[ntp:questions] Oddities in termination of cable from gps18.

David J Taylor david-taylor at blueyonder.co.uk.invalid
Thu Feb 23 17:47:03 UTC 2012

"unruh" <unruh at invalid.ca> wrote in message 
news:oxu1r.10154$I%6.2578 at newsfe16.iad...
> The Sure is only about a 2m cable, not 20m, and the TTL output on the
> Sure seems to work fine into both the parallel port ACK and the Serial
> port DCD interrupts. Ie, there seems to be no need to use the RS232
> converter. And I would rather not introduce delays into the chain unless
> I have to.

OK, so there are no problems for you, that's fine.

> What I still find weird is that all that the terminated line seems to do
> is to decrease the signal, while the unterminated line (open end) has
> the full 5V level with no ringing.

Essentially you have a resistive source, and you are loading it with a 
resistive termination.  Ohm's law should apply (except that they are pure 
resistances etc.).

> But let me get my assumptions on the table-- if one has a cable with an
> infinite resistance termination does the reflected signal have the same
> polarity (in the E field) as the incoming or opposite?

Same.  Short circuit would have the opposite, making the net signal zero.

> If it is the same polarity, then what I am seeing makes sense-- Say the
> output on the Garmin GPS has about a 400 ohm serial  output impedance, 
> so
> driving the line (100 ohms) gives a drop of voltage by about 1/5, but as
> the signal is reflected back with the same polarity, the effective
> impedance of the line seems to rise, finally leading to the full voltage
> developing on the line. If terminated by 100 ohm, the driver continues
> to see the line as a 100 ohm load giving an output voltage of about 1V
> rather than 5V. That also seems to fit with the structure I see on that
> rising edge of the pulse measured on the terminal end (initial step of
> about 1V on the unterminated line.  I'll try to see if I can get a
> photo.

So you are now taking about the GPS 18x LVC, whereas some of the earlier 
conversation was about the Sure.  No wonder we are confused!

As I'm sure you know, to get a feeling for the output impedance of the 
Garmin, watch the full-width signal on the 'scope and try different 
terminating resistors.  When the voltage is halved, that's the output 
impedance.  The line is virtually open circuit at DC, or for the 100/200 
millisecond pulse produced by the Garmin.  The line's impedance is 
irrelevant, except for the small time taken to transit the line (and any 
reflections).  Yes, with a 100 ohm line, terminated in 100 ohms, what you 
say about the pulse is exactly correct.

Without the termination, there will be a reflection at the receiver, that 
edge is reflected towards the transmitter, and because the line is not 
correctly terminated at the transmitter end, there will be another 
reflection there and so on.  Exactly how these appear will depend on the 
initial risetime of the signal, the bandwidth of the transmission line, 
and exactly what the terminating impedances are (they won't be purely 

You do need to consider how the receiver will respond to the signal - 
whether it has hysteresis or not, and any filtering on the signal from the 
pin you are connecting to, including any filtering inside the chip.  Chips 
dealing with outside world signals may well have low-pass filters, and 
overvoltage protection.

Whilst TTL levels will often work, using "proper" RS-232 drivers will 
likely slow the edges sufficiently that on shortish lengths of cables (say 
up to 10m) you won't see any of the effects we are talking about.


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