Weight vs Calories

Calories burnt = KG * KM * 1.036KG = your weight in kilograms

KM = distance run in kilometers

Healthy weight loss goal: Lose 0.5-2 kg/week

Example: Weight = 75kg; Distance run = 4km

Calories burnt = 310 (approx)

Calories burnt per week = 2170

1 kilogram = 7700 calories

Weight lost per week = 0.28 kg

Result: Need to improve.


Signal Integrity Principles

Each interconnect is a transmission line

Forget the word ground; Think return path.

Bandwidth of a signal is the highest sine wave frequency component.

All SI formulas are definitions or approximations.


Even a two inch long trace on a PCB can affect SI.


SI problems: Timing, Noise, EMI

Noise sources:

Signal on a net: Reflections, Distortions from Impedance discontinuities

Crosstalk: Mutual Inductance, Mutual (parasitic) capacitance

Rail collapse: Voltage drop in power/ground.

EMI: A component or entire system


Impedance discontinuities: Cross section, topology, added components(via, connector, etc.)


Ways to minimize cross talk (cross talk should kept as minimum as 5%)

1) Use uniform plane as return path.

2) Spacing traces farther apart.

3) Use guard traces.


Rail-Collapse noise (Rail-bounce, Ground bounce):

Because of the impedance of the power and ground distribution, a voltage drop will occur as the IC current switches. This voltage drop means the power and ground rails have collapsed from their nominal values.

Simultaneous switching noise (SSN) or simultaneous switching output (SSO): Whenever an output pin changes state, the current it draws from its ICs 0V and power rails eventually flows through the interconnections to the PCB’s 0V and power rails. The inductance in these paths causes voltage drops that make the IC’s internal 0V and power rails ‘bounce’ with respect to the PCB rails, according to the activity of the output ports

Ways to minimize:

1) Closely placed power-ground planes with thin dielectric

2) Low inductance decaps, on-chip decaps

3) Multiple short power, ground pins in packages


Sources of EMI

1) Conversion of some differential signal into a common signal

2) Ground bounce

Ways to minimize:

1) Use shielded cables

2) Use low-impedance connections


Rise time: Time taken to go from 10% to 90% of the final value.

Mostly rise time is about 10% of clock period.


Bandwidth = 0.35*[Rise Time]

For most microprocessor, ASIC based systems rise time is 7% of clock period.

Approximately bandwidth is five times the clock frequency.


An electron travels (in a copper wire) at a speed of 1cm/s.

Time to travel down a 6” of interconnect in FR4 is about 1ns.






U is unit of measure for vertical usable space, or height of racks, cabinets. 1U is equal to 1.75 inches (44.45mm).

Example : Form-factor of 3U cPCI board is 100mm x 160mm

Form-factor of 6U cPCI board is 100mm x 233.35mm

Note: 3U as per the definition is 3 x 44.45 mm = 133.35. But 160mm is defined in the cPCI specifications.

Transmission lines

Transverse Electromagnetic Waves: Propagation of energy in a transmission line takes place such that electric and magnetic fields transverse to one another and also to direction of propagation. The resultant wave is termed as TEM wave.

Consider a small section n of a parallel wire.

Length of this line is dx.

Voltage at input is V, at the other end is V+dV; Similarly current is I, I+dI.

Assume Primary Line Constants of the line are R,L,G and C

Note: w= 2.pi.f

For small dx, dI is zero.

Potential drop across the line is V – (V+dV) = R.dx.I + jwL.dx.I

à -V’ = I(R+jwL) ….. eq.1

Note: V’ = (d/dx)V

In a similar method, assuming dV is zero,

à -I’ = V(G+iwC) ….. eq.2

Differentiate eq.1,2

V”= V(R+jwL)(G+jwC) ….. eq.3

I” = I(R+jwL)(G+jwC) …..eq.4

Let (gamma)^2 = (R+jwL)(G+jwC)

Gamma = (alpha) + j(beta)

alpha is attenuation constant

beta is phase constant

gamma is propagation constant

Now eq.3,4 becomes

V” = V. (gamma)^2

I” = I. (gamma)^2

Solving the above equations,

V=A.exp(-gamma.x) + B.exp(gamma.x) …..eq.5

I= C.exp(-gamma.x) + D.exp(gamma.x) …..eq.6

Note: exp(x) = e^x

The first terms in eq.5,6 are called incident component(magnitude of V or I decreases from source towards load, whereas the second terms are called reflected component(magnitude of V or I decreases form load towards source)

Hypothetical infinite line: Voltage at distant end approaches zero(i.e no reflected component)

At x=0, V=Vs

Substitute in eq.5, Vs=A+B

At x=infinity, V=0

à B=0, V= Vs.exp(-gamma.x)




Where Z0=[(R+jwL)/(G+jwC)]^(1/2)

Z0 the input impedance of such infinite line is commonly referred to as Characteristic Impedance of the line. Z0 and propagation constant are termed as secondary constants (or coefficients) of the line

Line Terminated in a Load Impedance Zr:

At a distance x from the source, voltage and current are Vx, Ix.


Substitute the above after differentiating eq.5,6 and then simplifying,

Voltage, current at source are Vs=A+B, Is=(A-B)/Z0

Vx=Vs.cosh(gamma.x)-Is.Z0.sinh(gamma.x) …..eq.7

Ix=Is.cosh(gamma.x)-(Vs/Z0)sinh(gamma.x) …..eq.8

For load impedance Zr, length of transmission line is l.

Ix=Ir, Vx=Vr such that Vr=Ir.Zr

Substitute the above in eq.7,8, and solve for Vs/Is

Input Impedance Zin = Vs/Is = Z0.N/D

Where N = Zr.cosh(gamma.l) + Z0.sinh(gamma.l)

D= Z0.cosh(gamma.l) + Zr.sinh(gamma.l)

Line Terminated in Load Impedance Z0:

Zr=Z0 in N,D

à Zin=Z0

A line terminated in its characteristic impedance has input impedance equal to Z0. In such a line, there is no reflected component and at any point x distant from the signal source, the voltage and current are same as that for infinite length transmission line.

Low frequency transmission line:

R is very bigger than wL

G is very lesser than wC

Z0 = (R/jwC)^(1/2)

High frequency line:

R is very lesser than wL

G is very lesser than wC

Z0 = (L/C)^(1/2)

Calculation of Characteristic Impedance for given IO standard

Consider 3.3V CMOS, that provides drive capability of 24mA

Max value of VCC is 3.6V.

VOH is 2.4V.

Maximum allowed drop in the transmission line for proper operation is

VCCmax – VOH = 1.2V.

à Maximum allowed impedance of the line is 1.2/0.024 = 50 ohms