[sdiy] MIDI phantom power...over 5 pin MIDI connector ?

[rsdio at audiobanshee.com]

I forgot that you can also terminate a balanced pair by adding a resistance between the lines. In that case, rather than terminating to ground or power, the termination is referenced to the opposite signal voltage.

However, I still say that the resistors in the MIDI current loop are not terminating resistors. A terminating resistor would be in parallel to the opto-isolator diode, not in series.

The point of termination is to allow an alternate path for the signal to flow so that it doesn't bounce back from the receiver to the transmitter. Series resistance is just like having a longer cable - I don't see how it terminates.

Are you sure that the MIDI current loop is a "differential-mode signal?" It looks like a pure single-ended signal to me. What's unique is that the receiver is not independently grounded.

Brian


On Sep 11, 2015, at 9:15 AM, rburnett at richieburnett.co.uk wrote:
> I agree "mskala".  Technically the receiving end of a MIDI cable *IS* terminated for differential-mode signals (i.e. the wanted MIDI signal,) and it's un-terminated for common-mode signals (eg. noise picked up along the way.)  It's easy to see this because the load (the opto-coupler) is tied between the two MIDI lines and is therefore driven by the differential-mode signal.  But any common-mode signal has no way of getting back to ground at the MIDI input port, unless there's some RF filtering capacitors or something from the MIDI IN lines to the ground.
> 
> -Richie,
> 
> 
> On 2015-09-11 16:59, mskala at ansuz.sooke.bc.ca wrote:
>> A transmission line is two conductors.  The line has a characteristic
>> impedance, which will ordinarily be a pure resistance.  It is terminated
>> if, at the end of the line, there is a resistance (or something that looks
>> like a resistance) equal to the line's characteristic impedance.  If that
>> is the case, then at the start of the line, sources will see just the
>> line's characteristic impedance and not something more complicated.  With
>> a properly terminated line, signals will go from the start to the end and
>> then be absorbed nicely in the resistance.  In any other case, some part
>> of the signal will bounce off the end and propagate back to the start,
>> and the impedance seen by the source connected to the start of the line
>> may be something other than the line's characteristic impedance, and
>> other than a pure resistance.
>> Pretty often, especially if it's an unbalanced line, one of the conductors
>> will be connected to ground at some point.  But that has very little to do
>> with termination.  The termination is between the two conductors of the
>> line, nothing to do with DC ground.
>> At least, that's how termination works with the transmission lines used
>> in radio electronics.  Is this term used to mean something completely
>> different in the audio realm?  It wouldn't be the first time.
>> Are you just referring to a single conductor as a "line" and assuming
>> the other conductor is the unspecified path back through ground?  In that
>> case you'd certainly be right that such a path is necessary, but I
>> wouldn't call it a transmission line in the first place, and there'd be no
>> realistic hope of figuring out what the characteristic impedance actually
>> might be.