6.776 High Speed Communication Circuits and Systems Lecture 8 Broadband Amplifiers, Continued
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6.776 High Speed Communication Circuits and Systems Lecture 8 Broadband Amplifiers, Continued Massachusetts Institute of Technology March 1, 2005 Copyright © 2005 by Hae-Seung Lee and Michael H. Perrott The Issue of Velocity Saturation We often assume that MOS current is a quadratic function of Vgs: It can be shown, more generally -V corresponds to the saturation voltage at a given length, which we often refer to as ∆V dsat,l - In strong inversion below velocity saturation: which gives the quardatic equation above. H.-S. Lee & M.H. Perrott MIT OCW Velocity Saturation Continued It can be shown that - E : electric field (lateral) at which velocity saturation occurs - If then - If (V -V )/L approaches E in value, then the quadratic sat gs T sat equation is no longer valid - If then and the I-V characteristic becomes linear H.-S. Lee & M.H. Perrott MIT OCW Analytical Device Modeling in Velocity Saturation If L small (as in modern devices), than velocity saturation will impact us for even moderate values of Vgs-VT - Current increases linearly with V gs-VT! Transconductance in velocity saturation: - No longer a function of V gs- higer Vgs increases Id, but little increase in gm: wasted power H.-S. Lee & M.H. Perrott MIT OCW Example: Current Versus Voltage for 0.18µ Device Id Vgs M1 I versus V d 1.8µ W = L 0.18µ gs 1.4 1.2 Id (milliAmps) 1 0.8 strong Inversion (quadratic) 0.6 0.4 weak inversion velocity saturation 0.2 0 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Vgs (Volts) H.-S. Lee & M.H. Perrott MIT OCW Example: Gm Versus Voltage for 0.18µ Device Id Vgs M1 1.8µ W = L 0.18µ g versus V m gs 1 0.9 0.8 gm (milliAmps/Volts) 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0.4 H.-S. Lee & M.H. Perrott 0.6 0.8 1 1.2 V gs (Volts) 1.4 1.6 1.8 2 MIT OCW Example: Gm Versus Current Density for 0.18µ Device Id Vgs 1.8µ W = L 0.18µ Transconductance versus Current Density Transconductance (milliAmps/Volts) M1 1 0.8 0.6 0.4 0.2 0 0 H.-S. Lee & M.H. Perrott 100 200 300 400 500 Current Density (microAmps/micron) 600 700 MIT OCW How Do We Design the Amplifier? Highly inaccurate to assume square law behavior We will now introduce a numerical procedure based on the simulated gm curve of a transistor - A look at transconductance: - Observe that if we keep the current density (I - den=Id/W) constant, then gm scales directly with W This is independent of bias regime We can therefore relate gmof devices with different widths given that they have the same current density H.-S. Lee & M.H. Perrott MIT OCW A Numerical Design Procedure for Resistor Amp – Step 1 Vdd R R Vo+ Vo- αIbias Vin+ Ibias M1 M2 2Ibias M5 Two key equations - Set gain and swing (singleended) Vin- Equate (1) and (2) through R M6 Can we relate this formula to a gm curve taken from a device of width Wo? H.-S. Lee & M.H. Perrott MIT OCW A Numerical Design Procedure for Resistor Amp – Step 2 We now know: Substitute (2) into (1) The above expression allows us to design the resistor loaded amp based on the gm curve of a representative transistor of width Wo! H.-S. Lee & M.H. Perrott MIT OCW Example: Design for Swing of 1 V, Gain of 1 and 2 Assume L=0.18µ, use previous gm plot (Wo=1.8µ) Transconductance versus Current Density A=2 Transconductance (milliAmps/Volts) 1 A=1 gm(wo=1.8µ,Iden) 0.8 0.6 0.4 0.2 For gain of 1, current density = 250 µA/µm For gain of 2, current density = 115 µA/µm Note that current density reduced as gain increases! - f effectively t 0 0 100 200 300 400 500 Current Density - Iden (microAmps/micron) H.-S. Lee & M.H. Perrott 600 700 decreased MIT OCW Example (Continued) Knowledge of the current density allows us to design the amplifier - Recall - Free parameters are W, I bias, and R (L assumed to be fixed) Given Iden = 115 µA/µm (Swing = 1V, Gain = 2) - If we choose I bias = 300 µA Note that we could instead choose W or R, and then calculate the other parameters H.-S. Lee & M.H. Perrott MIT OCW How Do We Choose Ibias For High Bandwidth? Ctot = Cout+Cin+Cfixed Cin Cout Cin Amp Amp Cfixed Pick current density just below velocity saturation As you increase Ibias, the size of transistors also increases to keep a constant current density The size of Cin and Cout increases relative to Cfixed To achieve the highest bandwidth, size the devices (i.e., choose the value for Ibias), such that Cin+Cout dominates over Cfixed However, Cin+Cout=Cfixed is roughly the point of diminishing return because the bandwidth improvement becomes marginal while power and area continue to grow proportionally Thus, Cin+Cout=Cfixed is the most efficient point - H.-S. Lee & M.H. Perrott MIT OCW Resistor Loaded Amplifier (Unsilicided Poly) vout vin Vdd RL Id vin Vbias M1 vout Cfixed M2 gm1RL 1 Ctot = Cdb1+CRL/2 + Cgs2+KCov2 + Cfixed (+Cov1) slope = -20 dB/dec f gm1 2πCtot 1 2πRLCtot Miller multiplication factor We decided this was the fastest non-enhanced amplifier We will look at the following - Can we go faster? (i.e., can we enhance its bandwidth?) - Reduction of Miller effect on C - Shunt, series, and zero peaking - Distributed amplification gd H.-S. Lee & M.H. Perrott MIT OCW Miller Effect on Cgd Is Significant RL Id Zin C gd vin CL Cgs Vbias M1 Rs vout Cgd is quite significant compared to Cgs - In 0.18µ CMOS, C gd is about 45% the value of Cgs Input capacitance calculation - For 0.18µ CMOS, gain of 3, input cap is almost tripled over Cgs! H.-S. Lee & M.H. Perrott MIT OCW Reduction of Cgd Impact Using a Cascode Device RL Vbias2 M2 Zin C gd vin Vbias CL M1 Rs Cgs The cascode device lowers the gain seen by Cgd of M1(the total gain is the same as non-cascoded amp) vout For 0.18m CMOS and total gain of 3, impact of Cgd is reduced by 50%: Issue: cascoding lowers achievable voltage swing H.-S. Lee & M.H. Perrott MIT OCW Source-Coupled Amplifier (Unilateralization) RL vout Zin C gd vin Vbias M1 M2 2Ibias Remove impact of Miller effect by sending signal through source node rather than drain node -C Rs Cgd gd not Miller multiplied AND impact of Cgs cut in half! The bad news - Signal has to go through source node (C - Power consumption doubled sb H.-S. Lee & M.H. Perrott significant) MIT OCW Neutralization RL CN -1 Zin CL Consider canceling the effect of Cgd - Choose C = C - Charging of C N M1 vout Cgs Vbias Cgd Rs vin Id gd gd now provided by CN Benefit: Impact of Cgd reduced: :same as cascode H.-S. Lee & M.H. Perrott MIT OCW Neutralization, cont’d Issues: - What happens if C N is not precisely matched to Cgd? - Since the neutralization does not completely remove the - effect of Cgd, we can make CN slightly larger than Cgd to ‘over neutralize’ Over neutralization can reduce the effect of Cgs, but if CN is too large, the input capacitance is negative and can compromise stability. At high frequencies, this can lead to inductive input impedance How do we create the inverting amplifier? H.-S. Lee & M.H. Perrott MIT OCW Practical Implementation of Neutralization CN RL RL Cgd vin Vbias Rs CN Cgd Rs M2 M1 -vin 2Ibias Leverage differential signaling to create an inverted signal Only issue left is matching CN to Cgd - Often use lateral metal caps for C (or CMOS transistor) - If C too low, residual influence of C - If C too high, input impedance has inductive component N N gd N Causes peaking in frequency response Often evaluate acceptable level of peaking using eye diagrams H.-S. Lee & M.H. Perrott MIT OCW Shunt-peaked Amplifier Vdd Ld RL Id vin Vbias M1 vout Cfixed M2 Use inductor in load to extend bandwidth We can view impact of inductor in both time and frequency - Often implemented as a spiral inductor - In frequency: peaking of frequency response - In time: delay of changing current in R L Issue – can we extend bandwidth without significant peaking? H.-S. Lee & M.H. Perrott MIT OCW Shunt-peaked Amplifier - Analysis Vdd Small Signal Model Ld Zout vout RL Id vin Vbias M1 Ld vout Cfixed Expression for gain Parameterize with iin=gmvin M2 Ctot RL - Corresponds to ratio of RC to LR time constants H.-S. Lee & M.H. Perrott MIT OCW The Impact of Choosing Different Values of m – Part 1 Small Signal Model Parameterized gain expression Zout vout Comparison of new and old 3 dB frequencies Ld iin=gmvin Ctot RL Want to solve for ω2/ω1 H.-S. Lee & M.H. Perrott MIT OCW The Impact of Choosing Different Values of m – Part 2 From previous slide, we have After much algebra We see that m directly sets the amount of bandwidth extension! - Once m is chosen, inductor value is H.-S. Lee & M.H. Perrott MIT OCW Plot of Bandwidth Extension Versus m Bandwidth Extension (w /w ) Versus m 2 1 1.9 1.8 1.7 w1/w2 1.6 1.5 1.4 1.3 1.2 1.1 0 1 2 3 4 5 6 7 8 9 10 m Highest extension: ω2/ω1 = 1.85 at m ≈ 1.41 However, peaking occurs! H.-S. Lee & M.H. Perrott MIT OCW Plot of Transfer Function Versus m Maximum bandwidth: m = 1.41 (extension = 1.85) Maximally flat response: m = 2.41 (extension = 1.72) Best phase response: m = 3.1 (extension = 1.6) No inductor: m = infinity Eye diagrams often used to evaluate best m H.-S. Lee & M.H. Perrott Normalized Gain (dB) Normalized Gain for Shunt Peaked Amplifier For Different m Values 5 m=1.41 m=2.41 0 m=3.1 m=infinity -5 -10 -15 -20 -25 -30 -35 -40 -45 1/100 1/10 1 10 100 Normalized Frequency (Hz) MIT OCW Zero-peaked Common Source Amplifier vout vin Vdd RL Id M1 vout Cfixed M2 g mR L 1+gmRs zero 1 2πCsRs overall response vin Vbias Rs f Cs 1 2πCtotRL Inductors are expensive with respect to die area Can we instead achieve bandwidth extension with capacitor? - Idea: degenerate gain at low frequencies, remove degeneration at higher frequencies (i.e., create a zero) Issues: - Must increase R to keep same gain (lowers pole) Lowers achievable gate voltage bias (lowers device f ) H.-S. Lee & M.H. Perrott MIT OCW L t Zero-peaked Common Source Amplifier Analysis Vo gmvgs Vi Cgs ZL ZS Add Cgd to Ctot (as we did previously) Ignore the feed-forward effect of Cgd (It contributes high frequency zero of little consequence) Analysis shows H.-S. Lee & M.H. Perrott MIT OCW Zero-peaked Amplifier Analysis Continued Assuming ω<<ωT Transfer function can now be simplified to Adds a zero at 1/RsCs, but introduces a 2nd pole at Reduces low freq. gain to The 1st pole at 1/RLCtot can be cancelled by making RsCs=RLCtot, then the bandwidth is extended to the 2nd pole H.-S. Lee & M.H. Perrott MIT OCW Zero-peaked Amplifier Continued Pole-zero cancellation: Does it really help the bandwidth? If we designed the simple CS amplifier for the same gain, what would be the bandwidth? We need to first reduce RL to The bandwidth is then Same as zero-peaked amplifier! H.-S. Lee & M.H. Perrott MIT OCW Zero-peaked Amplifier Input impedance Input Impedance (ignoring Miller effect for now) Again, ω<<ωT Also, near the upper 3dB bandwidth, Negative resistance! The negative input resistance component can cause parasitic oscillation. The actual input impedance Zin,tot is the parallel connection between Zin and KCgd. H.-S. Lee & M.H. Perrott MIT OCW Back to Inductors – Shunt and Series Peaking Vdd vout vin RL -20 dB/dec L1 Id L2 vin Vbias M1 -40 dB/dec g mR L vout Cfixed f M2 1 2πCtotRL Combine shunt peaking with a series inductor - Bandwidth extension by converting to a second order filter response Can be designed for proper peaking Increases delay of amplifier Refer to Tom Lee’s book pp. 279-280 (2nd ed.) or 187-189 (1st ed.) H.-S. Lee & M.H. Perrott MIT OCW
