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Lecture 03 Basic Amplifier Stages / Cascode Amplifier This is a modified version of Dr Hesham’s Lectures Dr. Hesham A. Omran Analog IC Design ً ‫يل‬ِ ‫ل‬َ‫ق‬ ‫ا‬ ‫َّل‬ِ ‫إ‬ِ ‫م‬ْ‫ل‬ِ ‫ع‬ْ‫ل‬‫ا‬ َ ‫ن‬ِ ‫م‬ ْ‫م‬ُ ‫يت‬ِ ‫ت‬‫و‬ُ‫أ‬‫ا‬َ ‫م‬َ ‫و‬ 11 March 2022 8 ‫شعبان‬ 1443
Outline ❑ Recapping previous key results ❑ Basic amplifier operation ❑ Rin/out Shortcuts – Looking from drain – Looking from source ❑ GmRout method ❑ Basic amplifier topologies – Common Source (CS) – Common Gate (CG) – Common Drain (CD) – Source Follower (SF) ❑ Bipolar Junction Transistor (BJT) 06: Basic Amplifier Stages 2
MOSFET in Saturation ❑ The channel is pinched off if the difference between the gate and drain voltages is not sufficient to create an inversion layer 𝑉𝐺𝐷 ≤ 𝑉𝑇𝐻 𝑂𝑅 𝑉𝐷𝑆 ≥ 𝑉 𝑜𝑣 ❑ Square-law (long channel MOS) 𝐼𝐷 = 𝜇𝑛𝐶𝑜𝑥 2 𝑊 𝐿 ⋅ 𝑉 𝑜𝑣 2 1 + 𝜆𝑉𝐷𝑆 𝑉𝑇𝐻 = 𝑉𝑇𝐻0 + 𝛾 2Φ𝐹 + 𝑉𝑆𝐵 − 2Φ𝐹 06: Basic Amplifier Stages 3 n+ n+ G S D p-sub p+ B VGS>VTH VGD<VTH VDS>Vov VSB VGS>VTH VGD<VTH VDS>Vov
Regions of Operation Summary 06: Basic Amplifier Stages 4 OFF (Subthreshold) 𝑉𝐺𝑆 < 𝑉𝑇𝐻 ON 𝑉𝐺𝑆 > 𝑉𝑇𝐻 Triode 𝑉𝐷𝑆 < 𝑉 𝑜𝑣 Or 𝑉𝐺𝐷 > 𝑉𝑇𝐻 𝐼𝐷 = 𝜇𝐶𝑜𝑥 𝑊 𝐿 𝑉 𝑜𝑣𝑉𝐷𝑆 − 𝑉𝐷𝑆 2 2 Pinch-Off (Saturation) 𝑉𝐷𝑆 ≥ 𝑉 𝑜𝑣 Or 𝑉𝐺𝐷 ≤ 𝑉𝑇𝐻 𝐼𝐷 = 𝜇𝐶𝑜𝑥 2 𝑊 𝐿 𝑉 𝑜𝑣 2 1 + 𝜆𝑉𝐷𝑆
Low-Frequency Small-Signal Model 𝑔𝑚 = 𝜕𝐼𝐷 𝜕𝑉𝐺𝑆 = 𝜇𝐶𝑜𝑥 𝑊 𝐿 𝑉 𝑜𝑣 = 𝜇𝐶𝑜𝑥 𝑊 𝐿 ⋅ 2𝐼𝐷 = 2𝐼𝐷 𝑉 𝑜𝑣 𝑔𝑚𝑏 = 𝜂𝑔𝑚, 𝜂 ≈ 0.1 − 0.25 𝑟𝑜 = 1 𝜕𝐼𝐷/𝜕𝑉𝐷𝑆 = 1 𝜆𝐼𝐷 , 𝜆 ∝ 1 𝐿 06: Basic Amplifier Stages 5 gmvgs ro gmbvbs G D S B vgs vbs
Outline ❑ Recapping previous key results ❑ Basic amplifier operation ❑ Rin/out Shortcuts – Looking from drain – Looking from source ❑ Gm Rout method ❑ Basic amplifier topologies – Common Source (CS) – Common Gate (CG) – Common Drain (CD) – Source Follower (SF) ❑ Bipolar Junction Transistor (BJT) 06: Basic Amplifier Stages 6
Why Amplifiers? ❑ All the physical signals in the world around us are analog – Voice, light, temperature, pressure, etc. ❑ We (will) always need an “analog” interface circuit to connect between our physical world and our digital electronics 06: Basic Amplifier Stages 7 Amplifier A/D Digital processing and storage
Basic Amplifier Topologies Common Source (CS) Common Gate (CG) Common Drain (CD) Source Follower (SF) Voltage & current amplifier Current buffer Voltage buffer In 𝐺 𝑆 𝐺 Out 𝐷 𝐷 𝑆 06: Basic Amplifier Stages 8 RD vout vin RS RD vin RS vout RD vout vin RS
MOSFET CS Amplifier Example 06: Basic Amplifier Stages 9 [Sedra/Smith, 2015]
MOSFET CS Amplifier Example 06: Basic Amplifier Stages 10 [Sedra/Smith, 2015]
MOSFET CS Amplifier Example 06: Basic Amplifier Stages 11 [Sedra/Smith, 2015]
MOSFET CS Amplifier Example ❑ 𝑅𝑖𝑛 = 𝑣𝑖𝑛 𝑖𝑖𝑛 = 𝑣𝑖𝑛 𝑖𝑠𝑖𝑔 ❑ 𝑅𝑜𝑢𝑡 = 𝑣𝑥 𝑖𝑥 @ 𝑣𝑠𝑖𝑔 = 0 ❑ 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑔𝑎𝑖𝑛 = 𝐴𝑣 = 𝑣𝑜𝑢𝑡 𝑣𝑠𝑖𝑔 = 𝑣𝑖𝑛 𝑣𝑠𝑖𝑔 ⋅ 𝑣𝑜𝑢𝑡 𝑣𝑖𝑛 = 𝑅𝑖𝑛 𝑅𝑠𝑖𝑔+𝑅𝑖𝑛 ⋅ 𝑣𝑜𝑢𝑡 𝑣𝑖𝑛 ❑ 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑔𝑎𝑖𝑛 = 𝐴𝑖 = 𝑖𝑜𝑢𝑡 𝑖𝑠𝑖𝑔 = 𝑣𝑜𝑢𝑡/𝑅𝐿 𝑣𝑖𝑛/𝑅𝑖𝑛 = 𝑣𝑜𝑢𝑡 𝑣𝑖𝑛 ⋅ 𝑅𝑖𝑛 𝑅𝐿 06: Basic Amplifier Stages 12 Rsig vin RG1 RG2 RD vout M1 vsig Cc1 Cc2 RL Rout Rin isig iout
Large and Small Signal Analysis Large Signal Analysis Small Signal Analysis Model Large signal model Small signal model Linearity Non-linear Linear Simulation DC and transient analysis AC analysis Purpose Calculate bias point, signal swing, distortion, etc. Calculate 𝐴𝑣, 𝑅𝑖𝑛, 𝑅𝑜𝑢𝑡, 𝐵𝑊, 𝑒𝑡𝑐. VDC ✓ s.c. IDC ✓ o.c. Capacitor o.c. (in DC) 1/𝜔𝐶 Inductor s.c. (in DC) 𝜔𝐿 06: Basic Amplifier Stages 13
MOSFET Amplifier Analysis Steps 1. DC analysis • Coupling and bypass capacitors → open-circuit (o.c) • Calculate Q-point and check operation in saturation 𝑉𝐷𝑆 > 𝑉 𝑜𝑣 2. Calculate small signal parameters 𝑔𝑚, 𝑟𝑜 3. Draw the small signal equivalent circuit • DC voltage source → short-circuit (s.c.) • DC current source → open-circuit (o.c.) • Coupling and bypass capacitors → short-circuit (s.c) 4. Determine the amplifier parameters • Input resistance and output resistance • Voltage gain and current gain 06: Basic Amplifier Stages 14
Direct Analysis on Schematics ❑ No need to draw the small signal model every time ❑ Just remember we have two VCCSs and 𝑟𝑜 between D and S ❑ If G and B are ac connected, then 𝑣𝑏𝑠 = 𝑣𝑔𝑠 ➔ 𝑔𝑚 and 𝑔𝑚𝑏 add • 𝑔𝑚 → 𝑔𝑚 + 𝑔𝑚𝑏 06: Basic Amplifier Stages 15 gmvgs ro gmbvbs G D S vbs vgs B
Direct Analysis on Schematics ❑ No need to draw the small signal model every time ❑ Just remember we have two VCCSs and 𝑟𝑜 between D and S ❑ If G and B are ac connected, then 𝑣𝑏𝑠 = 𝑣𝑔𝑠 ➔ 𝑔𝑚 and 𝑔𝑚𝑏 add • 𝑔𝑚 → 𝑔𝑚 + 𝑔𝑚𝑏 06: Basic Amplifier Stages 16 gmvgs + gmbvbs ro B vgs S G D vbs
Intrinsic Gain 𝑣𝑜𝑢𝑡 = − 𝑔𝑚𝑣𝑖𝑛 𝑟𝑜 |𝐴𝑣| = 𝑣𝑜𝑢𝑡 𝑣𝑖𝑛 = 𝑔𝑚𝑟𝑜 ❑ 𝑔𝑚𝑟𝑜 is the max gain that can be obtained from a single transistor ❑ Common approximations that we usually use 𝑔𝑚𝑟𝑜 ≫ 1 𝑟𝑜 ≫ 1 𝑔𝑚 𝑔𝑚 + 1 𝑟𝑜 ≈ 𝑔𝑚 𝑟𝑜// 1 𝑔𝑚 ≈ 1 𝑔𝑚 06: Basic Amplifier Stages 17 vin vout
Outline ❑ Recapping previous key results ❑ Basic amplifier operation ❑ Rin/out Shortcuts • Looking from drain • Looking from source ❑ Gm Rout method ❑ Basic amplifier topologies • Common Source (CS) • Common Gate (CG) • Common Drain (CD) – Source Follower (SF) ❑ Bipolar Junction Transistor (BJT) 06: Basic Amplifier Stages 18
Rin/out Shortcuts ❑ Find equivalent impedance looking from Gate, Source, and Drain 06: Basic Amplifier Stages 19
Looking From Drain ❑ If G and B are ac connected, then 𝑣𝑏𝑠 = 𝑣𝑔𝑠 ➔ 𝑔𝑚 and 𝑔𝑚𝑏 add ❑ 𝑣𝑔𝑠 = −𝑖𝑥𝑅𝑆 and 𝑔𝑚𝑟𝑜 ≫ 1 ❑ Apply KCL at 𝐷 𝑖𝑥 = 𝑔𝑚 + 𝑔𝑚𝑏 −𝑖𝑥𝑅𝑆 + 𝑣𝑥 − 𝑖𝑥𝑅𝑆 𝑟𝑜 𝑅𝐿𝐹𝐷 = 𝑣𝑥 𝑖𝑥 = 𝑟𝑜 + 𝑔𝑚 + 𝑔𝑚𝑏 𝑟𝑜𝑅𝑆 + 𝑅𝑆 ≈ 𝑟𝑜 1 + 𝑔𝑚 + 𝑔𝑚𝑏 𝑅𝑆 06: Basic Amplifier Stages 20 RS vx RLFD ix ixRS ix ro vgs
Looking From Drain 𝑅𝐿𝐹𝐷 ≈ 𝑟𝑜 1 + 𝑔𝑚 + 𝑔𝑚𝑏 𝑅𝑆 ❑ Special case: 𝑅𝑆 = 0 (G and S ac s.c.) ➔ active load 𝑅𝐿𝐹𝐷 ≈ 𝑟𝑜 ❑ Drain is a high-impedance node (H.I.N.) 06: Basic Amplifier Stages 21 RS vx RLFD ix ixRS ix ro
Active Load (Source OFF) 06: Basic Amplifier Stages 22 ro ro
Quiz ❑ Assume M1 and M2 have the same 𝑔𝑚 and 𝑟𝑜, 𝑔𝑚𝑟𝑜 ≫ 1, and neglect body effect ❑ Find 𝑅𝑋 06: Basic Amplifier Stages 23 VB1 VB2 RX M2 M1 𝑅𝑋 = 𝑟𝑜1+𝑟𝑜2 + 𝑔𝑚2 + 𝑔𝑚𝑏2 𝑟𝑜1 𝑟𝑜2
Looking From Source ❑ 𝑣𝑔𝑠 = −𝑣𝑥, 𝑔𝑚 and 𝑔𝑚𝑏 add, and 𝑔𝑚𝑟𝑜 ≫ 1 ❑ Apply KCL at 𝑆 𝑖𝑥 = 𝑔𝑚 + 𝑔𝑚𝑏 𝑣𝑥 + 𝑣𝑥 − 𝑖𝑥𝑅𝐷 𝑟𝑜 𝑅𝐿𝐹𝑆 = 𝑣𝑥 𝑖𝑥 = 1 𝑔𝑚 + 𝑔𝑚𝑏 + 1 𝑟𝑜 1 + 𝑅𝐷 𝑟𝑜 ≈ 1 𝑔𝑚 + 𝑔𝑚𝑏 1 + 𝑅𝐷 𝑟𝑜 06: Basic Amplifier Stages 24 vx RLFS ix RD ixRD ix ro
Looking From Source 𝑅𝐿𝐹𝑆 ≈ 1 𝑔𝑚 + 𝑔𝑚𝑏 1 + 𝑅𝐷 𝑟𝑜 ❑ Special case: 𝑅𝐷 = 0 (G & D ac s.c.) ➔ diode connected 𝑅𝐿𝐹𝑆 ≈ 1 𝑔𝑚 + 𝑔𝑚𝑏 ❑ Source is a low impedance node (L.I.N.) 06: Basic Amplifier Stages 25 vx RLFS ix RD ixRD ix ro
Diode Connected (Source Absorption) ❑ Always in saturation (𝑉𝐷𝑆 = 𝑉𝐺𝑆 > 𝑉 𝑜𝑣) ❑ Body effect: 𝑔𝑚 → 𝑔𝑚 + 𝑔𝑚𝑏 (if G and B are ac connected) 06: Basic Amplifier Stages 26 1/gm 1/gm
Quiz ❑ Assume M1, M2, and M3 have the same 𝑔𝑚 and 𝑟𝑜, 𝑔𝑚𝑟𝑜 ≫ 1, and neglect body effect ❑ Find 𝑅𝑋 06: Basic Amplifier Stages 27 VB1 VB2 RX M2 M1 VB3 M3 𝑅𝑋 = 1 𝑔𝑚1 + 𝑔𝑚𝑏1 + 1 𝑟𝑜1 1 + 𝑅𝐿𝐹𝐷2 𝑟𝑜1 𝑅𝐿𝐹𝐷2 = 𝑟𝑜2+𝑟𝑜3 + 𝑔𝑚2 + 𝑔𝑚𝑏2 𝑟𝑜2 𝑟𝑜3
Rin/out Shortcuts Summary 06: Basic Amplifier Stages 28 1 𝑔𝑚 + 𝑔𝑚𝑏 1 + 𝑅𝐷 𝑟𝑜 L.I.N. 𝑟𝑜 1 + 𝑔𝑚 + 𝑔𝑚𝑏 𝑅𝑆 H.I.N. ∞ At low frequencies ONLY Note: Do not use shortcuts if a feedback element exists
Outline ❑ Recapping previous key results ❑ Basic amplifier operation ❑ Rin/out Shortcuts – Looking from drain – Looking from source ❑ Gm Rout method ❑ Basic amplifier topologies – Common Source (CS) – Common Gate (CG) – Common Drain (CD) – Source Follower (SF) ❑ Bipolar Junction Transistor (BJT) 06: Basic Amplifier Stages 29
Amplifier Model ❑ Rin/out 𝑅𝑖𝑛 = 𝑣𝑖𝑛 𝑖𝑖𝑛 𝑅𝑜𝑢𝑡 = 𝑣𝑥 𝑖𝑥 @ 𝑣𝑖𝑛 = 0 ❑ O.C. voltage gain (Thevenin model) 𝑣𝑜𝑢𝑡,𝑜𝑐 = 𝐴𝑣𝑣𝑖𝑛 𝐴𝑣 = 𝑣𝑜𝑢𝑡,𝑜𝑐 𝑣𝑖𝑛 06: Basic Amplifier Stages 30 Avvin Rout vin Rin vout iout iin
Amplifier Model ❑ Transconductance (Norton model) 𝑖𝑜𝑢𝑡,𝑠𝑐 = 𝐺𝑚𝑣𝑖𝑛 𝐺𝑚 = 𝑖𝑜𝑢𝑡,𝑠𝑐 𝑣𝑖𝑛 ❑ Thevenin  Norton 𝐴𝑣𝑣𝑖𝑛 = 𝐺𝑚𝑣𝑖𝑛 𝑅𝑜𝑢𝑡 𝐴𝑣 = 𝑣𝑜𝑢𝑡,𝑜𝑐 𝑣𝑖𝑛 = 𝐺𝑚𝑅𝑜𝑢𝑡 ❑ S.C. Current Gain 𝐴𝑖 = 𝑖𝑜𝑢𝑡,𝑠𝑐 𝑖𝑖𝑛 = 𝐺𝑚𝑅𝑖𝑛 06: Basic Amplifier Stages 31 Avvin Rout vin Rin vout iout iin vin Rin vout iout iin Gmvin Rout
Why GmRout? 𝑅𝑖𝑛 = 𝑣𝑖𝑛/𝑖𝑖𝑛 𝑅𝑜𝑢𝑡 = 𝑣𝑥/𝑖𝑥 @ 𝑣𝑖𝑛 = 0 𝐺𝑚 = 𝑖𝑜𝑢𝑡,𝑠𝑐/𝑣𝑖𝑛 𝐴𝑣 = 𝐺𝑚𝑅𝑜𝑢𝑡 𝐴𝑖 = 𝐺𝑚𝑅𝑖𝑛 ❑ Divide and conquer – Rout simplified: vin=0 – Gm simplified: vout=0 – We already need Rin/out – We can quickly and easily get Rin/out from the shortcuts 06: Basic Amplifier Stages 32 Avvin Rout vin Rin vout iout iin vin Rin vout iout iin Gmvin Rout
Outline ❑ Recapping previous key results ❑ Basic amplifier operation ❑ Rin/out Shortcuts – Looking from drain – Looking from source ❑ Gm Rout method ❑ Basic amplifier topologies – Common Source (CS) – Common Gate (CG) – Common Drain (CD) – Source Follower (SF) ❑ Bipolar Junction Transistor (BJT) 06: Basic Amplifier Stages 33
Basic Amplifier Topologies Common Source (CS) Common Gate (CG) Common Drain (CD) Source Follower (SF) Voltage & current amplifier Current buffer Voltage buffer In 𝐺 𝑆 𝐺 Out 𝐷 𝐷 𝑆 06: Basic Amplifier Stages 34 RD vout vin RS RD vin RS vout RD vout vin RS
Common Source (CS) ❑ Apply KCL at 𝑆 𝑮𝒎 = 𝒊𝒐𝒖𝒕,𝒔𝒄 𝒗𝒊𝒏 ≈ −𝒈𝒎 𝟏 + 𝒈𝒎 + 𝒈𝒎𝒃 𝑹𝑺 𝑅𝑜𝑢𝑡 ≈ 𝑅𝐷//𝑟𝑜 1 + 𝑔𝑚 + 𝑔𝑚𝑏 𝑅𝑆 𝐴𝑣 = 𝐺𝑚𝑅𝑜𝑢𝑡 ❑ If 𝑅𝐷 is ac o.c.: 𝐴𝑣 = −𝑔𝑚𝑟𝑜 ❑ If 𝑅𝐷 ≪ 𝑅𝐿𝐹𝐷: 𝐴𝑣 ≈ −𝒈𝒎𝑹𝑫 𝟏+ 𝒈𝒎+𝒈𝒎𝒃 𝑹𝑺 ❑ If 𝑅𝑆 = 0: 𝐴𝑣 = −𝑔𝑚 𝑅𝐷//𝑟𝑜 06: Basic Amplifier Stages 35 RD vout vin RS iout,sc -iout,scRS iout,sc ro 𝑖𝑜𝑢𝑡,𝑠𝑐 + 𝑔𝑚 𝑣𝑖𝑛 + 𝑖𝑜𝑢𝑡,𝑠𝑐𝑅𝑆 + 𝑔𝑚𝑏 𝑖𝑜𝑢𝑡,𝑠𝑐𝑅𝑆 + 𝑖𝑜𝑢𝑡,𝑠𝑐𝑅𝑆 𝑟𝑜 = 0 Shorted by o/p
Common Source (CS) 𝑮𝒎 = 𝒊𝒐𝒖𝒕,𝒔𝒄 𝒗𝒊𝒏 ≈ −𝒈𝒎 𝟏 + 𝒈𝒎 + 𝒈𝒎𝒃 𝑹𝑺 𝑅𝑜𝑢𝑡 ≈ 𝑅𝐷//𝑟𝑜 1 + 𝑔𝑚 + 𝑔𝑚𝑏 𝑅𝑆 𝐴𝑣 = 𝐺𝑚𝑅𝑜𝑢𝑡 ❑ If S and B are connected (for PMOS): 𝐴𝑣 ≈ −𝑅𝐷//𝑅𝐿𝐹𝐷 1 𝑔𝑚 + 𝑅𝑆 = − 𝐷𝑟𝑎𝑖𝑛 𝑅𝑒𝑠. 1 𝑔𝑚 + 𝑆𝑜𝑢𝑟𝑐𝑒 𝑅𝑒𝑠. ❑ If 𝑅𝑆 ≫ 1 𝑔𝑚 & 𝑅𝐷 ≪ 𝑅𝐿𝐹𝐷: 𝐴𝑣 ≈ −𝑅𝐷 𝑅𝑆 → 𝐿𝑖𝑛𝑒𝑎𝑟 ❑ 𝑅𝑆 reduces 𝐺𝑚 → Source degeneration – But improves linearity 06: Basic Amplifier Stages 36 RD vout vin RS iout,sc -iout,scRS iout,sc ro
Common Gate (CG) ❑ Apply KCL at 𝐷 𝑖𝑜𝑢𝑡,𝑠𝑐 + 𝑔𝑚 + 𝑔𝑚𝑏 −𝑣𝑖𝑛 − 𝑣𝑖𝑛 𝑟𝑜 = 0 𝑮𝒎 = 𝒊𝒐𝒖𝒕,𝒔𝒄 𝒗𝒊𝒏 ≈ 𝒈𝒎 + 𝒈𝒎𝒃 𝑅𝑜𝑢𝑡 ≈ 𝑅𝐷//𝑟𝑜 (𝒘𝒉𝒚? ) 𝐴𝑣 = 𝐺𝑚𝑅𝑜𝑢𝑡 ❑ If 𝑅𝐷 is ac o.c.: 𝐴𝑣 = 𝑔𝑚 + 𝑔𝑚𝑏 𝑟𝑜 ❑ If 𝑅𝐷 ≪ 𝑟𝑜: 𝐴𝑣 ≈ 𝑔𝑚 + 𝑔𝑚𝑏 𝑹𝑫 ❑ For PMOS connect S and B: 𝑔𝑚 + 𝑔𝑚𝑏 → 𝑔𝑚 ❑ Note that 𝐴𝑖 = 𝐺𝑚𝑅𝑖𝑛 <≈ 𝟏 (Current Buffer) 06: Basic Amplifier Stages 37 RD vout vin RS iout,sc ro
Common Drain (CD) – Source Follower ❑ Apply KCL at 𝐷 𝑖𝑜𝑢𝑡,𝑠𝑐 − 𝑔𝑚𝑣𝑖𝑛 + 𝑖𝑜𝑢𝑡,𝑠𝑐𝑅𝐷 𝑟𝑜 = 0 𝑮𝒎 = 𝒊𝒐𝒖𝒕,𝒔𝒄 𝒗𝒊𝒏 ≈ 𝒈𝒎 𝟏 + 𝑹𝑫/𝒓𝒐 𝑅𝑜𝑢𝑡 ≈ 𝑅𝑆// 1 𝑔𝑚 + 𝑔𝑚𝑏 1 + 𝑅𝐷 𝑟𝑜 𝐴𝑣 = 𝐺𝑚𝑅𝑜𝑢𝑡 ❑ If 𝑅𝑆 ≫ 𝑅𝐿𝐹𝑆: 𝐴𝑣 ≈ 𝑔𝑚 𝑔𝑚+𝑔𝑚𝑏 < 1 ❑ If 𝑆 and 𝐵 are connected (for PMOS): 𝐴𝑣 ≈ 𝟏 (Voltage buffer) 06: Basic Amplifier Stages 38 RD vout vin RS iout,sc -iout,scRD ro iout,sc
Summary of Basic Topologies CS CG CD (SF) Voltage & current amplifier Current buffer Voltage buffer Rin ∞ 𝑅𝑆// 1 𝑔𝑚 + 𝑔𝑚𝑏 1 + 𝑅𝐷 𝑟𝑜 ∞ Rout 𝑅𝐷//𝑟𝑜 1 + 𝑔𝑚 + 𝑔𝑚𝑏 𝑅𝑆 𝑅𝐷//𝑟𝑜 𝑅𝑆// 1 𝑔𝑚 + 𝑔𝑚𝑏 1 + 𝑅𝐷 𝑟𝑜 Gm −𝒈𝒎 𝟏 + 𝒈𝒎 + 𝒈𝒎𝒃 𝑹𝑺 𝒈𝒎 + 𝒈𝒎𝒃 𝒈𝒎 𝟏 + 𝑹𝑫/𝒓𝒐 06: Basic Amplifier Stages 39 RD vout vin RS RD vin RS vout RD vout vin RS
Quiz ❑ The circuit below shows a CS amplifier With diode-connected load ❑ Find the gain using Gm Rout (ignore body effect and CLM) – Express the gain in terms of (𝑊/𝐿)1 and (𝑊/𝐿)2 ❑ This is a “linear” CS amplifier 06: Basic Amplifier Stages 40 VB (W/L)1 (W/L)2 vin vout
Quiz ❑ The circuit below shows a complementary CS amplifier (inverter amp) ❑ Find the gain using Gm Rout Gm = - (gm1 + gm2) Rout = ro1 // ro2 06: Basic Amplifier Stages 41 M1 M2 vin vout
Output Signal Swing 𝑣𝑜𝑢𝑡,𝑚𝑎𝑥 = 𝑉𝐷𝐷 − 𝑉𝑜𝑣2 𝑣𝑜𝑢𝑡,𝑚𝑖𝑛 = 𝑉𝑜𝑣1 ❑ Output swing ≈ 𝑉𝐷𝐷 − 2𝑉 𝑜𝑣 06: Basic Amplifier Stages 42 VB vin vout M1 M2
Outline ❑ Recapping previous key results ❑ Basic amplifier operation ❑ Rin/out Shortcuts – Looking from drain – Looking from source ❑ Gm Rout method ❑ Basic amplifier topologies – Common Source (CS) – Common Gate (CG) – Common Drain (CD) – Source Follower (SF) ❑ Bipolar Junction Transistor (BJT) 06: Basic Amplifier Stages 43
Bipolar Junction Transistor (BJT) 06: Basic Amplifier Stages 44 Figure 5.26 (a) Basic common-emitter amplifier circuit. (b) Transfer characteristic of the circuit in (a). The amplifier is biased at a point Q, and a small voltage signal vi is superimposed on the dc bias voltage VBE. The resulting output signal vo appears superimposed on the dc collector voltage VCE. The amplitude of vo is larger than that of vi by the voltage gain Av. Common Emitter (CE)
BJT (Small Signal Model) 06: Basic Amplifier Stages 45 Π model: T model:
Rin/out Shortcuts (BJT) 06: Basic Amplifier Stages 46 Note: Do not use shortcuts if a feedback element exists
Lecture 3 part 2 Cascode Amplifiers Integrated Circuits Lab (ICL) Electronics and Communications Eng. Dept. Faculty of Engineering Ain Shams University Dr. Hesham A. Omran Analog IC Design ً ‫يل‬ِ ‫ل‬َ‫ق‬ ‫ا‬ ‫َّل‬ِ ‫إ‬ِ ‫م‬ْ‫ل‬ِ ‫ع‬ْ‫ل‬‫ا‬ َ ‫ن‬ِ ‫م‬ ْ‫م‬ُ ‫يت‬ِ ‫ت‬‫و‬ُ‫أ‬‫ا‬َ ‫م‬َ ‫و‬ 11 March 2022 8 ‫شعبان‬ 1443
Boosting Voltage Gain ❑ The max gain of a single transistor amplifier is the intrinsic gain (body effect may add). 𝐴𝑖 = 𝑔𝑚𝑟𝑜 ❑ 𝑔𝑚𝑟𝑜 is quite small for modern deep submicron technologies. – May be less than 10 ! – But we need high voltage gain to design an op-amp. ❑ How to boost the gain? – CS and CG stages provide voltage gain. 1. Use a cascade of CS stage: CS + CS + CS + … 2. Use a cascode amplifier: CS + CG + CG + … • Single cascode: CS + CG • Double cascode: CS + CG + CG 07: Cascode Amplifiers 48
Cascode ❑ CS + CG 07: Cascode Amplifiers 49 vout VB vin M1 M2 CS CG
Cascode Gain Using GmRout ❑ Transconductance is always related to the input transistor (VCCS) 𝑖𝑜𝑢𝑡,𝑠𝑐 ≈ −𝑔𝑚1𝑣𝑖𝑛 𝐺𝑚 = 𝑖𝑜𝑢𝑡,𝑠𝑐 𝑣𝑖𝑛 ≈ −𝑔𝑚1 Same Gm of CS 𝑅𝑜𝑢𝑡 ≈ 𝑟𝑜2 1 + 𝑔𝑚2 + 𝑔𝑚𝑏2 𝑟𝑜1 ≈ 𝑟𝑜2 𝑔𝑚2 + 𝑔𝑚𝑏2 𝑟𝑜1 Rout significantly boosted 𝐴𝑣 ≈ −𝑔𝑚1𝑟𝑜1 𝑔𝑚2 + 𝑔𝑚𝑏2 𝑟𝑜2 ❑ Assume all 𝑔𝑚 and 𝑟𝑜 are equal and neglect body effect 𝐴𝑣 = − 𝑔𝑚𝑟𝑜 2 07: Cascode Amplifiers 50 vout VB vin M1 M2 iout,sc iout,sc
Cascode as CS + CG 𝐶𝑆: 𝑣𝑥 𝑣𝑖𝑛 = −𝑔𝑚1 𝑟𝑜1//∞ 𝐶𝐺: 𝑣𝑜𝑢𝑡 𝑣𝑥 = 𝑔𝑚2 + 𝑔𝑚𝑏2 𝑟𝑜2 𝐴𝑣 = 𝑣𝑥 𝑣𝑖𝑛 ⋅ 𝑣𝑜𝑢𝑡 𝑣𝑥 ≈ −𝑔𝑚1𝑟𝑜1 𝑔𝑚2 + 𝑔𝑚𝑏2 𝑟𝑜2 ❑ Assume all 𝑔𝑚 and 𝑟𝑜 are equal and neglect body effect 𝐴𝑣 = − 𝑔𝑚𝑟𝑜 2 07: Cascode Amplifiers 51 vout VB vin M1 M2 CS CG vx
Quiz ❑ The circuit below shows a double cascode. ❑ Find the voltage gain. Assume all 𝑔𝑚 and 𝑟𝑜 are equal and neglect body effect. 07: Cascode Amplifiers 52 vout VB1 vin M1 M2 VB2 M3
What if 𝑹𝑫 is small? ❑ Is this cascode useful? – Not useful for boosting the gain – But useful for boosting the BW (more about this later) 07: Cascode Amplifiers 53 vout VB vin M1 M2 RD
Cascode Load ❑ If you want to keep the large Rout, you must use cascode load ❑ Assume all 𝑔𝑚 and 𝑟𝑜 are equal and neglect body effect 𝐴𝑣 = − 𝑔𝑚𝑟𝑜 2 2 ❑ Output swing ≈ 𝑉𝐷𝐷 − 4𝑉 𝑜𝑣 07: Cascode Amplifiers 54 vout VB1 vin M1 M2 VB2 VB3 M3 M4
Cascode Bias Voltage ❑ Keep M1 in sat 𝑉𝐵 > 𝑉𝐺𝑆2 + 𝑉𝑖𝑛,𝑚𝑎𝑥 − 𝑉𝑇𝐻1 𝑉𝑖𝑛,𝑚𝑎𝑥 < 𝑉𝑇𝐻1 + 𝑽𝑩 − 𝑉𝐺𝑆2 ❑ Keep M2 in sat 𝑉𝐵 < 𝑉𝑇𝐻2 + 𝑉𝑜𝑢𝑡,𝑚𝑖𝑛 𝑉𝑜𝑢𝑡,𝑚𝑖𝑛 > 𝑽𝑩 − 𝑉𝑇𝐻2 ❑ Increasing/decreasing 𝑉𝐵 either hurts 𝑽𝒊𝒏 range or 𝑽𝒐𝒖𝒕 range – Input and output ranges are coupled oppositely 07: Cascode Amplifiers 55 vout VB vin M1 M2 𝑉𝐵 < 𝑉𝑜𝑢𝑡,𝑚𝑖𝑛 − 𝑉𝑂𝑉2 + 𝑉𝐺𝑆2
Telescopic vs Folded Cascode ❑ Telescopic: CS + CG (both NMOS or both PMOS) – Both CS and CG use same bias current ❑ Folded: CS + CG (NMOS-PMOS combination) – The small signal current is folded up or down – Extra bias current is needed – Rout is lower (due to 𝐼𝐵1) – Why is it useful? 07: Cascode Amplifiers 56 vout VB vin M1 M2 CS CG vout VB M2 CS CG vin M1 IB1 IB2
Folded Cascode ❑ Input and output ranges are NOT coupled oppositely 𝑉𝑖𝑛,𝑚𝑖𝑛 > − 𝑉𝑇𝐻1 + 𝑽𝑩 − 𝑉𝐺𝑆2 𝑉𝑜𝑢𝑡,𝑚𝑖𝑛 > 𝑽𝑩 − 𝑉𝑇𝐻2 ❑ Choosing small 𝑽𝑩 extends both 𝑽𝒊𝒏 and 𝑽𝒐𝒖𝒕 ranges – Just bias 𝐼𝐵1 in saturation ❑ More on this point when we study operational transconductance amplifiers (OTAs) 07: Cascode Amplifiers 57 vout VB M2 CS CG vin M1 IB1 IB2
Quiz: Rout of Folded Cascode ❑ Assume all transistors have same gm and ro, and neglect body effect. Calculate Rout. 07: Cascode Amplifiers 58 VB2 M2 vin M1 VB1 Rout M3
Quiz: Rout of Folded Cascode ❑ Assume all transistors have same 𝑉 𝑜𝑣 and L, CS and CG have the same bias current, and neglect body effect. Calculate Rout. 07: Cascode Amplifiers 59 VB2 M2 vin M1 VB1 Rout M3
Quiz: Gain of Folded Cascode ❑ Calculate Av = Gm Rout. Assume all transistors have same gm and ro, and neglect body effect. 07: Cascode Amplifiers 60 VB2 M2 vin M1 VB1 M3 vout VB3 VB4 M3 M4 Note: Gm =gm1
Thank you!
References ❑ A. Sedra and K. Smith, “Microelectronic Circuits,” Oxford University Press, 7th ed., 2015 ❑ B. Razavi, “Fundamentals of Microelectronics,” Wiley, 2nd ed., 2014 ❑ B. Razavi, “Design Of Analog CMOS Integrated Circuit,” McGraw- Hill, 2nd ed., 2017 62 07: Cascode Amplifiers
Gain Boosting: Super Transistor ❑ 𝑔𝑚,𝑠𝑢𝑝𝑒𝑟 = 𝐴𝑣𝑔𝑚 ❑ 𝑟𝑜,𝑠𝑢𝑝𝑒𝑟 = 𝑟𝑜 07: Cascode Amplifiers 63 G,super S D Av G gmvgs =(Avgm)vgs,super ro gmbvbs G,super D S B vgs,super vbs
Gain Boosting: Super Transistor ❑ 𝐺𝑚 ≈ 𝑔𝑚,𝑠𝑢𝑝𝑒𝑟 1+𝑔𝑚,𝑠𝑢𝑝𝑒𝑟𝑅𝑠 ≈ 𝐴𝑠𝑔𝑚 1+𝐴𝑠𝑔𝑚𝑅𝑠 ≈ 1 𝑅𝑆 ❑ 𝑅𝑜𝑢𝑡 ≈ 𝑟𝑜 1 + 𝑔𝑚,𝑠𝑢𝑝𝑒𝑟𝑅𝑆 = 𝑟𝑜 1 + 𝐴𝑠𝑔𝑚𝑅𝑆 ❑ 𝐴𝑣 ≈ 𝐴𝑠𝑔𝑚𝑟𝑜 07: Cascode Amplifiers 64 As RS IB vout vin
Super Cascode ❑ A.k.a. regulated cascode or gain boosted cascode ❑ 𝐺𝑚 ≈ 𝑔𝑚1 ❑ 𝑅𝑜𝑢𝑡 = 𝑟𝑜2 1 + 𝑔𝑚2,𝑠𝑢𝑝𝑒𝑟𝑟𝑜1 = 𝑟𝑜2 1 + 𝐴𝑠𝑔𝑚2𝑟𝑜1 ❑ 𝐴𝑣 ≈ 𝐴𝑠 𝑔𝑚1𝑟𝑜1 𝑔𝑚2𝑟𝑜2 ❑ Gain is boosted while preserving headroom (no extra stacking) ❑ But more power and noise 07: Cascode Amplifiers 65 As IB vout vin VB M1 M2
Gain Boosting Implementation ❑ NMOS CS: headroom limitation – 𝑉𝑋 = 𝑉𝐺𝑆3 = 𝑉𝑇𝐻3 + 𝑉𝑜𝑣3 instead of 𝑉𝑜𝑣1 ❑ PMOS CS: M3 in triode – 𝑉𝐷𝐺 = 𝑉𝐺𝑆2 > 𝑉𝑇𝐻 ❑ Folded cascode: M4 provide level shift 07: Cascode Amplifiers 66 As IB1 vout vin M1 M2 M3 IB2 VX IB1 vout vin M1 M2 M3 IB2 VX As VDG>VTH IB1 vin M1 M2 M3 IB2 VX As M4 IB3 VB vout

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LEC 3 New basic_amplifiers.pdffffffffffffffffff

  • 1.
    Lecture 03 Basic AmplifierStages / Cascode Amplifier This is a modified version of Dr Hesham’s Lectures Dr. Hesham A. Omran Analog IC Design ً ‫يل‬ِ ‫ل‬َ‫ق‬ ‫ا‬ ‫َّل‬ِ ‫إ‬ِ ‫م‬ْ‫ل‬ِ ‫ع‬ْ‫ل‬‫ا‬ َ ‫ن‬ِ ‫م‬ ْ‫م‬ُ ‫يت‬ِ ‫ت‬‫و‬ُ‫أ‬‫ا‬َ ‫م‬َ ‫و‬ 11 March 2022 8 ‫شعبان‬ 1443
  • 2.
    Outline ❑ Recapping previouskey results ❑ Basic amplifier operation ❑ Rin/out Shortcuts – Looking from drain – Looking from source ❑ GmRout method ❑ Basic amplifier topologies – Common Source (CS) – Common Gate (CG) – Common Drain (CD) – Source Follower (SF) ❑ Bipolar Junction Transistor (BJT) 06: Basic Amplifier Stages 2
  • 3.
    MOSFET in Saturation ❑The channel is pinched off if the difference between the gate and drain voltages is not sufficient to create an inversion layer 𝑉𝐺𝐷 ≤ 𝑉𝑇𝐻 𝑂𝑅 𝑉𝐷𝑆 ≥ 𝑉 𝑜𝑣 ❑ Square-law (long channel MOS) 𝐼𝐷 = 𝜇𝑛𝐶𝑜𝑥 2 𝑊 𝐿 ⋅ 𝑉 𝑜𝑣 2 1 + 𝜆𝑉𝐷𝑆 𝑉𝑇𝐻 = 𝑉𝑇𝐻0 + 𝛾 2Φ𝐹 + 𝑉𝑆𝐵 − 2Φ𝐹 06: Basic Amplifier Stages 3 n+ n+ G S D p-sub p+ B VGS>VTH VGD<VTH VDS>Vov VSB VGS>VTH VGD<VTH VDS>Vov
  • 4.
    Regions of OperationSummary 06: Basic Amplifier Stages 4 OFF (Subthreshold) 𝑉𝐺𝑆 < 𝑉𝑇𝐻 ON 𝑉𝐺𝑆 > 𝑉𝑇𝐻 Triode 𝑉𝐷𝑆 < 𝑉 𝑜𝑣 Or 𝑉𝐺𝐷 > 𝑉𝑇𝐻 𝐼𝐷 = 𝜇𝐶𝑜𝑥 𝑊 𝐿 𝑉 𝑜𝑣𝑉𝐷𝑆 − 𝑉𝐷𝑆 2 2 Pinch-Off (Saturation) 𝑉𝐷𝑆 ≥ 𝑉 𝑜𝑣 Or 𝑉𝐺𝐷 ≤ 𝑉𝑇𝐻 𝐼𝐷 = 𝜇𝐶𝑜𝑥 2 𝑊 𝐿 𝑉 𝑜𝑣 2 1 + 𝜆𝑉𝐷𝑆
  • 5.
    Low-Frequency Small-Signal Model 𝑔𝑚= 𝜕𝐼𝐷 𝜕𝑉𝐺𝑆 = 𝜇𝐶𝑜𝑥 𝑊 𝐿 𝑉 𝑜𝑣 = 𝜇𝐶𝑜𝑥 𝑊 𝐿 ⋅ 2𝐼𝐷 = 2𝐼𝐷 𝑉 𝑜𝑣 𝑔𝑚𝑏 = 𝜂𝑔𝑚, 𝜂 ≈ 0.1 − 0.25 𝑟𝑜 = 1 𝜕𝐼𝐷/𝜕𝑉𝐷𝑆 = 1 𝜆𝐼𝐷 , 𝜆 ∝ 1 𝐿 06: Basic Amplifier Stages 5 gmvgs ro gmbvbs G D S B vgs vbs
  • 6.
    Outline ❑ Recapping previouskey results ❑ Basic amplifier operation ❑ Rin/out Shortcuts – Looking from drain – Looking from source ❑ Gm Rout method ❑ Basic amplifier topologies – Common Source (CS) – Common Gate (CG) – Common Drain (CD) – Source Follower (SF) ❑ Bipolar Junction Transistor (BJT) 06: Basic Amplifier Stages 6
  • 7.
    Why Amplifiers? ❑ Allthe physical signals in the world around us are analog – Voice, light, temperature, pressure, etc. ❑ We (will) always need an “analog” interface circuit to connect between our physical world and our digital electronics 06: Basic Amplifier Stages 7 Amplifier A/D Digital processing and storage
  • 8.
    Basic Amplifier Topologies CommonSource (CS) Common Gate (CG) Common Drain (CD) Source Follower (SF) Voltage & current amplifier Current buffer Voltage buffer In 𝐺 𝑆 𝐺 Out 𝐷 𝐷 𝑆 06: Basic Amplifier Stages 8 RD vout vin RS RD vin RS vout RD vout vin RS
  • 9.
    MOSFET CS AmplifierExample 06: Basic Amplifier Stages 9 [Sedra/Smith, 2015]
  • 10.
    MOSFET CS AmplifierExample 06: Basic Amplifier Stages 10 [Sedra/Smith, 2015]
  • 11.
    MOSFET CS AmplifierExample 06: Basic Amplifier Stages 11 [Sedra/Smith, 2015]
  • 12.
    MOSFET CS AmplifierExample ❑ 𝑅𝑖𝑛 = 𝑣𝑖𝑛 𝑖𝑖𝑛 = 𝑣𝑖𝑛 𝑖𝑠𝑖𝑔 ❑ 𝑅𝑜𝑢𝑡 = 𝑣𝑥 𝑖𝑥 @ 𝑣𝑠𝑖𝑔 = 0 ❑ 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑔𝑎𝑖𝑛 = 𝐴𝑣 = 𝑣𝑜𝑢𝑡 𝑣𝑠𝑖𝑔 = 𝑣𝑖𝑛 𝑣𝑠𝑖𝑔 ⋅ 𝑣𝑜𝑢𝑡 𝑣𝑖𝑛 = 𝑅𝑖𝑛 𝑅𝑠𝑖𝑔+𝑅𝑖𝑛 ⋅ 𝑣𝑜𝑢𝑡 𝑣𝑖𝑛 ❑ 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑔𝑎𝑖𝑛 = 𝐴𝑖 = 𝑖𝑜𝑢𝑡 𝑖𝑠𝑖𝑔 = 𝑣𝑜𝑢𝑡/𝑅𝐿 𝑣𝑖𝑛/𝑅𝑖𝑛 = 𝑣𝑜𝑢𝑡 𝑣𝑖𝑛 ⋅ 𝑅𝑖𝑛 𝑅𝐿 06: Basic Amplifier Stages 12 Rsig vin RG1 RG2 RD vout M1 vsig Cc1 Cc2 RL Rout Rin isig iout
  • 13.
    Large and SmallSignal Analysis Large Signal Analysis Small Signal Analysis Model Large signal model Small signal model Linearity Non-linear Linear Simulation DC and transient analysis AC analysis Purpose Calculate bias point, signal swing, distortion, etc. Calculate 𝐴𝑣, 𝑅𝑖𝑛, 𝑅𝑜𝑢𝑡, 𝐵𝑊, 𝑒𝑡𝑐. VDC ✓ s.c. IDC ✓ o.c. Capacitor o.c. (in DC) 1/𝜔𝐶 Inductor s.c. (in DC) 𝜔𝐿 06: Basic Amplifier Stages 13
  • 14.
    MOSFET Amplifier AnalysisSteps 1. DC analysis • Coupling and bypass capacitors → open-circuit (o.c) • Calculate Q-point and check operation in saturation 𝑉𝐷𝑆 > 𝑉 𝑜𝑣 2. Calculate small signal parameters 𝑔𝑚, 𝑟𝑜 3. Draw the small signal equivalent circuit • DC voltage source → short-circuit (s.c.) • DC current source → open-circuit (o.c.) • Coupling and bypass capacitors → short-circuit (s.c) 4. Determine the amplifier parameters • Input resistance and output resistance • Voltage gain and current gain 06: Basic Amplifier Stages 14
  • 15.
    Direct Analysis onSchematics ❑ No need to draw the small signal model every time ❑ Just remember we have two VCCSs and 𝑟𝑜 between D and S ❑ If G and B are ac connected, then 𝑣𝑏𝑠 = 𝑣𝑔𝑠 ➔ 𝑔𝑚 and 𝑔𝑚𝑏 add • 𝑔𝑚 → 𝑔𝑚 + 𝑔𝑚𝑏 06: Basic Amplifier Stages 15 gmvgs ro gmbvbs G D S vbs vgs B
  • 16.
    Direct Analysis onSchematics ❑ No need to draw the small signal model every time ❑ Just remember we have two VCCSs and 𝑟𝑜 between D and S ❑ If G and B are ac connected, then 𝑣𝑏𝑠 = 𝑣𝑔𝑠 ➔ 𝑔𝑚 and 𝑔𝑚𝑏 add • 𝑔𝑚 → 𝑔𝑚 + 𝑔𝑚𝑏 06: Basic Amplifier Stages 16 gmvgs + gmbvbs ro B vgs S G D vbs
  • 17.
    Intrinsic Gain 𝑣𝑜𝑢𝑡 =− 𝑔𝑚𝑣𝑖𝑛 𝑟𝑜 |𝐴𝑣| = 𝑣𝑜𝑢𝑡 𝑣𝑖𝑛 = 𝑔𝑚𝑟𝑜 ❑ 𝑔𝑚𝑟𝑜 is the max gain that can be obtained from a single transistor ❑ Common approximations that we usually use 𝑔𝑚𝑟𝑜 ≫ 1 𝑟𝑜 ≫ 1 𝑔𝑚 𝑔𝑚 + 1 𝑟𝑜 ≈ 𝑔𝑚 𝑟𝑜// 1 𝑔𝑚 ≈ 1 𝑔𝑚 06: Basic Amplifier Stages 17 vin vout
  • 18.
    Outline ❑ Recapping previouskey results ❑ Basic amplifier operation ❑ Rin/out Shortcuts • Looking from drain • Looking from source ❑ Gm Rout method ❑ Basic amplifier topologies • Common Source (CS) • Common Gate (CG) • Common Drain (CD) – Source Follower (SF) ❑ Bipolar Junction Transistor (BJT) 06: Basic Amplifier Stages 18
  • 19.
    Rin/out Shortcuts ❑ Findequivalent impedance looking from Gate, Source, and Drain 06: Basic Amplifier Stages 19
  • 20.
    Looking From Drain ❑If G and B are ac connected, then 𝑣𝑏𝑠 = 𝑣𝑔𝑠 ➔ 𝑔𝑚 and 𝑔𝑚𝑏 add ❑ 𝑣𝑔𝑠 = −𝑖𝑥𝑅𝑆 and 𝑔𝑚𝑟𝑜 ≫ 1 ❑ Apply KCL at 𝐷 𝑖𝑥 = 𝑔𝑚 + 𝑔𝑚𝑏 −𝑖𝑥𝑅𝑆 + 𝑣𝑥 − 𝑖𝑥𝑅𝑆 𝑟𝑜 𝑅𝐿𝐹𝐷 = 𝑣𝑥 𝑖𝑥 = 𝑟𝑜 + 𝑔𝑚 + 𝑔𝑚𝑏 𝑟𝑜𝑅𝑆 + 𝑅𝑆 ≈ 𝑟𝑜 1 + 𝑔𝑚 + 𝑔𝑚𝑏 𝑅𝑆 06: Basic Amplifier Stages 20 RS vx RLFD ix ixRS ix ro vgs
  • 21.
    Looking From Drain 𝑅𝐿𝐹𝐷≈ 𝑟𝑜 1 + 𝑔𝑚 + 𝑔𝑚𝑏 𝑅𝑆 ❑ Special case: 𝑅𝑆 = 0 (G and S ac s.c.) ➔ active load 𝑅𝐿𝐹𝐷 ≈ 𝑟𝑜 ❑ Drain is a high-impedance node (H.I.N.) 06: Basic Amplifier Stages 21 RS vx RLFD ix ixRS ix ro
  • 22.
    Active Load (SourceOFF) 06: Basic Amplifier Stages 22 ro ro
  • 23.
    Quiz ❑ Assume M1and M2 have the same 𝑔𝑚 and 𝑟𝑜, 𝑔𝑚𝑟𝑜 ≫ 1, and neglect body effect ❑ Find 𝑅𝑋 06: Basic Amplifier Stages 23 VB1 VB2 RX M2 M1 𝑅𝑋 = 𝑟𝑜1+𝑟𝑜2 + 𝑔𝑚2 + 𝑔𝑚𝑏2 𝑟𝑜1 𝑟𝑜2
  • 24.
    Looking From Source ❑𝑣𝑔𝑠 = −𝑣𝑥, 𝑔𝑚 and 𝑔𝑚𝑏 add, and 𝑔𝑚𝑟𝑜 ≫ 1 ❑ Apply KCL at 𝑆 𝑖𝑥 = 𝑔𝑚 + 𝑔𝑚𝑏 𝑣𝑥 + 𝑣𝑥 − 𝑖𝑥𝑅𝐷 𝑟𝑜 𝑅𝐿𝐹𝑆 = 𝑣𝑥 𝑖𝑥 = 1 𝑔𝑚 + 𝑔𝑚𝑏 + 1 𝑟𝑜 1 + 𝑅𝐷 𝑟𝑜 ≈ 1 𝑔𝑚 + 𝑔𝑚𝑏 1 + 𝑅𝐷 𝑟𝑜 06: Basic Amplifier Stages 24 vx RLFS ix RD ixRD ix ro
  • 25.
    Looking From Source 𝑅𝐿𝐹𝑆≈ 1 𝑔𝑚 + 𝑔𝑚𝑏 1 + 𝑅𝐷 𝑟𝑜 ❑ Special case: 𝑅𝐷 = 0 (G & D ac s.c.) ➔ diode connected 𝑅𝐿𝐹𝑆 ≈ 1 𝑔𝑚 + 𝑔𝑚𝑏 ❑ Source is a low impedance node (L.I.N.) 06: Basic Amplifier Stages 25 vx RLFS ix RD ixRD ix ro
  • 26.
    Diode Connected (SourceAbsorption) ❑ Always in saturation (𝑉𝐷𝑆 = 𝑉𝐺𝑆 > 𝑉 𝑜𝑣) ❑ Body effect: 𝑔𝑚 → 𝑔𝑚 + 𝑔𝑚𝑏 (if G and B are ac connected) 06: Basic Amplifier Stages 26 1/gm 1/gm
  • 27.
    Quiz ❑ Assume M1,M2, and M3 have the same 𝑔𝑚 and 𝑟𝑜, 𝑔𝑚𝑟𝑜 ≫ 1, and neglect body effect ❑ Find 𝑅𝑋 06: Basic Amplifier Stages 27 VB1 VB2 RX M2 M1 VB3 M3 𝑅𝑋 = 1 𝑔𝑚1 + 𝑔𝑚𝑏1 + 1 𝑟𝑜1 1 + 𝑅𝐿𝐹𝐷2 𝑟𝑜1 𝑅𝐿𝐹𝐷2 = 𝑟𝑜2+𝑟𝑜3 + 𝑔𝑚2 + 𝑔𝑚𝑏2 𝑟𝑜2 𝑟𝑜3
  • 28.
    Rin/out Shortcuts Summary 06:Basic Amplifier Stages 28 1 𝑔𝑚 + 𝑔𝑚𝑏 1 + 𝑅𝐷 𝑟𝑜 L.I.N. 𝑟𝑜 1 + 𝑔𝑚 + 𝑔𝑚𝑏 𝑅𝑆 H.I.N. ∞ At low frequencies ONLY Note: Do not use shortcuts if a feedback element exists
  • 29.
    Outline ❑ Recapping previouskey results ❑ Basic amplifier operation ❑ Rin/out Shortcuts – Looking from drain – Looking from source ❑ Gm Rout method ❑ Basic amplifier topologies – Common Source (CS) – Common Gate (CG) – Common Drain (CD) – Source Follower (SF) ❑ Bipolar Junction Transistor (BJT) 06: Basic Amplifier Stages 29
  • 30.
    Amplifier Model ❑ Rin/out 𝑅𝑖𝑛= 𝑣𝑖𝑛 𝑖𝑖𝑛 𝑅𝑜𝑢𝑡 = 𝑣𝑥 𝑖𝑥 @ 𝑣𝑖𝑛 = 0 ❑ O.C. voltage gain (Thevenin model) 𝑣𝑜𝑢𝑡,𝑜𝑐 = 𝐴𝑣𝑣𝑖𝑛 𝐴𝑣 = 𝑣𝑜𝑢𝑡,𝑜𝑐 𝑣𝑖𝑛 06: Basic Amplifier Stages 30 Avvin Rout vin Rin vout iout iin
  • 31.
    Amplifier Model ❑ Transconductance(Norton model) 𝑖𝑜𝑢𝑡,𝑠𝑐 = 𝐺𝑚𝑣𝑖𝑛 𝐺𝑚 = 𝑖𝑜𝑢𝑡,𝑠𝑐 𝑣𝑖𝑛 ❑ Thevenin  Norton 𝐴𝑣𝑣𝑖𝑛 = 𝐺𝑚𝑣𝑖𝑛 𝑅𝑜𝑢𝑡 𝐴𝑣 = 𝑣𝑜𝑢𝑡,𝑜𝑐 𝑣𝑖𝑛 = 𝐺𝑚𝑅𝑜𝑢𝑡 ❑ S.C. Current Gain 𝐴𝑖 = 𝑖𝑜𝑢𝑡,𝑠𝑐 𝑖𝑖𝑛 = 𝐺𝑚𝑅𝑖𝑛 06: Basic Amplifier Stages 31 Avvin Rout vin Rin vout iout iin vin Rin vout iout iin Gmvin Rout
  • 32.
    Why GmRout? 𝑅𝑖𝑛 =𝑣𝑖𝑛/𝑖𝑖𝑛 𝑅𝑜𝑢𝑡 = 𝑣𝑥/𝑖𝑥 @ 𝑣𝑖𝑛 = 0 𝐺𝑚 = 𝑖𝑜𝑢𝑡,𝑠𝑐/𝑣𝑖𝑛 𝐴𝑣 = 𝐺𝑚𝑅𝑜𝑢𝑡 𝐴𝑖 = 𝐺𝑚𝑅𝑖𝑛 ❑ Divide and conquer – Rout simplified: vin=0 – Gm simplified: vout=0 – We already need Rin/out – We can quickly and easily get Rin/out from the shortcuts 06: Basic Amplifier Stages 32 Avvin Rout vin Rin vout iout iin vin Rin vout iout iin Gmvin Rout
  • 33.
    Outline ❑ Recapping previouskey results ❑ Basic amplifier operation ❑ Rin/out Shortcuts – Looking from drain – Looking from source ❑ Gm Rout method ❑ Basic amplifier topologies – Common Source (CS) – Common Gate (CG) – Common Drain (CD) – Source Follower (SF) ❑ Bipolar Junction Transistor (BJT) 06: Basic Amplifier Stages 33
  • 34.
    Basic Amplifier Topologies CommonSource (CS) Common Gate (CG) Common Drain (CD) Source Follower (SF) Voltage & current amplifier Current buffer Voltage buffer In 𝐺 𝑆 𝐺 Out 𝐷 𝐷 𝑆 06: Basic Amplifier Stages 34 RD vout vin RS RD vin RS vout RD vout vin RS
  • 35.
    Common Source (CS) ❑Apply KCL at 𝑆 𝑮𝒎 = 𝒊𝒐𝒖𝒕,𝒔𝒄 𝒗𝒊𝒏 ≈ −𝒈𝒎 𝟏 + 𝒈𝒎 + 𝒈𝒎𝒃 𝑹𝑺 𝑅𝑜𝑢𝑡 ≈ 𝑅𝐷//𝑟𝑜 1 + 𝑔𝑚 + 𝑔𝑚𝑏 𝑅𝑆 𝐴𝑣 = 𝐺𝑚𝑅𝑜𝑢𝑡 ❑ If 𝑅𝐷 is ac o.c.: 𝐴𝑣 = −𝑔𝑚𝑟𝑜 ❑ If 𝑅𝐷 ≪ 𝑅𝐿𝐹𝐷: 𝐴𝑣 ≈ −𝒈𝒎𝑹𝑫 𝟏+ 𝒈𝒎+𝒈𝒎𝒃 𝑹𝑺 ❑ If 𝑅𝑆 = 0: 𝐴𝑣 = −𝑔𝑚 𝑅𝐷//𝑟𝑜 06: Basic Amplifier Stages 35 RD vout vin RS iout,sc -iout,scRS iout,sc ro 𝑖𝑜𝑢𝑡,𝑠𝑐 + 𝑔𝑚 𝑣𝑖𝑛 + 𝑖𝑜𝑢𝑡,𝑠𝑐𝑅𝑆 + 𝑔𝑚𝑏 𝑖𝑜𝑢𝑡,𝑠𝑐𝑅𝑆 + 𝑖𝑜𝑢𝑡,𝑠𝑐𝑅𝑆 𝑟𝑜 = 0 Shorted by o/p
  • 36.
    Common Source (CS) 𝑮𝒎= 𝒊𝒐𝒖𝒕,𝒔𝒄 𝒗𝒊𝒏 ≈ −𝒈𝒎 𝟏 + 𝒈𝒎 + 𝒈𝒎𝒃 𝑹𝑺 𝑅𝑜𝑢𝑡 ≈ 𝑅𝐷//𝑟𝑜 1 + 𝑔𝑚 + 𝑔𝑚𝑏 𝑅𝑆 𝐴𝑣 = 𝐺𝑚𝑅𝑜𝑢𝑡 ❑ If S and B are connected (for PMOS): 𝐴𝑣 ≈ −𝑅𝐷//𝑅𝐿𝐹𝐷 1 𝑔𝑚 + 𝑅𝑆 = − 𝐷𝑟𝑎𝑖𝑛 𝑅𝑒𝑠. 1 𝑔𝑚 + 𝑆𝑜𝑢𝑟𝑐𝑒 𝑅𝑒𝑠. ❑ If 𝑅𝑆 ≫ 1 𝑔𝑚 & 𝑅𝐷 ≪ 𝑅𝐿𝐹𝐷: 𝐴𝑣 ≈ −𝑅𝐷 𝑅𝑆 → 𝐿𝑖𝑛𝑒𝑎𝑟 ❑ 𝑅𝑆 reduces 𝐺𝑚 → Source degeneration – But improves linearity 06: Basic Amplifier Stages 36 RD vout vin RS iout,sc -iout,scRS iout,sc ro
  • 37.
    Common Gate (CG) ❑Apply KCL at 𝐷 𝑖𝑜𝑢𝑡,𝑠𝑐 + 𝑔𝑚 + 𝑔𝑚𝑏 −𝑣𝑖𝑛 − 𝑣𝑖𝑛 𝑟𝑜 = 0 𝑮𝒎 = 𝒊𝒐𝒖𝒕,𝒔𝒄 𝒗𝒊𝒏 ≈ 𝒈𝒎 + 𝒈𝒎𝒃 𝑅𝑜𝑢𝑡 ≈ 𝑅𝐷//𝑟𝑜 (𝒘𝒉𝒚? ) 𝐴𝑣 = 𝐺𝑚𝑅𝑜𝑢𝑡 ❑ If 𝑅𝐷 is ac o.c.: 𝐴𝑣 = 𝑔𝑚 + 𝑔𝑚𝑏 𝑟𝑜 ❑ If 𝑅𝐷 ≪ 𝑟𝑜: 𝐴𝑣 ≈ 𝑔𝑚 + 𝑔𝑚𝑏 𝑹𝑫 ❑ For PMOS connect S and B: 𝑔𝑚 + 𝑔𝑚𝑏 → 𝑔𝑚 ❑ Note that 𝐴𝑖 = 𝐺𝑚𝑅𝑖𝑛 <≈ 𝟏 (Current Buffer) 06: Basic Amplifier Stages 37 RD vout vin RS iout,sc ro
  • 38.
    Common Drain (CD)– Source Follower ❑ Apply KCL at 𝐷 𝑖𝑜𝑢𝑡,𝑠𝑐 − 𝑔𝑚𝑣𝑖𝑛 + 𝑖𝑜𝑢𝑡,𝑠𝑐𝑅𝐷 𝑟𝑜 = 0 𝑮𝒎 = 𝒊𝒐𝒖𝒕,𝒔𝒄 𝒗𝒊𝒏 ≈ 𝒈𝒎 𝟏 + 𝑹𝑫/𝒓𝒐 𝑅𝑜𝑢𝑡 ≈ 𝑅𝑆// 1 𝑔𝑚 + 𝑔𝑚𝑏 1 + 𝑅𝐷 𝑟𝑜 𝐴𝑣 = 𝐺𝑚𝑅𝑜𝑢𝑡 ❑ If 𝑅𝑆 ≫ 𝑅𝐿𝐹𝑆: 𝐴𝑣 ≈ 𝑔𝑚 𝑔𝑚+𝑔𝑚𝑏 < 1 ❑ If 𝑆 and 𝐵 are connected (for PMOS): 𝐴𝑣 ≈ 𝟏 (Voltage buffer) 06: Basic Amplifier Stages 38 RD vout vin RS iout,sc -iout,scRD ro iout,sc
  • 39.
    Summary of BasicTopologies CS CG CD (SF) Voltage & current amplifier Current buffer Voltage buffer Rin ∞ 𝑅𝑆// 1 𝑔𝑚 + 𝑔𝑚𝑏 1 + 𝑅𝐷 𝑟𝑜 ∞ Rout 𝑅𝐷//𝑟𝑜 1 + 𝑔𝑚 + 𝑔𝑚𝑏 𝑅𝑆 𝑅𝐷//𝑟𝑜 𝑅𝑆// 1 𝑔𝑚 + 𝑔𝑚𝑏 1 + 𝑅𝐷 𝑟𝑜 Gm −𝒈𝒎 𝟏 + 𝒈𝒎 + 𝒈𝒎𝒃 𝑹𝑺 𝒈𝒎 + 𝒈𝒎𝒃 𝒈𝒎 𝟏 + 𝑹𝑫/𝒓𝒐 06: Basic Amplifier Stages 39 RD vout vin RS RD vin RS vout RD vout vin RS
  • 40.
    Quiz ❑ The circuitbelow shows a CS amplifier With diode-connected load ❑ Find the gain using Gm Rout (ignore body effect and CLM) – Express the gain in terms of (𝑊/𝐿)1 and (𝑊/𝐿)2 ❑ This is a “linear” CS amplifier 06: Basic Amplifier Stages 40 VB (W/L)1 (W/L)2 vin vout
  • 41.
    Quiz ❑ The circuitbelow shows a complementary CS amplifier (inverter amp) ❑ Find the gain using Gm Rout Gm = - (gm1 + gm2) Rout = ro1 // ro2 06: Basic Amplifier Stages 41 M1 M2 vin vout
  • 42.
    Output Signal Swing 𝑣𝑜𝑢𝑡,𝑚𝑎𝑥= 𝑉𝐷𝐷 − 𝑉𝑜𝑣2 𝑣𝑜𝑢𝑡,𝑚𝑖𝑛 = 𝑉𝑜𝑣1 ❑ Output swing ≈ 𝑉𝐷𝐷 − 2𝑉 𝑜𝑣 06: Basic Amplifier Stages 42 VB vin vout M1 M2
  • 43.
    Outline ❑ Recapping previouskey results ❑ Basic amplifier operation ❑ Rin/out Shortcuts – Looking from drain – Looking from source ❑ Gm Rout method ❑ Basic amplifier topologies – Common Source (CS) – Common Gate (CG) – Common Drain (CD) – Source Follower (SF) ❑ Bipolar Junction Transistor (BJT) 06: Basic Amplifier Stages 43
  • 44.
    Bipolar Junction Transistor(BJT) 06: Basic Amplifier Stages 44 Figure 5.26 (a) Basic common-emitter amplifier circuit. (b) Transfer characteristic of the circuit in (a). The amplifier is biased at a point Q, and a small voltage signal vi is superimposed on the dc bias voltage VBE. The resulting output signal vo appears superimposed on the dc collector voltage VCE. The amplitude of vo is larger than that of vi by the voltage gain Av. Common Emitter (CE)
  • 45.
    BJT (Small SignalModel) 06: Basic Amplifier Stages 45 Π model: T model:
  • 46.
    Rin/out Shortcuts (BJT) 06:Basic Amplifier Stages 46 Note: Do not use shortcuts if a feedback element exists
  • 47.
    Lecture 3 part2 Cascode Amplifiers Integrated Circuits Lab (ICL) Electronics and Communications Eng. Dept. Faculty of Engineering Ain Shams University Dr. Hesham A. Omran Analog IC Design ً ‫يل‬ِ ‫ل‬َ‫ق‬ ‫ا‬ ‫َّل‬ِ ‫إ‬ِ ‫م‬ْ‫ل‬ِ ‫ع‬ْ‫ل‬‫ا‬ َ ‫ن‬ِ ‫م‬ ْ‫م‬ُ ‫يت‬ِ ‫ت‬‫و‬ُ‫أ‬‫ا‬َ ‫م‬َ ‫و‬ 11 March 2022 8 ‫شعبان‬ 1443
  • 48.
    Boosting Voltage Gain ❑The max gain of a single transistor amplifier is the intrinsic gain (body effect may add). 𝐴𝑖 = 𝑔𝑚𝑟𝑜 ❑ 𝑔𝑚𝑟𝑜 is quite small for modern deep submicron technologies. – May be less than 10 ! – But we need high voltage gain to design an op-amp. ❑ How to boost the gain? – CS and CG stages provide voltage gain. 1. Use a cascade of CS stage: CS + CS + CS + … 2. Use a cascode amplifier: CS + CG + CG + … • Single cascode: CS + CG • Double cascode: CS + CG + CG 07: Cascode Amplifiers 48
  • 49.
    Cascode ❑ CS +CG 07: Cascode Amplifiers 49 vout VB vin M1 M2 CS CG
  • 50.
    Cascode Gain UsingGmRout ❑ Transconductance is always related to the input transistor (VCCS) 𝑖𝑜𝑢𝑡,𝑠𝑐 ≈ −𝑔𝑚1𝑣𝑖𝑛 𝐺𝑚 = 𝑖𝑜𝑢𝑡,𝑠𝑐 𝑣𝑖𝑛 ≈ −𝑔𝑚1 Same Gm of CS 𝑅𝑜𝑢𝑡 ≈ 𝑟𝑜2 1 + 𝑔𝑚2 + 𝑔𝑚𝑏2 𝑟𝑜1 ≈ 𝑟𝑜2 𝑔𝑚2 + 𝑔𝑚𝑏2 𝑟𝑜1 Rout significantly boosted 𝐴𝑣 ≈ −𝑔𝑚1𝑟𝑜1 𝑔𝑚2 + 𝑔𝑚𝑏2 𝑟𝑜2 ❑ Assume all 𝑔𝑚 and 𝑟𝑜 are equal and neglect body effect 𝐴𝑣 = − 𝑔𝑚𝑟𝑜 2 07: Cascode Amplifiers 50 vout VB vin M1 M2 iout,sc iout,sc
  • 51.
    Cascode as CS+ CG 𝐶𝑆: 𝑣𝑥 𝑣𝑖𝑛 = −𝑔𝑚1 𝑟𝑜1//∞ 𝐶𝐺: 𝑣𝑜𝑢𝑡 𝑣𝑥 = 𝑔𝑚2 + 𝑔𝑚𝑏2 𝑟𝑜2 𝐴𝑣 = 𝑣𝑥 𝑣𝑖𝑛 ⋅ 𝑣𝑜𝑢𝑡 𝑣𝑥 ≈ −𝑔𝑚1𝑟𝑜1 𝑔𝑚2 + 𝑔𝑚𝑏2 𝑟𝑜2 ❑ Assume all 𝑔𝑚 and 𝑟𝑜 are equal and neglect body effect 𝐴𝑣 = − 𝑔𝑚𝑟𝑜 2 07: Cascode Amplifiers 51 vout VB vin M1 M2 CS CG vx
  • 52.
    Quiz ❑ The circuitbelow shows a double cascode. ❑ Find the voltage gain. Assume all 𝑔𝑚 and 𝑟𝑜 are equal and neglect body effect. 07: Cascode Amplifiers 52 vout VB1 vin M1 M2 VB2 M3
  • 53.
    What if 𝑹𝑫is small? ❑ Is this cascode useful? – Not useful for boosting the gain – But useful for boosting the BW (more about this later) 07: Cascode Amplifiers 53 vout VB vin M1 M2 RD
  • 54.
    Cascode Load ❑ Ifyou want to keep the large Rout, you must use cascode load ❑ Assume all 𝑔𝑚 and 𝑟𝑜 are equal and neglect body effect 𝐴𝑣 = − 𝑔𝑚𝑟𝑜 2 2 ❑ Output swing ≈ 𝑉𝐷𝐷 − 4𝑉 𝑜𝑣 07: Cascode Amplifiers 54 vout VB1 vin M1 M2 VB2 VB3 M3 M4
  • 55.
    Cascode Bias Voltage ❑Keep M1 in sat 𝑉𝐵 > 𝑉𝐺𝑆2 + 𝑉𝑖𝑛,𝑚𝑎𝑥 − 𝑉𝑇𝐻1 𝑉𝑖𝑛,𝑚𝑎𝑥 < 𝑉𝑇𝐻1 + 𝑽𝑩 − 𝑉𝐺𝑆2 ❑ Keep M2 in sat 𝑉𝐵 < 𝑉𝑇𝐻2 + 𝑉𝑜𝑢𝑡,𝑚𝑖𝑛 𝑉𝑜𝑢𝑡,𝑚𝑖𝑛 > 𝑽𝑩 − 𝑉𝑇𝐻2 ❑ Increasing/decreasing 𝑉𝐵 either hurts 𝑽𝒊𝒏 range or 𝑽𝒐𝒖𝒕 range – Input and output ranges are coupled oppositely 07: Cascode Amplifiers 55 vout VB vin M1 M2 𝑉𝐵 < 𝑉𝑜𝑢𝑡,𝑚𝑖𝑛 − 𝑉𝑂𝑉2 + 𝑉𝐺𝑆2
  • 56.
    Telescopic vs FoldedCascode ❑ Telescopic: CS + CG (both NMOS or both PMOS) – Both CS and CG use same bias current ❑ Folded: CS + CG (NMOS-PMOS combination) – The small signal current is folded up or down – Extra bias current is needed – Rout is lower (due to 𝐼𝐵1) – Why is it useful? 07: Cascode Amplifiers 56 vout VB vin M1 M2 CS CG vout VB M2 CS CG vin M1 IB1 IB2
  • 57.
    Folded Cascode ❑ Inputand output ranges are NOT coupled oppositely 𝑉𝑖𝑛,𝑚𝑖𝑛 > − 𝑉𝑇𝐻1 + 𝑽𝑩 − 𝑉𝐺𝑆2 𝑉𝑜𝑢𝑡,𝑚𝑖𝑛 > 𝑽𝑩 − 𝑉𝑇𝐻2 ❑ Choosing small 𝑽𝑩 extends both 𝑽𝒊𝒏 and 𝑽𝒐𝒖𝒕 ranges – Just bias 𝐼𝐵1 in saturation ❑ More on this point when we study operational transconductance amplifiers (OTAs) 07: Cascode Amplifiers 57 vout VB M2 CS CG vin M1 IB1 IB2
  • 58.
    Quiz: Rout ofFolded Cascode ❑ Assume all transistors have same gm and ro, and neglect body effect. Calculate Rout. 07: Cascode Amplifiers 58 VB2 M2 vin M1 VB1 Rout M3
  • 59.
    Quiz: Rout ofFolded Cascode ❑ Assume all transistors have same 𝑉 𝑜𝑣 and L, CS and CG have the same bias current, and neglect body effect. Calculate Rout. 07: Cascode Amplifiers 59 VB2 M2 vin M1 VB1 Rout M3
  • 60.
    Quiz: Gain ofFolded Cascode ❑ Calculate Av = Gm Rout. Assume all transistors have same gm and ro, and neglect body effect. 07: Cascode Amplifiers 60 VB2 M2 vin M1 VB1 M3 vout VB3 VB4 M3 M4 Note: Gm =gm1
  • 61.
  • 62.
    References ❑ A. Sedraand K. Smith, “Microelectronic Circuits,” Oxford University Press, 7th ed., 2015 ❑ B. Razavi, “Fundamentals of Microelectronics,” Wiley, 2nd ed., 2014 ❑ B. Razavi, “Design Of Analog CMOS Integrated Circuit,” McGraw- Hill, 2nd ed., 2017 62 07: Cascode Amplifiers
  • 63.
    Gain Boosting: SuperTransistor ❑ 𝑔𝑚,𝑠𝑢𝑝𝑒𝑟 = 𝐴𝑣𝑔𝑚 ❑ 𝑟𝑜,𝑠𝑢𝑝𝑒𝑟 = 𝑟𝑜 07: Cascode Amplifiers 63 G,super S D Av G gmvgs =(Avgm)vgs,super ro gmbvbs G,super D S B vgs,super vbs
  • 64.
    Gain Boosting: SuperTransistor ❑ 𝐺𝑚 ≈ 𝑔𝑚,𝑠𝑢𝑝𝑒𝑟 1+𝑔𝑚,𝑠𝑢𝑝𝑒𝑟𝑅𝑠 ≈ 𝐴𝑠𝑔𝑚 1+𝐴𝑠𝑔𝑚𝑅𝑠 ≈ 1 𝑅𝑆 ❑ 𝑅𝑜𝑢𝑡 ≈ 𝑟𝑜 1 + 𝑔𝑚,𝑠𝑢𝑝𝑒𝑟𝑅𝑆 = 𝑟𝑜 1 + 𝐴𝑠𝑔𝑚𝑅𝑆 ❑ 𝐴𝑣 ≈ 𝐴𝑠𝑔𝑚𝑟𝑜 07: Cascode Amplifiers 64 As RS IB vout vin
  • 65.
    Super Cascode ❑ A.k.a.regulated cascode or gain boosted cascode ❑ 𝐺𝑚 ≈ 𝑔𝑚1 ❑ 𝑅𝑜𝑢𝑡 = 𝑟𝑜2 1 + 𝑔𝑚2,𝑠𝑢𝑝𝑒𝑟𝑟𝑜1 = 𝑟𝑜2 1 + 𝐴𝑠𝑔𝑚2𝑟𝑜1 ❑ 𝐴𝑣 ≈ 𝐴𝑠 𝑔𝑚1𝑟𝑜1 𝑔𝑚2𝑟𝑜2 ❑ Gain is boosted while preserving headroom (no extra stacking) ❑ But more power and noise 07: Cascode Amplifiers 65 As IB vout vin VB M1 M2
  • 66.
    Gain Boosting Implementation ❑NMOS CS: headroom limitation – 𝑉𝑋 = 𝑉𝐺𝑆3 = 𝑉𝑇𝐻3 + 𝑉𝑜𝑣3 instead of 𝑉𝑜𝑣1 ❑ PMOS CS: M3 in triode – 𝑉𝐷𝐺 = 𝑉𝐺𝑆2 > 𝑉𝑇𝐻 ❑ Folded cascode: M4 provide level shift 07: Cascode Amplifiers 66 As IB1 vout vin M1 M2 M3 IB2 VX IB1 vout vin M1 M2 M3 IB2 VX As VDG>VTH IB1 vin M1 M2 M3 IB2 VX As M4 IB3 VB vout