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This is a front panel for a piano-keyboard (Kurzweil SP3X), with buttons and button-LEDs. The LEDs did not work at all, and I traced it to a shorted capacitor (C2), which I removed and confirmed it was shorted. Is it ok to simply leave this (C2) open, not replacing the cap? It's just for convenience as I don't have tools at my current location. For learning it would be nice if you explained why it's ok or not.

Mostly there will be only 4-6 LEDs active on right panel. But during startup all LEDs light up simultaneously for about 0.5sec, as seen here at 12sec-mark: https://www.youtube.com/watch?v=szlCpS0GEhA

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    \$\begingroup\$ For anyone searching, C2 is in the center left of the first schematics image. \$\endgroup\$ – Anders Petersson Nov 12 '19 at 11:49
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The LEDs are connected in a matrix that is scanned at a rate high enough to fool the eye into thinking that they are on continuously. The transistors Q1-Q10 are turned on one at a time, and for each time slot, any or all of Q11-Q18 are turned on to light up some of the LEDs in the corresponding column.

R16 feeds the collectors of Q11-Q18, so without C2 the current through it would vary with how many of them are turned on at any given time. This variable voltage drop across R16 would cause variations in the brightness of the LEDs. C2 filters out those variations so that R16 only "sees" the average current for the entire display, eliminating those brightness variations.

It also prevents those current variations from affecting the rest of the system — especially any sensitive audio circuits.

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  • \$\begingroup\$ "scanned at a rate" ..I'm curious of the component(s) that gave this away? Or is it an assumption as simply a proper way to feed the circuit? (Added input schematic above). \$\endgroup\$ – bretddog Nov 14 '19 at 11:59
  • \$\begingroup\$ When LEDs are connected in a matrix like that, the only way to control them independently is to scan them. In order to avoid visible flicker, each LED must be "refreshed" a few hundred times per second or more. \$\endgroup\$ – Dave Tweed Nov 14 '19 at 12:30
  • \$\begingroup\$ Yes of course. I see that now.. :) \$\endgroup\$ – bretddog Nov 14 '19 at 13:45
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C2 and R16 appear to be designed to isolate the power rail from LED switching transients.

The question is why would they do that? I doubt the transients from switching a few LEDs would have a significant impact on a typical digital power rail.

Maybe the power rail is shared with some analog audio circuitry (or fed from a rail that runs analog audio) and there was a percived need to avoid injecting audible spikes.

Another possibility is that this was done to reduce electromagnetic emissions from the cable connecting the display board to the rest of the system or maybe to reduce crosstalk issues in that cable. However the capacitor doesn't seem the right type for that, a big electrolytic is more likely aimed at noise in the audio band.

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The designer obviously thought it a good idea. Its function would appear to be a filter removing high frequency switching noise. My guess is that it would not damage anything to remove it, but you might end up with flickering display lights.

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  • \$\begingroup\$ "HF switching noise", what components would that potentially influence? I also noticed now after assembly/testing that a very quick tap on the buttons makes them not respond, so that would perhaps be a purpose of C2? Btw that resistor R16 I measured to 50ohm, not 4.7 as this schematic shows. \$\endgroup\$ – bretddog Nov 12 '19 at 11:59
  • \$\begingroup\$ @bretddog We can only guess...it is possible that the traces (particularly ground traces) for the display components are routed so that they're distributed from the ground-side of C2. This may help to isolate ground noise from getting back to critical audio ground-paths. Peter's answer suggested this too. \$\endgroup\$ – glen_geek Nov 12 '19 at 13:22

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