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SINGLE-STAGE AMPLIFIERS EMT451/4 AMPLIFIER : DEFINITIONS Amplification the process of increasing an ac signal’s power Amplifier The circuit that amplifies Two Amplifier properties that are of interest to us: Gain Impedance AMPLIFICATION? Essential function in integrated circuits Signal too small: to drive load Overcome noise of subsequent stage Provide the right level to the subsequent circuit (e.g. logic level) Analyze large-signal & small-signal characteristics of single-stage amplifiers (CS, CG, CD, cascode) Goal: develop intuitive technique and fundamental models Need to be able to simplify analysis yet maintain acceptable accuracy Approach: simple model -> add second order phenomena (e.g. body effect, channel-length modulation) TRADE-OFFS in ANALOG DESIGN AMPLIFIER GAIN Amplifier Ratio of output signal to input signal Ratio < 1: attenuator Ratio = 1: buffer Ratio > 1: amplifier 3 types of gains associated with an amplifier Voltage gain Current gain Power gain VOLTAGE GAIN Defined as the ratio of ac output voltage to ac input voltage Or, the mathematical expression: VOut AV V In CURRENT GAIN Defined as the ratio of ac output current to ac input current Mathematically, expressed as: I Out AI I In AMPLIFIER IMPEDANCE When a signal (current or voltage) is fed into the input, a portion of it will not get through the amplifier. This is due to external resistance effects. 2 types of impedances associated with an amplifier: Input impedance Output impedance SINGLE-STAGE AMPLIFIER Single-stage amplifiers are used in virtually every op-amp design By replacing a passive load (resistor) with a MOS transistor (called an active load), minimize chip area Active load : produce higher values of resistance higher gain Types of active load : G-D load and current source load. COMMON-SOURCE STAGE with resistive load Commonsource stage Equivalent circuit in deep triode region Input-output characteristic Small-signal model for saturation region ANALYSIS If Vin is low (below threshold) transistor M1 is OFF For Vin not-too-much-above threshold, transistor M1 is in Saturation, and Vout decreases. Vout 1 W 2 VDD RD nCox Vin VTH 2 L For (Vin>Vin1) – M1 is in Triode Mode. Vout VDD RD 1 W 2 nCox 2Vin VTH Vout Vout 2 L If Vin is high enough to drive M1 into deep triode region, Vout VDD << 2(Vin – VTH) V Vout VDD Ron Ron RD out 1 nCox W RD Vin VTH L Voltage gain Vout W Av RD nCox Vin VTH g m RD Vin L Av 2nC W ox L VRD ID Design tradeoffs Gain is determined by 3 factors: W/L, RD , DC voltage and ID If current and RD are kept constant, an increase of W/L increases the gain, but it also increases the gate capacitance – lower bandwidth If ID and W/L are kept constant, and we increase RD, then VDS becomes smaller. Operating Point gets closer to the borderline of Triode Mode. It means – less “swing” (room for the amplified signal) If ID decreases and W/L and VRD are kept constant, then we must increase RD Large RD consumes too much space, increases the noise level and slows the amplifier down (time constant with input capacitance of next stage). COMMON-SOURCE tradeoffs Av gm ro || RD Larger RD values increase the influence of channel-length modulation (ro term begins to strongly affect the gain) Common Source Maximum Gain Av gm ro || RD Av gmr o " intri nsi c gain" COMMON-SOURCE STAGE with diode-connected load “Diode-Connected” MOSFET is only a name. In BJT if Base and Collector are short-circuited, then the BJT acts exactly as a diode. We want to replace RD with a MOSFET that will operate like a small-signal resistor. V1 VX I X Rd VX g mVX ro 1 1 || ro gm gm Common Source with Diode-Connected NMOS Load (Body Effect included) Vx (gm gmb)V x Ix ro Vx 1 1 || ro Ix gm gmb gm gmb CS with Diode-Connected NMOS Load Voltage Gain Calculation Av gm1 1 gm1 1 gm2 gmb2 gm2 1 2nCOX (W / L)1 I D1 1 Av 2nCOX (W / L) 2 I D 2 1 (W / L)1 1 Av (W / L)2 1 CS with Diode-Connected NMOS Load Voltage Gain Calculation (W / L)1 1 Av (W / L)2 1 If variations of η with the output voltage are neglected, the gain is independent of the bias currents and voltages (as long as M1 stays in Saturation) The point: Gain does not depend on Vin Amplifier is relatively linear (even for large signal analysis!) W W 0.5n COX (Vin VTH1 ) 2 0.5n COX (VDD Vout VTH 2 ) 2 L 1 L 2 W W (Vin VTH1 ) (VDD Vout VTH 2 ) L 1 L 2 Assumptions made along the way: W W (Vin VTH1 ) (VDD Vout VTH 2 ) L 1 L 2 We neglected channel-length modulation effect We neglected VTH voltage dependent variations We assumed that two transistors are matching in terms of COX. Comment about “cutting off” If I1 is made smaller and smaller, what happens to Vout? It should be equal to VDD at the end of the current reduction process, or is it? Comment about “cutting off” (cont’d) If I1 is made smaller and smaller, VGS gets closer and closer to VTH. Very near I1=0, if we neglect sub-threshold conduction, we should have VGS≈VTH2, and therefore Vout≈VDD-VTH2 ! Cutoff conflict resolved: In reality, sub-threshold conduction, at a very low current, eventually brings Vout to the value VDD. Output node capacitance slows down this transition. In high-speed switching, sometimes indeed Vout doesn’t make it to VDD Back to large-signal behavior of the CS amplifier with diode-connected NMOS load: W W (Vin VTH1 ) (VDD Vout VTH 2 ) L 1 L 2 Large signal behavior of CS amplifier with diode-connected load When Vin<VTH1 (but near), we have the above “imperfect cutoff” effect. Then we have a, more or less, linear region. When Vin exceeds Vout+VTH1 amplifier enters the Triode mode, and becomes nonlinear. CS with Diode-Connected PMOS Load Voltage Gain n (W / L)1 Av p (W / L) 2 No body effect! Numerical Example Say that we want the voltage gain to be 10. Then: Av n (W / L)1 10 p (W / L) 2 n (W / L)1 100 p (W / L) 2 Example continued n (W / L)1 100 p (W / L) 2 Typically n 2 p Example continued (W / L)1 50 (W / L) 2 Need a “strong” input device, and a “weak” load device. Large dimension ratios lead to either a larger input capacitance (if we make input device very wide (W/L)1>>1) or to a larger output capacitance (if we make the load device very narrow (W/L)2<<1). Latter option (narrow load) is preferred from bandwidth considerations. CS with Diode-Connected Load Swing Issues I D1 I D2 , W W 2 n (VGS1 VTH1) p (VGS2 VTH2) 2 L 1 L 2 n (W / L)1 | VGS2 VTH2 | Av p (W / L) 2 (VGS1 VTH1) This implies substantial voltage swing constraint. Why? Numerical Example to illustrate the swing problems Assume for instance VDD=3V Let’s assume that VGS1-VTH1=200mV (arbitrary selection, consistent with current selection) Assume also that |VTH2|=0.7V (for PMOS load VTH2=-0.7V) For a gain of 10, we now need |VGS2|=2.7V (for PMOS load VGS2<-2.7V) Therefore because VGD2=0: VDS2=VGS2=-2.7V). Now VDS1=VDD-VSD2<3-2.7=0.3V, and recall that VDS1>VGS1-VTH1=0.2V. Not much room left for the amplified signal vds1. DC Q-Point of the Amplifier VG1 determines the current and the voltage VGS2 If we neglect the effect of the transistors’ λ, the error in predicting the Q-point solution may be large! CS with Diode-Connected Load –How do we take into account λ? 1 Av gm1 ( || ro1 || ro 2 ) gm2 gmb2 1 Av g m1 ( || ro1 || ro 2 ) gm2 Example as Introduction to Current Source Load M1 is biased to be in Saturation and have a current of I1. A current source of IS=0.75I1 is hooked up in parallel to the load – does this addition ease up the amplifier’s swing problems? Example as Introduction to Current Source Load Now ID2=I1/4. Therefore (from the ratio of the two transconductances with different currents): 4 n (W / L)1 Av p (W / L) 2 Example: Swing Issues I D1 4I D2 , W W 2 n (VGS1 VTH1) 4 p (VGS2 VTH2) 2 L 1 L 2 Av 4 | VGS2 VTH2 | (VGS1 VTH1) It sure helps in terms of swing. COMMON-SOURCE STAGE with current-source load How can Vin change the current of M1 if I1 is constant? W I D1 0.5n COX (Vin VTH1 ) 2 (1 Vout ) I1 L 1 As Vin increases, Vout must decrease Simple Implementation: Current Source obtained from M2 in Saturation CS with Current Source Load Av g m ( ro1 || ro2) DC Conditions W I D1 0.5 n C OX (Vin VTH1 ) 2 (1 1Vout ) I D 2 L 1 W 0.5 p C OX (Vb VDD VTH 2 ) 2 (1 2 [Vout VDD ]) L 2 {(W/L)1,VG1} and {(W/L)2,Vb} need to be more or less consistent, if we wish to avoid too much dependence on λ values. Need DC feedback to fix better the DC Vout Swing Considerations W I D1 0.5 n C OX (Vin VTH1 ) 2 (1 1Vout ) I D 2 L 1 W 0.5 p C OX (Vb VDD VTH 2 ) 2 (1 2 [Vout VDD ]) L 2 We can make |VDS2|>|VGS2-VTH2| small (say a few hundreds of mV), if we compensate by making W2 wider. How do we make the gain large? Av g m ( ro1 || ro2) Recall: λ is inversely proportional to the channel length L. To make λ values smaller (so that ro be larger) need to increase L. In order to keep the same current, need to increase W by the same proportion as the L increase. CS Amplifier with Current-Source Load Gains Typical gains that such an amplifier can achieve are in the range of -10 to -100. To achieve similar gains with a RD load would require much larger VDD values. For low-gain and high-frequency applications, RD load may be preferred because of its smaller parasitic capacitance (compared to a MOSFET load) Numerical Example Let W/L for both transistors be W/L = 100µm / 1.6µm Let µnCox=90µA/V2, µpCox=30µA/V2 Bias current is ID=100µA Numerical Example (Cont’d) Let ro1=8000L/ID and ro2=12000L/ID where L is in µm and ID is in mA. What is the gain of this stage? Numerical Example (Cont’d) g m1 2 nCOX (W / L)1 I D 1.06mA / V ro1 8000 1.6 / 0.1 128K ro 2 12000 1.6 / 0.1 192 K AV g m1 (ro1 || ro 2 ) 81.4 How does L influence the gain? Av gm ro1 || ro2 Assuming ro2 large, ox D W 1 Av g m ro1 2 nC I L 1 I D How does L influence the gain? ox D W 1 Av g m ro1 2 nC I L 1 1 I D As L1 increases the gain increases, because λ1 depends on L1 more strongly than gm1 does! As ID increases the gain decreases. Increasing L2 while keeping W2 constant increases ro2 and the gain, but |VDS2| necessary to keep M2 in Saturation increases. COMMON-SOURCE STAGE with triode load Av g m1RON 2 RON 2 1 pCox WL (VDD Vb VTH 2) 2 How should Vb be chosen? M2 must conduct: Vb-VDD≤VTH2, or Vb≤VDD+VTH2 M2 must be in Triode Mode: Vout-VDD≤Vb-VDD-VTH2, or Vb≥Vout+VTH2 M2 must be “deep inside” Triode Mode: 2(Vb-VDDVTH2)>>Vout-VDD, or: Vb>>VDD/2+VTH2+Vout/2 Also Vb and (W/L)2 determine the desired value of Ron2 It’s not easy at all to determine (and implement) a working value for Vb CS Amplifiers with Triode Region load is rarely used COMMON-SOURCE STAGE with source degeneration gm Gm 1 g m RS Av Gm RD Summary of key formulas, neglecting channel-length modulation and body effect g m RD Av 1 g m RS CS Amplifier with Source Degeneration – Formula Explained V1 Vin I D RS Vin g mV1 RS Vin V1 1 g m RS Vout g mV1 RD g m RD Av 1 g m RS Idea is the same as that of adding emitter resistor to a BJT CE amplifier g m RD AV 1 g m RS With RS=0, the gain strongly depends on gm, and as we recall gm depends on the input’s amplitude, temperature and other effects. If gmRS>>1, then AV≈-RD/RS CS Amplifier with Source Degeneration Key Formulas with λ and Body-Effect Included gm Gm R S [1 (gm gmb )RS ] ro ROU T [1 (gm gmb )ro ]RS ro Av Gm ( RD || ROUT ) Derivation is similar to that of the simplified case – generalized transconductance I It doesn’t matter what RD is because we treat ID as “input” I D g mV1 g mbVbs I ro g mV1 g mbVX I D RS ro I D RS g m (Vin I D RS ) g mb ( I D RS ) ro Gm g m ro ID Vin RS [1 ( g m g mb ) RS ]ro Linearizing effect of RS: gm RD RD Av 1 gm RS 1/ gm RS Linearizing effect of RS: gm RD RD Av 1 gm RS 1/ gm RS For low current levels 1/gm>>RS and therefore Gm≈gm. For very large Vin, if transistor is still in Saturation, Gm approaches 1/RS. Estimating Gain by Inspection gm RD RD Av 1 gm RS 1/ gm RS Denominator: Resistance seen the Source path, “looking up” from ground towards Source. Numerator: Resistance seen at Drain. Example to demonstrate method: Note that M2 is “diode-connected”, thus acting like a resistor 1/gm2 AV=-RD/(1/gm1+1/gm2) RS effect on CS Output Resistance I X g mV1 g mbVbs I ro g m ( I X RS ) g mb I X RS I ro VX ro ( I X ( g m g mb ) RS I X ) I X RS CS Output Resistance ROU T [1 (gm gmb )ro ]RS ro ROU T ro' ro [1 (gm gmb )RS ] RS causes a significant increase in the output resistance of the amplifier CS Amplifier with Source Degeneration Gain Formula with λ and Body-Effect Included gm Gm R S [1 (gm gmb )RS ] ro ROU T [1 (gm gmb )ro ]RS ro Av Gm ( RD || ROUT ) Derivation of Gain Formula Vout I ro ( g mV1 g mbVbs ) RD Vout RS RS [ g m (Vin Vout ) g mbVout ) RD RD RD Derivation of Gain Formula Vout I ro ro Vout RS RD Vout RS RS RS ro [ g m (Vin Vout ) g mbVout ]ro Vout RD RD RD RD Now we can relate Vout to Vin - formula results General Networks Result Voltage gain in any linear circuit equals –GmRout Gm is circuit transconductance when output is shorted to ground Rout is output resistance when circuit’s input voltage is set to zero. General Networks Result –GmRout formula is useful if Gm and Rout can be determined by inspection. Proof: Rout is Norton equivalent. Vout=-IoutRout Gm=Iout/Vin, which leads to the result.