ampliﬁer Frequency Responses Summary of frequency response of single-stages: CE/CS: suffers from Miller effect CC/CD: “wideband” -- see Section 10.5 CB/CG: “wideband” -- see Section 10.6 (wideband means that the stage operates to near the frequency limit of the device ... fT)s How to ﬁnd the Bode plot for a general multistage ampliﬁer? can’t handle n poles and m zeroes analytically --> SPICE develop analytical tool for an important special case: * no zeroes * exactly one “dominant” pole (ω1 << ω2, ω3, ... , ωn)V-----o---u--t = -(--1----+------j--(--ω------⁄---ω----1---)--)---(--1-----+-----j--(---ω---A--⁄--o-ω----2---)--)---(--…-------)--(--1-----+-----j--(---ω-----⁄---ω----n---)---)-Vin(the example shows a voltage gain ... it could be Iout/Vin or Vout/Iin ) EE 105 Spring 1997 Lecture 23

Finding the Dominant Poles Multiplying out the denominator: V-----o---u--t = ------------------------------------------A----o------------------------------------------ Vin 1 + b1 jω + b2( jω)2 + … + bn( jω)n The coefﬁcient b1 originates from the sum of jω/ωi factors -- n ∑b1 = ω--1---1- + ω--1---2- + … + -ω-1---n- = i ω--1---i ≈ ω--1---1- Therefore, if we can estimate the linear coefﬁcient b1 in the demoninator polynomial, we can estimate of the dominant poleProcedure: see P. R. Gray and R. G. Meyer, Analysis and Design of Analog Integrated Circuits, 3rd ed., Wiley, 1994, pp. 502-504. 1. Find circuit equations with current sources driving each capacitor 2. Denominator polynomial is determinant of the matrix of coefﬁcients 3. b1 term comes from a sum of terms, each of which has the form: RTj Cj where Cj is the jth capacitor and RTj is the Thévenin resistance across the jth capacitor terminals (with all capacitors open-circuited) EE 105 Spring 1997 Lecture 23

Open-Circuit Time Constantss The dominant pole of the system can be estimated by: --1--- n –1 n –1 b1 j RTjC j 1 τ j ,≈∑ ∑ω1 = = where τj = RTj Cj is the open-circuit time constant for capacitor Cjs This technique is valuable because it estimates the contribution of each capacitor to the dominant pole frequency separately ... which enables the designer to understand what part of a complicated circuit is responsible for limiting the bandwidth of the ampliﬁer. EE 105 Spring 1997 Lecture 23

Example: Revisit CE Ampliﬁers Small-signal model: RS CµVs + rπ + gmVπ ro roc + − Vπ Cπ RL Vout − −s Apply procedure to each capacitor separately 1. Cπ’s Thévenin resistance is found by inspection as the resistance across its terminals with all capacitors open-circuited: RTπ = RS rπ = Rin′ --> τCπo = RTπCπ 2. Cµ’s Thévenin resistance is not obvious --> must use test source and network analysis EE 105 Spring 1997 Lecture 23

Time Constant for Cµs Circuit for ﬁnding RTµ − vt + + it vπR'in = RS rπ − gmvπ R'out = ro roc RLvπ is given by: vπ = –it(Rs rπ) = –itRin′vo is given by:vo = –ioRout′ = (it – gmvπ)Rout′ = it(gmRin′ + 1)Rout′vt is given by: vt = vo – vπ = it((1 + gmRin′)Rout′ + Rin′)solving for RTµ = vt / it RTµ = Rin′ + Rout′ + gmRin′Rout′τCµo = RTµCµ = (Rin′ + Rout′ + gmRin′Rout′)Cµ EE 105 Spring 1997 Lecture 23

Estimate of Dominant Pole for CE Ampliﬁers Estimate dominant pole as inverse of sum of open-circuit time constants ω1–1 = ( RTπCπ + RTµCµ) = Rin′Cπ + ( Rin′ + Rout′ + gmRin′Rout′)Cµ inspection --> identical to “exact” analysis (which also assumed ω1 « ω2 )s Advantage of open-circuit time constants: general technique Example: include Ccs and estimate its effect on ω1 EE 105 Spring 1997 Lecture 23

Multistage Ampliﬁer Frequency Responses Applying the open-circuit time constant technique to ﬁnd the dominant pole frequency -- use CS/CB cascode as an example V+ iSUP iOUT RS Q2 + M1 RL vOUT −Vs + −VBIAS + − V−s Systematic approach: 1. two-port small-signal models for each stage (not the device models!) 2. carefully add capacitances across the appropriate nodes of two-port models, which may not correspond to the familiar device conﬁguation for some models EE 105 Spring 1997 Lecture 23

Two-Port Model for Cascode s The base-collector capacitor Cµ2 is located between the output of the CB stage (the collector of Q2) and small-signal ground (the base of Q2) RS Cgd1 I2 Iout Cgs1 + +Vs + Vgs1 gm1Vgs1 ro1 1/gm2 Cπ2 −I2 βo2ro2 Cµ2 RL Vout − − − We have omitted Cdb1, which would be in parallel with Cπ2 at the output of the CS stage, and Ccs2 which would be in parallel with Cµ2. In addition, the current supply transistor will contribute additional capacitance to the output node. s Time constants τCgs1o = RSCgs1 τCgd1o = (Rin′ + Rout′ + gm1Rin′Rout′)Cgd1 where Rin′ = RS and Rout′ = ro1 g----m1----2- ≈ ----1----- gm2 Since the output resistance is only 1/gm2, the Thévenin resistance for Cgd1 is not magniﬁed (i.e., the Miller effect is minimal): τCgd1o = RS + ----1----- + g-g---mm----21- Cgd1 ≈ RS(1 + gm1 ⁄ gm2)Cgd1 gm2 RS EE 105 Spring 1997 Lecture 23

Cascode Frequency Response (cont.)s The base-emitter capacitor of Q2 has a time constant of τCπ2o = g----m1----2- Cπ2 s The base-collector capacitor of Q2 has a time constant of τCµ2o = (βo2ro2 roc RL)Cµ2 ≈ RLCµ2s Applying the theorem, the dominant pole of the cascode is approximately ω3–1db ≈ τCgs1o + τCgd1o + τCπ2o + τCµ2oω–31db ≈ RSCgs1 + RS(1 + gm1 ⁄ gm2)Cgd1 + g----m1----2- Cπ2 + RLCµ2 EE 105 Spring 1997 Lecture 23

Gain-Bandwidth Products A useful metric of an ampliﬁer’s frequency response is the product of the low- frequency gain |Avo| and the 3 dB frequency ω3dB For the cascode, the gain is |Avo| = |-gm1RL| and the gain-bandwidth product isAvo ω3dB ≈ -------------------------------------------------------------g----m----1---R----L--------------------------------------------------------------- g----m1----2- RSCgs1 + RS(1 + gm1 ⁄ gm2)Cgd1 + Cπ2 + RLCµ2s If the voltage source resistance is small, then Avo ω3dB ≈ (---C----π---2----⁄---g-g--m-m---2-1---+R----L-R----L---C-----µ---2---)-which has the same form as the common-base gain-bandwidth product (andwhich is much greater than the Miller-degraded common-source) EE 105 Spring 1997 Lecture 23

EE 105 Spring 1997 Lecture 23

# The Miller Approximation - prenhall.com

##
**Description: ** EE 105 Spring 1997 Lecture 23 The Miller Approximation The “exact” analysis is not particularly helpful for gaining insight into the frequency response ...

### Read the Text Version

No Text Content!

- 1 - 14

Pages: