[ntp:questions] Oddities in termination of cable from gps18.
unruh at invalid.ca
Thu Feb 23 21:43:15 UTC 2012
On 2012-02-23, David J Taylor <david-taylor at blueyonder.co.uk.invalid> wrote:
> "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,
>> 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
> 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!
Nope, I was always talking about the GPS 18 LVC ( not 18x either).
> 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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