EE348L Lecture 1. EE348L Lecture 1. Complex Numbers, KCL, KVL, Impedance,Steady State Sinusoidal Analysis. Motivation

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1 EE348L Lecture 1 Complex Numbers, KCL, KVL, Impedance,Steady State Sinusoidal Analysis 1 EE348L Lecture 1 Motivation

2 Example CMOS 10Gb/s amplifier Differential in,differential out, 5 stage dccoupled,broadband amplifier Output driver compensates for 14 db of loss in external world Receiver has 1mV sensitivity. Circuit is capable of independently matching input and output impedances to / 10ohms (differential) Same performance across process corners (75), 0100C, /10% VDD 3 Transconducta nce pair with resistor load Dummy devices for matching Dccoupled cascade Similar to above, but with dualloop feedback to boost bandwidth Baising circuit not shown Basic amplifier cells

3 Basic amplifier cell 1 out out in in 5 v gs (s) Linear model of basic amplifier Cell 1 g m v gs (s) out out in in i o (s) i o (s) r o R L C L v gs (s) C gs C gs g m v gs (s) r o R L C L

4 EE348L Lecture 1 Complex Numbers 7 Complex Numbers A = a j b ; B = c jd ; j = Sqrt(1) A B = (ac) j (b d) A x B = ac bd j (ad bc) A/B = (ajb ) x (c jd)/((cjd) x (cjd)) = A x B * /(c 2 d 2 ) ; where B* is complex conjugate of B

5 Complex Numbers a j b is realimaginary (rectangular) notation of A Magnitude Phase representation is polar (vector) representation of A in complex plane = re jθ = A angle(a). jy y=b r θ x=a re jθ = r ( cos(θ) j sin(θ) ) r = sqrt ( a 2 b 2 ) ; θ = tan 1 (b/a) x 9 Why do we care? Link between phasor and sine/cosine functions Consider rotating vector of length r describing a circle. Let the rate of rotation be ω radians/s (degrees of arc covered per second). Trace the projection of vector on x axis as a function of time. We get r cos(ωt). ω is angular frequency (radians/s) = 2πf, where f is frequency = 1/T, where T is time period. y r θ x t=0 t=t/4 t=t/2 t=3t/4

6 EE348L Lecture 1 Transforms 11 Why work with sinusoidal excitations? Magic of Fourier Transform Arbitrary time domain signals can be decomposed into constituent sine and cosine functions of ωt. ω is angular frequency (radians/s) = 2πf, where f is frequency = 1/T, where T is time period. If we know how to analyze the behavior of a circuit for sine or cosine functions, we are done. Algebra is cumbersome with sine and cosine functions. So we use exponentials. Have nice properties (derivative of an exponent is an exponent) etc.

7 Laplace transform Utility of Laplace Transform Reduce integrodifferential equations to algebraic equations. E.g.: Time domain description of RLC circuit response is secondorder differential equation. Time domain response of a linear circuit is described by convolution in time domain. Very painful. Using Laplace transform, response of a linear, timeinvariant circuit is simply the product of the transforms of circuit and input y(s) = x(s)h(s) ; H(s) is transfer function! 13 Transfer function sdomain ratio of the output to the input when all initial conditions are zero. Capacitors have no charge before signal is applied (t=0) Inductors have no current before signal is applied (t=0) What is s? s is complex frequency. H(s)=H(jω) is the steady state response of system to sinusoidal input of frequency ω; j=sqrt(1).

8 Poles of Transfer Function Laplace transform of exp(at) = 1/(sa), a is a complex number. Re(a) > 0 => exponentially increasing signal in time domain Re(a) < 0 => exponentially decaying signal in time domain Re(a) = 0 => oscillating signal in time domain Typically for linear timeinvariant (RLC) circuits, H(s) = P(s)/Q(s) ; P(s) and Q(s) are polynomial functions of s. 15 Now for H(jω)! In phasor land, sin(ωt) and cosine(ωt) functions are represented by Im(exp(jωt)) and Re(exp(jωt)) This is why we substitute s=jω to get the steady state response of linear timeinvariant circuits to sinusoidal excitations. Given V o (s) = H(s)V i (s) and sinusoidal excitation in time domain V i exp(jωt), V o = H(jω)V i where V o exp(jωt) is the steady state response of the system to V i exp(jωt). H(jw) called frequency response (function)

9 EE348L Lecture 1 Laws, Ideal sources 17 Ohm s law: V = I Z The Laws KVL : In any closed loop of interconnected elements in a circuit, Σ V i = 0 walk around a loop, get back to the same potential energy is conserved KCL : at any node in a circuit, Σ I i = 0, I i is current in branch i connected to node. What is this a restatement of? Have to maintain consistent polarity of currents and voltages! In general, V, I, and Z are complex numbers.

10 Ideal Sources Ideal Voltage Source V 1 : Always produce potential V 1 across terminals. Zero resistance in series. Delivers any amount of current. V/ I =? Ideal Current Source I 1 : Always sink I 1 across terminals. Infinite resistance in parallel. Sustains any amount of potential difference V across its terminals! V/ I =? I 1 V V 1 I 19 V 1 V 1 V 2 V 2 Algebra of Ideal sources I 1 I 1 I 2 I 2

11 V 1 I 1 Algebra of Ideal Sources I 1 V 1 I 1 V 1 21 EE348L Lecture 1 Impedance

12 Z = V/ I Impedance Small V in response to large I = small Z Large V in response to small I = Large Z In general, Z is a complex number = Z angle(z) 23 Resistor, Inductor, Capacitor Impedance of Resistor = V/I = R Impedance of Capacitor = 1/sC I = C dv/dt V=V o sin(ωt)=> I=ωCV o cos(ωt) => I = ωcv o sin(ωt90) I leads V by 90 degrees (π/2 radians) V=V o exp(jωt)=> I=jωCV o exp(jωt); j is anticlockwise rotation by 90 degrees (π/2 radians) in complex plane. Impedance of Inductor = sl V = L di/dt I=I o sin(ωt)=> V=ωLΙ o cos(ωt) => V = ωlι o sin(ωt90) V leads I by 90 degrees (π/2 radians) C R Z(ω) = V/I = 1/jωC Note: Derivative causes dependent variable to lead L Z(ω) = V/I = jωl

13 EE348L Lecture 1 Equivalent Impedance 25 Equivalent Impedance R1 R2 R1R2 Z1 Z2 Z1Z2 Z is arbitrary complex impedance R1 or Z1 R1 R2 = (R1 R2)/(R1R2) Z1 Z2 = (Z1 Z2)/(Z1Z2) R2 or Z2

14 V i (t) or V i (s) or V i (jω) Potential Divider R1 R2 V o (t) or V o (s) or V o (jω) What is V o (t)/v i (t)? V o (s)/v i (s)? V o (jw)/v i (jw)? 27 V i (t) or V i (s) or V i (jω) Potential Divider R1 R2 V o (t) or V o (s) or V o (jω) V o (t)/v i (t) = V o (s)/v i (s) = V o (jw)/v i (jw) = R2/(R1R2)

15 Complex Potential Divider V i (s) or V i (jω) Z1 Z2 V o (s) or V o (jω) V o (s)/v i (s) = V o (jw)/v i (jw) = Z2/(Z1Z2) Z can be R, L or C network 29 I i (t) or I i (s) or I i (jω) Current Divider R1 R2 I 2 (t) or I 2 (s) or I 2 (jω) I 2 (t)/i i (t) = I 2 (s)/i i (s) = I 2 (jw)/i i (jw) = R1/(R1R2) I 1 (t)/i i (t) = I 1 (s)/i i (s) = I 1 (jw)/i i (jw) = R2/(R1R2) Note: I 1 (t) I 2 (t) = I i (t) etc. (OK)

16 I i (s) or I i (jω) Complex Current Divider Z1 Z2 I 2 (s) or I 2 (jω) I 2 (s)/i i (s) = I 2 (jω)/i i (jω) = R1/(R1R2) I 1 (s)/i i (s) = I 1 (jω)/i i (jω) = R2/(R1R2) Note: I 1 (jω) I 2 (jω) = I i (jω) etc. (OK) Z can be R, L or C network 31 EE348L Lecture 1 RC Circuits, gainphase response, bode plots, decibel, power

17 Voltage driven series RC Circuit Complex voltage divider (quick solution) KVL (slower solution) H(S) = V o (s) / V in (s) = 1/jωC / (R 1/jωC) = 1/(1jωRC) Define ω o = 1/RC; H(s) = 1/(1 jω/ ω o ) V in (s) R V o (s) For ω >> ω o ; H(jω) ~= 0 For ω << ω o ;H(jω) ~= 1 For ω = ω o ; H(jω) = 1/(1j) H(jω o ) = 1/(1j) = 1/ 2 /_H(jω o )= tan 1 (1)= 45 deg. At ω = ω o, Power = 0.5 *Power at DC. Note that 10*log10(0.5) ~= 3dB. Hence ω o called 3dB Bandwidth; occurs at pole for 1 st order circuit C 33 Gain and Phase Response (sketch) H(jω) ω = ω o H(jω) 0 ω = ω o Note: Output Phase lags behind that of input! ω ω

18 The decibel scale To design, measure and talk about practical amplifiers, we need lots of dynamic range to be represented on one piece of paper, say from 10 9 to orders of magnitude. Cannot do this with linear scale. Need log scale. Alexander Graham Bell defined Power Ratio in decibels = 0.1 Bel = 10 log 10 (P1/P2); P2 is often taken as reference power. For transfer functions, P2= H(jω) ω=0 H(jω) ω=0 *. 35 Gain Phase Plot in db scale For ω >> ω o ; H(jω) = large negative db value For ω << ω o ;H(jω) ~= 1 = 0dB For ω = ω o ; H(jω) = 1/(1j) H(jω o ) = 1/(1j) = 1/ 2 H(jω o ) * = 1/(1j) = 1/ 2 Power = ½ = 10log 10 (1/2)= 3dB /_H(jω o ) = tan 1 (1)= 45 deg. For ω >= ω o, consider ω 1 and 10ω 1, which are 1 decade apart. 10log 10 (P(10ω 1 )/P(ω 1 )) = 10log 10 ( (1(ω o /ω 1 ) 2 )/ 1100(ω o /ω 1 ) 2 )) ~= 10log 10 (1/10) = 20dB. Single Pole Circuits have 20dB/decade slope in gain plot

19 45 db( H(jω) ) H(jω) Gain and Phase Response in db (sketch) Bode plot: /3dB frequency is break point ω = ω o ω = ω o 3dB Slope is 20dB/decade ω ω 37 Bode plots Approximation for quick analysis. Start with db value at ω=0. Break with 20dB/decade for pole, 20dB/decade for zero. Zeros and poles less than a decade away interact! Pole causes phase to become negative, zero causes phase to become positive.

20 Back of the envelope For the single pole RC filter, What value of ω results in 1% loss of power? What value of ω results in 5% loss of power? What value of ω results in 10% loss of power? 10% power loss output/input = 0.9 = 0.046dB What value of ω results in 50% loss of power? 50% power loss output/input = 0.5 = 3dB! What happens when we look at voltage instead of power? 50% voltage loss 20 log 10 (output/input) = 6 db! 39 Current driven parallel RC Circuit V=Parallel impedance x I (quick solution) KCL slower solution, current divider about the same. H(S) = V o (s) / I in (s) =? I in (s) R C For ω >> ω o ; H(jω) ~=? For ω << ω o ;H(jω) ~=? For ω = ω o ; H(jω) =? H(jω o ) =? /_H(jω o )=? V o (s)

21 Current driven parallel RC Circuit V=Parallel impedance x I (quick solution) KCL slower solution, current divider about the same. H(S) = V o (s) / I in (s) = R/jωC / (R 1/jωC) = R/(1jωRC) Define ω o = 1/RC; H(s) = R/(1 jω/ ω o ) I in (s) For ω >> ω o ; H(jω) ~= 0 For ω << ω o ;H(jω) ~= R For ω = ω o ; H(jω) = R/(1j) H(jω o ) = R/(1j) = R/ 2 /_H(jω o )= tan 1 (1)= 45 deg. At ω = ω o, Power = 0.5 *Power at DC. Note that 10*log10(0.5) ~= 3dB. Hence ω o called 3dB Bandwidth; occurs at pole for 1 st order circuit R C V o (s) 41 Current driven series RC Circuit V=? R H(S) = V o (s) / I in (s) =? I in (s) What is ω o? V o (s) C

22 V in (s) R V o (s) Relevance C I in (s) Conversion of voltage driven series RC circuit to current driven series RC circuit eliminates RC pole! This speeds up the circuit. Practical example of circuit design. Need to understand basic topologies presented so far Circuit design is more of the same subtle in many cases. R V o (s) C 43