cautery@montac.com said:

It's a 1 Hz repetition rate, but the bandwidth depends upon the rise time.
If the rise time is ballpark of 1 ns, the bandwidth will be ballpark of 1
GHz.  So, yes, you want low inductance.  That includes the power to the chip
as well as the resistors.  Surface mount is your friend.  So are ground/power
planes.

How good is your scope?

The FIVE is 6-1.  The one is for isolation.  The 5 is the rest of the
package.  You might as well use them all as drivers.  You don't want to use
them for another signal (even if it is supposed to be identical) or you will
get minor crosstalk when you do things like plug or unplug a cable.

I don't know why you are saying "gate impedance".  That's over on the input
side.  I would have said "output impedance" or "driver impedance" of the chip
or section.

I'm not sure why they picked 100 ohms.  Assume the net source impedance is 25
ohms.  Suppose the far end is terminated with 50 ohms.  There won't be any
reflections so the source value doesn't matter.  25 ohms will provide a
higher voltage at the far end than 50.  If you have a CMOS driver and a CMOS
receiver, 1/2 the voltage at the receiver is nasty.  It might be OK if you
have HT type receivers.

I would suggest a bit of lab work.  What are you going to use on the far end?

Lots of gear has 1000 ohms rather than 50 so a 50 ohm source impedance takes
care of the reflections and leaves (almost) the full voltage at the receiver.

cautery@montac.com said:

More than 250 divided by 5 and rounded up a bit for the output impedance of
the chip will be more than 50 ohms.  I'd do some experiments.

--
These are my opinions.  I hate spam.

Read the spec sheet on the part you are using for a driver.
The size of the resistor on the paralleled driver side will be set by the
maximum current of the devices.

Decide how conservative a design you want.
Do you want it to drive into a shorted load and survive?
[If each output is rated at 20 mA, in a 5 Volt system, then R=E/I, = 250
Ohms]

Or do you only want it to work into a 50 Ohm load (to ground)?

Or perhaps only a Thevenin load? 50 Ohm load, but 100 Ohms up to +V, and
100 Ohms to ground.

Remember that 50 Ohms in a 5 Volt system will draw 100 mA on the 'High'
which can be a lot of current
for a modern IC.

--- Graham

==

On Sat, Jun 17, 2017 at 4:01 AM, Hal Murray hmurray@megapathdsl.net wrote:

On 6/17/2017 4:01 AM, Hal Murray wrote:

Not nearly good enough, as I don't have one.  Have to borrow.  Thank you
for pointing me in the right direction on bandwidth...

Yes, I got this part... At least I generally get something.  ;-)

Apologize, was parroting the term used in the archived Time Nuts post I
found referencing the need to address the "driver impedance" in
selecting series resistors when trying to set the output impedance.

OK... I'll have to try and study this some more, because for a simpleton
like me, and impedance mismatch between the PPS out port on the Nortel
and the device to which it is connected of 25:50 ohms, reduces to 1:2,
flipped around is a 2:1 VSWR, which means a fairly significant power
loss...  Maybe it doesn't matter much with a short cable.. but the cable
is also tiny/lossy.

Yes, I intend to do extensive testing.  I was just trying to trim away
some of the waste up front... time, materials, et al.  :-)

Yep... gonna need some more instruments asap...  I may just have to wing
it and accept "good enough" until I can round up the rest of the
required instrumentation.

Thanks for your assistance and not brutalizing me for my lack of
knowledge.  I'm trying to learn.

• C. Autery