First and Second Order Circuits. Claudio Talarico, Gonzaga University Spring 2015

Transcription

1 First and Second Order Circuits Claudio Talarico, Gonzaga University Spring 2015

2 Capacitors and Inductors intuition: bucket of charge q = Cv i = C dv dt Resist change of voltage DC open circuit Store voltage (charge) Energy stored = 0.5 C v(t) 2 i(t) intuition: water hose λ = Li v = L di dt Resist change of current DC short circuit Store current (magnetic flux) Energy stored = 0.5 L i(t) 2 C +" v(t) " i(t) + v(t) EE406 Introduction to IC Design 2

3 Characterization of an LTI system s behavior Techniques commonly used to characterize an LTI system: 1. Observe the response of the system when excited by a step input (time domain response) Assumption: x in (t) is causal (i.e. x(t)=0 for t < 0) x in (t) L.T.I. h(t) x out (t) x out = t 0 x in (τ )h(t τ )dτ x in (t)* h(t) 2. Observe the response of the system when excited by sinusoidal inputs (frequency response) X out (s) = X in (s) H(s) EE406 Introduction to IC Design 3

4 Frequency Response The merit of frequency-domain analysis is that it is easier than time domain analysis: L[x(t)] = 0 e st x(t)dt = X(s) The transfer function of any of the LTI circuits we consider Are rational with m n Are real valued coefficients a j and b i Have poles and zeros that are either real or complex conjugated Furthermore, if the system is stable All denominator coefficients are positive The real part of all poles are negative One sided Laplace Transform (assumption: x(t) is causal or is made causal by multiplying it by u(t)) H(s) = a 0 + a 1 s a m sm 1+ b 1 s b n s n = K (s +ω z1)...(s +ω zm) (s +ω p1 )...(s +ω pn ) = = K (s z 1)...(s z m ) (s p 1 )...(s p n ) with K a m b n root form mathematicians style EE406 Introduction to IC Design 4

5 Frequency Response! s # 1+ " H(s) = a 0! # " ω z1 ω p1 $! &...# % " $! s &...# 1+ % " ω zm ω pn $ & % $ & %! s $! # 1 &... 1 s $ # & " z = a 1 % " z m % 0! 1 s $ # " p 1 % &...! 1 s $ # " p & % n EE style NOTE : p i = ω pi z i = ω zi (poles) (zeros) EE406 Introduction to IC Design 5

6 Magnitude and Phase (1) When an LTI system is exited with a sinusoid the output is a sinusoid of the same frequency. The magnitude of the output is equal to the input magnitude multiplied by the magnitude response ( H(jω in ) ). The phase difference between the output and input sinusoid is equal to the phase response (ϕ=phase[h(jω in )]) x in (t) = A in cos(ωt) = A in e jω t + e jω t F X in ( jω) = F[x in (t)] 2 unit circle IM axis θ 1e jθ = cosθ +jsinθ RE axis H( jω) = H( jω) e jωt 0 X out ( jω) = X in ( jω) H( jω) cosθ = e jθ + e jθ 2 Euler s formula EE406 Introduction to IC Design 6

7 Magnitude and Phase (2) X out ( jω) = X in ( jω) H( jω) = X in ( jω) H( jω) e jωt 0 F -1 Time Shift Property: F[x(t t 0 )] = X(f) e j2πft0 x out (t) = H( jω in ) x in (t + t 0 ) = = A in H( jω in ) cos[ω(t + t 0 )] = = A in H( jω in ) cos(ωt +ωt 0 ) EE406 Introduction to IC Design 7

8 First order circuits A first order transfer function has a first order denominator H(s) = A 0 ω p First order low pass transfer function. This is the most commonly encountered transfer function in electronic circuits H(s) = A 0 ω z ω p General first order transfer function. EE406 Introduction to IC Design 8

9 Step Response of first order circuits (1) Case 1: First order low pass transfer function H(s) = A 0 ω p x in (t) = A in u(t) X in (s) = A in s X out (s) = A in s A 0 ω p = A in A 0! x out (t) = A in A 0 u(t) " # 1 e t/τ % & " 1 s 1 % $ ' # $ s +ω p &' with τ =1/ω p EE406 Introduction to IC Design 9

10 Step Response of first order circuits (2) Case 2: General first order transfer function x in (t) = A in u(t) X in (s) = A in s H(s) = A 0 ω z ω p X out (s) = A ina 0 s ω z ω p! ( " x out (t) = A in A 0 u(t) 1 1 ω p * $ )* # ω z % 'e t/τ & + -,- where τ =1/ω p EE406 Introduction to IC Design 10

11 Step Response of first order circuits (3) Notice x out (t) short term and long term behavior x out (0+) = A in A 0 ω p ω z x out ( ) = A in A 0 The short term and long term behavior can also be verified using the Laplace transform x out (0+) = lim s s X out (s) = lim s s A in s H(s) = A ina 0 ω p ω z x out ( ) = lim s 0 s X out (s) = lim s 0 s A in s H(s) = A ina 0 EE406 Introduction to IC Design 11

12 Equation for step response to any first order circuit x out (t) = x out ( ) [ x out ( ) x out (0+)] e t/τ where τ =1/ω p Steady response Transitory response EE406 Introduction to IC Design 12

13 Example #1 R = 1KΩ, C= 1µF. Input is a 0.5V step at time 0 Source: Carusone, Johns and Mar2n H(s) = 1 RC τ = RC =1µs ω p = 1 τ =1Mrad / s f = ω p 3dB 2π 159KHz V out (t) = 0.5 ( 1 e t/τ )u(t) EE406 Introduction to IC Design 13

14 Example #2 (1) R1 = 2KΩ, R2=10KΩ C1= 5pF. C2=10pF Input is a 2V step at time 0 Source: Carusone, Johns and Mar2n H(s 0) = R 2 R 1 + R 2 A 0 ; H(s ) = C 1 C 1 + C 2 A τ p = (R 1 R 2 ) (C 1 C 2 ) = R R 1 2 (C 1 + C 2 ) R 1 + R 2 " % R $ H(s) = 2 R 1 C ' $ 1 R 1 + R 2 $ R R ' 1 2 (C 1 + C 2 )' # $ R 1 + R 2 &' By inspection τ z = R 1 C 1 EE406 Introduction to IC Design 14

15 V out (t) = Example #2 (2) V out (0+)+ $ % V out ( ) V out (0+) 1 e t/τ p 0 fort 0 ( ) & ' fort > 0 V out ( ) V out (t) = A in R 2 R 1 + R 2 A in C 1 C 1 + C 2 = V out (0+) EE406 Introduction to IC Design 15

16 Example #3 Consider an amplifier having a small signal transfer function approximately given by A(s) = A 0 ω p A 0 = 1 x 10 5 ω p = 1 x 10 3 rad/s Find approx. unity gain BW and phase shift at the unity gain frequency since A 0 >> 1: A(s) A 0 s ω p = A 0 ω p s A( jω) A 0ω p jω A 0 ω p jω u =1 ω u A 0 ω p " Phase[A( jω u )] Phase A 0ω p % $ ' = 90! # jω u & EE406 Introduction to IC Design 16

17 Second-order low pass Transfer Function H(s) = a 0 1+ b 1 s + b 2 s 2 = a 0 Q + s2 2 b 1 1 Q ; b Interesting cases: - Poles are real - one of the poles is dominant - Poles are complex ω 3dB 1 b 1 = τ j b 1 ( ) EE406 Introduction to IC Design 17

18 Poles Location Roots of the denominator of the transfer function: Complex Conjugate poles (overshooting in step response) Q + s2 2 = 0 Poles location with Q> 0.5 jω for Q > 0.5 p 1,2 = 2Q ( 1 j 4Q2 1) = ω R jω I - For Q = (Φ=45 ), the -3dB frequency is (Maximally Flat Magnitude or Butterworth Response) - For Q > the frequency response has peaking Real poles (no overshoot in the step response) for Q 0.5 p 1,2 = 2Q 1 1 4Q2 ( ) Φ = Source: Carusone σ EE406 Introduction to IC Design 18

19 Frequency Response 10 H(s) 5 H(s) = 1 Q + s2 2 Magnitude [db] Q=0.5 Q=1/sqrt(2) Q=1 Q= / 0 EE406 Introduction to IC Design 19

20 Step Response Underdamped Q > 0.5 Source: Carusone Q=0.5 Overdamped Q < 0.5 time Ringing for Q > 0.5 The case Q=0.5 is called maximally damped response (fastest settling without any overshoot) EE406 Introduction to IC Design 20

21 Widely- Spaced Real Poles H(s) = a 0 1+ b 1 s + b 2 s 2 = a 0 Q + s2 2 b 1 1 Q ; b Real poles occurs when Q 0.5: for Q 0.5 p 1,2 = ω ( 0 2Q 1 1 ) 4Q2 Real poles widely-spaced (that is one of the poles is dominant) implies: p 1 2Q 2Q 1 4Q2 << p 2 2Q + 2Q! 1 4Q2 0 << 2 1 4Q 2 0 << 1 4Q 2 0 <<1 4Q 2 Q 2 << 1 4 b 2 b 1 2 << 1 4 EE406 Introduction to IC Design 21

22 Widely-Spaced Real Poles a 0 H(s) = 1+ b 1 s + b 2 s = a 0 2 " 1 s % $ # p 1 & ' " $ 1 s # p 2 % ' & = a 0 1 s p 1 s p 2 + s2 p 1 p 2 a 0 1 s p 1 + s2 p 1 p 2 p 1 1 b 1 p 2 1 p 1 b 2 = b 1 b 2 This means that in order to estimate the -3dB bandwidth of the circuit, all we need to know is b 1! H(s) a 0 1 s p 1 ω -3dB p 1 1 b 1 ZVTC method: b 1 = τ j ω 3dB 1 = 1 b 1 τ j EE406 Introduction to IC Design 22

23 Example: Series RLC circuit (1) v in (t) = V I u(t) R L +" " C i(t) +" v out (t) " V out (s) V in (s) = 1 sc R + sl + 1 sc = 1 RC + s 2 LC 1 Q + s2 2 2 = 1 LC ; Q = 1 RC = L / C R Z 0 R EE406 Introduction to IC Design 23

24 Example - Series RLC: Poles location (2) Q + s2 ω = 0 s2 + s 2 0 Q +ω 2 0 = 0 s 1,2 = Q $ ω & 0 % Q 2 ' ) ( s 1,2 = ω ( 0 2Q 1 1 ) 4Q2 2Q α Two possible cases: For Q ½ real poles: s 1,2 = ω ( 0 2Q 1 1 ) 4Q2 For Q > ½ complex poles α ( 1 1 ) 4Q2 s 1,2 = ω ( 0 2Q 1 j 4Q2 1) α 1 j 4Q2 1 ( ) EE406 Introduction to IC Design 24

25 V out (s) V in (s) = 1 sc R + sl + 1 sc 2 = 1 LC ; Q = 1 RC = Example series RLC = 1 RC + s 2 LC 1 L / C R Z 0 R ; 2Q α Q + s2 2 C = 1 L ω ; L = C 1 Q = = L RC R 1 α = 2Q = R 2L EE406 Introduction to IC Design 25

26 Example series RLC s 2 + s Q +ω 2 0 = 0 s 2 + s 2α +ω 2 0 = 0 s 1,2 = α α 2 2 For Q ½ real poles: s 1,2 = α α 2 2 For Q > ½ complex poles! # " for α $ <<1 the polesarewidelyspaced& % IM s 1,2 = α j 2 α 2 = α jω n ω n = natural (damped) frequency = resonant frequency ω n ω p1 X X ω p2 α= /2Q RE EE406 Introduction to IC Design 26

27 v out (t) Example series RLC: Step response time ζ α / >1 (a) Overdamped Q < 0.5 (α > ) ζ α / =1 (b) Critically damped Q=0.5 (α = ) ζ α / <1 (c) Underdamped Q > 0.5 (α < ) ω n = ringing frequency α = damping factor (rate of decay) = resonance frequency ζ = Damping ratio EE406 Introduction to IC Design 27

28 Example RLC series: Quality Factor Q For a system under sinusoidal excitation at a frequency ω, the most fundamental definition for Q is: energy stored Q = ω average power dissipated At the resonant frequency, the current through the network is simply V in /R. Energy in such a network sloshes back and forth between the inductance and the inductor, with a constant sum. The peak inductor current at resonance is I pk =V pk /R, so the energy stored by the network can be computed as: E stored = 1 2 L I 2 pk dimensionless The average power dissipated in the resistor at resonance is: P avg = 1 2 R I 2 pk E Q = ω stored 0 = 1 P avg LC L R = L / C R = Z 0 R Z 0 = Characteristic impedance of the network At resonance Z C = Z L = L=( C) 1 = L / C Z 0 EE406 Introduction to IC Design 28

29 Example series RLC: Ringing and Q Since Q is a measure of the rate of energy loss, one expect a higher Q to be associated with more persistent ringing than a lower Q. IM R large α large Q small Φ= ω n α= RE Q=0.5 à Φ = 0 Q=0.707 à Φ = 45 Q=1 à Φ = 60 Q=10 à Φ 87 Ringing doesn t last long and its excursion is small The decaying envelope is proportional to: e αt = e t 2Q Rule of thumb: Q is roughly equal to the number of cycles of ringing. Frequency of the ringing oscillations: 1/f n = T n = 2π/ω n EE406 Introduction to IC Design 29

30 Example RLC series by intuition (1) We can predict the behavior of the circuit without solving pages of differential equations or Laplace transforms. All we need to know is the characteristics equation (denominator of H(s)) and the initial conditions v I (t) = V I u(t) +" " i(t) L R +" C v(t) " Assume the values of R,L,C are such that > α (underdamping). Initial conditions: i(0) = I 0 v(0) = +V 0 The voltage across the capacitor cannot jump: v(0+)=v(0)=+v 0 The current through the inductor cannot jump: i(0+)=i(0)= I 0 The output voltage starts at V 0, it ends at v( )=V I times before settling at V I and it rings about Q EE406 Introduction to IC Design 30

31 Example RLC series using intuition (2) The only question left is to decide if v(t) will start off shooting down or up? But, we know that i(0+) is negative à this means the current flows from the capacitor toward the inductor à which means the capacitor must be discharging à the voltage across the capacitor must be dropping v(t) Ringing stops after Q! cycles V I V I? v (0) i(0) 0 is negative so v(t) must drop π! 2! ω n period of ringing t EE406 Introduction to IC Design 31

32 Example RLC series using intuition (3) In practice for the under damped case it useful to compute two parameters Overshoot Settling time " OS = exp π % $ α' # ω n & Normalized overshoot = % overshoot w.r.t. final value t s 1 α ln # ε ω & n % ( ε is the % error that we are willing $ ' to tolerate w.r.t. the ideal final value EE406 Introduction to IC Design 32

33 First order vs. Second order circuits Behavior First order circuits introduce exponential behavior Second order circuits introduce sinusoidal and exponential behavior combined Fortunately we will not need to go on analyzing 3 rd, 4 th, 5 th and so on circuits because they are not going to introduce fundamentally new behavior EE406 Introduction to IC Design 33