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Chapter 11 Frequency Response 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9 Fundamental Concepts High-Frequency Models of Transistors Analysis Procedure Frequency Response of CE and CS Stages Frequency Response of CB and CG Stages Frequency Response of Followers Frequency Response of Cascode Stage Frequency Response of Differential Pairs Additional Examples 1 Chapter Outline CH 11 Frequency Response 2 High Frequency Roll-off of Amplifier As frequency of operation increases, the gain of amplifier decreases. This chapter analyzes this problem. CH 11 Frequency Response 3 Example: Human Voice I Natural Voice Telephone System Natural human voice spans a frequency range from 20Hz to 20KHz, however conventional telephone system passes frequencies from 400Hz to 3.5KHz. Therefore phone conversation differs from face-to-face conversation. CH 11 Frequency Response 4 Example: Human Voice II Path traveled by the human voice to the voice recorder Mouth Air Recorder Path traveled by the human voice to the human ear Mouth Air Ear Skull Since the paths are different, the results will also be different. CH 11 Frequency Response 5 Example: Video Signal High Bandwidth Low Bandwidth Video signals without sufficient bandwidth become fuzzy as they fail to abruptly change the contrast of pictures from complete white into complete black. CH 11 Frequency Response 6 Gain Roll-off: Simple Low-pass Filter In this simple example, as frequency increases the impedance of C1 decreases and the voltage divider consists of C1 and R1 attenuates Vin to a greater extent at the output. CH 11 Frequency Response 7 Gain Roll-off: Common Source Vout 1 g mVin RD || C s L The capacitive load, CL, is the culprit for gain roll-off since at high frequency, it will “steal” away some signal current and shunt it to ground. CH 11 Frequency Response 8 Frequency Response of the CS Stage Vout g m RD Vin RD2 C L2 2 1 At low frequency, the capacitor is effectively open and the gain is flat. As frequency increases, the capacitor tends to a short and the gain starts to decrease. A special frequency is ω=1/(RDCL), where the gain drops by 3dB. CH 11 Frequency Response 9 Example: Figure of Merit F .O.M . 1 VT VCC C L This metric quantifies a circuit’s gain, bandwidth, and power dissipation. In the bipolar case, low temperature, supply, and load capacitance mark a superior figure of merit. CH 11 Frequency Response 10 Example: Relationship between Frequency Response and Step Response H s j 1 R12C12 2 1 t Vout t V0 1 exp u t R1C1 The relationship is such that as R1C1 increases, the bandwidth drops and the step response becomes slower. CH 11 Frequency Response 11 Bode Plot s s 1 1 z1 z 2 H ( s) A0 s s 1 1 p1 p2 When we hit a zero, ωzj, the Bode magnitude rises with a slope of +20dB/dec. When we hit a pole, ωpj, the Bode magnitude falls with a slope of -20dB/dec CH 11 Frequency Response 12 Example: Bode Plot p1 1 RD C L The circuit only has one pole (no zero) at 1/(RDCL), so the slope drops from 0 to -20dB/dec as we pass ωp1. CH 11 Frequency Response 13 Pole Identification Example I p1 p2 1 RS Cin Vout Vin CH 11 Frequency Response 1 1 RD C L g m RD 2 p21 1 2 p2 2 14 Pole Identification Example II p1 1 1 RS || Cin gm CH 11 Frequency Response p2 1 RD C L 15 Circuit with Floating Capacitor The pole of a circuit is computed by finding the effective resistance and capacitance from a node to GROUND. The circuit above creates a problem since neither terminal of CF is grounded. CH 11 Frequency Response 16 Miller’s Theorem ZF Z1 1 Av ZF Z2 1 1 / Av If Av is the gain from node 1 to 2, then a floating impedance ZF can be converted to two grounded impedances Z1 and Z2. CH 11 Frequency Response 17 Miller Multiplication With Miller’s theorem, we can separate the floating capacitor. However, the input capacitor is larger than the original floating capacitor. We call this Miller multiplication. CH 11 Frequency Response 18 Example: Miller Theorem 1 in RS 1 g m RD C F CH 11 Frequency Response out 1 1 C F RD 1 g m RD 19 High-Pass Filter Response Vout Vin R1C1 R12C1212 1 The voltage division between a resistor and a capacitor can be configured such that the gain at low frequency is reduced. CH 11 Frequency Response 20 Example: Audio Amplifier Ci 79.6nF CL 39.8nF Ri 100K g m 1 / 200 In order to successfully pass audio band frequencies (20 Hz-20 KHz), large input and output capacitances are needed. CH 11 Frequency Response 21 Capacitive Coupling vs. Direct Coupling Capacitive Coupling Direct Coupling Capacitive coupling, also known as AC coupling, passes AC signals from Y to X while blocking DC contents. This technique allows independent bias conditions between stages. Direct coupling does not. CH 11 Frequency Response 22 Typical Frequency Response Lower Corner CH 11 Frequency Response Upper Corner 23 High-Frequency Bipolar Model C Cb C je At high frequency, capacitive effects come into play. Cb represents the base charge, whereas C and Cje are the junction capacitances. CH 11 Frequency Response 24 High-Frequency Model of Integrated Bipolar Transistor Since an integrated bipolar circuit is fabricated on top of a substrate, another junction capacitance exists between the collector and substrate, namely CCS. CH 11 Frequency Response 25 Example: Capacitance Identification CH 11 Frequency Response 26 MOS Intrinsic Capacitances For a MOS, there exist oxide capacitance from gate to channel, junction capacitances from source/drain to substrate, and overlap capacitance from gate to source/drain. CH 11 Frequency Response 27 Gate Oxide Capacitance Partition and Full Model The gate oxide capacitance is often partitioned between source and drain. In saturation, C2 ~ Cgate, and C1 ~ 0. They are in parallel with the overlap capacitance to form CGS and CGD. CH 11 Frequency Response 28 Example: Capacitance Identification CH 11 Frequency Response 29 Transit Frequency gm 2f T CGS gm 2f T C Transit frequency, fT, is defined as the frequency where the current gain from input to output drops to 1. CH 11 Frequency Response 30 Example: Transit Frequency Calculation 2fT 3 n VGS VTH 2 2L L 65nm VGS VTH 100m V n 400cm2 /(V .s ) fT 226GHz CH 11 Frequency Response 31 Analysis Summary The frequency response refers to the magnitude of the transfer function. Bode’s approximation simplifies the plotting of the frequency response if poles and zeros are known. In general, it is possible to associate a pole with each node in the signal path. Miller’s theorem helps to decompose floating capacitors into grounded elements. Bipolar and MOS devices exhibit various capacitances that limit the speed of circuits. CH 11 Frequency Response 32 High Frequency Circuit Analysis Procedure Determine which capacitor impact the low-frequency region of the response and calculate the low-frequency pole (neglect transistor capacitance). Calculate the midband gain by replacing the capacitors with short circuits (neglect transistor capacitance). Include transistor capacitances. Merge capacitors connected to AC grounds and omit those that play no role in the circuit. Determine the high-frequency poles and zeros. Plot the frequency response using Bode’s rules or exact analysis. CH 11 Frequency Response 33 Frequency Response of CS Stage with Bypassed Degeneration Vout g m RD RS Cb s 1 s VX RS Cb s g m RS 1 In order to increase the midband gain, a capacitor Cb is placed in parallel with Rs. The pole frequency must be well below the lowest signal frequency to avoid the effect of degeneration. CH 11 Frequency Response 34 Unified Model for CE and CS Stages CH 11 Frequency Response 35 Unified Model Using Miller’s Theorem CH 11 Frequency Response 36 Example: CE Stage RS 200 I C 1m A 100 C 100 fF C 20 fF CCS 30 fF p,in 2 516MHz p,out 2 1.59GHz The input pole is the bottleneck for speed. CH 11 Frequency Response 37 Example: Half Width CS Stage W 2X p ,in p ,out CH 11 Frequency Response 1 C g R C RS in 1 m L XY 2 2 2 1 Cout 2 C XY RL 1 2 g R 2 m L 38 Direct Analysis of CE and CS Stages gm | z | C XY 1 | p1 | 1 g m RL C XY RThev RThevCin RL C XY Cout 1 g m RL C XY RThev RThevCin RL C XY Cout | p 2 | RThev RL Cin C XY CoutC XY Cin Cout Direct analysis yields different pole locations and an extra zero. CH 11 Frequency Response 39 Example: CE and CS Direct Analysis p1 1 1 g m1 rO1 || rO 2 C XY RS RS Cin rO1 || rO 2 (C XY Cout ) 1 g m1 rO1 || rO 2 C XY RS RS Cin rO1 || rO 2 (C XY Cout ) p2 RS rO1 || rO 2 Cin C XY CoutC XY Cin Cout CH 11 Frequency Response 40 Example: Comparison Between Different Methods RS 200 CGS 250 fF CGD 80 fF C DB 100 fF g m 150 1 0 RL 2 K Dominant Pole Miller’s Exact p,in 2 571MHz p,in 2 264MHz p,in 2 249MHz p,out 2 428MHz p,out 2 4.53GHz p,out 2 4.79GHz CH 11 Frequency Response 41 Input Impedance of CE and CS Stages 1 1 Z in || r Z in CGS 1 g m RD CGD s C 1 g m RC C s CH 11 Frequency Response 42 Low Frequency Response of CB and CG Stages Vout g m RC Ci s s 1 g m RS Ci s g m Vin As with CE and CS stages, the use of capacitive coupling leads to low-frequency roll-off in CB and CG stages (although a CB stage is shown above, a CG stage is similar). CH 11 Frequency Response 43 Frequency Response of CB Stage p, X 1 1 RS || C X gm C X C p ,Y rO CH 11 Frequency Response 1 RL CY CY C CCS 44 Frequency Response of CG Stage 1 p , Xr O 1 RS || C X gm C X CGS CSB p,Y rO 1 RL CY CY CGD C DB Similar to a CB stage, the input pole is on the order of fT, so rarely a speed bottleneck. CH 11 Frequency Response 45 Example: CG Stage Pole Identification p, X 1 1 RS || C SB1 CGD1 g m1 CH 11 Frequency Response p ,Y 1 1 C DB1 CGD1 CGS 2 C DB2 g m2 46 Example: Frequency Response of CG Stage RS 200 CGS 250 fF CGD 80 fF C DB 100 fF g m 150 p , X 2 5.31GHz 0 p ,Y 2 442MHz 1 Rd 2 K CH 11 Frequency Response 47 Emitter and Source Followers The following will discuss the frequency response of emitter and source followers using direct analysis. Emitter follower is treated first and source follower is derived easily by allowing r to go to infinity. CH 11 Frequency Response 48 Direct Analysis of Emitter Follower Vout Vin C 1 s gm 2 as bs 1 CH 11 Frequency Response RS C C C C L C C L a gm C R S b RS C 1 gm r CL gm 49 Direct Analysis of Source Follower Stage Vout Vin CGS 1 s gm 2 as bs 1 CH 11 Frequency Response RS CGDCGS CGDC SB CGS C SB a gm CGD C SB b RS CGD gm 50 Example: Frequency Response of Source Follower RS 200 C L 100 fF CGS 250 fF CGD 80 fF C DB 100 fF g m 150 1 0 CH 11 Frequency Response p1 2 1.79GHz j 2.57GHz p 2 2 1.79GHz j 2.57GHz 51 Example: Source Follower Vout Vin CGS 1 s gm 2 as bs 1 RS CGD1CGS1 (CGD1 CGS1 )(C SB1 CGD2 C DB2 ) a g m1 b RS CGD1 CGD1 C SB1 C GD 2 C DB 2 g m1 CH 11 Frequency Response 52 Input Capacitance of Emitter/Source Follower rO C / CGS Cin C / CGD 1 g m RL CH 11 Frequency Response 53 Example: Source Follower Input Capacitance 1 Cin CGD1 CGS1 1 g m1 rO1 || rO 2 CH 11 Frequency Response 54 Output Impedance of Emitter Follower VX RS r C s r RS IX r C s 1 CH 11 Frequency Response 55 Output Impedance of Source Follower V X RS CGS s 1 I X CGS s g m CH 11 Frequency Response 56 Active Inductor The plot above shows the output impedance of emitter and source followers. Since a follower’s primary duty is to lower the driving impedance (RS>1/gm), the “active inductor” characteristic on the right is usually observed. CH 11 Frequency Response 57 Example: Output Impedance rO VX rO1 || rO 2 CGS 3 s 1 IX CGS 3 s g m3 CH 11 Frequency Response 58 Frequency Response of Cascode Stage Av, XY g m1 1 g m2 C x 2C XY For cascode stages, there are three poles and Miller multiplication is smaller than in the CE/CS stage. CH 11 Frequency Response 59 Poles of Bipolar Cascode p, X 1 RS || r 1 C 1 2C1 p ,out CH 11 Frequency Response p ,Y 1 1 CCS1 C 2 2C1 g m2 1 RL CCS 2 C 2 60 Poles of MOS Cascode p, X 1 g m1 CGD1 RS CGS1 1 g m2 p ,Y CH 11 Frequency Response p ,out 1 RL C DB 2 CGD2 1 1 g m2 g m2 CGD1 C DB1 CGS 2 1 g m1 61 Example: Frequency Response of Cascode RS 200 CGS 250 fF CGD 80 fF C DB 100 fF g m 150 1 p , X 2 1.95GHz 0 p ,Y 2 1.73GHz RL 2 K p ,out 2 442MHz CH 11 Frequency Response 62 MOS Cascode Example p, X 1 g m1 CGD1 RS CGS1 1 g m2 p ,Y 1 C DB1 CGS 2 CH 11 Frequency Response g m2 p ,out 1 RL C DB 2 CGD2 1 g m2 CGD1 CGD3 C DB3 1 g m1 63 I/O Impedance of Bipolar Cascode 1 Z in r 1 || C 1 2C1 s CH 11 Frequency Response Z out 1 RL || C 2 CCS 2 s 64 I/O Impedance of MOS Cascode 1 Z in g m1 CGD1 s CGS1 1 g m2 CH 11 Frequency Response Z out 1 RL || CGD2 C DB2 s 65 Bipolar Differential Pair Frequency Response Half Circuit Since bipolar differential pair can be analyzed using halfcircuit, its transfer function, I/O impedances, locations of poles/zeros are the same as that of the half circuit’s. CH 11 Frequency Response 66 MOS Differential Pair Frequency Response Half Circuit Since MOS differential pair can be analyzed using halfcircuit, its transfer function, I/O impedances, locations of poles/zeros are the same as that of the half circuit’s. CH 11 Frequency Response 67 Example: MOS Differential Pair p, X p ,Y p ,out CH 11 Frequency Response 1 RS [CGS1 (1 g m1 / g m 3 )CGD1 ] 1 g m3 CGD1 C DB1 CGS 3 1 g m1 1 RL C DB3 CGD3 1 g m3 68 Common Mode Frequency Response Vout g R R C 1 m D SS SS VCM RSS CSS s 2 g m RSS 1 Css will lower the total impedance between point P to ground at high frequency, leading to higher CM gain which degrades the CM rejection ratio. CH 11 Frequency Response 69 Tail Node Capacitance Contribution Source-Body Capacitance of M1, M2 and M3 Gate-Drain Capacitance of M3 CH 11 Frequency Response 70 Example: Capacitive Coupling Rin2 RB 2 || r 2 1RE L1 1 2 542Hz r 1 || RB1 C1 CH 11 Frequency Response L 2 1 22.9 Hz RC Rin2 C2 71 Example: IC Amplifier – Low Frequency Design Rin2 CH 11 Frequency Response RF 1 Av 2 L1 g m1RS1 1 2 42.4MHz RS1C1 L 2 1 2 6.92MHz RD1 Rin2 C2 72 Example: IC Amplifier – Midband Design vX g m1 RD1 || Rin2 3.77 vin CH 11 Frequency Response 73 Example: IC Amplifier – High Frequency Design p1 2 (308 MHz) p 2 2 (2.15 GHz) p3 1 RL 2 (1.15CGD 2 C DB 2 ) 2 (1.21 GHz) CH 11 Frequency Response 74