1

EE348L Lecture 1

Complex Numbers, KCL, KVL, Impedance,Steady State Sinusoidal

Analysis

2

EE348L Lecture 1

Motivation

3

Prof

esso

r M

adha

van

EE34

8L L

ectu

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lides

, Spr

ing

2005 Example CMOS 10Gb/s amplifier

Differential in,differential out, 5 stage dc-coupled,broadband amplifierOutput driver compensates for 14 dB of loss in external worldReceiver has 1mV sensitivity.Circuit is capable of independently matching input and output impedances to +/- 10ohms (differential)Same performance across process corners (75), 0-100C, +/-10% VDD

4

Prof

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r M

adha

van

EE34

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2005 Basic amplifier cells

Transconductance pair with resistor loadDummy devices for matchingDc-coupled cascade

Similar to above, but with dual-loop feedback to boost bandwidth

Baising circuit not shown

5

Prof

esso

r M

adha

van

EE34

8L L

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2005 Basic amplifier cell 1

out+

in+ in-out-

6

Prof

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r M

adha

van

EE34

8L L

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2005 Linear model of basic amplifier Cell 1

out+

in+ in-out-

io(s)

rogmvgs(s)

Cgs

vgs(s) RL CL

io(s)

rogmvgs(s)

Cgs

vgs(s) RL CL

7

EE348L Lecture 1

Complex Numbers

8

Prof

esso

r M

adha

van

EE34

8L L

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lides

, Spr

ing

2005 Complex Numbers

A = a + j b ; B = c +jd ; j = Sqrt(-1)A + B = (a+c) + j (b + d)A x B = ac – bd + j (ad + bc)A/B = (a+jb ) x (c – jd)/((c+jd) x (c-jd)) = A x B*/(c2 + d2) ; where B* is complex conjugate of B

9

Prof

esso

r M

adha

van

EE34

8L L

ectu

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lides

, Spr

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2005 Complex Numbers

a + j b is real-imaginary (rectangular) notation of AMagnitude Phase representation is polar (vector) representation of A in complex plane = rejθ =

|A| angle(A).

θx

jy

r

x=a

y=b

rejθ = r ( cos(θ) + j sin(θ) )

r = sqrt ( a2 + b2) ; θ = tan-1(b/a)

10

Prof

esso

r M

adha

van

EE34

8L L

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lides

, Spr

ing

2005 Why do we care ?

Link between phasor and sine/cosine functionsConsider 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.

θ

x

y

r

t=0

t=T/4

t=T/2

t=3T/4

11

EE348L Lecture 1

Transforms

12

Prof

esso

r M

adha

van

EE34

8L L

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lides

, Spr

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2005 Why work with sinusoidal excitations ?

Magic of Fourier Transform -

Arbitrary time domain signals can be decomposed into constituent sine and cosine functions of ωt. ωis 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.

13

Prof

esso

r M

adha

van

EE34

8L L

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lides

, Spr

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2005 Laplace transform

Utility of Laplace Transform -

Reduce integro-differential equations to algebraic equations. E.g.: Time domain description of RLC circuit response is second-order 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, time-invariant circuit is simply the product of the transforms of circuit and input y(s) = x(s)H(s) ; H(s) is transfer function !

14

Prof

esso

r M

adha

van

EE34

8L L

ectu

re S

lides

, Spr

ing

2005 Transfer function

s-domain 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).

15

Prof

esso

r M

adha

van

EE34

8L L

ectu

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lides

, Spr

ing

2005 Poles of Transfer Function

Laplace transform of exp(-at) = 1/(s+a), a is a complex number.Re(a) > 0 => exponentially increasing signal in time domainRe(a) < 0 => exponentially decaying signal in time domainRe(a) = 0 => oscillating signal in time domainTypically for linear time-invariant (RLC) circuits, H(s) = P(s)/Q(s) ; P(s) and Q(s) are polynomial functions of s.

16

Prof

esso

r M

adha

van

EE34

8L L

ectu

re S

lides

, Spr

ing

2005 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 time-invariant circuits to sinusoidal excitations.

Given Vo(s) = H(s)Vi(s) and sinusoidal excitation in time domain Viexp(jωt), Vo = H(jω)Vi where Voexp(jωt) is the steady state response of the system to Viexp(jωt).

H(jw) called frequency response (function)

17

EE348L Lecture 1

Laws, Ideal sources

18

Prof

esso

r M

adha

van

EE34

8L L

ectu

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lides

, Spr

ing

2005 The “Laws”

Ohm’s law: V = I Z

KVL : In any closed loop of interconnected elements in a circuit, Σ Vi = 0 walk around a loop, get back to the same potential energy is conserved

KCL : at any node in a circuit, Σ Ii = 0, Ii 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.

19

Prof

esso

r M

adha

van

EE34

8L L

ectu

re S

lides

, Spr

ing

2005 Ideal Sources

Ideal Voltage Source “V1”: Always produce potential “V1” across terminals. Zero resistance in series. Delivers any amount of current. ∆V/∆I = ?

Ideal Current Source “I1”: Always sink “I1” across terminals. Infinite resistance in parallel. Sustains any amount of potential difference V across its terminals ! ∆V/∆I = ?

+-

V1 I

I1 +

-V

20

Prof

esso

r M

adha

van

EE34

8L L

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lides

, Spr

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2005 Algebra of Ideal sources

+-

V1

+- V2

I2I1

+-

V1+-

V2

I2

I1

21

Prof

esso

r M

adha

van

EE34

8L L

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lides

, Spr

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2005 Algebra of Ideal Sources

I1

+-

V1

+-

V1I1 +

-

V1

I1

22

EE348L Lecture 1

Impedance

23

Prof

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r M

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EE34

8L L

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2005 Impedance

Z = ∆V/∆I

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)

24

Prof

esso

r M

adha

van

EE34

8L L

ectu

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lides

, Spr

ing

2005 Resistor, Inductor, Capacitor

Impedance of Resistor = V/I = R

Impedance of Capacitor = 1/sCI = C dV/dtV=Vo sin(ωt)=> I=ωCVo cos(ωt) => I = ωCVo sin(ωt+90) I leads V by 90 degrees (π/2 radians)V=Vo exp(jωt)=> I=jωCVoexp(jωt); j is anticlockwise rotation by 90 degrees (π/2 radians) in complex plane.

Impedance of Inductor = sL V = L dI/dtI=Io sin(ωt)=> V=ωLΙo cos(ωt) => V = ωLΙo sin(ωt+90) V leads I by 90 degrees (π/2 radians)

R

C

L

Z(ω) = V/I = 1/jωC

Note: Derivative causes dependent variable to lead

Z(ω) = V/I = jωL

25

EE348L Lecture 1

Equivalent Impedance

26

Prof

esso

r M

adha

van

EE34

8L L

ectu

re S

lides

, Spr

ing

2005 Equivalent Impedance

R1 R2 R1+R2

Z1 Z2 Z1+Z2

Z is arbitrary complex impedance

R1 or Z1

R2 or Z2

R1 || R2 = (R1 R2)/(R1+R2)

Z1 || Z2 = (Z1 Z2)/(Z1+Z2)

27

Prof

esso

r M

adha

van

EE34

8L L

ectu

re S

lides

, Spr

ing

2005 Potential Divider

R1

R2

+

-

+

-

Vi(t) or

Vo(t) orVo(s) orVo(jω)

Vi(s) or

Vi(jω)

What is Vo(t)/Vi(t) ? Vo(s)/Vi(s) ? Vo(jw)/Vi(jw) ?

28

Prof

esso

r M

adha

van

EE34

8L L

ectu

re S

lides

, Spr

ing

2005 Potential Divider

R1

R2

+

-

+

-

Vi(t) or

Vo(t) orVo(s) orVo(jω)

Vi(s) or

Vi(jω)

Vo(t)/Vi(t) = Vo(s)/Vi(s) = Vo(jw)/Vi(jw) = R2/(R1+R2)

29

Prof

esso

r M

adha

van

EE34

8L L

ectu

re S

lides

, Spr

ing

2005 Complex Potential Divider

Z1

Z2

+

-

+

-

Vo(s) orVo(jω)

Vi(s) or

Vi(jω)

Vo(s)/Vi(s) = Vo(jw)/Vi(jw) = Z2/(Z1+Z2)

Z can be R, L or C network

30

Prof

esso

r M

adha

van

EE34

8L L

ectu

re S

lides

, Spr

ing

2005 Current Divider

R1 R2 I2(t) orI2(s) orI2(jω)

Ii(s) orIi(jω)

I2(t)/Ii(t) = I2(s)/Ii(s) = I2(jw)/Ii(jw) = R1/(R1+R2)

I1(t)/Ii(t) = I 1(s)/Ii(s) = I1(jw)/Ii(jw) = R2/(R1+R2)

Note: I1(t) + I2(t) = Ii(t) etc. (OK)

Ii(t) or

31

Prof

esso

r M

adha

van

EE34

8L L

ectu

re S

lides

, Spr

ing

2005 Complex Current Divider

Z1 Z2 I2(s) orI2(jω)

Ii(s) orIi(jω)

I2(s)/Ii(s) = I2(jω)/Ii(jω) = R1/(R1+R2)

I 1(s)/Ii(s) = I1(jω)/Ii(jω) = R2/(R1+R2)

Note: I1(jω) + I2(jω) = Ii(jω) etc. (OK)

Z can be R, L or C network

32

EE348L Lecture 1

RC Circuits, gain-phase response, bode plots, decibel , power

33

Prof

esso

r M

adha

van

EE34

8L L

ectu

re S

lides

, Spr

ing

2005 Voltage driven series RC Circuit

Complex voltage divider (quick solution)

KVL (slower solution)

H(S) = Vo(s) / Vin(s) = 1/jωC / (R + 1/jωC) = 1/(1+jωRC)

Define ωo = 1/RC; H(s) = 1/(1 + jω/ ωo)

R

C+-

Vin(s)

Vo(s)

For ω >> ωo; H(jω) ~= 0

For ω << ωo; H(jω) ~= 1

For ω = ωo; H(jω) = 1/(1+j)

|H(jωo)|=|1/(1+j)| = 1/√2

/_H(jωo)= -tan-1(1)= -45 deg.

At ω = ωo, Power = 0.5 *Power at DC. Note that 10*log10(0.5) ~= -3dB. Hence ωo called –3dB Bandwidth;

occurs at pole for 1st order circuit

34

Prof

esso

r M

adha

van

EE34

8L L

ectu

re S

lides

, Spr

ing

2005 Gain and Phase Response (sketch)

ω

ω

|H(jω)|

H(jω)

ω = ωo

ω = ωo

10.707

0

-45

-90

Note: Output Phase lags behind that of input !

35

Prof

esso

r M

adha

van

EE34

8L L

ectu

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lides

, Spr

ing

2005 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 106 – 15 orders of magnitude. Cannot do this with linear scale.

Need log scale.

Alexander Graham Bell defined Power Ratio in deciBels = 0.1 Bel = 10 log10(P1/P2); P2 is often taken as reference power.

For transfer functions, P2= |H(jω)|ω=0| |H(jω)|ω=0|* .

36

Prof

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r M

adha

van

EE34

8L L

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lides

, Spr

ing

2005 Gain Phase Plot in dB scale

For ω >> ωo; H(jω) = large negative dB value

For ω << ωo; H(jω) ~= 1 = 0dB

For ω = ωo; H(jω) = 1/(1+j)

|H(jωo)|=|1/(1+j)| = 1/√2

|H(jωo)|*=|1/(1-j)| = 1/√2

Power = ½ = 10log10(1/2)= -3dB

/_H(jωo) = -tan-1(1)= -45 deg.

For ω >= ωo, consider ω1 and 10ω1, which are 1 decade apart.

10log10(P(10ω1)/P(ω1)) = 10log10(√(1+(ωo/ω1)2)/ 1+100(ωo/ω1)2)) ~= 10log10(1/10) = -20dB.

Single Pole Circuits have –20dB/decade slope in gain plot

37

Prof

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r M

adha

van

EE34

8L L

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lides

, Spr

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2005 Gain and Phase Response in dB (sketch)

ω

ω

dB(|H

(jω

)|)

ω = ωo

-3dB

-45

H(j

ω)

ω = ωo

Slope is -20dB/decade

Bode plot: +/-3dB frequencyis break point

38

Prof

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r M

adha

van

EE34

8L L

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lides

, Spr

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2005 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.

39

Prof

esso

r M

adha

van

EE34

8L L

ectu

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lides

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2005 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 log10(output/input) = -6 dB!

40

Prof

esso

r M

adha

van

EE34

8L L

ectu

re S

lides

, Spr

ing

2005 Current driven parallel RC Circuit

V=Parallel impedance x I (quick solution)

KCL slower solution, current divider about the same.

H(S) = Vo(s) / Iin(s) = ?For ω >> ωo; H(jω) ~= ?

For ω << ωo; H(jω) ~= ?

For ω = ωo; H(jω) = ?

|H(jωo)|=?

/_H(jωo)= ?

R CIin(s)

Vo(s)

41

Prof

esso

r M

adha

van

EE34

8L L

ectu

re S

lides

, Spr

ing

2005 Current driven parallel RC Circuit

V=Parallel impedance x I (quick solution)

KCL slower solution, current divider about the same.

H(S) = Vo(s) / Iin(s) = R/jωC/ (R + 1/jωC) = R/(1+jωRC)

Define ωo = 1/RC; H(s) = R/(1 + jω/ ωo)

For ω >> ωo; H(jω) ~= 0

For ω << ωo; H(jω) ~= R

For ω = ωo; H(jω) = R/(1+j)

|H(jωo)|=|R/(1+j)| = R/√2

/_H(jωo)= -tan-1(1)= -45 deg.

At ω = ωo, Power = 0.5 *Power at DC. Note that 10*log10(0.5) ~= -3dB. Hence ωo called –3dB Bandwidth;

occurs at pole for 1st order circuit

R CIin(s)

Vo(s)

42

Prof

esso

r M

adha

van

EE34

8L L

ectu

re S

lides

, Spr

ing

2005 Current driven series RC Circuit

V= ?

H(S) = Vo(s) / Iin(s) = ?

What is ωo ?

R

CIin(s)

Vo(s)

43

Prof

esso

r M

adha

van

EE34

8L L

ectu

re S

lides

, Spr

ing

2005 Relevance

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 farCircuit design is more of the same – subtle in many cases.

R

CIin(s)

Vo(s)R

C+-

Vin(s)

Vo(s)