[hsdd] High-Speed Digital Design Newsletter - Law of Product Development
- From: "Dr. Howard Johnson" <howie03@xxxxxxxxxx>
- To: <hsdd@xxxxxxxxxxxxx>
- Date: Mon, 3 Oct 2005 10:06:52 -0700
LAW OF PRODUCT DEVELOPMENT
HIGH-SPEED DIGITAL DESIGN ? online newsletter ?
Vol. 8 Issue 06
My next web cast happens tomorrow!
"Jitter Effects in Modern System Design",
see registration:
www.xilinx.com/events/webcasts/100405_signalintegrity.htm
Live Web cast
Date: 10-04-05
Time: 11:00AM PDT; 2:00EDT; 18:00GMT
Location: Your Desktop
I feel lucky to be back home this week after
aborting a trip to Houston just as Hurricane Rita
bore down on the city. I escaped just days before
the brunt of the storm arrived. Unless you have
lived through such an event, you cannot imagine the
surreal panic spreading slowly through a city as
people began to realize the enormity of the
situation unfolding around them.
I did a presentation at Rice University on Tuesday,
but then cancelled all my classes at H.P. scheduled
for Thursday and Friday when it became apparent that
no one would be at work on those days--too busy
buying flashlights and bottled water.
Hot, humid, still air hung like a wet blanket across
my shoulders as I made to way slowly, painstakingly,
across the metropolitan area on Wednesday afternoon,
headed north for Dallas. Why didn't I fly? All
flights from Houston were booked. The drive to
Dallas (my hometown), usually a 3-1/2 hour jaunt,
took 14 hours. Police blocked entrances to the
interstate to "keep traffic flowing", although you
could hardly call 6-lane bumper-to-bumper traffic a
"flow". People who exited the freeway found gas
stations out of gas, stores out of food and water,
and everyone out of patience. I am glad I fueled up
early that morning while the city was still
functioning. Once at the Dallas airport, my flight
finally left, but 8 hours late. Compared to the
suffering of others, my story seems inconsequential.
Just a few days lost.
If you live in the gulf region, or have family
there, I extend to you my sympathy and best wishes
for a speedy recovery from the incalculable damage
done by this season's storms.
I would like to thank my attentive staff for quickly
recognizing the seriousness of the situation and
finding a flight from Dallas that got me out of
there. THANK YOU JENNIFER AND LIZ.
______________________________________________________
LAW OF PRODUCT DEVELOPMENT
Mehran Abbasi writes: [edited]
I am trying to connect a source at 5V to a load at
3.3V using a 50-ohm transmission line. What kind of
termination can I use which does this attenuation?
______________________________________________________
Dr. Johnson replies:
Achieving both attenuation and termination requires
two resistors. At the source end of your connection,
install a series resistor of approximately 22 ohms.
At the destination end, install a shunt termination
resistance of 50 ohms. These two components, in
conjunction with the transmission line, form a
perfect attenuator with a ratio of 50/(50+22).
Adjust the 22-ohm resistor to account for the series
resistance of your driver, in order to perfect the
received signal size.
Best regards,
Dr. Howard Johnson
______________________________________________________
Dr. Johnson,
Your answer was very simple and great, but why does
one bother using a PI or T networks for attenuation?
Thanks,
Mehran Abbasi
_____________________________________________________
Dear Mehran,
It is an immutable law of product development that
the more independent requirements you place on a
circuit, the more complex the circuit must become to
accommodate those new requirements. For example, if
you said, "attenuate this signal", I could just
stick a 100 meg-ohm resistor in series with your
driver--that makes plenty of attenuation.
If you say you want a specific amount of
attenuation, like a ratio of 2/3, then I must get
more details about the source and load impedances in
the circuit so I can derive the correct value of
resistance.
If you want a specific amount of attenuation AND
ALSO the property of termination (i.e., a specific
input impedance at the load, or at the source, or
both), then I have to add complexity to meet that
new requirement. The circuit I supplied has two
resistors, representing your two requirements:
specific attenuation and termination.
"PI" and "T" networks have three components, and are
thus *more* than capable of meeting your
requirements. With a PI or T network, I can design a
circuit that presents a controlled impedance on the
left side, a controlled impedance on the right side,
and also a specific degree of attenuation (within
practical limits).
Your requirements can be met using only two parts,
because all you asked for is a controlled impedance
at one point (at the end termination) and a specific
attenuation.
Other examples of two-element resistive networks and
their application to digital design appear in my
article "Matching Pads":
www.sigcon.com/Pubs/edn/matchingpads.htm
The networks in that article couple two transmission
lines having different impedances. The networks have
only two requirements (at most), namely that the
network present a specific controlled impedance on
one side, or the other, or both. The exact degree of
attenuation remains unspecified. In that article I
present networks that produce the minimum
attenuation given the requirements. Controlling the
impedance on only one side or the other requires
only one resistor, but controlling the impedance on
both sides requires two resistors.
In that article the attenuation remains unspecified,
so the two impedance requirements can be met with
(at most) just two components.
Suppose I change the problem to impose three
requirements: (1) controlled impedance for signals
encountering the net from left side, (2) controlled
impedance for signals encountering the net from the
right side, and (3) a specific value of attenuation.
In this case, you need three components to solve the
constraint equations, resulting in a "T" or "PI"
network.
Radio engineers use "T" and "PI" attenuating
networks a lot. The networks are usually designed
for controlled impedances equal to 50 ohms as viewed
from both sides, and a specific degree of
attenuation. Such a network can be "dropped in"
anywhere within a 50-ohm circuit.
Does this answer your question?
Best regards,
Dr. Howard Johnson
______________________________________________________
Dear Dr. Johnson,
One more question.
Suppose I use either a "PI" or "T" resistive
attenuation at end of a 50-ohm transmission line. Do
I still need a 50-ohm parallel resistor at the input
of device or a series termination at the output of
source?
Kind regards,
Mehran Abbasi
______________________________________________________
Dear Mehran,
I suppose a lot of projects go this way--the
requirements do not become clear until you are
almost finished with the design.
You seem to really want to use a "PI" or "T"
network. That would be overkill for the requirements
you have articulated to me, namely:
(1) Some kind of termination,
(2) Located at the end of the line
(3) With a gain of 2/3.
Assume the driver is on the left, leading to a
receiver on the right side of your diagram.
To form a 50-ohm end termination with a gain of 2/3
you connect the right end of the transmission line
to a 17-ohm resistor, and the far end of the
resistor to your receiver. Then shunt a second
resistor (33 ohms) between the receiver input and
ground.
The resulting two-element network looks like either
a "T" missing it's right arm, or a "PI" missing its
left leg, depending on which way you look at it.
This circuit produces a nominal 50-ohm end
termination. I believe the circuit I first suggested
is superior for two reasons:
(a) The first circuit draws less current from the
driver (reducing power, radiation, and
crosstalk).
(b) The first circuit provides some nominal
amount of termination at the source. If the
receiver has a lot of parasitic capacitance, the
source resistor helps damp the reflections that
result from that capacitance. In that way, it
works somewhat like a both-ends terminated setup.
This time I am going to publish my remarks in this
newsletter and hope that perhaps, finally, I have
gotten to the root of your question. If you need
more help, find an electrical engineer local to your
area that can explain to you how electronic circuits
behave, and craft a solution that will satisfy you.
Best regards,
Dr. Howard Johnson
______________________________________________________
We have been extraordinarily busy this season doing
private classes; join us at our next upcoming public
seminars in San Jose the week of January 30, 2006. A
full schedule of cities and dates appears at:
www.sigcon.com
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