# characteristic impedance

• posted

characteristic impedance of a line is square root of L/C. It does not depend on the length. But the losses occuring in resistance, no matter how small, is dependent on the length. Any literature to read on how the length affects the charactristic impedance. When doing circuit analysis, will this charactreritic impedance be added to the series connecting element? I have a device that says input impedance 50ohm, i am not using it. But to use other part of the device i have terminate it with 50ohm. How can i do it? can i just connect a 50ohm characterisitc impedance cable and leave it? thanks in advance

• posted

Zo is sqr(L/C), to a close approximation, provided the following is true:

1. The frequencies of interest are well above the skin effect knee frequency. Above this frequency, R(s) is fairly constant (but not completely).

1. Rs is negligible (per unit length) relative to Zo.

As noted, you should read up on transmission line theory and filter theory (which is where that equation comes from). For most circuits, provided the above are true, you can simply terminate to a resistor, although many circuits designed for high frequency have terminators internally. Some do, some don't.

This is something you can spend years on and still not understand completely (there are many things in this field which have been found empirically).

Note that to get a decent 'feel' for characteristic impedance (and transmission line theory in general), you should be comfortable analysing in the frequency domain and dealing with S-parameters (there's some more reading).

Cheers

PeteS

• posted

Something like that.

Most texts leave out the resistance as negligible, but I've seen it figured in - just don't remember where.

Read up on VSWR, reflection coefficient, return loss, ... If a 50 ohm source is connected to a 50 ohm line terminated with 50 ohms, you'd ignore the line impedance, but the line loss is another story. There's different kinds of analysis, so exactly how you do it is another story.

No. You'd want a 50 ohm resistor. Whether or not you use the line is up to you, but it would need to be terminated. You can buy terminators that mate to different connectors. Good connections like that keep water out and RF leakage down. You should probably terminate the input whether or not you use it.

And your secret device isn't worth keeping secret, so if this doesn't answer your questions, you'll have to spill the beans.

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Best Regards,
Mike```
• posted

I read in sci.electronics.design that Krish wrote (in ) about 'characteristic impedance', on Sun, 27 Mar 2005:

The full expression for characteristic impedance is (use Courier font): ________ /R + jwL Zo = /--------- \/ G + jwC

where R = series resistance per unit length L = inductance per unit length G = shunt conductance (dielectric loss) per unit length C = capacitance per unit length

No. Use a 50 ohm resistor.

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Regards, John Woodgate, OOO - Own Opinions Only.
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• posted

In real coaxial cable, the effect of resistance can be specified as a certain amount of dB power-loss per foot of cable. For example, if the cable has 10 dB of loss per foot, then 5 feet of cable would absorb 50 dB of the total power.

Another effect to keep in mind is that as the frequency goes up, the dB losses per foot usually go up. So if a particular cable has 10 dB of loss per foot at 500 MHz, it may have 50 dB of loss per foot at 1.5 GHz. Note that I'm just making these numbers up.

So, to answer your question, this power loss doesn't really effect the characteristic impedance of the cable. To a first approximation, the loss can be modeled separately from the characteristic impedance.

Note that the above discussion assumes you have a resistive load at the end of the transmission line whose impedance is the same as the transmission line's impedance. For example, if it is a 50 Ohm transmission line, you should have a 50 Ohm resistor (or 49.9) at the end as a load.

And to answer your other question, the best way to terminate a device at frequencies up to UHF is to use a small surface-mount resistor (49.9 Ohms in this case) as close as possible to the device. If you can't get close to the device, for some reason, use a 50 Ohm cable with a 49.9 Ohm resistor at the end of it. You can also buy 50 Ohm terminators for cables and connectors. If the device has some kind of an RF connector on it, it may be easier to use a terminator for it.

If the frequencies are down around 100 MHz or less, then you don't really have to be all that close to the device or do anything fancy. Just put any type of 50 Ohm resistor across the terminals somehow.

--Mac

• posted

in theory yes, a VERY long cable will look like 50 Ohms due to it's losses ...

in practice, what you want is a termination plug

just by a 50 Ohm termination plug to connect to your source.

mark

• posted

It will indeed look like 50 ohms - but this is nothing to do with its losses. It is simply that power is propagated rather than dissipated. Provided none is reflected back while you are measuring, the cable will appear to be a 50 ohm resistor.

d

Pearce Consulting

• posted

A bit of thinking is required. Apart what has been said, there is DC termination and there is AC termination. The AC termination being an additional cap in series with the 50 ohm. It all depends whether a DC is required or not.

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