terminator use

I'm confused about the use of terminators. I'm sending pulse signals from some 93 ohm output equipment through 93 ohm coax to my scope which is 50 ohm. I have a t at the scope. Should I be puting a 93 ohm terminator on the other end of the t or a 50ohm? Is there a rule of thumb where to place terminators? thanks jk

Reply to
entropy429
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The terminator? He's currently governor of California :-)

It's rather simple: Connect a 43ohms resistor from the center conductor of the coax to the scope input. If you have to do this a lot and the cable has BNC it might pay off to grind off the corner of a 90 degree BNC adapter until you've just separated the center. Solder the 43ohms in series there and close it up so no dirt and grime gets in. Use a chip resistor or at least one that is low in inductance.

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Regards, Joerg

http://www.analogconsultants.com/
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Reply to
Joerg

If the generator is really a 93 ohm source, you don't need a terminator at the far end, no matter what impedance the scope is. Your system is "source terminated" and any reflections from the scope will be re-absorbed when they bounce back into the generator.

But if you want to soak up any reflections at the scope (say, the generator isn't truly a 93 ohm source, and you don't want stuff sloshing around) you need a series resistor of 43 ohms between the end of the coax and the 50 ohm scope input... not a tee.

John

Reply to
John Larkin

Hi John. Dont we need a transmission line of 1/4 wavelength of the freq of interest before we start worring about termination?

Reply to
BobG

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If you have a 50 ohm scope, then putting a 93 ohm resistor in parallel
with its input (which is what you\'re doing if you\'re using a tee) will
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Reply to
John Fields

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No.

Sourcing and terminating a line with its characteristic impedance will
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Reply to
John Fields

Does source termination only refer to periodic signal or does it also apply to one shot signals also?? What is source termination? I haven't heard that before,( shows you my level) :) jf

Reply to
jfisher864

"Source termination" can be used when you have a driver, a transmission line, and a single high-impedance load at the very end of that transmission line. It works with any type of signal. The need for termination depends upon your situation. Giving this subject proper treatment, here, would take too much time and effort.

The idea for source termination is that if your driver will put out V into an open, then when you connect the transmission line (with same characterstic impedance as the driver's net source impedance) then V/2 will be "launched" into the transmission line (toward the load). When the singal hits the open end of the transmission line then V will be present at the end of the line (into your high-impedance load) and V/2 will be reflected back to the source. When the reflection arrives back at its source, it will be completely absorbed by the driver's source impedance.

So, the advantages of source termination are that it is simple to implement, and you get the entire open-circuit drive voltage at the load. The drawbacks are that if you have any discontinuities along the transmission line (other than the intentional one at the end) you will get multiple reflections and (relatively) poor signal integrity at the load, and with source termination you cannot daisy chain loads because the ones not at the end will "see" multiple copies of the signal.

Bob

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Reply to
BobW

Reply to
John Fields

Is the scope terminated at 50 ohms, or is it just a 50 ohm BNC jack into the usual high-impedance scope input?

If it's the high-impedance input and you _really_ need the line terminated, then use a 93 ohm termination. You can roll your own with a

91-ohm resistor, that'll be plenty close enough.

If the scope really does present a 50 ohm termination, then you have plenty of suggestions already.

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Tim Wescott
Wescott Design Services
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Reply to
Tim Wescott

Any signal.

If the generator behaves like a real 93 ohm source....

___________ | | | 93r gen | ------------------------ any | |--| 93r coax |----------load | | |------------------------| | | |--+ -----------+ ___________

Then, for any signal waveform and for any coax length, the waveform at the load is exactly the same as if there was no coax present....

___________ | | | gen | | |------load | | | | |-------+ ___________

except for the coax time delay, of course.

That's "source termination", namely driving an X ohm coax from an X ohm generator.

That's handy in fast logic systems. If gate A drives a 50 ohm trace that runs to gate B, and you end terminate at B with 50 ohms to ground, gate A sees a 50 ohm load, AC and DC, and that's hard to drive. But if you put a series 50 ohm resistor at A, before the trace, and don't terminate at B, it works just as well, except that driver gate A now sees a 100 ohm AC load and an infinite DC load, which saves a lot of power.

John

Reply to
John Larkin

If a line is 1/4 wavelength, and the far end is open, it looks like a dead short to a sinewave drive. Numbers like 1/20 are closer to invisible. It really depends on the impedances and how much precision you want.

John

Reply to
John Larkin

I expected at least 1 bad John Conner joke out of this thread (did I miss it?). Instead I actually learned something useful. Boy, this newsgroup sure has gone downhill! LOL

Jim

Reply to
James Beck

--------------------------------+ | | / But there's no reflection! \\ / OF COURSE. I'M A TERMINATOR. \\ | | gnd

Reply to
John Larkin

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