Coax cable - capacitive load

Question

I still have an understanding problem with impedance matched coaxial cable and its behaviour.

We do impedance matching to prevent reflections.

How does my signal source see the whole cable? At a certain frequency it sees only 50 Ohms. How do I calculate that frequency? What does my signal source see at lower frequencies? A capacitive load, depending of the length of the cable?

I already read many articles and posts, but it is something which is still not clear to me.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

how does my signal source see the whole cable?

The characteristic impedance (\$Z_0\$) of any transmission line be it coax or twisted pair (screened or unscreened) is determined by: -

$$Z_0 = \sqrt{\dfrac{R + j\omega L}{G + j\omega C}}$$

Where R, L, G and C are the resistance, inductance, conductance and capacitance of the actual line per unit length. Because it's per unit length, we can choose any length we want to measure those numbers and get the same answer.

At a certain frequency it sees only 50 Ohm ?

If we assume the frequency is quite high (about 1 MHz or more), the above equation can be simplified because \$\omega L\$ and \$\omega C\$ dominate over R and G hence we get: -

$$Z_0 = \sqrt{\dfrac{j\omega L}{j\omega C}} = \sqrt{\dfrac{L}{C}}$$

So, a typical value for L might be 250 nH per metre and C might be 100 pF per metre and this gives: -

$$Z_0 = \sqrt{2500} = 50 \text{ ohms}$$

Above approximately 1 MHz, the characteristic impedance is resistive at some fixed value (quite often 50 ohms) up to the GHz area when other things happen.

How do I calculate that frequency?

It's usually around 1 MHz but, you have the full formula and if you wanted to know how things shape up below 1 MHz, the formula tends to become this for any practical cable with negligible conductance (G) : -

$$Z_0 = \sqrt{\dfrac{R}{j\omega C}}$$

That formula dominates for the majority of the audio spectrum such as a cable like this: -

Picture taken from this wiki site and please note that there is an error in the x axis - it should say "300 k" and not "3 M".

What does my signal source sees at lower frequencies? A Capacitive Load, depending of the length of the cable?

Not quite, the ratio of R to \$j\omega C\$ is easy to understand but the square root of it (and in particular the "j" term) implies a phase angle of 45 degrees.

So, we do Impedance matching to prevent reflections.

Yes, but at low frequencies this is usually pointless because the length of the cable is normally so short compared to the wavelengths of the (audio) signal that reflections are trivial.

What wiki says about coaxial cable R, L, G and C: -

New circuit disclosed by the OP

A picture has appeared showing a differential to single-ended driver with a 50 ohm termination to a length of coax. The coax is terminated in 50 ohm. Given that there is no explanation, I have the following to add: -

• There doesn't need to be a termination resistor at both ends for this to work. It's quite reasonable to use a series driver termination and have the coax far-end open-circuit terminated.
• The advantage of a single driver series termination is that any reflections from the open-circuit load become dissipated at the driver end AND, more importantly, the voltage seen at the receiver is exactly what is transmitted i.e. there is no loss.
• A termination at both ends is missing a trick and the output voltage is 50% of the driver signal voltage.
• Coax cable screens must be grounded, and for best quality, this should be done at both ends BUT....
• If it can't be done at both ends then ground at the sending end and use a differential receiver to adequately avoid noise pick-up AND...
• Grounding should be done using an impedance equal to the impedance used in feeding the inner conductor. This significantly improves EMI susceptibility issues.

This would be my recommendation based on the limited information supplied by the OP: -

But, it might work adequately with 50 ohm on the inner and a hard ground at the shield at the driver end. It’s still unclear what the question addition is all about.

Answer 2

So, we do Impedance matching to prevent reflections.

Usually, yes. Actually, we do impedance matching for maximum power transfer; you can measure how much power is not transferred by how much is reflected.

So, no reflections is a symptom of what we wanted to achieve: that the maximum power is transmitted through the transmission line

how does my signal source see the whole cable?

Not at all.

If it's matched, it only sees the wave impedance at the point of entry into the transmission line.

At a certain frequency it sees only 50 Ohm ?!

Again, yes, if the transmission line has a 50 Ω wave impedance at any particular frequency, that's what it sees.

How do I calculate that frequency?

This question makes no sense – you're the one with a system that works at a specific frequency, and you match your system and the transmission line for that specific frequency.

What does my signal source sees at lower frequencies?

Some other impedance.

A Capacitive Load, depending of the length of the cable?

Capacitive, ohmic, inductive: Depends on what the actual transmission line wave impedance and length are; there's no general statement, aside from saying:

A matching is the only way the effective wave impedance seen by the source doesn't depend on cable length.

Note that "matched" doesn't mean "real-valued" impedance. If your source has a complex impedance, then the matched impedance is the complex conjugate of that – and still complex.

Answer 3

how does my signal source see the whole cable?

Let's start with simple, lossless, uniform, coax. There are more complicated systems that other people will be able to tell you about.

If the cable is terminated in its characteristic impedance, which we'll assume is 50 Ω, then the cable input looks like 50 Ω. The important thing here is that for lossless cable, that's true for any waveform, at all frequencies including DC, and for any length of cable.

^{simulate this circuit – Schematic created using CircuitLab}

Here are two ways we might describe a transmission line line as 'terminated' (there are others). In your comments, you talk about using 3m of RG316 coax cable. This length has a loop resistance of about 1 Ω, while the cable has an impedance tolerance of about 2 Ω.

In the upper diagram, the opamp will see a load of about 25 Ω. This will be essentially resistive from DC to 100 MHz. Any capacitive or inductive component will be insignificant. Very few common amplifiers will be able to drive this without trouble.

In the lower diagram, the opamp will see a load of about 100 Ω from DC to 100 MHz, with the same caveats.

We use slightly different models to explain time domain and frequency domain behaviour. Time domain involves steps and impulses in time, which have a very wide frequency spectrum. Frequency domain descriptions tend to use single frequencies, which focus on long term behaviour and ignore the initial transient. Remember they are both true, and any apparent conflict is one of language and the model's domain.

If we throw a step input at the cable, so a broad range of frequencies including very high frequencies, the cable input looks like 50 Ω initially. It looks like 50 Ω for as long as it takes the step wave to reach the far end of the cable. If it finds 50 Ω there, it doesn't get reflected, and the cable input continues to look like 50 Ω indefinitely.

If the step finds an open circuit, then it gets reflected voltage in phase, current in antiphase, and when the step reaches the input again, the voltage doubles and the input current drops to zero. At low frequencies in the frequency domain, when the cable transit time is very short compared to the period of the signal, this open circuit behaviour looks capacitive at the input. With a short circuit on the output, the low frequency input behaviour looks inductive.

With a long line, the behaviour is more interesting. If the output is open circuit, then at a frequency for which the line is a quarter wavelength long, the input actually looks like a short circuit. This special line length is much used in filters and other components as this impedance transformation is so useful. However, whatever the frequency in the frequency domain, an open circuited 50 Ω line will never look like 50 Ω at the input, only short, open, inductive or capacitive depending on the length and frequency.

With a lossy line, the behaviour is more complicated. Once the series resistance becomes a significant fraction of the impedance, it can no longer be ignored. This is complicated by the fact that at RF, the skin depth effect increases the effective resistance.

Answer 4

The impedance of a cable comes from the concept of an ideal cable (no resistance, no dielectric loss) with infinite length and constant properties over its length.

If you power such cable w/ DC, you will see a constant and finite current as the voltage change propagates along the cable and charges the capacity of the cable along. The cable is effectively a resistor.

You can power the same cable w/ AC - the cable will show the same resistance that emerges by the same mechanism.

If you cut the cable at some point and replace the rest of the infinite part with a resistor with the same resistance as the infinite cable, nothing will change from the power source point of view. The input side of the cable will still behave as the same resistor.

Answer 5

The cable is a frequency dependent impedance. Its impedance depends on its termination and its electrical length. At very low frequencies where the cable is electrically short, it seen as a capacitance, because the series inductance is just simply too small at very low frequencies to make the cable a distributed system. The frequency where it will show its nominal characteristic impedance is where the impedance of the series inductance starts to dominate over the series resistive impedance.