Every source has an effective resistance, defined by it's voltage to current ratio. R=V/I Measuring may be a bit tricky, but that is what it is.

 

From the below data sheet info, you can determine that the high output resistance is typically about (5.5V-4.86)V / 24mA = 26.7Ω, and the low output impedance is 0.36V / 24mA = 15Ω.
So that should have only a small effect on the parallel impedance of the 5 outputs.

Edit: Redid the calculation to use 5.5V, not 5V.

 

Another question related to this topic
I cannot quite understand how the scope sees 2.5V when an incident wave of 5V is supplied to one end of the coax and the other end is left open. I can understand that this is a voltage divider (50ohms output impedance of the step generator and 50ohms of the coax cable), but since the other end of the line is open, how is this voltage divider being created?

 

The characteristic impedance of the coax is distributed along the length of the cable, not lumped at the open end. What you have not asked about is: "what happens when the wavefront reaches the open end and encounters the impedance discontinuity"?

 

But what i cannot understand is how there is voltage divider if the other end of the conductor is not grounded?

The signal doesn't see the open until the time it takes to reaches the end of the coax.
Before that, the signal sees the 50 ohm impedance of the coax.

When the signal reaches the end it jumps to 5V, which then travels back down the line until it reaches the near end where the signal at the generator 50Ω load then goes to 5V.
This is shown in my post #16 simulation.

 

But what i cannot understand is how there is voltage divider if the other end of the conductor is not grounded?
When open ended, the equivalent circuit will be like the one on top, correct? A circuit wired this way will give you 5V as there is no ground reference.

View attachment 268227

Let me repeat this slowly. Circuits with transmission lines ARE NOT EQUIVALENT to circuits with lumped parameters. Lumped parameter analysis only works on circuits that are "small" with respect the wavelength of the signals in them. The rising edge of a step function has considerable high frequency content (equivalent to very short wavelengths) as would be evident from a Fourier analysis. For some period of time after the rising edge of the step function, what is at the other end is completely unknown and unknowable. For that period of time the unterminated coax behaves as if it was terminated with its characteristic impedance which is equal to the distributed impedance along the cable.

 

The signal doesn't see the open until the time it takes to reaches the end of the coax.
Before that, the signal sees the 50 ohm impedance of the coax.

When the signal reaches the end it jumps to 5V, which then travels back down the line until it reaches the near end where the signal at the generator 50Ω load then goes to 5V.
This is shown in my post #16 simulation.

The simulation in Post #16 shows another feature of transmission lines and impedance discontinuities. An incident wave can do all or part of each of three things at a discontinuity: it can be transmitted, it can be reflected, or it can pass through. In our case of an unterminated line substantially all of the incident energy will be reflected back to the source. In practice this is normally a situation you want to avoid, because depending on the power levels the source may be damaged by the returning signal power. The situation you normally want is for the signal power to be transmitted through or be absorbed by the discontinuity.

 

Really, it IS a lot more complicated than one might guess based on a DC analysis.

 

Really, it IS a lot more complicated than one might guess based on a DC analysis.

Systems governed by partial differential equations can have many surprising and counter intuitive properties.

 

I've been watching this channel for years

The way this works is reflections.

There is 2 types of circuits - Imagine a bathtub slowly getting filled up: the water height is pretty much the same at every point. The other type is where I get a bucket of water and throw it in one side. The wave goes down and bounces back.

You might also be interested to know that w2aew channel has more related videos on this subject, and you actually don't need any circuit to generate a sharp edge...

 

But what i cannot understand is how there is voltage divider if the other end of the conductor is not grounded?

It is not a divider. It is a subtracter.

ak

 

Today I connected a capacitance meter (UNI-T UT601) to one end of the coax. I got a reading of 3.15nF. The datasheet states "Capacity Core/Shield = 100pF/m". The cable is 32.75m long. If 3.15nF is divided by 32.75m, one gets 96pF.
Lets say that we have a coax with unknown Capacity Core/Shield, is this a reliable way of how to determine the capacitance?

 

The readings appear to be close to what the specification would predict, it appears, within 5%. Allowing for measurement errors plus production variability, this appears to be a reliable method.
I do wonder about the effect of the surroundings on the results, though.

 

It is not a divider. It is a subtracter.

How so?
For the leading edge of the pulse, the source output impedance acts as a voltage divider with the coax characteristic impedance.

 

I do wonder about the effect of the surroundings on the results, though.

Experimented with this back when I was building a TV station. Three cable groups: multiple cables in a bundle with wire ties, multiple cables together but unbundled, laying in an overhead raceway, and individual cables not near anything. All cables were one of two types of precision 75 ohm video coax. Lengths varied between 50' and 250'. The capacitance measurement method worked pretty well, always less than 5% error. The errors were both short and long, so the average over one 30-cable bundle was close to zero. We found no significant accuracy differences among the three groups.

Because we had a brand new 500 MHz Tek 7000 mainframe scope, we also whipped up a TDR circuit with Schottky TTL (this was 1972 or 73). Due to the limits of the display, the effective accuracy got worse as the cables got longer. However, it did find a kink in one cable that we suspected. The TDR gave us an approximate location, and the problem was about 4 feet from that. Saved hours of semi-dangerous crawling around in a suspended raceway. Not bad for just fooling around on the night shift.

ak

 

TDR operation can result in reflective waves that are also related to a cable's velocity factor.
https://portalvhds963slh4m3fqg2.blob.core.windows.net/megger-products/TDR_AN58_EN_V02.pdf

Practical use:

with this short introduction you can go further and compare a known good cable with an unknown cable of the same length.
Unlike a check on the copper the cable's dielectric might be faulty but also the distance to fault can be found.