Simple transmission line question

ts a 5% tolerance too. Just pretend you don't see the 10mV overshoot, it's just monitor front end parasitics ....

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Say If you had a long transmission line could you send a pulse down it, with no source termination, and then switch in the termination for the return pulse? Waste energy only when you need to. (Assuming the pulse width is less than the length of the T-line.)

George H.

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Sure, All you gotta do is make sure the voltage at the source end of the line doesn't change when you insert the resistor.

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I guess you could drive the line with current, turn off the current, then terminate.

Anyone want to write-up what waveform you'd see ?:-) ...Jim Thompson

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Ok, but to what voltage do you terminate?

If you start with a discharged line and stuff some electrons into it, the terminal condition will be V=Q/C. The voltage on the line may never go to zero. The logical line voltage is whatever the voltage was at the source end of the line when you decided to insert said termination. And the at the line to which you tie the resistor and the current into the line at that instant all determine the voltage on the other end of the resistor required to keep the line voltage constant. Simple in theory, but rather difficult if you don't know what's on the other end or the elapsed time or the width of the pulse or...or... Doing it all dynamically likely takes WAY more energy than you'd save.

If you pick the right voltage, there's one reflection at the end that gets absorbed at the source as the resistor current goes to zero.

"dave" "Phil Allison"

So the wave travels at 200,000,000 m/s.

... Phil

a 5% tolerance too. Just pretend you don't see the 10mV overshoot, it's just monitor front end parasitics ....

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Transmission lines have a mind of their own ;-)

Take ANY line, no source termination, I care not what the far-end termination, hit it with a current pulse, say 1 Amp, for a time shorter than the out and back transit time... what voltage will you FIRST see at the input to the line?

Answer: A STEP of voltage of value (1A * Zo)

What you see after that will depend on the far-end termination. ...Jim Thompson

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I'm toying around with just such a design. Imagine a 50 ohm pullup resistor from Vcc and a fast transistor that can pull it to ground. The transistor collector drives a coaxial cable, through a series capacitor, out to the load.

If I turn the transistor on hard, for some 10s of picoseconds, it launches a pulse of -Vcc down the cable. By the time any reflections return, the transistor is off and the 50 ohm terminator is back in place.

a 5% tolerance too. Just pretend you don't see the 10mV overshoot, it's just monitor front end parasitics ....

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It doesn't even have to be a short pulse. Just apply a 1 Amp step of current and watch the fun >:-} ...Jim Thompson

tolerance too. Just pretend you don't see the 10mV overshoot, it's just monitor front end parasitics ....

Try two 100 ohm resistors, "Y" from cable to ground, about 60 degrees?

o 450

gets a 5% tolerance too. Just pretend you don't see the 10mV overshoot, it' s just monitor front end parasitics ....

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I was thinking you could do it either way. 50 ohms to ground with a current source or added in series with a voltage source. Hey maybe just a diode in parallel with 50 ohms driven by a voltage source... ohh time to play with Jeroen's spice file a bit!

George H.

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so 450

gets a 5% tolerance too. Just pretend you don't see the 10mV overshoot, it 's just monitor front end parasitics ....

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Fun! I don't have any use for such a scheme, but if you had an unknown or variable load at the end of the coax it might be a nice trick.

George H.

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wave

Do I multiply 75 X 0.87 = 62.5?

gets a 5% tolerance too. Just pretend you don't see the 10mV overshoot, it's just monitor front end parasitics ....

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Where do you get that "10s of picoseconds" transistor? ...Jim Thompson

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Typical coax is around 0.66 or thereabouts...

gets a 5% tolerance too. Just pretend you don't see the 10mV overshoot, it's just monitor front end parasitics ....

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The load in the case at hand would be a lithium niobate e/o modulator, nominally

gets a 5% tolerance too. Just pretend you don't see the 10mV overshoot, it's just monitor front end parasitics ....

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Hey, George, Jim is asking you where to get those transistors. You'd better answer him or he might get violent.

I use Times Microwave LMR series. (Or BuryFlex if I don't have to pass code.) Velocity [sub] Propagation depends on the dielectric constant of the spacer between the outer surface of the inner conductor and the inner surface of the outer conductor. I did not have to look that up. I grew up reading Reference Data for Radio Engineers (Bill Orr.et al), and have some of it memorized.

There is always a reflection from the unterminated end. The 50 ohm source termination absorbs the reflected energy.

A hi-Z 'scope input isn't an open circuit, BTW, there's capacitance to consider.

My Gould pulse generator's 50V-into-50 ohms output module has a source impedance of about 7 ohms. I made a 43 ohm source terminator (Caddock 10W precision resistors) for that. Minimum ringing from an open cable.

That's not coax, that's pipe >:-}

I didn't have to look it up either. My mind-built-in rule-of-thumb (*) says typical coax looks like 1.5x air length.

I remember a Bill Orr from my Motorola days, '62-'70, applications engineer... probably the same guy.

(*) "Rule-of-thumb", dates from old English law... the size of the stick you can legally beat your wife with is the same as the diameter of your thumb ;-) ...Jim Thompson