Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
28 June 2020 ِ وما أُوتِيتم ن ا ْلعِْلِم إِاَّل قَلِ ًيل م ََ َ ُْ 1441 ذو القعدة7 Analog IC Design Lecture 15 Negative Feedback Dr. Hesham A. Omran Integrated Circuits Laboratory (ICL) Electronics and Electrical Communications Eng. (EECE) Dept. Faculty of Engineering Ain Shams University Outline ❑ ❑ ❑ ❑ ❑ ❑ ❑ Recapping previous key results General feedback system Loop gain Why negative feedback? Stability of feedback system Root locus and Bode plot Phase and gain margin 15: Negative Feedback 2 Outline ❑ ❑ ❑ ❑ ❑ ❑ ❑ Recapping previous key results General feedback system Loop gain Why negative feedback? Stability of feedback system Root locus and Bode plot Phase and gain margin 15: Negative Feedback 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 ≥ 𝑉𝑜𝑣 ❑ Square-law (long channel MOS) VSG > |VTH| 𝐼𝐷 = 𝜇𝑛 𝐶𝑜𝑥 𝑊 2 𝐿 2 1 + 𝜆𝑉 ⋅ 𝑉𝑜𝑣 𝐷𝑆 VSD > |Vov| 𝑉𝑆𝐵 ↑ ⇒ 𝑉𝑇𝐻 ↑ B VSB VGS>VTH VDG < |VTH| G VGD<VTH S D VGD < VTH VDS > Vov p+ n+ n+ p-sub 15: Negative Feedback VDS>Vov VGS > VTH 4 Regions of Operation Summary OFF (Subthreshold) 𝑉𝐺𝑆 < 𝑉𝑇𝐻 Triode 𝑉𝐷𝑆 < 𝑉𝑜𝑣 Or 𝑉𝐺𝐷 > 𝑉𝑇𝐻 2 𝑊 𝑉𝐷𝑆 𝐼𝐷 = 𝜇𝐶𝑜𝑥 𝑉 𝑉 − 𝐿 𝑜𝑣 𝐷𝑆 2 15: Negative Feedback ON 𝑉𝐺𝑆 > 𝑉𝑇𝐻 Pinch-Off (Saturation) 𝑉𝐷𝑆 ≥ 𝑉𝑜𝑣 Or 𝑉𝐺𝐷 ≤ 𝑉𝑇𝐻 𝜇𝐶𝑜𝑥 𝑊 2 𝐼𝐷 = 𝑉 1 + 𝜆𝑉𝐷𝑆 2 𝐿 𝑜𝑣 5 High Frequency Small Signal Model 𝑔𝑚 = 𝜕𝐼𝐷 𝜕𝑉𝐺𝑆 = 𝜇𝐶𝑜𝑥 𝑊 𝑉 𝐿 𝑜𝑣 𝑔𝑚𝑏 = 𝜂𝑔𝑚 𝑟𝑜 = 1 𝜕𝐼𝐷 /𝜕𝑉𝐷𝑆 = 𝑉𝐴 𝐼𝐷 = = 𝜇𝐶𝑜𝑥 𝑊 𝐿 ⋅ 2𝐼𝐷 = 2𝐼𝐷 𝑉𝑜𝑣 𝜂 ≈ 0.1 − 0.25 1 𝜆𝐼𝐷 𝑉𝐴 ∝ 𝐿 ↔ 𝜆 ∝ 𝐶𝑔𝑏 ≈ 0 1 𝐿 𝐶𝑔𝑠 ≫ 𝐶𝑔𝑑 𝑉𝐷𝑆 ↑ 𝑉𝐴 ↑ 𝐶𝑠𝑏 > 𝐶𝑑𝑏 Cgd G Cgb D Cgs Csb gmvgs gmbvbs ro Cdb S B 15: Negative Feedback 6 Rin/out Shortcuts Summary 𝑟𝑜 1 + 𝑔𝑚 + 𝑔𝑚𝑏 𝑅𝑆 H.I.N. ∞ At low frequencies ONLY 15: Negative Feedback 1 𝑅𝐷 1+ 𝑔𝑚 + 𝑔𝑚𝑏 𝑟𝑜 L.I.N. 7 Summary of Basic Topologies CS CG CD (SF) RD RD RD vout vout vin vin vin RS RS Voltage & current amplifier Voltage amplifier Current buffer 𝑅𝑆 || RS Voltage buffer Current amplifier 1 𝑅𝐷 1+ 𝑔𝑚 + 𝑔𝑚𝑏 𝑟𝑜 Rin ∞ Rout 𝑅𝐷 ||𝑟𝑜 1 + 𝑔𝑚 + 𝑔𝑚𝑏 𝑅𝑆 𝑅𝐷 ||𝑟𝑜 Gm −𝒈𝒎 𝟏 + 𝒈𝒎 + 𝒈𝒎𝒃 𝑹𝑺 𝒈𝒎 + 𝒈𝒎𝒃 15: Negative Feedback vout ∞ 𝑅𝑆 || 1 𝑅𝐷 1+ 𝑔𝑚 + 𝑔𝑚𝑏 𝑟𝑜 𝒈𝒎 𝟏 + 𝑹𝑫 /𝒓𝒐 8 Differential Amplifier Pseudo Diff Amp 𝑨𝒗𝒅 Diff Pair (w/ ideal CS) −𝑔𝑚 𝑅𝐷 −𝑔𝑚 𝑅𝐷 −𝑔𝑚 𝑅𝐷 𝑨𝒗𝑪𝑴 −𝑔𝑚 𝑅𝐷 0 −𝑔𝑚 𝑅𝐷 1 + 2 𝑔𝑚 + 𝑔𝑚𝑏 𝑅𝑆𝑆 𝑨𝒗𝒅 /𝑨𝒗𝑪𝑴 1 ∞ 2 𝑔𝑚 + 𝑔𝑚𝑏 𝑅𝑆𝑆 ≫1 𝐴𝑣𝐶𝑀2𝑑 𝑣𝑜𝑑 Δ𝑅𝐷 Δ𝑔𝑚 𝑅𝐷 = ≈ + 𝑣𝑖𝐶𝑀 2𝑅𝑆𝑆 2𝑔𝑚1,2 𝑅𝑆𝑆 𝐶𝑀𝑅𝑅 = 15: Negative Feedback Diff Pair (w/ 𝑹𝑺𝑺 ) 𝐴𝑣𝑑 𝐴𝑣𝐶𝑀2𝑑 9 Op-Amp ❑ An op-amp is simply a high gain differential amplifier ▪ The gain can be increased by using cascodes and multi-stage amplification ❑ The diff amp is a key block in many analog and RF circuits ▪ DEEP understanding of diff amp is ESSENTIAL M3 Vin+ Vin- Vout Vout Vin+ M1 VB 15: Negative Feedback M4 M2 Vin- M5 10 Op-Amp vs OTA ❑ In short, an OTA is an op-amp without an output stage (buffer) ❑ Some designers just use op-amp name and symbol for both Rout OTA LOW HIGH Rout vin Model Op-amp iin Rin Avvin iout vout iout vin iin Rin Gmvin vout Rout Diff input, SE output Fully diff 15: Negative Feedback 11 ∗ V-star 𝑽 ❑ V-star 𝑉 ∗ is inspired by 𝑉𝑜𝑣 but calculated from actual simulation data 2𝐼𝐷 2𝐼𝐷 2 ∗ 𝑔𝑚 = ∗ ↔ 𝑉 = = 𝑉 𝑔𝑚 𝑔𝑚 /𝐼𝐷 ❑ Figures-of-merit in terms of 𝑉 ∗ 2𝐼𝐷 1 2 𝑔𝑚 𝑟𝑜 = ∗ ⋅ = ∗ 𝑉 𝜆𝐼𝐷 𝜆𝑉 𝑔𝑚 1 2𝐼𝐷 1 𝑓𝑇 = = ⋅ ∗ ⋅ 2𝜋𝐶𝑔𝑔 2𝜋 𝑉 𝐶𝑔𝑔 𝑔𝑚 2 = ∗ 𝐼𝐷 𝑉 ❑ The boundary between weak and strong inversion (𝑛 = 1.2 → 1.5) 𝑉𝑜𝑣 𝑆𝐼 = 𝑉 ∗ 𝑊𝐼 = 2𝑛𝑉𝑇 ≈ 60 → 80𝑚𝑉 15: Negative Feedback 12 Outline ❑ ❑ ❑ ❑ ❑ ❑ ❑ Recapping previous key results General feedback system Loop gain Why negative feedback? Stability of feedback system Root locus and Bode plot Phase and gain margin 15: Negative Feedback 13 General Feedback System Vin 15: Negative Feedback + _+ Verr Feedforward (Actuator) AOL Vfb Feedback (Sensor) β Vout 14 General Feedback System ❑ Error signal = 𝑉𝑒𝑟𝑟 = 𝑉𝑖𝑛 − 𝑉𝑓𝑏 ❑ Open loop (OL) gain = 𝐴𝑂𝐿 = ❑ Feedback factor = 𝛽 = 𝑉𝑜𝑢𝑡 𝑉𝑒𝑟𝑟 𝑉𝑓𝑏 𝑉𝑜𝑢𝑡 ❑ Closed loop (CL) gain = 𝐴𝐶𝐿 = Vin 15: Negative Feedback ≫1 + _+ 𝑉𝑜𝑢𝑡 𝑉𝑖𝑛 Verr Feedforward (Actuator) AOL Vfb Feedback (Sensor) β Vout 15 Closed-loop Gain 𝑉𝑜𝑢𝑡 = 𝐴𝑂𝐿 𝑉𝑖𝑛 − 𝑉𝑓𝑏 = 𝐴𝑂𝐿 𝑉𝑖𝑛 − 𝛽𝑉𝑜𝑢𝑡 𝑉𝑜𝑢𝑡 𝐴𝑂𝐿 𝐴𝑂𝐿 𝐴𝐶𝐿 = = = 𝑉𝑖𝑛 1 + 𝛽𝐴𝑂𝐿 1 + 𝐿𝐺 ❑ Loop gain = 𝐿𝐺 = 𝛽𝐴𝑂𝐿 ≫ 1 1 𝐴𝐶𝐿 ≈ 𝛽 ❑ Closed-loop gain is independent of open-loop gain! Vin 15: Negative Feedback + _+ Verr Feedforward (Actuator) AOL Vfb Feedback (Sensor) β Vout 16 Error Signal 𝑉𝑒𝑟𝑟 = 𝑉𝑖𝑛 − 𝑉𝑓𝑏 = 𝑉𝑖𝑛 − 𝛽𝑉𝑜𝑢𝑡 = 𝑉𝑖𝑛 − 𝛽𝐴𝑂𝐿 𝑉𝑒𝑟𝑟 𝑉𝑖𝑛 𝑉𝑖𝑛 𝑉𝑒𝑟𝑟 = = 1 + 𝛽𝐴𝑂𝐿 1 + 𝐿𝐺 ❑ Loop gain = 𝐿𝐺 = 𝛽𝐴𝑂𝐿 ≫ 1 𝑉𝑖𝑛 𝑉𝑒𝑟𝑟 = →0 1 + 𝐿𝐺 ❑ Negative feedback loop works to minimize the error signal Vin 15: Negative Feedback + _+ Verr Feedforward (Actuator) AOL Vfb Feedback (Sensor) β Vout 17 Feedback Example ❑ Op-amp functions: (1) subtraction and (2) amplification ❑ The network 𝑅1 and 𝑅2 functions: (1) sensing the output voltage and (2) providing a 𝑅2 feedback factor 𝛽 = 𝑅1 +𝑅2 𝐴𝐶𝐿 Vin + 𝑉𝑜𝑢𝑡 𝐴𝑂𝐿 = = = 𝑉𝑖𝑛 1 + 𝛽𝐴𝑂𝐿 1 + _+ Verr Vfb AOL β 𝐴𝑂𝐿 𝑅1 + 𝑅2 𝑅1 ≈ =1+ 𝑅2 𝑅2 𝑅2 ⋅ 𝐴𝑂𝐿 𝑅1 + 𝑅2 Vout Vin Verr AOL Vfb Vout R1 β R2 15: Negative Feedback 18 Outline ❑ ❑ ❑ ❑ ❑ ❑ ❑ Recapping previous key results General feedback system Loop gain Why negative feedback? Stability of feedback system Root locus and Bode plot Phase and gain margin 15: Negative Feedback 19 Loop Gain ❑ Deactivate the input → Break the loop → Apply a test source → Calculate the gain around the loop ❑ Loop gain = 𝐿𝐺 = − 𝑉𝑜𝑢𝑡 𝑉𝑡 = 𝛽𝐴𝑂𝐿 ❑ A.k.a. loop transmission, return ratio … ❑ But biasing/loading changes when we break the loop! ▪ Make sure dc biasing is properly set ▪ Add a dummy load Vin = 0 + + Verr Vout AOL _ Vfb 15: Negative Feedback β Vt 20 Loop Gain ❑ Modern circuit simulators can compute the loop gain without explicitly breaking the loop ▪ Use stability (STB) analysis ▪ Insert a 0V dc voltage source or iprobe in the loop • Polarity matters for Eldo, but not for Spectre ▪ Loop gain = 𝐿𝐺 = 𝛽𝐴𝑂𝐿 is calculated by the simulator Vin + + Verr AOL Vout _ V=0 Vfb 15: Negative Feedback β 21 Outline ❑ ❑ ❑ ❑ ❑ ❑ ❑ Recapping previous key results General feedback system Loop gain Why negative feedback? Stability of feedback system Root locus and Bode plot Phase and gain margin 15: Negative Feedback 22 Why Negative Feedback? ❑ We use a very high gain amplifier 𝐴𝑂𝐿 , but end up with a much smaller gain 𝐴𝐶𝐿 𝐴𝑂𝐿 1 = ≈ 1+𝛽𝐴𝑂𝐿 𝛽 ❑ We can design high gain amplifiers, but we really do not need all that gain ❑ High gain is the balance that we use to buy other useful properties ❑ Negative feedback useful properties 1. Gain Desensitization → Stable, linear, and accurate gain 2. Bandwidth Extension 3. Modification of I/O Impedances 15: Negative Feedback 23 Gain Desensitization 𝐴𝐶𝐿 𝐴𝑂𝐿 1 = = 1 + 𝛽𝐴𝑂𝐿 𝛽 1 1 𝛽𝐴𝑂𝐿 1 1 1 1 ≈ 1− = 1− 𝛽 𝛽𝐴𝑂𝐿 𝛽 𝐿𝐺 +1 ❑ Static gain error 𝐴𝐶𝐿,𝑖𝑑𝑒𝑎𝑙 − 𝐴𝐶𝐿,𝑎𝑐𝑡𝑢𝑎𝑙 1 1 𝜖𝑠 = ≈ = 𝐴𝐶𝐿,𝑖𝑑𝑒𝑎𝑙 𝛽𝐴𝑂𝐿 𝐿𝐺 ❑ 𝐴𝑂𝐿 varies due to PVT, load, and input signal variations ❑ 𝐴𝐶𝐿 almost independent of 𝐴𝑂𝐿 (if 𝐿𝐺 ≫ 1) ▪ Independent of PVT: stable and robust ▪ Independent of load: stable and robust ▪ Independent of input level: linear 15: Negative Feedback 24 Gain Desensitization ❑ In IC design, we cannot control absolute values due to PVT, load, and input signal variations ❑ But we can precisely control ratios of MATCHED components 𝐴𝐶𝐿 𝑌 𝐴𝑂𝐿 = = = 𝑋 1 + 𝛽𝐴𝑂𝐿 1 + ❑ 𝑅1 = 3𝑅 𝑎𝑛𝑑 𝑅2 = 𝑅 ⇒ 𝐴𝐶𝐿 𝐴𝑂𝐿 𝑅1 + 𝑅2 𝑅1 ≈ =1+ 𝑅2 𝑅2 𝑅2 ⋅ 𝐴𝑂𝐿 𝑅1 + 𝑅2 = 4 ⇒ Stable, linear, and accurate Vin Verr AOL Vfb Vout R1 β R2 15: Negative Feedback [Razavi, 2017] 25 Bandwidth Extension ❑ Assume the op-amp (OL) is a first order system 𝐴𝑂𝐿 𝑠 = 𝐴𝑂𝐿𝑜 𝑠 1+ 𝜔𝑝,𝑂𝐿 ❑ The CL transfer function is also a first order system 𝐴𝐶𝐿 𝐴𝑂𝐿𝑜 𝐴𝑂𝐿 𝑠 𝐴𝐶𝐿𝑜 1 + 𝛽𝐴𝑂𝐿𝑜 𝑠 = = = 𝑠 𝑠 1 + 𝛽𝐴𝑂𝐿 𝑠 1+ 1+ 𝜔𝑝,𝐶𝐿 1 + 𝛽𝐴𝑂𝐿𝑜 𝜔𝑝,𝑂𝐿 ▪ But the pole is at a much higher frequency 𝜔𝑝,𝐶𝐿 = 1 + 𝐿𝐺𝑜 𝜔𝑃,𝑂𝐿 ❑ CL DC gain reduced by 1 + 𝐿𝐺𝑜 ❑ CL bandwidth extended by 1 + 𝐿𝐺𝑜 ❑ GBW (and UGF) remains constant 15: Negative Feedback 26 Bandwidth Extension ❑ CL DC gain reduced by 1 + 𝐿𝐺𝑜 ❑ CL bandwidth extended by 1 + 𝐿𝐺𝑜 ❑ GBW (and UGF) remains constant 𝐴𝐶𝐿 𝐴𝑂𝐿𝑜 𝐴𝐶𝐿𝑜 1 + 𝐿𝐺𝑜 𝑠 = = 𝑠 𝑠 1+ 1+ 𝜔𝑝,𝐶𝐿 1 + 𝐿𝐺𝑜 𝜔𝑝,𝑂𝐿 𝐴𝑂𝐿 𝐴𝑂𝐿𝑜 1 + 𝛽𝐴𝑂𝐿𝑜 𝜔𝑝,𝑂𝐿 𝜔𝑢 𝜔𝑝,𝑂𝐿 1 + 𝛽𝐴𝑂𝐿𝑜 15: Negative Feedback 27 Modification of I/O Impedances ❑ Shunt sensing/mixing → R decreases ❑ Series sensing/mixing → R increases 𝑨𝑶𝑳 𝑨𝑶𝑳 𝜷 𝜷 𝑨𝑶𝑳 𝑨𝑶𝑳 𝜷 15: Negative Feedback 𝜷 [Razavi, 2017] 28 The Price We Pay ❑ Negative feedback useful properties 1. Gain Desensitization → Stable, linear, and accurate gain 2. Bandwidth Extension 3. Modification of I/O Impedances ❑ The price we pay to buy these useful properties 1. Gain reduction 2. The risk of instability 15: Negative Feedback 29 Outline ❑ ❑ ❑ ❑ ❑ ❑ ❑ Recapping previous key results General feedback system Loop gain Why negative feedback? Stability of feedback system Root locus and Bode plot Phase and gain margin 15: Negative Feedback 30 Stability of Feedback System 𝐴𝐶𝐿 𝑉𝑜𝑢𝑡 𝑠 𝐴𝑂𝐿 𝑠 𝑠 = = 𝑉𝑖𝑛 𝑠 1 + 𝛽𝐴𝑂𝐿 𝑠 ❑ Barkhausen’s Oscillation Criteria 𝛽𝐴𝑂𝐿 𝑠 = 1 ∠𝛽𝐴𝑂𝐿 𝑠 = −180 Vin + + _ Verr Vfb 15: Negative Feedback AOL Vout β 31 Stable vs Unstable System: Pole-Zero Plot Laplace domain 1 𝑠−𝑎 15: Negative Feedback Time domain 𝑒 𝑎𝑡 [Razavi, 2017] 32 Stable vs Unstable System: Bode Plot ❑ Gain crossover frequency (GX): @ 𝛽𝐴𝑂𝐿 𝑠 = 1 ▪ Same as 𝜔𝑢 ❑ Phase crossover frequency (PX): @ ∠𝛽𝐴𝑂𝐿 𝑠 = −180 ❑ For a stable system: GX < PX 𝟐𝟎 log 𝜷𝑨𝑶𝑳 𝝎 𝟐𝟎 log 𝜷𝑨𝑶𝑳 𝝎 GX 𝟎 𝒅𝑩 ∠𝜷𝑨𝑶𝑳 𝝎 −𝟏𝟖𝟎𝒐 15: Negative Feedback PX 𝝎 𝝎 𝟎 𝒅𝑩 ∠𝜷𝑨𝑶𝑳 𝝎 GX 𝝎 PX 𝝎 −𝟏𝟖𝟎𝒐 33 Effect of Feedback Factor 𝜷 ❑ We assume 𝛽 is independent of frequency ▪ ∠𝛽𝐴𝑂𝐿 is independent of 𝛽 → PX is independent of 𝛽 ❑ Increasing 𝛽 shifts mag up → GX increases → bad for stability ❑ Worst-case stability corresponds to 𝛽 = 1 → 𝛽𝐴𝑂𝐿 = 𝐴𝑂𝐿 ▪ Unity-gain feedback → buffer → smallest CL gain 𝟐𝟎 log 𝜷𝑨𝑶𝑳 𝝎 𝟎 𝒅𝑩 ∠𝜷𝑨𝑶𝑳 𝝎 𝝎 PX 𝝎 −𝟏𝟖𝟎𝒐 15: Negative Feedback 34 Outline ❑ ❑ ❑ ❑ ❑ ❑ ❑ Recapping previous key results General feedback system Loop gain Why negative feedback? Stability of feedback system Root locus and Bode plot Phase and gain margin 15: Negative Feedback 35 Single-Pole System: Root Locus ❑ Note that 𝛽 does not affect poles of 𝐿𝐺 (assuming 𝛽 is real!) ▪ But it affects poles of 𝐴𝐶𝐿 : Roots of the characteristic equation 1 + 𝛽𝐴𝑂𝐿 ❑ The locus exists on real axis to the left of an odd number of poles and zeros. ❑ The locus starts at the open-loop poles and ends at the open-loop zeros or at infinity. ❑ For first-order system, pole always in LHP: Unconditionally stable jω β↑ β=0 σ 15: Negative Feedback 36 Single-Pole System: Bode Plot 𝐴𝐶𝐿 𝐴𝑂𝐿𝑜 𝐴𝑂𝐿 𝑠 𝐴𝐶𝐿𝑜 1 + 𝛽𝐴𝑂𝐿𝑜 𝑠 = = = 𝑠 𝑠 1 + 𝛽𝐴𝑂𝐿 𝑠 1+ 1+ 𝜔𝑝,𝐶𝐿 1 + 𝛽𝐴𝑂𝐿𝑜 𝜔𝑝,𝑂𝐿 𝟐𝟎 log 𝜷𝑨𝑶𝑳 𝝎 ∠𝜷𝑨𝑶𝑳 𝝎 15: Negative Feedback [Razavi, 2017] 37 Two-Pole System: Root Locus ❑ Poles always in LHP: Unconditionally stable 𝐴𝐶𝐿𝑜 𝐴𝐶𝐿𝑜 𝐴𝐶𝐿 𝑠 = = 2 1 𝑠 𝑠 𝑠 𝑠2 1+ + 1 + 2𝜁 + 𝑄 𝜔𝑜 𝜔𝑜2 𝜔𝑜 𝜔𝑜2 ▪ 𝜁 > 1 (𝑄 < 0.5): Overdamped (real and distinct CL poles) ▪ 𝜁 = 1 (𝑄 = 0.5): Critical damped (real and equal CL poles) ▪ 𝜁 < 1 (𝑄 > 0.5): Underdamped (complex conjugate CL poles) 15: Negative Feedback [Sedra/Smith, 2015] & [Razavi, 2017] 38 Two-Pole System: Bode Plot ❑ Phase shift always < 180𝑜 ▪ Unconditionally stable ❑ 𝜁 > 1 (𝑄 < 0.5): Overdamped ❑ 𝜁 = 1 (𝑄 = 0.5): Critical damped ❑ 𝜁 < 1 (𝑄 > 0.5): Underdamped ▪ Overshoot in step response 𝟐𝟎 log 𝜷𝑨𝑶𝑳 𝝎 ❑ 𝜁 < 1/ 2 = 0.707 ▪ 𝑄 > 1/ 2 = 0.707 ▪ Peaking in frequency response ▪ Ringing in step response 15: Negative Feedback ∠𝜷𝑨𝑶𝑳 𝝎 𝑨𝑪𝑳 𝝎 [Razavi, 2017] 39 Three-Pole System: Root Locus ❑ Poles cross 𝑗𝜔 axis at a specific value of 𝛽 = 𝛽𝑐𝑟𝑖𝑡 ▪ The system becomes unstable ❑ Pole are NOT always in LHP ▪ Conditionally stable: 𝛽 < 𝛽𝑐𝑟𝑖𝑡 jω β↑ σ 15: Negative Feedback 40 Three-Pole System: Bode Plot ❑ Instability (oscillation) condition can be satisfied 𝟐𝟎 log 𝜷𝑨𝑶𝑳 𝝎 ∠𝜷𝑨𝑶𝑳 𝝎 𝑨𝑪𝑳 𝝎 15: Negative Feedback [Razavi, 2017] 41 Stability Summary ❑ First-order system ▪ Unconditionally stable ❑ Third order system ▪ Conditionally stable: Set 𝛽 < 𝛽𝑐𝑟𝑖𝑡 ❑ Second-order system ▪ Unconditionally stable ▪ But may suffer from CL peaking/ringing if close to oscillation condition ▪ How much margin is needed? 15: Negative Feedback 42 Outline ❑ ❑ ❑ ❑ ❑ ❑ ❑ Recapping previous key results General feedback system Loop gain Why negative feedback? Stability of feedback system Root locus and Bode plot Phase and gain margin 15: Negative Feedback 43 Phase and Gain Margin 𝑃𝑀 = 180𝑜 − ∠𝛽𝐴𝑂𝐿 𝐺𝑋 = 180𝑜 − tan−1 𝜔𝑢 𝜔𝑢 −1 − tan 𝜔𝑝1 𝜔𝑝2 𝐺𝑀 = 0 − 20 𝑙𝑜𝑔 𝛽𝐴𝑂𝐿 𝑃𝑋 𝟐𝟎 log 𝜷𝑨𝑶𝑳 𝝎 GX 𝟎 𝒅𝑩 GM 𝝎 ∠𝜷𝑨𝑶𝑳 𝝎 PX 𝝎 −𝟏𝟖𝟎𝒐 PM 15: Negative Feedback 44 Phase Margin (PM) 𝑃𝑀 = 90𝑜 − tan−1 𝜔𝑢 𝜔𝑝2 PM > 0 → stable But low PM means: → frequency domain peaking → time domain ringing 𝜷𝑨𝑶𝑳 𝝎 ∠𝜷𝑨𝑶𝑳 𝝎 𝑨𝑪𝑳 𝝎 Frequency domain peaking → noise amplification Time domain ringing → poor settling time 15: Negative Feedback [Razavi, 2017] 45 Phase Margin (PM) 𝑃𝑀 = 90𝑜 − tan−1 𝜔𝑢 𝜔𝑝2 PM > 0 → stable But low PM means: → frequency domain peaking → time domain ringing 𝜷𝑨𝑶𝑳 𝝎 ∠𝜷𝑨𝑶𝑳 𝝎 𝑨𝑪𝑳 𝝎 Frequency domain peaking → noise amplification Time domain ringing → poor settling time 15: Negative Feedback [Razavi, 2017] 46 Phase Margin: Ultimate GBW ❑ If 𝜔𝑝2 = 𝜔𝑢 : PM = 45o ▪ Typically inadequate (peaking/ringing) ❑ Thus 𝜔𝑝2 should be > 𝜔𝑢 → 𝜔𝑝1 ≪ 𝜔𝑢 < 𝜔𝑝2 ▪ 𝜔𝑝1 defines OL BW and 𝜔𝑝2 defines ultimate GBW (max CL BW) 𝟐𝟎 log 𝜷𝑨𝑶𝑳 𝝎 Frequency domain peaking → noise amplification Time domain ringing → poor settling time GX 𝝎 ∠𝜷𝑨𝑶𝑳 𝝎 𝝎 PM 15: Negative Feedback 47 Asymptotic vs Actual UGF ❑ For 𝜔 < 𝜔𝑢 the Bode plot is similar to a 1st order system ❑ For a true 1st order system: GBW = UGF = 𝜔𝑢 = 𝜔𝑢1 ❑ But the 2nd pole causes some bending below the asymptote ▪ Actual 𝜔𝑢 slightly < 𝜔𝑢1 : UGF < GBW 15: Negative Feedback [Jespers & Murmann, 2017] 48 Optimum Phase Margin (Freq Response) ❑ Maximum CL BW without peaking occurs at 𝜁 = 𝑄 = 0.707 ▪ Maximally flat response 15: Negative Feedback 49 Optimum Phase Margin (Freq Response) ❑ Maximum CL BW without peaking occurs at 𝜁 = 𝑄 = 0.707 ▪ Maximally flat response ▪ 𝜔𝑝2 = 2𝜔𝑢1 and 𝑃𝑀 ≈ 65𝑜 15: Negative Feedback [Jespers & Murmann, 2017] 50 Optimum Phase Margin (Freq Response) ❑ Maximum CL BW without peaking occurs at 𝜁 = 𝑄 = 0.707 ▪ Maximally flat response ▪ 𝜔𝑝2 = 2𝜔𝑢1 and 𝑃𝑀 ≈ 65𝑜 ❑ But 𝜁 < 1: Underdamped system ▪ Overshoot exists in transient response 15: Negative Feedback 51 Optimum Phase Margin (Tran Response) ❑ Fastest settling without overshoot occurs at 𝜁 = 1 (𝑄 = 0.5) ▪ Critical damped system ▪ 𝜔𝑝2 = 4𝜔𝑢1 and 𝑃𝑀 ≈ 76𝑜 15: Negative Feedback [Jespers & Murmann, 2017] 52 Second-Order System Speed-up ❑ First order system corresponds to 𝜔𝑝2 /𝜔𝑢1 → ∞ ▪ Too much overdamping ❑ The settling of 2nd order system with optimum PM is faster ▪ Critical damped system is faster than overdamped system ❑ But we must take some extra margin to account for variations 15: Negative Feedback [Jespers & Murmann, 2017] 53 Thank you! 15: Negative Feedback 54 References ❑ A. Sedra and K. Smith, “Microelectronic Circuits,” Oxford University Press, 7th ed., 2015 ❑ B. Razavi, “Design Of Analog CMOS Integrated Circuit,” 2nd ed., McGraw-Hill, 2017. ❑ T. C. Carusone, D. Johns, and K. W. Martin, “Analog Integrated Circuit Design,” 2nd ed., Wiley, 2012. ❑ P. Jespers and B. Murmann, Systematic Design of Analog CMOS Circuits Using PreComputed Lookup Tables, Cambridge University Press, 2017. 15: Negative Feedback 55 Bandwidth Extension ❑ Cascade of feedback amplifiers provides the same gain and a much faster response ▪ But power consumption and area doubled 15: Negative Feedback [Razavi, 2017] 56