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?
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.
Dr. Howard Johnson
Your answer was very simple and great, but why does one bother using a PI or T networks for attenuation?
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-million-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 in the middle (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".
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:
- Controlled impedance for signals encountering the net from left side,
- Controlled impedance for signals encountering the net from the right side, and
- A specific value of attenuation.
Given those requirements, you will need three components to solve the constraint equations. Good choices would include a "T" network or a "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 without affecting anything but the attenuation.
Does that answer your question?
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?
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:
- Some kind of termination,
- Located at the end of the line,
- 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 R1, and the other end of R1 to your receiver. Then shunt a second resistor (R2=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 three reasons:
- The first circuit draws less current from the driver (reducing power, radiation, and crosstalk).
- 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.
- It's unusual to find a driver that can feed a 50-ohm load connected striaght to ground. With my first circuit, the driver doesn't do that. The load the driver sees is 50+22=72 ohms.
- You could probably improve the drive-current characteristics by changing the 50-ohm termination at the end into a two-element split termination. That would reduce the I[OH] required from the driver at the expense of not being able to drive all the way down to zero volts, which is OK because you only need to get down the V[IL] anyway.
I am going to publish these remarks with the hope that perhaps, finally, I have gotten to the root of your question.
Dr. Howard Johnson
OK, I'm re-editing this collection of articles and noticed that I left something out of my reply. Sorry, Mehran.
Store-bought attenuators are designed to be connected into a circuit with a specific impedance on both the source and load sides. Violate that rule and the attenuator may not work properly.
If Mehran uses a store-bought 50-ohm "T" or "PI" attenuator then, to make it work properly, he will need to provide an additional 50-ohm termination at either the source or the load.