EE105 – Fall 2015 Microelectronic Devices and Circuits

Prof. Ming C. Wu [email protected] 511 Sutardja Dai Hall (SDH)

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LTI: Linear Time-­Invariant System

• System islinear (studied thoroughly in 16AB):

• System istime invariant: – There is no “clock” or time reference – The transfer function is not a function of time – It does not matter when you apply the input. The transfer function is going to be the same …

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1 Linear Systems

• Continuous time linear systems have a lot in common with finite dimensional linear systems we studied in 16AB: – Linearity:

– Basis Vectors à basis functions:

– Superposition:

– Matrix Representation à Integral representation:

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Linear Systems (cont)

• Eigenvectors à eigenfunctions

• Orthonormal basis

• Eigenfunction expansion

• Operators acting on eigenfunction expansion

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2 LTI Systems

• Since most periodic (non-­periodic) signals can be decomposed into a summation (integration) of sinusoids via Fourier Series (Transform), the response of a LTI system to virtually any input is characterized by the frequency response of the system:

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Example: Low Pass Filter (LPF)

Phase shift • Input signal: vs (t) =Vs cos(wt) v (t) = K ×V cos(wt +f) • We know that: o
s V0 Amp scale

v0 (t) = vs (t) - i(t)R dv i(t) = C 0 dt dv v (t) = v (t) - RC 0 0 s dt dv v (t) = v (t) +t 0 s 0 dt

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3 Complex Exponential

(z) ℑ

y z = x + jy

z| φ = z | ̸ = 1 jφ θ j | e j̸z jφ e z = z e = me | | | φ (z) x ℜ

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The Rotating Complex Exponential

• So the complex exponential is nothing but a point tracing out a unit circle on the complex plane:

eix = cos x + isin x

eiwt + e-iwt eiwt e-iwt 2

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4 Magic: Turn Diff Eq into Algebraic Eq

• Integration and differentiation are trivial with complex numbers:

d 1 eiwt = iweiwt eiwt dt = eiwt dt ò iw

• Any ODE is now trivial algebraic manipulations … in fact, we’ll show that you don’t even need to directly derive the ODE by using phasors • The key is to observe that the current/voltage relation for any element can be derived for complex exponential excitation

2-9 University of California,

Complex Exponential is Powerful

• To find steady state response we can excite the system with a complex exponential Mag Response

iwt LTI System w i(wt+f) e H H( ) e Phase Response • At any frequency, the system response is characterized by a single complex number H: H(w) f =
H(w)

• This is not surprising since a sinusoid is a sum of complex exponentials (and because of linearity!) eiwt - e-iwt eiwt + e-iwt sin wt = coswt = 2i 2

• From this perspective, the complex exponential is even more fundamental

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5 Solving LPF with Phasors

• Let’s excite the system with a complex exp:

dv0 vs (t) = v0 (t) +t dt use j to avoid confusion jwt vs (t) = Vse j(wt+f ) jwt vo (t) = V0 e = V0e

real complex

jwt jwt jwt Vse = V0e +t × jw ×V0e

Vs =V0 (1+ jw ×t ) V 1 0 = Easy!!! Vs (1+ jw ×t )

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Magnitude and Phase Response

• The system is characterized by the complex function V 1 H (w) = 0 = Vs (1+ jw ×t )

• The magnitude and phase response match our previous calculation:

V 1 H (w) = 0 = 2 Vs 1+ (wt )

H (w) = - tan -1 wt

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6 Why did it work?

• Again, the system is linear:

y = L(x1 + x2 ) = L(x1) +L(x2 )

• To find the response to a sinusoid, we can find the response to and and sum the results: 𝒊𝒊𝝎𝝎𝝎𝝎 �𝒊𝒊𝝎𝝎𝝎𝝎 𝒆𝒆 LTI System𝒆𝒆 w +f iwt w i( t 1 ) e H H( ) e

LTI System -w +f -iwt -w i( t 2 ) e H H( )e

iwt -iwt iwt -iwt e + e LTI System H(w)e + H(-w)e 2 H 2

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(cont.)

• Since the input is real, the output has to be real: H(w)eiwt + H(-w)e-iwt y(t) = 2 • That means the second term is the conjugate of the first: H(-w) = H(w) (even function)
H(-w) = -
H(w) = -f (odd function)

• Therefore the output is: H (w) y(t) = (ei(wt+f ) + e-i(wt+f ) ) 2 = H (w) cos(wt +f)

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7 “Proof” for Linear Systems

• For an arbitrary linear circuit (L,C,R,M, and dependent sources), decompose it into linear-­ sub operators, like multiplication by constants, time derivatives, or integrals: d d 2 y = L(x) = ax + b x + b x +
+ x + x + x +
1 dt 2 dt 2 ò òò òòò • For a complex exponential input x this simplifies to:

d d 2 y = L(e jwt ) = ae jwt + b e jwt + b e jwt +
+ c e jwt + c e jwt +
1 dt 2 dt 2 1 ò 2 òò e jwt e jwt y = ae jwt + b jwe jwt + b ( jw)2 e jwt +
+ c + c +
1 2 1 jw 2 ( jw)2 æ c c ö = = jwt ç + w + w 2 + + 1 + 2 + ÷ y Hx e ça b1 j b2 ( j )
2
÷ è jw ( jw) ø

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“Proof” (cont.)

• Notice that the output is also a complex exp times a complex number: æ c c ö = = jwt ç + w + w 2 + + 1 + 2 + ÷ y Hx e ça b1 j b2 ( j )
2
÷ è jw ( jw) ø

• The amplitude of the output is the magnitude of the complex number and the phase of the output is the phase of the complex number æ c c ö = = jwt ç + w + w 2 + + 1 + 2 + ÷ y Hx e ça b1 j b2 ( j )
2
÷ è jw ( jw) ø y = e jwt H (w) e jH (w) Re[ y] = H (w) cos(wt+  H (w))

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8 Phasors

• With our new confidence in complex numbers, we go full steam ahead and work directly with them … we can even drop the time factor since it will cancel out of the equations. 𝒊𝒊𝝎𝝎𝝎𝝎 𝒆𝒆 ~ jf1 • Excite system with phasor a : V1 = V1e

~ jf2 • Response will also be phasor: V2 = V2e • For those with a Linear System background, we’re going to work in the frequency domain – This is the Laplace domain with s = jw

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Capacitor I-­V Phasor Relation

• Find the Phasor relation for current and voltage in a cap: + jωt ic (t) = Ice dvC (t) v (t) ic (t) = C jωt C ic (t) dt vc (t) = Vce _

ω d ω I e j t = C [V e j t ] c dt c d ω ω CV e j t = jω CV e j t c dt c jωt jωt Ice = jω CVce

Ic = jω CVc

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9 Inductor I-­V Phasor Relation

l Find the Phasor relation for current and voltage in an inductor:

jωt + di(t) i(t) = Ie v(t) = L dt v(t) = Ve jωt v(t) i(t) _ ω d ω Ve j t = L [Ie j t ] dt d ω ω LI e j t = jω LIe j t dt Ve jωt = jω LIe jωt V = jω L I

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Impede the Currents !

• Suppose that the“ input” is defined as the voltage of a terminal pair (port) and the “output” is defined as the current into the port:

+ w w +f v(t) = Ve j t = V e j( t v ) Arbitrary LTI v(t) i(t) w j(wt+f ) Circuit i(t) = Ie j t = I e i –

• The impedance Z is defined as the ratio of the phasor voltage to phasor current (“self” transfer function)

V V f -f Z(w) = H (w) = = e j( v i ) I I

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10 Admit the Currents!

• Suppose that the“ input” is defined as the current of a terminal pair (port) and the “output” is defined as the voltage into the port:

+ w w +f v(t) = Ve j t = V e j( t v ) Arbitrary LTI v(t) i(t) w j(wt+f ) Circuit i(t) = Ie j t = I e i –

• The admittance Y is defined as the ratio of the phasor current to phasor voltage (“self” transfer function) I I f -f Y(w) = H(w) = = e j( i v ) V V

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Voltage and Current Gain

• The voltage (current) gain is just the voltage (current) transfer function from one port to another port: + + v (t) i (t) Arbitrary LTI i (t) v (t) 1 1 Circuit 2 2 – –

V2 V2 j(f2 -f1 ) Gv (w) = = e V1 V1

I2 I2 j(f2 -f1 ) Gi (w) = = e I1 I1 – If , the circuit has voltage (current) gain – If , the circuit has loss or attenuation 𝑮𝑮 > 𝟏𝟏 𝑮𝑮 < 𝟏𝟏 2-22

11 Transimpedance/admittance

• Current/voltage gain are unit-­less quantities • Sometimes we are interested in the transfer of voltage to current or vice versa

+ + v (t) i (t) Arbitrary LTI i (t) v (t) 1 1 Circuit 2 2 – –

V V f -f J(w) = 2 = 2 e j( 2 1 ) [W] I1 I1

I I f -f K(w) = 2 = 2 e j( 2 1 ) [S] V1 V1

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Direct Calculation ofH (no DEs)

• To directly calculate the transfer function (impedance, trans-­impedance, etc) we can generalize the circuit analysis concept from the “real” domain to the “phasor” domain • With the concept of impedance (admittance), we can now directly analyze a circuit without explicitly writing down any differential equations • Use KVL, KCL, mesh analysis, loop analysis, or node analysis where inductors and capacitors are treated as complex resistors

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12 LPF Example: Again!

• Instead of setting up the DE in the time-­domain, let’s do it directly in the frequency domain • Treat the capacitor as an imaginary “resistance” or impedance:

→

time domain “real” circuit frequency domain “phasor” circuit

• We know the impedances: 1 Z = R Z = R C jwC 2-25

Bode Plots

• Simply the log-­log plot of the magnitude and phase response of a circuit (impedance, transimpedance, gain, …) • Gives insight into the behavior of a circuit as a function of frequency • The “ log” expands the scale so that breakpoints in the transfer function are clearly delineated • In EECS 140, Bode plots are used“ to compensate” circuits in feedback loops

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13 Frequency Response of Low-­Pass Filters

1 jωC 1 1 T(ω) = = = 1 + ω + ω ω R + 1 j RC 1 j / 0 jωC 1 ω0 = RC 1 T(ω) = 2 1+(ω /ω0 ) −1 ∠T(ω) = −tan (ω /ω0 )

ω3dB = ω0 [rad/sec]

ω0 f3dB = [Hz] 2π

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Frequency Response of High-­Pass Filters

R 1 1 T(ω) = = = 1 1 − ω ω R + 1+ 1 j 0 / jωC jωRC 1 ω0 = RC 1 T(ω) = 2 1+(ω0 /ω) −1 ∠T(ω) = tan (ω0 /ω)

ω3dB = ω0 [rad/sec]

ω0 f3dB = [Hz] 2π

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14 Example: High-­Pass Filter

• Using the voltage divider rule: L jw jwL H (w) = = R + w L R j L 1+ jw R jwt H (w) = 1+ jwt jwt w ® ¥ H ® =1 jwt 0 w ® 0 H ® = 0 1+ 0 1 j 1 w = H = = t 1+ j 2

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HPF Magnitude Bode Plot

• Recall that log of product is the sum of log jwt 1 H (w) = = jwt + dB + wt dB + wt 1 j dB 1 j dB

wt j dB Increase by 20 dB/decade wt = Þ wt = Equals unity at breakpoint 1 j dB 0dB

40 dB 20 dB 20 dB

0 dB 0.1 1 10 100

-­20 dB 2-30

15 HPF Bode Plot (dissection)

• The second term can be further dissected: 1 = 0dB - 1+ jwt + wt dB 1 j dB 1 .1/t 1/t 10 /t w << 0 dB 0 dB t 1 20 dB w >> -­20 dB/dec -­20 dB t 1 -­40 dB w = ~ -­ 3dB t -­60 dB

-­ 3dB 1 w = 2-31 t

Composite Plot

• Composite is simply the sum of each component: 1 jwt + dB + wt jwt High frequency ~ 0 dB Gain 1 j dB dB

0 dB .1/t 1/t 10 /t Low frequency attenuation

-­20 dB 1 + wt 1 j dB -­40 dB

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16 Approximate versus Actual Plot

• Approximate curve accurate away from breakpoint • At breakpoint there is a 3 dB error

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HPF Phase Plot

• Phase can be naturally decomposed as well: jwt 1 p
H(w) =
=
jwt+
= - tan -1 wt 1+ jwt 1+ jwt 2

• First term is simply a constant phase of 90 degrees • The second term is the arctan function • Estimate arctan function: 1 w << t 1 w = t 45 Actual curve 1 0 w >> t 2-34

17 Power Flow

• The instantaneous power flow into any element is the product of the voltage and current: P(t) = i(t)v(t) • For a periodic excitation, the average power is: = t t t Pav òi( )v( )d T • In terms of sinusoids we have = ω +ϕ ω +ϕ τ Pav ∫ I cos( t i ) V cos( t v )d T = ⋅ ω ϕ − ω ϕ ⋅ ω ϕ − ω ϕ τ I V ∫ (cos t cos i sin t sin i ) (cos t cos v sin t sin v )d T = ⋅ τ 2 ω ϕ ϕ + 2 ω ϕ ϕ + ω ω I V ∫ d cos t cos i cos v sin t sin i sin v csin t cos t T I ⋅ V I ⋅ V = (cosϕ cosϕ + sinϕ sinϕ ) = cos(ϕ −ϕ ) 2 i v i v 2 i v 2-35

Power Flow with Phasors

I × V P = cos(f -f ) av 2 i v

Power Factor

• Note that if , then 𝝅𝝅 𝑰𝑰 �𝑽𝑽 𝝅𝝅 • Important: Power𝝓𝝓𝒊𝒊 − 𝝓𝝓 𝒗𝒗 is= a𝟐𝟐 non-­linear𝑷𝑷 𝒂𝒂 function𝒂𝒂 = 𝟐𝟐 𝒄𝒄𝒄𝒄𝒄𝒄 so𝟐𝟐 we= 𝟎𝟎 can’t simply take the real part of the product of the phasors: P ¹ Re[I ×V ]

• From our previous calculation: I × V 1 1 P = cos(f -f ) = Re[I ×V * ] = Re[I * ×V ] 2 i v 2 2

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18 More Power to You!

• In terms of the circuit impedance we have: 2 1 1 V V P = Re[I ×V * ] = Re[ ×V * ] = Re[Z -1] 2 2 Z 2 2 * 2 2 V Z V * V = Re[ 2 ] = 2 Re[Z ] = 2 Re[Z] 2 Z 2 Z 2 Z

• Check the result for a real impedance (resistor) • Also, in terms of current: 2 1 1 I P = Re[I * ×V ] = Re[I * × I × Z] = Re[Z] 2 2 2

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Summary

• Complex exponentials are eigen-­functions of LTI systems – Steady-­state response of LCR circuits are LTI systems – Phasor analysis allows us to treat all LCR circuits as simple “resistive” circuits by using the concept of impedance (admittance) • Frequency response allows us to completely characterize a system – Any input can be decomposed into either a continuum or discrete sum of frequency components – The transfer function is usually plotted in the log-­log domain (Bode plot) – magnitude and phase – Location of poles/zeros is key

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