**From:** Scott McMorrow (*scott@vasthorizons.com*)

**Date:** Fri Apr 07 2000 - 18:00:22 PDT

**Next message:**Bob Lewandowski: "Re: [SI-LIST] : Terminator location with larger BGA's"**Previous message:**D. C. Sessions: "Re: [SI-LIST] : Terminator location with larger BGA's"**In reply to:**mcgiffb@ttc.com: "[SI-LIST] : Terminator location with larger BGA's"

Bill,

Resistors always attenuate a signal. As the resistor is placed

further away from the driver (not just the package pin) there is a finite

time delay before the driver "sees" the effect of the termination resistor.

Unfortunately, part of the time delay is actually burried inside of the

BGA package itself. As the delay between the driver and resistor

is increased reflections will occur. At 1/10 the rise time these reflections

are small enough to be generally neglibible.

When the delay between the driver and resistor is increased to 1/2

the rise time, or 95ps in your case, the driver will not see the effect

of the resistor until a complete edge transition. In this case, the driver

is driving a pure transmission line at whatever the impedance of that

line is ... and (this is important) it is the transmission line that is driving

the resistor. One can now use transmission line theory and compute

the reflection coefficient at the resistor boundary.

For the case where the delay is greater than 1/2 Trise:

Given that the trace on both sides of the resistor has the same characteristic

impedance Z then the circuit looks something like this:

Driver ----T-line1 ---- Resistor ---- T-line2 -- Receiver

Zsource Z R Z Open

The driver will launch a wavefront into the transmission line which

is Vout = Vdd * (Z/(Zsource+Z)

That wavefront will reach the T-line1/Resistor/T-line2 boundary where

some of the wave will be reflected, some of the wave will be attenuated,

and some of the wave will be transmitted down T-line2. The reflection

coeffiecient at the boundary is computed as follows:

Tau = ((R+Z) - Z)/ ((R+Z) +Z) = R/(R+2Z) (i.e. the resistor lumps into the transmission line impedance)

thus, the reflection coefficient for a zero ohm resistor is 0. (no reflection)

and the reflection coeffiecient for an infinite resistance is 1. (complete reflection)

In between, some of the energy is reflected back down to the driver.

The reflection coefficient can be used to compute the voltage reflected back

down the line to the driver as follows:

Vreflected = Vout * Tau = Vout * (R/(R+2Z))

The voltage at the driver side of the resistor is now:

Vresistor = Vout + Vreflected = Vout + Vout * Tau

= Vout (1 + Tau)

(1 + Tau) is known as the transmission coefficient, T.

The voltage launched on the far side of the resistor is

now:

Far side voltage = Vresistor * (Z/(R + Z))

= Vout *(1+(R/(R+2Z)) * (Z/(R+Z))

At the end of the line the Reflection coefficient is 1 (for an open line),

so the transmission coefficient is now 2. Thus the end of line

voltage is:

Vendofline = 2 * Vout *(1+(R/(R+2Z)) * (Z/(R+Z)))

and referenced back to Vdd

Vendofline = 2 * (Vdd * (Z/(Zsource+Z)) *(1+(R/(R+2Z))) * (Z/(R+Z))

Now lets try some numbers:

Case1:

Zsource = 0 (a perfect driver)

Z = 50

R = 0

Vdd = 3.3 V

Vendofline = 6.6 V ( 3.3 Volts of overshoot)

this is the normal case of a perfect driver driving a perfect transmission line

and terminated at the end by an open circuit

Case2:

Zsource = 0 (a perfect driver)

Z = 50

R = 50

Vdd = 3.3 V

In this case, if the resistor were at the driver there would be a perfect

source impedance match to the T-line and no overshoot would occur.

However, with the resistor placed 1/2 risetime away from the driver

the voltage seen at the receiver is:

Vendofline = 2 * ( 3.3 * (50/(0+50)) * (1 + (50/(50 + 100))) * (50/(50+50))

= 6.6 * 1 * (1 + 1/3) * 1/2)

= 6.6 * 4/6

= 4.4 V (or 1.1 Volts of overshoot.)

Thus, a series resistor placed 1/2 a rise time (or more) away from a driver

is effective, but not as effective as a resistor placed at the driver. In general,

if the resistor is placed at the driver it is 100% effective as a series terminator

and if it is placed 1/2 a rise time away or greater it's effectiveness is

reduced by 33%. Between those two points the effectiveness of a resistor

is changed linearly from 100% to 67%.

Please understand that this analytical formulation is a first approximation.

There will be additional reflections that will complicate the behavior of the

circuit. However, as a first approximation one can use it to make some

general observations:

1) The effectiveness of a resistor as a series terminator is reduced as

it is moved away from the driver. This reduction can be up to 33% at

1/2 a risetime away.

2) One could compensate for the reduction in performance by increasing

the value of the series resistance.

3) Compared to a perfectly terminated system the amount of overshoot

seen at a single perfect receiver due to the delay from the driver to

the series resistor can be calculated as follows:

Overshoot percentage = 33 * (delay from driver to resistor/ (risetime/2))

for delays from 0 to risetime/2.

Beyond one half risetime the overshoot does not increase.

Fortunately, rather than performing calculations such as these, one can

use one of the many good signal integrity simulators out there in the world

to create a test circuit which will analyze these sorts of situations. Then you

can adjust the resistor accordingly. An increase in the distance of the

series resistor from the driver will not kill you if you adjust the resistance

accordingly.

regards,

scott

-- Scott McMorrow Principal Engineer SiQual, Signal Quality Engineering 18735 SW Boones Ferry Road Tualatin, OR 97062-3090 (503) 885-1231 http://www.siqual.commcgiffb@ttc.com wrote:

> We are using some 484 pin BGA's (FPGA's), which have edge rates of about 170ps. > Using a series termination, it would have to be placed within 1/10 of the > transition length according to Howard Johnson. On FR4, the transition length > would be .98, making the distance from source to termination .098. It is not > possible to get all of the required termination's within that distance. > > Aside from slowing down the edges (not desirable), waiting longer for things to > settle, or going with buried resistance, are there any other solutions for > placement of termination's? How are other people terminating large, high pin > count devices? > > Thanks > > Bill McGiffin > > **** To unsubscribe from si-list or si-list-digest: send e-mail to majordomo@silab.eng.sun.com. In the BODY of message put: UNSUBSCRIBE si-list or UNSUBSCRIBE si-list-digest, for more help, put HELP. > si-list archives are accessible at http://www.qsl.net/wb6tpu > ****

**** To unsubscribe from si-list or si-list-digest: send e-mail to majordomo@silab.eng.sun.com. In the BODY of message put: UNSUBSCRIBE si-list or UNSUBSCRIBE si-list-digest, for more help, put HELP. si-list archives are accessible at http://www.qsl.net/wb6tpu ****

**Next message:**Bob Lewandowski: "Re: [SI-LIST] : Terminator location with larger BGA's"**Previous message:**D. C. Sessions: "Re: [SI-LIST] : Terminator location with larger BGA's"**In reply to:**mcgiffb@ttc.com: "[SI-LIST] : Terminator location with larger BGA's"

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