I was wandering -
If I want to use Midronome's analog clock out (one channel)
and put it through a passive multiplier, will the signal weaken that much
so it would render the strength of the pulses unusable,
or there's enough signal strength for that?
Unfortunately, I'm unable right now to try this in practice.
Thank you!
analog clock out and a passive multiplier
Re: analog clock out and a passive multiplier
Hi 
As I wrote in the email, it's a matter of impedance.
Short answer: it will work fine.
Technical explanation:
---------
You want the input impedance Zin to be "much higher" than the output impedance Zout, so the voltage is maintained. The received voltage will be:
When you multiply the clock passively (with a cable splitter for example), all the input impedances are in parallel and therefore are getting smaller. For example, 2 identical Zin in parallel gives Zin/2.
The ANLG outputs on the Midronome have Zout=1kOhm. Most analog clock inputs will have Zin=50kOhm (or more).
So even if you would passively multiply 10 of these, the total Zin=50k/10=5k. That would mean the voltage goes from 5V to 5V*5k/(5k+1k) = 4.2V. That would still work perfectly fine. Multiply 50 of them, and the total Zin=50k/50=1k. That would mean a voltage of 2.5V. Not ideal, but this actually still has a good chance of working.
---------
And remember that the Midronome has two ANLG outputs: the left and the right channel - and they can each have their own speed
Enjoy!
Simon
As I wrote in the email, it's a matter of impedance.
Short answer: it will work fine.
Technical explanation:
---------
You want the input impedance Zin to be "much higher" than the output impedance Zout, so the voltage is maintained. The received voltage will be:
Code: Select all
5V * Zin / (Zin+Zout)The ANLG outputs on the Midronome have Zout=1kOhm. Most analog clock inputs will have Zin=50kOhm (or more).
So even if you would passively multiply 10 of these, the total Zin=50k/10=5k. That would mean the voltage goes from 5V to 5V*5k/(5k+1k) = 4.2V. That would still work perfectly fine. Multiply 50 of them, and the total Zin=50k/50=1k. That would mean a voltage of 2.5V. Not ideal, but this actually still has a good chance of working.
---------
And remember that the Midronome has two ANLG outputs: the left and the right channel - and they can each have their own speed
Enjoy!
Simon
Re: analog clock out and a passive multiplier
Thank you for the reply!
I didn't know that the impedance is the main issue here,
but after your elaboration, now it makes total sense.
I am aware that there are two analog clocks,
that's why I mentioned ''one channel''.
And that's yet another glorious thing about Midronome!
I don't know what else anyone could want from a master clock
(maybe a USB host option, and a TRS MIDI Out; but can't think of anything else).
You managed to cover everything needed for a total synchronization of any setup!
I didn't know that the impedance is the main issue here,
but after your elaboration, now it makes total sense.
I am aware that there are two analog clocks,
that's why I mentioned ''one channel''.
And that's yet another glorious thing about Midronome!
I don't know what else anyone could want from a master clock
(maybe a USB host option, and a TRS MIDI Out; but can't think of anything else).
You managed to cover everything needed for a total synchronization of any setup!