I am loathe to admit my ignorance in a public forum, but I do want to learn
something so I have to ask why I want a gain of two? I am not trying to fool
anyone here, its not something I know a lot about and would rather not be
sitting in the back room of this place in a month, trying to figure out why
what I did does not work. I would prefer to look silly up front and get it
right, if you know what I mean
No, its just that I have never video distribution before, and will listen to
any and all advice before commiting to a design that is flawed from the
begining. Some would probably tell me not to do it at all, but there is enough
help and wisdom in the world that I will give it a try. I need the money, and
I am sure I can figure it out as I go.
Thank you for the artwork. It came across fine once I switched the fonts. I
appreciate you taking the time to draw it up and explain it.
Much obliged to you,
John
Well, I have to repeat, I'm not a video guy. Most of what I know about
video I learned from datasheets. ;-)
Anyway, I guess I'll address this by giving a quick review of transmission
line basics.
Typical transmission lines have a characteristic impedance. For radio and
microwave equipment this is usually 50 Ohms. For video it is usually 75
Ohms. Other values are also possible. Twisted pair might have around 80
or 100 Ohms or 150 Ohms.
But lets stick with video cable which is 75 Ohms.
What does the characteristic impedance mean? It means that if the 75
Ohm cable were infinitely long, it would look exactly like a 75 Ohm
resistor as far as the driving buffer (your buffer) is concerned.
But with finite length cables, the signal eventually reaches the end of
the cable and might reflect back towards the source. This can lead to some
messy interference on the cable, but there is a simple way to avoid it.
You just put a 75 Ohm resistor at the far end of the cable. As far as I
know, all video equipment will be designed this way. That is, it will look
like a 75 Ohm resistor to whatever drives it. Because of the way
transmission lines work, a cable terminated with a resistor of the same
characteristic impedance looks just like the resistor, as far as the
driver is concerned. It doesn't matter if the cable is 1 foot or 1000
feet. It will present a 75 Ohm load to the driver.
So, now you know that your buffer needs to drive a 75 Ohm load. That
still doesn't explain why you need a gain of two instead of one.
Sometimes the resistor (or whatever) at the end of the cable is not
exactly 75 Ohms. When this happens, a little bit of energy will reflect
back towards the source. When it gets to the source, depending on the
impedance presented by the source, some of that energy will reflect back
towards the load again. If you don't have the series 75 Ohm resistor,
virtually all of the energy will be reflected back toward the load, and
this can lead to interference. But WITH the 75 Ohm resistor, almost all of
the energy will be absorbed. This will lead to the cleanest signals,
especially if long cables are involved.
There is another reason to have the 75 Ohm resistor, too. When the buffer
is not attached to a load, the op-amp will still see a small capacitance
as a load. This capacitance can cause the op-amp to oscillate and possibly
overheat. The series resistor will isolate this tiny capacitance and keep
that from happening. The series resistor will also provide pretty
bullet-proof short circuit protection.
So now that we have this series 75 Ohm resistor, we need a gain of two
because the load at the end of the cable and the series resistor will
share the voltage equally. That is why you "need" a gain of 2. If you
could be totally sure that no one would ever short circuit it, leave it
open, or connect it to a load that was way off of 75 Ohms, you could drop
the series resistor, and stick with a gain of 1. But I would strongly
recommend that you include the series 75 Ohm resistor.
As a general rule, whenever you have a long cable, you will have the best
signal integrity if both ends of the cable are terminated in the
characteristic impedance of the cable.
Oh, since the op-amp is a voltage source, we want a series resistor to
terminate it (just like I showed in my ascii art diagram). If we had a
current source (this is rare, AFAIK, in these types of circuits), we would
want a shunt termination, just like we use at the load end.
So there are a couple of main things you should take away from this:
1) You want a series termination for each 75 Ohm load you drive.
2) You want to make sure that whatever drives your board from the outside
world sees your board as a 75 Ohm load.
3) You will need to keep your input and output signal traces as short as
practical, and, if possible, choose a trace width that will give you 75
Ohm impedance for the trace. (There are equations for this.)
If you have some video signals on the board which originate on the board
and do not leave it, you probably don't need to worry about impedance
as much for those signals.
I hope this helps!
--Mac