EE105 – Fall 2015 Microelectronic Devices and Circuits Prof. Ming C. Wu [email protected] 511 Sutardja Dai Hall (SDH) 2-1 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 … 2-2 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: 2-3 Linear Systems (cont) • Eigenvectors à eigenfunctions • Orthonormal basis • Eigenfunction expansion • Operators acting on eigenfunction expansion 2-4 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: 2-5 Example: Low Pass Filter (LPF) Phase shift • Input signal: vs (t) =Vs cos(wt) v (t) = K ×V cos(wt +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 2-6 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 2-7 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 eiwt + e-iwt eiwt e-iwt 2 2-8 4 Magic: Turn Diff Eq into Algebraic Eq • Integration and differentiation are trivial with complex numbers: d 1 eiwt = iweiwt eiwt dt = eiwt 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 iwt LTI System w i(wt+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!) eiwt - e-iwt eiwt + e-iwt sin wt = coswt = 2i 2 • From this perspective, the complex exponential is even more fundamental 2-10 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 jwt vs (t) = Vse j(wt+f ) jwt vo (t) = V0 e = V0e real complex jwt jwt jwt Vse = V0e +t × jw ×V0e Vs =V0 (1+ jw ×t ) V 1 0 = Easy!!! Vs (1+ jw ×t ) 2-11 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+ (wt ) H (w) = - tan -1 wt 2-12 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: � LTI System w +f iwt w i( t 1 ) e H H( ) e LTI System -w +f -iwt -w i( t 2 ) e H H( )e iwt -iwt iwt -iwt e + e LTI System H(w)e + H(-w)e 2 H 2 2-13 (cont.) • Since the input is real, the output has to be real: H(w)eiwt + H(-w)e-iwt 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(wt+f ) + e-i(wt+f ) ) 2 = H (w) cos(wt +f) 2-14 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 òò òòò • For a complex exponential input x this simplifies to: d d 2 y = L(e jwt ) = ae jwt + b e jwt + b e jwt ++ c e jwt + c e jwt + 1 dt 2 dt 2 1 ò 2 òò e jwt e jwt y = ae jwt + b jwe jwt + b ( jw)2 e jwt ++ c + c + 1 2 1 jw 2 ( jw)2 æ c c ö = = jwt ç + w + w 2 + + 1 + 2 + ÷ y Hx e ça b1 j b2 ( j ) 2 ÷ è jw ( jw) ø 2-15 “Proof” (cont.) • Notice that the output is also a complex exp times a complex number: æ c c ö = = jwt ç + w + w 2 + + 1 + 2 + ÷ y Hx e ça 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 = = jwt ç + w + w 2 + + 1 + 2 + ÷ y Hx e ça b1 j b2 ( j ) 2 ÷ è jw ( jw) ø y = e jwt H (w) e jH (w) Re[ y] = H (w) cos(wt+ H (w)) 2-16 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. ~ jf1 • Excite system with phasor a : V1 = V1e ~ jf2 • 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 2-17 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 2-18 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 2-19 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(wt+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 2-20 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(wt+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 2-21 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(f2 -f1 ) Gv (w) = = e V1 V1 I2 I2 j(f2 -f1 ) Gi (w) = = e I1 I1 – If , the circuit has voltage (current) gain – If , the circuit has loss or attenuation > < 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 2-23 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 2-24 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 jwC 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 2-26 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 2-27 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 2-28 14 Example: High-­Pass Filter • Using the voltage divider rule: L jw jwL H (w) = = R + w L R j L 1+ jw R jwt H (w) = 1+ jwt jwt w ® ¥ H ® =1 jwt 0 w ® 0 H ® = 0 1+ 0 1 j 1 w = H = = t 1+ j 2 2-29 HPF Magnitude Bode Plot • Recall that log of product is the sum of log jwt 1 H (w) = = jwt + dB + wt dB + wt 1 j dB 1 j dB wt j dB Increase by 20 dB/decade wt = Þ wt = 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+ jwt + wt 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 jwt + dB + wt jwt High frequency ~ 0 dB Gain 1 j dB dB 0 dB .1/t 1/t