EECS 105 Fall 1998Lecture 29Multistage Amplifier Frequency Response■Summary of frequency response of single-stages:CE/CS: suffers from Miller effectCC/CD: “wideband” -- see Section 10.5CB/CG: “wideband” -- see Section 10.6(wideband means that the stage operates to near the frequencylimit of the device ... fT)■How to find the Bode plot for a general multistage amplifier?can’t handle n poles and m zeroes analytically --> SPICEdevelop analytical tool for an important special case:* no zeroes* exactly one “dominant” pole (ω1 << ω2, ω3, ... , ωn)(the example shows a voltage gain ... it could be Iout/Vin or Vout/Iin )VoutVin-----------Ao1 j ωω1⁄()+()1 j ωω2⁄()+()…()1 j ωωn⁄()+()------------------------------------------------------------------------------------------------------------------------=29EECS 105 Fall 1998Lecture 29Finding the Dominant Pole■Multiplying out the denominator:The coefficient b1 originates from the sum of jω/ωi factors --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 capacitor2. Denominator polynomial is determinant of the matrix of coefficients3. b1 term comes from a sum of terms, each of which has the form:RTj Cjwhere Cj is the jth capacitor and RTj is the Thévenin resistance across the jth capacitor terminals (with all capacitors open-circuited)VoutVin-----------Ao1 b1jω b2jω()2… bnjω()n++ ++--------------------------------------------------------------------------------------=b11ω1------1ω2------…1ωn------+++1ωi-----in∑1ω1------≈==EECS 105 Fall 1998Lecture 29Open-Circuit Time Constants■The dominant pole of the system can be estimated by:,where τj = RTj Cj is the open-circuit time constant for capacitor Cj■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.ω11b1-----≈RTjCjjn∑1–τj1n∑1–==EECS 105 Fall 1998Lecture 29Example: Revisit CE Amplifier■Small-signal model: ■Apply procedure to each capacitor separately1. Cπ’s Thévenin resistance is found by inspection as the resistance across its terminals with all capacitors open-circuited: --> 2. Cµ’s Thévenin resistance is not obvious --> must use test source and network analysis+−VsVπRSCµCπrπgmVπ+−+−ro rocRLVoutRTπRSrπRin′==τCπoRTπCπ=EECS 105 Fall 1998Lecture 29Time Constant for Cµ■Circuit for finding RTµ vπ is given by:vo is given by:vt is given by:solving for RTµ = vt / it +−R'out = ro roc RL R'in = RS rπ itvπ−+vtgmvπvπitRsrπ()– itRin′–==voi–oRout′ itgmvπ–()Rout′ itgmRin′1+()Rout′== =vtvovπ– it1 gmRin′+()Rout′ Rin′+()==RTµRin′ Rout′ gmRin′Rout′++=τCµoRTµCµRin′ Rout′ gmRin′Rout′++()Cµ==EECS 105 Fall 1998Lecture 29Estimate of Dominant Pole for CE Amplifier■Estimate dominant pole as inverse of sum of open-circuit time constantsinspection --> identical to “exact” analysis (which also assumed )■Advantage of open-circuit time constants: general techniqueExample: include Ccs and estimate its effect on ω1 ω11–RTπCπRTµCµ+()Rin′CπRin′ Rout′ gmRin′Rout′++()Cµ+==ω1ω2«EECS 105 Fall 1998Lecture 29Multistage Amplifier Frequency Response■Applying the open-circuit time constant technique to find the dominant pole frequency -- use CS/CB cascode as an example■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 modelsV+V−RLiSUPiOUTRSM1Q2VsVBIAS+−+−+−vOUTEECS 105 Fall 1998Lecture 29Two-Port Model for Cascode ■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)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.■Time constantswhere and Since the output resistance is only 1/gm2, the Thévenin resistance for Cgd1 is not magnified (i.e., the Miller effect is minimal):gm1Vgs1Cπ2Cgd1Cgs11/gm2Cµ2βo2ro2−I2RLVsro1IoutI2+−+−VoutRS+−Vgs1τCgs1oRSCgs1=τCgd1oRin′Rout′gm1Rin′Rout′++()Cgd1=Rin′RS= Rout′ro11gm2---------1gm2---------≈=τCgd1oRS1gm2---------gm1gm2---------RS++Cgd1= RS1 gm1gm2⁄+()Cgd1≈EECS 105 Fall 1998Lecture 29Cascode Frequency Response (cont.)■The base-emitter capacitor of Q2 has a time constant of■The base-collector capacitor of Q2 has a time constant of■Applying the theorem, the dominant pole of the cascode is approximatelyτCπ2o1gm2---------Cπ2=τCµ2oβo2ro2rocRL()Cµ2RLCµ2≈=ω3db1–τCgs1oτCgd1oτCπ2oτCµ2o+++≈ω3db1–RSCgs1RS1 gm1gm2⁄+()Cgd11gm2---------Cπ2RLCµ2+++≈EECS 105 Fall 1998Lecture 29Gain-Bandwidth Product■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 is■If the voltage source resistance is small, thenwhich has the same form as the common-base gain-bandwidth product (and which is much greater than the Miller-degraded common-source)Avoω3dBgm1RLRSCgs1RS1 gm1gm2⁄+()Cgd11gm2---------Cπ2RLCµ2+++------------------------------------------------------------------------------------------------------------------------------------------≈Avoω3dBgm1RLCπ2gm2⁄
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