Circuit Intuitions Ali Sheikholeslami Dept. of Elec. & Comp. Engineering University of Toronto, Canada ali@ece.utoronto.ca sponsored by SSCS Distinguished Lecture Program August 23, 2019 University of California, San Diego Ali Sheikholeslami Circuit Intuitions 1 of 39
Snapshots from Magazine Ali Sheikholeslami Circuit Intuitions 2 of 39
Outline Why Circuit Intuitions? o Overview of Articles Series o Looking into a Node o Use of Thevenin and Norton Equivalent Circuits n Ali Sheikholeslami Circuit Intuitions 3 of 39
Why Circuit Intuitions? Turning circuit design/analysis into a fun game! o Gain intimate understanding of how circuits behave o Making things simple, obvious! o A gateway to innovation! o Ali Sheikholeslami Circuit Intuitions 4 of 39
Circuit Intuitions Series Methods of Analysis o Looking into a Node n Source Degeneration; Bandwidth Extension n Miller’s Theorem; Miller’s Approximation n Fundamental Concepts o Process Variation and Pelgrom’s Law n Why Sinusoids? Reinventing the Wheels, Random Walk n Capacitor Analogy (3 articles) n Norton and Thevenin Equivalent Circuits (3 articles) n Special Circuits o Negative Cap.; Chopper Amp.; Capacitor as a Resistor n Ali Sheikholeslami Circuit Intuitions 5 of 39
Focus of this Talk Methods of Analysis o Looking into a Node n Source Degeneration; Bandwidth Extension n Miller’s Theorem; Miller’s Approximation n Fundamental Concepts o Process Variation and Pelgrom’s Law n Why Sinusoids? Reinventing the Wheels n Capacitor Analogy (3 articles) n Norton and Thevenin Equivalent Circuits (3 articles) n Special Circuits o Negative Cap.; Chopper Amp.; Capacitor as a Resistor n Ali Sheikholeslami Circuit Intuitions 6 of 39
Linear Time Invariant (LTI) Circuits Output is a linear combination of inputs! o Simple Case: If there are no storage elements (C or L) n 𝑧 = ℎ $ 𝑦 $ + ℎ ' 𝑦 ' + ℎ ( 𝑦 ( + ⋯ Superposition holds! o 𝑧 = 𝑧 $ + 𝑧 ' + 𝑧 ( + ⋯ Output is sum of contributions from individual inputs! o Thevenin/Norton Theorem: consequence of superposition o Ali Sheikholeslami Circuit Intuitions 7 of 39
Thevenin/Norton Theorems Ali Sheikholeslami Circuit Intuitions 8 of 39
An Intuitive Proof of Thevenin 𝑤 +,- 𝑢 = 𝑤 +/ 𝑢 − 𝑆 23 𝑗 5 (𝑢) Ali Sheikholeslami Circuit Intuitions 9 of 39
An Intuitive Proof of Norton 𝑗 5 𝑢 = 𝑗 8/ 𝑢 − 𝑤 +,- (𝑢) 𝑆 23 ⁄ Ali Sheikholeslami Circuit Intuitions 10 of 39
Small-Signal Model of NMOS/PMOS D S G B G B S D Assumption: All transistors are in Saturation Region! g m is the short-circuit transconductance of the transistor o g mb is additional g m due to non-zero v bs o r o is the transistor output resistance o Ali Sheikholeslami Circuit Intuitions 11 of 39
Circuits with More Transistors v out v out M 3 M 4 R L v out r o M 2 g m v gs g mb v bs M 2 M 1 v d1 v in v d5 v d1 r o M 1 g m v gs g mb v bs M 5 v in o Analysis becomes too cumbersome very quickly Practice of KVL/KCL, leaves no room for intuition o Ali Sheikholeslami Circuit Intuitions 12 of 39
Basic Premise o Assume low-frequency analysis in this talk! o That is, capacitors do not show up! o In small-signal, transistors behave like LTI systems Superposition holds! o Every circuit has a Thevenin/Norton equivalent circuit o Open-circuit voltage source (v oc ) in series with R eq n Short-circuit current source (i sc ) in parallel with R eq n When there is no input signal, they are just resistors! n We rely on Thevenin/Norton equivalent circuits ONLY! o Ali Sheikholeslami Circuit Intuitions 13 of 39
“Library Elements” o We call these “library elements” o Commonly-Used Configurations Looking into the gate and the drain n Looking into the source n Diode-Connected transistor n Looking into the drain with source degeneration n Looking into the source with load at the drain n Thevenin/Norton Equivalent looking into the drain n Thevenin/Norton Equivalent looking into the source n Let us build this library; one element at a time o Ali Sheikholeslami Circuit Intuitions 14 of 39
Looking into the Gate ① g g r o g m v gs R eq =? R eq g mb v bs R eq = ∞ o Looking into the gate, we see infinite resistance Ali Sheikholeslami Circuit Intuitions 15 of 39
Looking into the Drain ② g g d d r o r o R eq R eq =? g m v gs g mb v bs R eq = r o Looking into the drain, while gate and source are o grounded, we see r o ! Ali Sheikholeslami Circuit Intuitions 16 of 39
Looking into the Source ③ g r o r o r o 1/g me -g mb v s -g m v s g me v s R eq s s s R eq = r o || 1/g me R eq =? g me = g m + g mb = 1.1-1.2 g m o Looking into the source, while gate and drain are o grounded, we see a resistor whose value is 𝑆 23 = 𝑠 + ||1/ F2 ≈ 1/ F2 Ali Sheikholeslami Circuit Intuitions 17 of 39
Diode-Connected Transistor ④ g g m v in r o 1/g m r o v in v in R eq R eq = r o || 1/g m R eq =? o A diode-connected transistor is a resistor whose value is 𝑆 23 = 𝑠 + ||1/ F ≈ 1/ F Ali Sheikholeslami Circuit Intuitions 18 of 39
Looking into the Drain with R S ⑤ R eq =? g me v s g me r o v s r o g me r o R S R eq r o R S s R S r o R S v s R S R S R eq = R s + r o + g me r o R s Also known as Source Degeneration o Effective way to increase the output impedance o Multiplying the source resistance by g me r o o Ali Sheikholeslami Circuit Intuitions 19 of 39
Looking into the Source with R D ⑥ R D -g me r o v s R D r o R D g me v s r o R D R D g me r o v s r o R eq R eq =? v s v s v s 𝑆 K + 𝑠 + 𝑆 23 = 1 + F2 𝑠 + If R D <<r o , then 𝑆 23 ≈ 1/ F2 o R D is divided by (1+g me r o ) and appears at the source o Ali Sheikholeslami Circuit Intuitions 20 of 39
Looking into the Drain with Input Two Methods: ⑦ Find the Thevenin Equivalent Circuit: o i sc open-circuit voltage v oc in series with R eq n Find Norton Equivalent Circuit: o v oc v in short-circuit current i sc in parallel with R eq n Note that v oc =i sc x R eq o R S In this circuit, easier to find v oc first 𝑤 +/ = − F 𝑠 + 𝑤 LM v s =0, no body effect, v oc =-g m r o v in à 𝑗 8/ = 𝑤 +/ /𝑆 23 i sc =v oc /R eq à Norton more intuitive due to high R eq Ali Sheikholeslami Circuit Intuitions 21 of 39
Looking into the Source with Input Find Norton Equivalent Circuit first: o ⑧ short-circuit current i sc in parallel with R eq n R D v in Shorting output to ground à v s =0 o v oc i sc i sc R D Use current division à U V 𝑗 8/ = F 𝑤 LM à 𝑗 𝑡𝑑 = 𝑛 𝑤 𝑗𝑜 𝑠 𝑝 g m v in r o U V WX Y 𝑠 𝑝 + 𝑆 K Given small R eq , Thevenin à 𝑤 𝑝𝑑 = 𝑗 𝑡𝑑 𝑆 𝑓𝑟 is more intuitive i sc Ali Sheikholeslami Circuit Intuitions 22 of 39
Putting It All Together (1 to 4) ① ② ③ ④ R eq R eq R eq R eq R eq = ∞ R eq = r o R eq = r o || 1/g me R eq = r o || 1/g m Looking into one terminal while the other two grounded o Looking into the drain of a diode-connected transistor o No additional resistor in the circuit o Ali Sheikholeslami Circuit Intuitions 23 of 39
Putting It All Together (5 to 8) ⑤ ⑥ ⑦ ⑧ i sc R D v in v oc v in R D R eq v oc i sc R S R S R eq 𝑗 𝑡𝑑 = 𝑛 𝑤 𝑗𝑜 𝑠 𝑝 𝑆 K + 𝑠 𝑤 +/ = − F 𝑠 + 𝑤 LM R eq = R s + r o + 𝑆 23 = 𝑠 𝑝 + 𝑆 K 𝑗 8/ = 𝑤 +/ /𝑆 23 + g me r o R s 1 + F2 𝑠 + 𝑤 𝑝𝑑 = 𝑗 𝑡𝑑 𝑆 𝑓𝑟 Adding resistors and input to the circuit o Circuit Intuitions 24 of 39 Ali Sheikholeslami
Finding Voltage of “any” Node Method 1: Use Norton equivalent circuit o Short the node to ground; find i sc o Find R eq (using library elements) o Multiply the two: v out = i sc x R eq Method 2: Use Thevenin equivalent circuit o Open the load; find v oc o Find R eq (same as in Method 1) o Use voltage divider rule to find v out Ali Sheikholeslami Circuit Intuitions 25 of 39
Example: Common-Source Amplifier R L Finding v out : Use i sc x R eq : v out à i sc = - g m v in i sc R eq à R eq = R L || r o à v out = - g m ( R L || r o ) v in v in o What is the intuition here? How to increase the gain? How to increase v out ? n Output voltage comes from i sc x R eq n à either increase i sc or R eq n How can we increase these two? o Ali Sheikholeslami Circuit Intuitions 26 of 39
Example: Cascode Circuit To find the output voltage v out Use i sc x R eq : o M 2 R L Find out how i sc and R eq change o compared to those for CS amplifier v d1 Gain intuition on why the gain o M 1 changes from one to the other. v in Also, find intermediate voltages o (such as vd1) for added insight Ali Sheikholeslami Circuit Intuitions 27 of 39
Cascode Circuit: Find any Voltage (1) Method 1: Finding v d1 : Use Thevenin: v out v out ② à R eq1 = r o1 M 2 M 2 ⑦ à v oc1 = - g m r o1 v in R L R L ⑥ à R eq2 = (r o2 + R L )/ (1+ g me2 r o2 ) à v d1 = - g m r o1 v in (r o1 || R eq2 ) R eq2 v d1 v oc1 , R eq1 v d1 Finding v out : Use R eq2 to find the load current first : R eq1 M 1 à i L = v d1 /R eq2 v oc1 v in à v out = i L R L = v d1 R L /R eq2 Ali Sheikholeslami Circuit Intuitions 28 of 39
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