RF question about coax

[vk6zgo]

Although it is a sine wave that is fed to the LED,the LED is a diode,so it will  create harmonics.
It is doubtful that any of these will radiate,but bear in mind,they do exist.
As I said before,the coax is extremely short,compared to a wavelength at 2.3MHz,& even at those harmonics which might be generated.

PAL video cables carry frequency components at up to 5MHz &,if incorrectly terminated do not radiate.
(OK,before somebody says it,they are terminated correctly at the source end).

I feel you are over thinking this thing,& I would suggest,just go with your original idea,& see if you have any significant radiation.
Buy a cheap all band portable radio,& wave the thing around near the coax.Even though these things are junk radios,they are more than sensitive enough to show up any problems.
I know it doesn't sound very "EE like",but then,I'm just a Tech!

VK6ZGO

[jahonen]

Tried out the series resistor with the LED in LTSpice. Seems to work quite nicely, certainly much better than with no termination at all. Slight undershoot is visible (mismatch) but I think that it is in acceptable limits here.

Regards,
Janne

[SoftwareSamurai]

Tried out the series resistor with the LED in LTSpice. Seems to work quite nicely, certainly much better than with no termination at all. Slight undershoot is visible (mismatch) but I think that it is in acceptable limits here.

Very interesting! I'll have to try that!

So is that 47 ohms + LED resistance = 50 ohms?

Thanks Janne!

[jahonen]

Very interesting! I'll have to try that!

So is that 47 ohms + LED resistance = 50 ohms?

Thanks Janne!

Yes, that's basically it. Of course, LED "resistance" is not constant so termination is never perfect. But good enough, certainly. I assume that you have relatively high bias current and small frequency modulated sine on top of that?

Regards,
Janne

[SoftwareSamurai]

Yes, that's basically it. Of course, LED "resistance" is not constant so termination is never perfect. But good enough, certainly.

Makes sense. Got it.

I assume that you have relatively high bias current and small frequency modulated sine on top of that?

The "current version" circuit that I posted earlier is the one I need to modify for moving the LED to a "movable" platform. The op amp together with the transistor provide the driving current for the LED, which seems to work well.

The signal is an FM sine, 2.3 MHz with a +/-75 kHz modulation.

The signal itself comes out of an IC, to a 22n cap, then a 2k4 resistor, then to the op amp positive. The op amp positive input is pulled up by a two-resistor voltage divider:
+Vcc -- 24k -+- 2k7 -- GND.

[elCap]

Very interesting thread!

As understand it, it's only the LED that will be outside of the main board, would an optical fibre work?

[SoftwareSamurai]

As understand it, it's only the LED that will be outside of the main board, would an optical fibre work?

Perhaps so.

The LED I'm using has high-intensity and a fairly wide FOV. For my project, that's important because it needs to be movable (within certain limits), yet still received by the receiver board. Fiber optic cable is pretty good at transporting light, but only in a narrow FOV (AFAIK). I'd have to add a lens to it in order to regain the wide FOV.

But that is a good question!

[ejeffrey]

How big of a FOV do you need?  Cheap plastic optical fibers typically have a fairly high NA / wide FOV.  A 60 degree cone is pretty typical.  Glass fibers and those used for high speed communication (ethernet, not TOSLINK) are typically smaller, on the order of 10-20 degrees.

[SoftwareSamurai]

How big of a FOV do you need?

The LED I'm planning on using has an angle of half-intensity of +/- 60 degrees.
Still, your idea of using a fiber optic cable does have merit. I'll have to think about that.

[IanB]

This thread is full of discussion about 50 ohms. What exactly is the "magic" to 50 ohms? In a world full of numbers, why is 50 ohms special? Could someone explain?

[SoftwareSamurai]

This thread is full of discussion about 50 ohms. What exactly is the "magic" to 50 ohms? In a world full of numbers, why is 50 ohms special? Could someone explain?

I was considering the different ways I could relocate an LED off of a main board, but keep it connected. One idea I had was to use a micro-coax cable. (e.g. 0.81mm). Those types of coax cables are typically 50 ohm cables, as opposed to, say, CATV cable that's typically 75 ohm. So the question turned into how best attach an LED to a meter-long micro-coax cable, but still match the impedance of the cable. If you simply connect the LED across the coax, it won't work very well due to impedance mis-match. Janne suggested adding a 47 ohm resistor in series with the LED (which has a little resistance itself) in order to better balance the coax impedance.

As for "why 50 ohms", there's a pretty good write up at Wikipedia (http://en.wikipedia.org/wiki/Coaxial_cable), under "Choice of impedance". In a nutshell, and typical of engineers, it's the best compromise between "power handling capability and attenuation".

[amspire]

This thread is full of discussion about 50 ohms. What exactly is the "magic" to 50 ohms? In a world full of numbers, why is 50 ohms special? Could someone explain?

50 ohms is a common impedance for coax cables, but there are many other impedances around like 75 ohm coax and 100 ohm twisted pair.

In a way it is pretty "magic" and "special" because what is actually happening in these cables is amazing.

As you probably know, when two wires are near each other, these is a capacitance between them, so a very long cable has lots of capacitance.  This is a bad thing as capacitance tends to attenuate AC.

All wire also has a self inductance, so the longer the wires, the more inductive impedance which again attenuates signals.

Now if you have a pair of wires that are separated with a constant spacing, such as in coax or twisted pair, then at one particular ratio of voltage to current, the self inductance exactly counters the effect of the capacitance.  So no attenuation except for resistive and dielectric dissipative losses, even for very high frequencies.  This explains why you are able to send a 1GHz signal down a 10 meter coax cable with little attenuation, even though from the apparent capacitance, there should be massive signal loss.  This one particular ratio of voltage to current is the cable impedance and so if it is terminated at the far end by a resistor of the same impedance, a signal can travel down the cable with almost no deformation or attenuation.

What is actually happening is that each microscopic slice of the cable with its inductance and capacitance  is temporarily storing the energy from the previous slice, and then passes all the energy on to the next slice at somethings a bit less then the speed of light. So each microscopic slice is actually a bit like a memory, as each different slice could have a different voltage.  This sequence of differing voltages get faithfully passed from one slice to the next down the cable.

As an example, analogue oscilloscopes need to delay waveforms in the the channels by something like 5 - 100nS so that they can have the trigger firing before the edge you want to look at reaches the screen.  The way they used to do this was to have a long coiled length of 50 ohm coax. If the sweep rate was set to 1 nSec per division, the complete waveform you see on the screen was stored at one point in the delay cable.

Why it is magical is that with discrete inductors and capacitance, their impedance only cancels out exactly at one frequency. With the case of the distributed inductance and capacitance in a coax cable, it surprisingly cancels out at every frequency.

This is incredibly lucky for us. Without this, the phone system wouldn't work and it would be very hard to get the signal from your TV antenna to your TV.

Richard