CMOS Cascode Transconductance Amplifier Basic topology. Current Supply Topology p-channel cascode current supply is an obvious solution Current supply must have a very high source resistance r oc since otherwise it will limit the output resistance of the amplifier need to design a totem pole voltage supply to generate V G2, V G3, and V G4 EE 05 Fall 2000 Page Week 4 EE 05 Fall 2000 Page 2 Week 4
Totem Pole Voltage Reference Match device sizes of M 2B, M 3B, and M 4B to M 2, M 3, and M 4 Complete Transconductance Amplifier V BIAS : user must supply a very precise DC voltage V BIAS.2 V so that the CS/CG cascode is biased so that it is in the high gain region CS-CG two-port parameters: G m = g m Output resistance: R out = r o2 ( + g m2 r o ) r o3 ( + g m3 r o4 ) Output swing: V OUT(max) = V D4 - V SD3(sat) = V DD - V SG4B - V SG3B + V SG3 - V SD3(sat) V OUT(max) = 5 V -.5 V -.5 V +.5 V - 0.5 V = 3.0 V V OUT(min) = V D + V DS2(sat) = V G2 - V GS2 + V DS2(sat) = 2 V -.4 V + 0.4 V = V EE 05 Fall 2000 Page 3 Week 4 EE 05 Fall 2000 Page 4 Week 4
Multistage Voltage Amplifier Example to understand a complicated circuit Eliminating Current and Voltage Sources Replace current and voltage sources with symbols V DD = 5 V -I D5 -I D6 Q 4 M 3 V B2 Q 2 v OUT + v s _ V + BIAS _ R S M I D0 0 MOSFETs, 3 BJTs, resistor... must identify building blocks Step-by-step approach to identifying the important transistors --. replace all transistor current sources and voltage sources by their symbols -- look for diodes and current mirrors! (M 5, M 6 /M 6B, M 7 /M 7B, and M 0 and Q 2B are part of current sources or a totem pole voltage reference.) 2. for the (few) remaining transistors, identify the type and use two-port smallsignal models to understand the circuit s operation. (For the above amplifier, the remaining transistors are M, Q 2, M 3, and Q 4.) EE 05 Fall 2000 Page 5 Week 4 EE 05 Fall 2000 Page 6 Week 4
Identifying Amplifier Stages n-channel MOSFET M has its source grounded --> common source npn BJT Q 2 has its gate tied to a voltage source (from totem pole string of diode-connected transistors) --> common base p-channel MOSFET M 3 has its drain connected to ground --> common drain npn BJT Q 4 has its collector tied to the positive supply --> common collector Voltage amplifier is a cascade of two-port models: Cascode Stage Output Resistance Cascode input stage output resistance determines gain V DD = 5 V -I D6 R out,cb/cs V B2 Q 2 CS CB CD CC R S R S2 M R in R out(cs/cb) R out V + BIAS _. common source/common base with cascoded current-source supply: very high output resistance R out (CS/CB) --> can get extremely high output resistance, with a transconductance equal to that of the CS stage 2. common drain: no loading on previous stage since infinite input resistance 3. common collector: low output resistance Output resistance: note that R S2 = r o >> r π2 R out,cb = ( β o2 r o2 ) r oc6 = ( β o r o2 ) ( r o6 ( + g m6 r o7 )) EE 05 Fall 2000 Page 7 Week 4 EE 05 Fall 2000 Page 8 Week 4
CS-CB-CD-CC Two-Port Parameters Since CC and CD stages have unity gain (approximately), we can quickly estimate the voltage gain by finding v in3 /v in where v in3 is the input to the CD stage Voltage gain: DC Bias and Output Swing Assuming all n-channel devices have V GS =.5 V and p-channel devices have V SG =.5 V, we can find all the node voltages... we also assume that V BIAS has been adjusted such that the circuit is in the high-gain region A v ( g m )R out,cb = g m (( β o r o2 ) ( r o6 ( + g m6 r o7 ))) Output resistance: source resistance of CC output stage is relatively small, since it preceded by a CD stage. R R out R out,cc --------- S,CC = + ----------------- = --------- + ------------------ g m4 β o4 g m4 g m3 β o4 Output swing: must consider the limited swing of previous stages (back to cascode) since the the CD/CC output stages are DC level shifters EE 05 Fall 2000 Page 9 Week 4 EE 05 Fall 2000 Page 0 Week 4
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 ) Finding the Dominant Pole Multiplying out the denominator: V out A ---------- = -------------------------------------------------------------------------------------- o V in + b jω + b 2 ( jω) 2 + + b n ( jω) n The coefficient b originates from the sum of jω/ω i factors -- 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 A ---------- = ------------------------------------------------------------------------------------------------------------------------ o V in ( + 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 ) b = ------ + ------ + + ------ = ω ω 2 ω n n ----- ------ ω i 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. 502-504.. 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 Fall 2000 Page Week 4 EE 05 Fall 2000 Page 2 Week 4
Open-Circuit Time Constants The dominant pole of the system can be estimated by: Small-signal model: Example: Revisit CE Amplifier ω ----- b n n R Tj C j = = τ j, 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. 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 Cπ = R o Tπ C π 2. C µ s Thévenin resistance is not obvious --> must use test source and network analysis EE 05 Fall 2000 Page 3 Week 4 EE 05 Fall 2000 Page 4 Week 4
Time Constant for C µ Estimate of Dominant Pole for CE Amplifier Circuit for finding R Tµ 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 v π is given by: v o is given by: v π = i t ( R s r π ) = i t R in 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 Cµ = R o Tµ C µ = ( R in + R out + g m R in R out )C µ EE 05 Fall 2000 Page 5 Week 4 EE 05 Fall 2000 Page 6 Week 4
Multistage Amplifier Frequency Response Applying the open-circuit time constant technique to find the dominant pole frequency -- use CS/CB cascode as an example 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 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 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): τ Cgdo R g S --------- m = + + --------- RS Cgd R g m2 g S ( + g m g m2 )C gd m2 EE 05 Fall 2000 Page 7 Week 4 EE 05 Fall 2000 Page 8 Week 4
Cascode Frequency Response (cont.) The base-emitter capacitor of Q 2 has a time constant of τ Cπ2o = --------- Cπ2 g 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 m R L and the gain-bandwidth product is 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 ω 3db τ Cgso + τ Cgdo + τ Cπ + τ 2o Cµ2o R S C gs + R S ( + g m g m2 )C gd + --------- g Cπ2 + R L C µ2 m2 g A vo ω m R L 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 A vo ω m 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) EE 05 Fall 2000 Page 9 Week 4 EE 05 Fall 2000 Page 20 Week 4