Multistage Amplifier Frequency Response

Transcription

1 Multistage Amplifier Frequency Response * Summary of frequency response of single-stages: CE/CS: suffers from Miller effect CC/CD: wideband -- see Section 0.5 CB/CG: wideband -- see Section 0.6 (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 << ω 2, ω 3,..., ω n ) V out V in = A o ( + j( ω ω ))( + j( ω ω 2 ))( )( + j( ω ω n )) (the example shows a voltage gain... it could be I out /V in or V out /I in ) EE 05 Spring 2000 Page Week 5, Lecture 35

2 Finding the Dominant Pole * Multiplying out the denominator: V out V in = A o + b jω + b 2 ( jω) b n ( jω) n The coefficient b originates from the sum of jω/ω i factors -- b = = ω ω 2 ω n n i ω i ω Therefore, if we can estimate the linear coefficient b 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, 994, pp Find circuit equations with current sources driving each capacitor 2. Denominator polynomial is determinant of the matrix of coefficients 3. b 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) EE 05 Spring 2000 Page 2 Week 5, Lecture 35

3 Open-Circuit Time Constants * The dominant pole of the system can be estimated by: ω b n n = R Tj C j = τj, j 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. EE 05 Spring 2000 Page 3 Week 5, Lecture 35

4 Example: Revisit CE Amplifier * Small-signal model: * Apply procedure to each capacitor separately. 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 EE 05 Spring 2000 Page 4 Week 5, Lecture 35

5 Time Constant for C µ * Circuit for finding R Tµ 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 + )R out v t is given by: solving for R Tµ = v t / i t v t = v o v π = i t (( + 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 EE 05 Spring 2000 Page 5 Week 5, Lecture 35

6 Estimate of Dominant Pole for CE Amplifier * Estimate dominant pole as inverse of sum of open-circuit time constants ω = ( 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 «ω 2 ) * Advantage of open-circuit time constants: general technique Example: include C cs and estimate its effect on ω EE 05 Spring 2000 Page 6 Week 5, Lecture 35

7 Multistage 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:. 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 05 Spring 2000 Page 7 Week 5, Lecture 35

8 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 ) We have omitted C db, 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 Cgso = R S C gs τ Cgdo = ( R in + R out + g m R in R out )C gd where R in = R S and R out r o = g m2 g m2 Since the output resistance is only /g m2, the Thévenin resistance for C gd is not magnified (i.e., the Miller effect is minimal): g τ Cgdo R S m = RS Cgd R g m2 g S ( + g m g m2 )C gd m2 EE 05 Spring 2000 Page 8 Week 5, Lecture 35

9 Cascode Frequency Response (cont.) * The base-emitter capacitor of Q 2 has a time constant of 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 ω 3db τ Cgso + τ Cgdo + τ Cπ2o + τ Cµ2o ω 3db R S C gs + R S ( + g m g m2 )C gd Cπ2 + R g L C µ2 m2 EE 05 Spring 2000 Page 9 Week 5, Lecture 35

10 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 m R L and the gain-bandwidth product is g m R L A vo ω 3dB R S C gs + R S ( + g m g m2 )C gd Cπ2 + R g L C µ2 m2 * If the voltage source resistance is small, then g m R L A vo ω 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) EE 05 Spring 2000 Page 0 Week 5, Lecture 35