On Saturday 10 October 2009, Roland Jollivet wrote:
>A twisted pair, or any long piece of wire coming out of the back of a PC,
> is first and foremost, a transmission line, and needs to be terminated
> properly. This has nothing to do with the DC bias. The impedance of such a
> line is generally 50R to 75R.
>
>In the case of a printer port line, which is unbalanced, the line might
> have a 1k pullup, to satisfy DC conditions, but also needs a signal(AC)
> termination, usually a RC combination, and is often of low value, maybe a
> 100R and small capacitor.
>
>In the case of a proper balanced line, there is usually a single low value
>resistor (50R) across the line as it enters the differntial input of the
>amplifier, but thats not the case here.
>
>Trying to satisfy both AC and DC requirements with a single resistor, may
>either overload the port line if it is too low, or allow signal reflections
>if it too high.
>
Or too low.

School time.  I hope I don't bore everybody...

This is why the scsi disk terminations are as complex as they are.  The 
impedance of that ribbon cable, with every other wire grounded, which is the 
case for scsi setups not labeled as LVDS, is nominally 120 to 122 ohms.

The resistive terminator pack has a ground on one end, and a 5 volt 
connection on the other, and there are pairs of resistors wired between the 5 
volts and ground, whose job it is to both protect the driver from over 
current (it is a wired 'or' circuit).  The resistors are connected at the 
junction of the 2 resistors of each pair, to one of the scsi data or control 
lines.  The idea was to have  a resting voltage of about 3 volts when nothing 
is pulling to line to a logic zero.  To achieve this AND emulate the 120 ohm 
cable impedance, thereby giving a perfect match, the two resistors are a 330 
ohm to the ground, and a 220 ohm connected to the 5 volts.  The formula for 
parallel resistors is 1/r=(1/r1 + 1/r2), so r=132 ohms.  This is generally 
close enough, and cannot in std resistor values (read cheap) be improved.  
But the ideal setup is whats called 'active' terminations, where a fairly 
high powered op-amp is biased to have a very low impedance output, quite well 
nailed to 3.0 volts.

You then connect a 120 ohm resistor from the op-amp output to the signal 
lines, one resistor per signaling line.
 
To get to what we are doing here, this active termination might be just the 
ticket but is needlessly costly.

Another tidbit of info:  Most of these receiver chips with differential 
inputs, need have only one input connected to a bias point of about 1/2 vcc, 
or 2.5 volts.  1 volt either way will not normally effect them.  Their inputs 
are very very high impedance, and the line terminating resistor will complete 
the circuit to establish the same bias, within a very few pico-volts on the 
other input.

This, since we are only talking one pair of lines per signal to worry about, 
also means you do not need the complexity of the op-amp.  A pair of 560 ohm 
resistors in series from the 5 volt supply rail to ground, with a bypass of 
about 10uf from the junction to ground, can be used as the bias point for 
several sets of differential receivers at the same time.  If there are any 
interactions (I'd be surprised) then isolate each such circuit pair with a 
10k resistor.  There won't be enough drop across the resistor to effect the 
bias because the input current the chips need is usually in pico-amps.

This means you are then free to use whatever that wire needs as its 
termination to be very well matched and echo free.

Most of the two wire microphone and general purpose mylar shielded audio wire 
we use miles of in broadcasting, has an impedance of about 60 ohms.  That, 
when used in a  600 ohm circuit, causes several db of loss at 15 kilohertz in 
100 feet of it, and many techs have never grokked that when they can't make a 
Proof of Performance & can't find the loss of treble.  They don't subscribe 
to the broadcast engineering magazines that discuss this at about 5 year 
intervals trying to educate them about it either.

>The line should be terminated with a pull-up according to the power output
>of the driver chip, then a scope used to select the correct RC combination.
>The RC connects the line to ground. Use a 100R resistor, and start with cap
>order of 50pF and increase it until you see a nice square wave.

If the rest of it is done right, this trial and error procedure & capacitance 
is not required.  It can be a bandaid that bites you under the right 
conditions.  If, because of extreme noise conditions, you need to add a 
capacitor to slow the response, put it directly across the differential 
receivers input pins.  But using the cable we use is much the better idea, 
its mylar shielding is excellent, 50db (or more) quieter than just a braided 
shield alone, and its probably cheaper than the braid only shielded junk. 

Oh oh, I hear the recess bell, and its beer thirty to boot. :)

>Roland
>
>
>
>
>2009/10/9 Steve Blackmore <st...@pilotltd.net>
>
>> On Thu, 08 Oct 2009 21:34:12 -0400, you wrote:
>> >If 120 ohms from the inputs to +5 is too much of a load, try 470 or 510
>>
>> ohms.
>>
>> > Any pull up action should work against the noise.  If that's where the
>>
>> noise is attacking,
>>
>> >some is better than none, and even 1K ohm might be enough to hold the
>>
>> noise at bay.
>>
>> >Of course it's easy for me to talk, when I have a parts bin with
>> > assorted
>>
>> resistors handy.
>>
>> >|  Hi Jon - tried 120 Ohm across the inputs - it stopped the line
>> >| receiver working? The pull ups on the outputs helped marginally but
>> >| it's much better than it was and is quite usable. I'll order some
>> >| better encoder cable and will try and rewire.
>>
>> The 120 Ohm are not pull-ups, they are to balance the twisted pair.
>>
>> Steve Blackmore
>> --
>>
>>
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