The Miller Approximation

The Miller Approximation The exact analysis is not particularly helpful for gaining insight into the frequency response... consider the effect of C µ on the input only I t C µ V t g m V t R'out = r o r oc R L V out neglect the feedforward current I µ in comparison with g m V π... a good approximation I t = (V t - V o ) / Z µ V o = - g m V t R L / (R L R out ) = A vcµ V t where A vcµ is the low frequency voltage gain across C µ I t = V t (1- A v ) / Z µ Z eff = V t / I t = Z µ / (1 - A v ) 1 Z eff ------------- 1 -------------------- 1 = = jωc µ 1 A --------------------------------------------- = vcµ jω( C µ ( 1 A vcµ )) 1 --------------- j ω C M C M = ( 1 A vcµ )C µ is the Miller capacitor

Generalized Miller Approximation An impedance Z connected across an amplifier with voltage gain A vz can be replaced by an impedance to ground... multiplied by (1-A vz ) Z V i A vo V o V i Z eff A vo V o Z eff = Z/(1 A vo ) Common-emitter and common-source: A vz = large and negative for C µ or C gd --> capacitance at the input is magnified Common-collector and common-drain: A vz 1 --> capacitance at the input due to C π or C gs is greatly reduced

Voltage Gain vs. Frequency for CE Amplifier Using the Miller Approximation The Miller capacitance is lumped together with C π, which results in a singlepole low-pass RC filter at the input R S V s r π V π C π C M g m V π R' out V out C M = C µ (1 g m R' out ) Transfer function has one pole and no zero after Miller approximation: 1 ω 3dB = ( r π R S )( C π C M ) 1 ω 3dB = ( r π R S )[ C π ( 1 g m r o r oc R L )C µ ] 1 1 ω 3dB ω 1 from the exact analysis (final term R out C µ is missing)

Multistage Amplifier Frequency Response 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... f T ) 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 (ω 1 << ω 2, ω 3,..., ω n ) V out ---------- V in = A o ------------------------------------------------------------------------------------------------------------------------- ( 1 j ( ω ω 1 ))( 1 j( ω ω 2 ))( )( 1 j( ω ω n )) (the example shows a voltage gain... it could be I out /V in or V out /I in )

Finding the Dominant Pole Multiplying out the denominator: V out ---------- V in = A o ---------------------------------------------------------------------------------------- 1 b 1 j ω b 2 ( j ω ) 2 b n ( jω) n The coefficient b 1 originates from the sum of jω/ω i factors -- 1 1 b 1 = ------ ω ------ 1 1 ω 2 ω ------ = n n i 1 1 ---- ------ ω i ω 1 Therefore, if we can estimate the linear coefficient b 1 in the demoninator polynomial, we can estimate of the dominant pole Procedure: see P. R. Gray and R. G. Meyer, Analysis and Design of Analog Integrated Circuits, 3 rd 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. b 1 term comes from a sum of terms, each of which has the form: R Tj C j where C j is the j th capacitor and R Tj is the Thévenin resistance across the j th capacitor terminals (with all capacitors open-circuited)

Open-Circuit Time Constants The dominant pole of the system can be estimated by: 1 ω 1 ----- b 1 n 1 n R Tj C j 1 = = τ j, j 1 where τ j = R Tj C j is the open-circuit time constant for capacitor C j 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.

Example: Revisit CE Amplifier Small-signal model: R S C µ V s r π V π C π g m V π r o r oc R L V out Apply procedure to each capacitor separately 1. C π s Thévenin resistance is found by inspection as the resistance across its terminals with all capacitors open-circuited: R Tπ = R S r π = R in --> τ Cπo = R Tπ C π 2. C µ s Thévenin resistance is not obvious --> must use test source and network analysis

Time Constant for C µ Circuit for finding R Tµ v t i t R' in = R S r π v π g m v π R' out = r o r oc R L v π is given by: v π = i t ( R s r π ) = i t R in v o is given by: v o = i o R out = ( i t g m v π )R out = i t ( g m R in 1)R out v t is given by: solving for R Tµ = v t / i t v t = v o v π = i t (( 1 g m R in )R out R in ) R Tµ = R in R out g m R in R out τ Cµo = R Tµ C µ = ( R in R out g m R in R out )C µ

Estimate of Dominant Pole for CE Amplifier Estimate dominant pole as inverse of sum of open-circuit time constants 1 ω 1 = ( R Tπ C π R Tµ C µ ) = R in C π ( R in R out g m R in R out )C µ inspection --> identical to exact analysis (which also assumed ω 1 «ω 2 ) Advantage of open-circuit time constants: general technique Example: include C cs and estimate its effect on ω 1

Multistage Amplifier Frequency Response Applying the open-circuit time constant technique to find the dominant pole frequency -- use CS/CB cascode as an example V i SUP i OUT Q 2 R S M 1 R L v OUT V s V BIAS V 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

Two-Port Model for Cascode The base-collector capacitor C µ2 is located between the output of the CB stage (the collector of Q 2 ) and small-signal ground (the base of Q 2 ) R S C gd1 I 2 I out V s C gs1 V gs1 g m1 V gs1 r o1 1/g m2 C π2 I 2 β o2 r o2 C µ2 R L V out We have omitted C db1, which would be in parallel with C π2 at the output of the CS stage, and C cs2 which would be in parallel with C µ2. In addition, the current supply transistor will contribute additional capacitance to the output node. Time constants τ Cgs1o = R S C gs1 τ Cgd1o = ( R in R out g m1 R in R out )C gd1 where R in = R S and R out r 1 o1 --------- 1 = --------- g m2 g m2 Since the output resistance is only 1/g m2, the Thévenin resistance for C gd1 is not magnified (i.e., the Miller effect is minimal): 1 g m 1 τ C R gd1o S --------- --------- = RS Cgd1 R g m2 g S ( 1 g m1 g m2 )C gd1 m 2

Cascode Frequency Response (cont.) The base-emitter capacitor of Q 2 has a time constant of 1 τ Cπ2o = --------- Cπ2 g m2 The base-collector capacitor of Q 2 has a time constant of τ Cµ2o = ( β o2 r o2 r oc R )C L µ2 R L C µ2 Applying the theorem, the dominant pole of the cascode is approximately 1 ω 3db τ Cgs1o τ Cgd1o τ Cπ2o τ Cµ2o 1 ω 3db 1 R S C gs1 R S ( 1 g m1 g m2 )C gd1 --------- Cπ2 R g L C µ2 m2

Gain-Bandwidth Product A useful metric of an amplifier s frequency response is the product of the lowfrequency gain A vo and the 3 db frequency ω 3dB For the cascode, the gain is A vo = -g m1 R L and the gain-bandwidth product is g m1 R L A vo ω 3dB ----------------------------------------------------------------------------------------------------------------------------------------- 1 R S C gs1 R S ( 1 g m1 g m2 )C gd1 --------- Cπ2 R g L C µ2 m2 If the voltage source resistance is small, then A vo ω g m1 R L 3dB ( --------------------------------------------------- C π2 g m2 R L C ) µ2 which has the same form as the common-base gain-bandwidth product (and which is much greater than the Miller-degraded common-source)