[time-nuts] pulse height
shalimr9 at gmail.com
shalimr9 at gmail.com
Tue Nov 27 14:58:07 UTC 2012
Said, I agree with all you say. My web page does not intend to show what is best, simply what is. Many people are puzzled by the issues of line impedance. Best practices have been discussed at length on this list, but always theoretically. I thought a few scope pictures and explanations would be useful.
I did not think it would raise such an issue.
Didier
Sent from my Droid Razr 4G LTE wireless tracker.
-----Original Message-----
From: SAIDJACK at aol.com
To: time-nuts at febo.com
Sent: Mon, 26 Nov 2012 4:01 PM
Subject: Re: [time-nuts] pulse height
Hi Didier,
yes, if you put a 50 Ohm termination at the far end all looks good, but you
are still driving a 91mA DC current through the cable during the high
times, and that will have rippling effects on the driver board by loading the
5V power supply down with a 1Hz period.
And if you forget to switch on the 50 Ohms end-termination, you get ~10V
as shown in your plots, and you might just blow up your expensive instrument
:) One reason I don't like fast edges being driven by << 50 Ohms series
resistance.
Also, if you use 50 Ohms series termination, AND 50 Ohms end-termination,
you still get a 2.5V pulse, enough voltage to cleanly switch most of
today's logic inputs (either TTL or 3.3V LVCMOS).
There are just so many things wrong with the 5 Ohms termination, for
example what happens if you short that output to ground? What happens if you
feed a parasitic pulse into the line, say from a nearby lightning strike or
EMI or ESD event etc? With "proper" 50 Ohms series termination the pulse
should simply be absorbed if it is not unreasonably high and the resistor is
large enough. With 5 Ohms, the driver will likely fly off the PCB..
In terms of the incident wave switching issue that Hal mentioned, the wave
will stay at about 2.5V for a while, then go to 5V, - again enough voltage
to switch TTL and LVCMOS logic inputs cleanly. But then again it is never a
good idea to daisy-chain inputs via BNC-T's anyway's.
bye,
Said
In a message dated 11/26/2012 13:11:21 Pacific Standard Time,
shalimr9 at gmail.com writes:
Said,
I agree. I intended to complete the page by doing more tests, but the
interesting point of the demonstration is that it is sufficient to match the
cable at the far end, and in doing so, you preserve the full amplitude of the
pulse. If you put 45 ohms in series and terminate in 50 ohms at the other
end, you end up with half the signal. However you do not need to do that.
That was the reason why I wrote the page in the first place. I will try to
clarify that when I get a chance.
Didier
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