The formula for channel loading on a standard optical DFB laser (Philips Diamond Transport series) is as follows:
The required RF input level, and the +/- 3dB AGC window centered around it, changes with the number of channels loaded as shown in the following formula:
-10*log(# of channels/77) = adjustment to input level.
For 77 channels, the input level is +15dBmV and the AGC window is +12dBmV to + 18dBmV. An RF signal outside of this window cannot be adjusted to the reference point using AGC.
For 40 channels, the input level will be -10*log(40/77) = 2.8 higher or 18dB. The AGC window is +15dBmV to +21dBmV. An RF signal outside of this window cannot be adjusted to the reference point using AGC.
This was taken from their catalog spec's. I believe that this basis is a good rule to follow since the formula is based on optical transmitter power.
Good luck!!!
> I could really use that channel loading formula you were going to post for me. Your help is appreciated.
>
> Re: Channel loading * CORRECTION *
>
> by Buffalo Chips, 4/24/2002 10:35:14 PM , IP: 63.26.236.166
>
> --------------------------------------------------------------------------------
> I gave you the wrong formula by mistake. I will put the correct one on ASAP. The rest of the information is correct though. Sorry, I must have had a bad day that day...lol.
>
> > With a standard DFB optical transmitter you would need to find out the manufacturer's reccomended input level at full (100%) channel loading.
> >
> > Then,
> >
> > input level = 10log(# actual channel's) - mfg. rec. level
> >
> > I hope that this helps. Also remember that whenever you use half (50%) of the maximum channel loading (100%), you have a 3dB difference in overall input. This means you could increase the input by 3dB.
> >
> > EXAMPLE: You have a DFB laser with a maximum input 15dBmV with 80 NTSC channels per the manufacturer spec's.
> >
> > In reality, you only have 40 active NTSC channels. Since you have half channel loading (50%), you should increase input by 3dB to 18dBmV for better c/n performance.
> >
> > If you have 20 channels, ditto. Another 3dB difference to 21dBmV input.
> >
> > Just like a 2-way splitter. You split the input in half, you lose 3.01dB in theory. Reality you lose 3.4 - 4.0dB depending on the quality and frequency of the splitter. In optical input power, you are much closer to that 3.01dB theory number than with coaxial splitters.
> >
> > Good luck!
> >
> >
> > > Help! What is the formula for figuring a power level based on channel loading,e.g. I am allowed a maximum RF input level of +60dBmV on a device based on 1 NTSC channel loading and have 118 analog channels, what is my maximum allowable input level? THIS IS JUST AN EXAMPLE, I AM LOOKING FOR FORMULA.
> > > Thanks for your help.
>
