Help with Casio keyboard not working

[LJ89]

If you want to. It probably doesn't matter much, but if you wanted to re-measure some, just do these:
5) the right lead of the 22Ω resistor (which is below LC7881), which should be the same as 3) above,
6) the left lead of the 22Ω resistor in 5) above.

To ensure you are measuring the 22Ω resistor, this is the red-red-black one that is just below the row of C101 green capacitors and just to the right of the 2 electrolytic capacitors. My instruction were not all that clear when I read them again. Thanks.

yes you were right I did measure the wrong resistor. I measured the resistor that was just bellow pin 1 of LC7881 (which I have circled green in the attached photo)

When I measured the resistor that the yellow arrow points to I got
5) 1.24v
6)1.24v

One odd the I did notice is when I was kind of poking around with my finger my finger touched the first resistor I measured ( the green circle ) and I believe at the same time it had contact with pin 1 of LC7881 (due to them being so close) and the keyboard was making music (due to the demo button being engaged) and it wasn't 100% clear and loud, but it was the most i have heard out of it since it has broke. Does that mean anything?

thanks

 

[LJ89]

Sir LJ89 . . . . .

GAMES ON !

Now, just especially for you, is your . . . . . winning secrets . . . . . playbook . . . . .

Now . . .for us to find out . . . . . WHY the units AVDD +5 volt supply line is now being SO abysmally low ?
Is it because of excess power pull and loading of one of the 3 circuits served, or is it the AVDD +5 volt portion within the regulator IC, more in depth testing should reveal.

Using the markups of my REFERENCES right half , note the initial derivation of that supply from LA/CA5668's pin 6 and is initially being marked up with an ORANGE flow line, then it has to transition to a bare wire jumper as the YELLOW line, then it feeds into a 33 ohm (bands of red-red-black) resistor and then on that resistors other side, grabs some filtering and bypass decoupling, provided by the BLUE E-cap , being just physically above.. . to the top.
Those two parts junction and then flow straight down, to make a color flow line transition, to then being of a RED color coding.

The first circuit encounter will then be at the TURQOISE hexagonal marker to feed thru a 47 ohm (bands of yel-viol-black) resistor and then on that resistors other side, it grabs some filtering and bypass decoupling, being provided by the BLUE E-cap that is just physically to the right of pin 8 of the units M5218 I.C..
Now, going back to the RED supply line, since it then takes a very round about routing to ALSO then finally end up at about the almost same identical place and then has to use a YELLOW jumper wire to drop down and continue a RED downward path.

It then encounters a TURQOISE hexagonal marker to then feed thru a 33 ohm (bands of red-red-black) resistor, then the RED line on the other end of the resistor receives filtering and bypass decoupling, from the BLUE E-cap just physically above, and to the right.
The RED line also branches to the left to feed pin 4, then loops down to feed pin10 and uses a YELLOW jumper wire to then continue as a RED line and connect into pin 14.
A final connection is made from the RED line buss via a directional steering diode. One of its connection is just to the right of pin 4 then up and out into pin 2 . . . AND this junctilon also gets some filtering and bypass decoupling by a BLUE E-cap, with its + being connected into I.C.pin 2's right buss.

ALL ABOARD . . .LAST STOP ! . . .as the end of the line, is now coming up.

Follow the RED line down to the final TURQOISE hexagonal marker and its feeding off into a 10 ohm (bands of brown-black-black) *** resistor, then a short RED line flow path downward until another YELLOW wire jumper. *** ( Correct me on that resistor value, if being found different . . . .as it wuz, vewy-vewy hards to TRY to reeds ! )
Exit into a RED line until it makes a dual junction , one being into two pins of the u /Processor and another route into some filtering and bypass decoupling by the positive of a BLUE E-cap, down at the junctions bottom and with the E-caps grounding lead being on far left.

UPDATES . . . . you said . ..

I did desolder the 6V220μF capacitor that you marked with the red square. And I fired up the keyboard with battery power. There was no change, still no sound.

Well, that is telling us, THAT specific E-cap is neither dead shorted nor is it having excess internal leakage. . . .Mamma mia . . . .that's a guuud !

I was under the assumption that the capacitor needs to be soldered on the board for the keyboard to work?

BASE QUALIFIER . . . . . with you using battery power with its PURE DC output; it is being equivalent to a SUPER capacitor bank, with only the slight potential possibility of some squeaks-squawks-warbles or low frequency motor boating in the audio.
BUT that would at,least let us know that the unit is now being quasi operational.

You Said . . . .
and I was not sure what to do with the GREEN square markup...
Disregard that, after my looking at all of the quasi -similar schematics and the very best close ups of your top board pics, I see that the units 3 pin u/Processor reset I.C. is getting its supply power from a DIFFERENT +5VDC source

HOWYOUCANNOWDOITTOITINAMETHODDRIVENANDLOGICALMANNER . . . . .
THE GAME PLAN . . . .

IF you still have two E-caps off the board, no need to reinstall, as we now KNOW that they were not the root cause of the resultant low supply voltage faults.

First go to the TURQOISE hexagonal markers and you will see that each will have those 10- - - - - 47 ohm supply resistors leading off to their sections being supplied.
We will be wanting to lift one lead of each of those, so that they will then disconnect from their respective circuits being supplied.

CAUTIONARY ALERT . . . .
I see that all of the innumerable bare jumper wires have been "stitched" to the boards, preliminary to other components. THEN all of the small round / tubular ceramic capacitors and resistors have been " stapled " in.
This precludes any potential component fall out, prior to laminar flow soldering processing.
You need to give cautious care to working with those resistors fragile leads and their end caps.
Since I now know of your doing soldering, but not of your degree of prowess.
Your soldering iron always needs to be tinned to the degree of perfection, with its working area of the tip looking like its chrome plated shiny !
A fresh drop of rosin fluxed wire solder, added just at the instant of tip / to / work junction contact REALLY enhances an optimal thermal heat transfer.
On a resistor you would want to unfold and straighten one of the resistors lead wires end wrap around.
Heat up the reistors solder blob and quickly slip in a Swis s s s s s s h Army knife blade or an X-acto knifes blade into the foldover side and easily lever the wire up to being straight up and it then no longer having a folded over clinch action / shape . Then stop and get a LARGE sewing needle / thin and pointed scribe / or / probably, your very smallest jewelers screwdriver.
That will go over to the component side of the board and be inserted within the resistor (that was just straightened out on its foil side) leads right angle end loop to the pcb. Then you reheat the solder blob and very slowly and easily lever that lead up and out ( NO MAKEE BWEAKEE ! ) until it just clears outwardly from its board hole.
Move on to the other hexagonal TURQOISE designated supply resistors in taking one resistor end out of its circuit.
When all are floating, power up and meter for ~ +5 VDC voltage at the point where the RED . . . . . AVDD 5+ supply connects to the 22 ohm resistor (the resistor has its other end being connected to POWER LA/CA 5668 I.C pin 6 via a YELLOW jumper wire link).
Then see if that voltage is now up to and around the specified 5V, if so drop power .

Another test would be to have the unit unpowered and since you now have all of the 3 supply resistors loose from their circuitry, then you do an ohms test.
With your meter being set up exactly like when you measured the 22 ohm resistor and got its 22 ohms reading. This time the meter negative is on STAR ground and the + probe goes to where you disconnected the feed resistor fromon each circuit to be tested.
Maybe on one of them we will luck out and read an abnormally low ohms reading for that circuit.
If no concrete evidence by resistive testing, then on to doing a power up testing.

POWER UP TESTING . . . .
And I'd expect that the FIRST circuit that you then want to re establish hooking up with its floating resistor lead, is being that bottom 10 ohm resistor to the . . . . . GAME ENDING . . . u/ Processor and then just . . .PRAY! . . . that the then to be retested AVDD +5 supply is still being up to that +5 VDC now normal level.
Otherwise, the only other thing on that circuits RED power line is being that circuits BLUE E-cap.

Then you move on to the next two remaining circuits and connect in their other 2 supply resistors one at a time and test and then be fully expecting one of them to reinstate the abnormal voltage pull down condition.
It should then be the leaky ? or shorted ? BLUE E-cap associated with that sole circuit or the IC itself is actually being defective within the chip.
To confirm that, you solder suck away or use rosin flux soaked copper wick / braid to absorb away any and ALL of the pins surrounding solder and leave the IC pin floating free of there being any solder connectivity.
To be able to retest and see if that IC was at fault.
Just pin 8 on the cheap M5218 is used . . . and of course, its associated BLUE E-cap.. ( Please . .let it be one of them ! )

On the LC7881C it could be its associated E-cap, or shorted steering diode coupling from pin 4 to 2. Or pins 2-4-10 or 14 faulted inside the I.C.



UPCOMING . . . . . later

With your earlier comments on the sensitivity / derived speaker noise in your final attempt in evaluating a tone output from the instrument to the audio amps inputs that fed to the speakers . I believe that when touching meter probes and getting speaker noise was indicative of your LA4127 functioning.

For investigating an uber simplified testing of the audio output stage of that unit.
But . . . need to know . . . .
Do you have any components parts supply- donor(s) , like old audio equipment, radio or TV of 1970-1990 with their old style wire leaded discrete component builds usage ?

Thaaaaaaaaaaasssit . . . . .


SELECT CIRCUIT REFERENCING . . . . .

https://i.ibb.co/vvZpFfx/Casio-5-AVDD-Supply-Power-Flow-Path.png

73's de Edd . . . . .

View attachment 51321


.

Sir 73's de Edd,

I have some preliminary results as per your request!

At the first blue hexagon I removed one lead of the 47ohm resistor. I checked the voltage at pin 6 of CA5668 and at where AVDD 5+ supply connects to the 22 ohm resistor. They both read 1.64 v

Than at the second blue hexagon I removed one lead of the 22ohm resistor. I checked the voltage at pin 6 of CA5668 and at where AVDD 5+ supply connects to the 22 ohm resistor. They both read 0 v.

I found this kind of odd... I could not turn on the keyboard at all (no red light for power) with wall wart or batteries... So I resoldered the floating pin of the 22ohm resistor back onto the PCB. And now the keyboard would turn on and I checked the voltage at pin 6 of CA5668 and at where AVDD 5+ supply connects to the 22 ohm resistor. They both read 1.64 v.

So I continued to the next hexagon and removed one pin of the 10ohm resistor. I checked the voltage at pin 6 of CA5668 and at where AVDD 5+ supply connects to the 22 ohm resistor. They both read 1.8 v.
I tried pressing some buttons and you could hear some very staticky, garbled, tones that you can hear without putting your ear to the speakers.

So at this point I have one lead of the first hexagon 47ohm resistor floating and one lead of the third hexagon 10ohm resistor floating. The second hexagon 22ohm resistor is resoldered to the board.

Let me know what you think doctor.

Should I just go ahead with the other tests you requested in your post, regarding setting my meter to OHMS. let me know!

Thank you doctor, looking forward to hearing from you.

lj89

 

[73's de Edd]

Sires LJ89 et vbplayer . . . . . .

There is an increasing crescendo of voicing coming upwards from your board, in the *** RED arrow designated sector shouting
. . . . it's me . . . . . . IT"S ME . . . . . *** IT"S ME ! ***
And one of the two likely options, is being the M5218 I.C., BUT, just HOPE that its being that circuits associated E-cap . . . . dead shorted.

With all of your researched voltages to the different stages . . . as compiled in the ref below . . . .initially, in a no power applied state, do an ohms test by
using + probe to that stages E--cap + and the meters black ground lead to star ground.
If shorted, pull it off the board and you might then reconnect any disconnected feeder resistors, so that you can then power up the unit again to retest the AVDD 5+ supply line and confirm that it is now staying up to a +5' ish volt level now, with power getting to all sections..


Woe is you ! . . .SHOULD IT THEN HAPPEN TO BE THE UNITS M5218 I.C. . . instead.

USE THIS PROCEDURE TO PULL THAT M5218 I. C., OFF THE BOARD . . . . .
( if you are not being adequately adept in electro-mechanical-skillset coordination of totally floating its pin 8 of ANY solder presence, thereby leaving its pin 8 FLOATING of any electrical circuit connectivity. )

HOWTODOITTOIT . . . . . ( waiting to see if the related circuitry's E-cap was bad . . . . thus not needing to do this)


Re REFERENCING . . . . . .
https://i.ibb.co/Tcp6rwJ/Casio-CT638-Diagnostic-Op-Cit.png

73's de Edd . . . . .




.

 

[Martaine2005]

Loving your first ever pic Edd. I feel like we could ALL be related.
Apparently, we had millions of siblings that didn't make it.

Martin

 

[vbplayer]

yes you were right I did measure the wrong resistor. I measured the resistor that was just bellow pin 1 of LC7881 (which I have circled green in the attached photo)

When I measured the resistor that the yellow arrow points to I got
5) 1.24v
6)1.24v

One odd the I did notice is when I was kind of poking around with my finger my finger touched the first resistor I measured ( the green circle ) and I believe at the same time it had contact with pin 1 of LC7881 (due to them being so close) and the keyboard was making music (due to the demo button being engaged) and it wasn't 100% clear and loud, but it was the most i have heard out of it since it has broke. Does that mean anything?

thanks
View attachment 51353

Cool. So those measurements make much more sense (i.e. what I would have expected with the 1.24V showing up @ pin #6 of CA5668).

As for your finger touching both the first resistor and pin #1 of LC7881, not sure what happened here. It is a good sign of life after death though. Pin #1 is one of the outputs of LC7881 (most likely the Left channel) and pin #20 is the other output (most likely the Right channel). If your finger contacted pin #1 and the resistor lead closest to that pin, I don't think anything should have happened as pin #1 should be connected to this resistor lead by a trace. And likewise, if your finger contacted pin #1 and the resistor lead farthest from that pin, I don't think anything should have happened here because from your previous measurements on each side of that resistor (i.e. 0.55V from Post #71), no current is flowing through this resistor. Perhaps somehow you created some kind of short against some other contact that caused the sound clip. Maybe 73's de Edd could chime in on this encounter with a more enlightening explanation.

 

[vbplayer]

So it sounds like 73's de Edd has a great plan. His highest suspicion is the electrolytic capacitor just under M5218 (or to the right of this IC if looking at Edd's diagram from Post #74). Test that first in-circuit (again with power off, at least one battery removed and shorting the + and - lead to discharge the capacitor). Do the continuity test and see if this capacitor charges quickly (i.e. should be some number reading on meter before it quickly goes to OL). If it doesn't do this, remove the capacitor and do continuity test off-board. Then, if capacitor seems OK, desolder pin #8 of M5218 (which is where this IC gets it's 5ish volt power) to disconnect ("float") this pin from the PCB (you need to get all the solder out of the hole and ensure the pin is in the middle of the hole not touching any sides of the hole). After desoldering this pin, the best way to ensure it is isolated ("floating") from the board trace is to check for any resistance from pin #8 to where that electrolytic capacitor's positive (+) lead was before your removed it. Make sure you resolder the resistor leads you had desoldered before (i.e. the 47Ω and 22Ω resistors which feed AVDD voltage to M5218 and LC7881, respectively). Now, put your battery (ies) back in, power on your keyboard and see if you are reading 5Vish volts DC at pin #6 of CA5668. If you are, check pin #10 of LC7881 too (which should also be 5Vish volts DC). If you get these proper voltages, then most likely M5218 is damaged and needs to be replaced. Let us know what happens with these tests or if you have any questions.

 

[LJ89]

So it sounds like 73's de Edd has a great plan. His highest suspicion is the electrolytic capacitor just under M5218 (or to the right of this IC if looking at Edd's diagram from Post #74). Test that first in-circuit (again with power off, at least one battery removed and shorting the + and - lead to discharge the capacitor). Do the continuity test and see if this capacitor charges quickly (i.e. should be some number reading on meter before it quickly goes to OL). If it doesn't do this, remove the capacitor and do continuity test off-board. Then, if capacitor seems OK, desolder pin #8 of M5218 (which is where this IC gets it's 5ish volt power) to disconnect ("float") this pin from the PCB (you need to get all the solder out of the hole and ensure the pin is in the middle of the hole not touching any sides of the hole). After desoldering this pin, the best way to ensure it is isolated ("floating") from the board trace is to check for any resistance from pin #8 to where that electrolytic capacitor's positive (+) lead was before your removed it. Make sure you resolder the resistor leads you had desoldered before (i.e. the 47Ω and 22Ω resistors which feed AVDD voltage to M5218 and LC7881, respectively). Now, put your battery (ies) back in, power on your keyboard and see if you are reading 5Vish volts DC at pin #6 of CA5668. If you are, check pin #10 of LC7881 too (which should also be 5Vish volts DC). If you get these proper voltages, then most likely M5218 is damaged and needs to be replaced. Let us know what happens with these tests or if you have any questions.

Thank you very much for the reply! Much appreciated. I will get to this asap and hopefully get some numbers by the end of the day.

By the way how is your keyboard going ?

Thanks

 

[vbplayer]

Thanks for asking how my keyboard is going.

Unfortunately, I am having a difficult time understanding where I need to go. My situation, although similar to yours, is a bit different at this point. The main difference is my LA5668N (same as your CA5668) has 0V on pin #6 now (as I think it is damaged). I have one more good LA5668N, but am reluctant to put it on the PCB as this may become damaged too (I have already been through 2 ICs).

I am trying to determine a way of using a separate AVDD (+5ish volt) power source that I can somehow protect against getting potentially shorted and damaged ) due to suspected bad components in this pathway (like M5218, LC7881 and the myriad of resistors, capacitors and diodes). Then, I can take some of the steps you have to isolate circuitry which will hopefully lead to the problem area. But, in working through your problem and learning several things from 73's de Edd, I'm hopeful to be able to fix my keyboard someday.

I have checked many resistors and electrolytic capacitors in-circuit on my PCB (as you have via continuity testing), but have yet to discover any problems. Visually, everything looks OK (no swollen caps, burned resistors, bad solder joints, etc.). It's a bit frustrating as I have been working on my keyboard for several months (on and off). But, I have learned a lot from this forum. So, therefore, I am hopeful.

If anyone, like 73's de Edd or some other expert, knows a simple way I can create a separate 5ish volt source that protects itself, that would be helpful. Currently, I don't have any regulated 5V wall warts or an adjustable power supply. I was wondering if I could just put some 1.5VDC batteries together (like 3 AAs, Cs or Ds to make 4.5VDC) and use this for testing. But, I don't know if this really is regulated enough or not. Or, if I have to put some resistors and capacitors on the output of this supply to filter noise. And then, whether I need to put some type of fuse after this supply to protect the batteries from overheating if there is a short. More questions than answers right now, but that's part of learning.

I really want to make sure we get your problem solved, or at least headed towards resolution before asking to many questions about my problem.

 

[LJ89]

Sires LJ89 et vbplayer . . . . . .

There is an increasing crescendo of voicing coming upwards from your board, in the *** RED arrow designated sector shouting
. . . . it's me . . . . . . IT"S ME . . . . . *** IT"S ME ! ***
And one of the two likely options, is being the M5218 I.C., BUT, just HOPE that its being that circuits associated E-cap . . . . dead shorted.

With all of your researched voltages to the different stages . . . as compiled in the ref below . . . .initially, in a no power applied state, do an ohms test by
using + probe to that stages E--cap + and the meters black ground lead to star ground.
If shorted, pull it off the board and you might then reconnect any disconnected feeder resistors, so that you can then power up the unit again to retest the AVDD 5+ supply line and confirm that it is now staying up to a +5' ish volt level now, with power getting to all sections..


Woe is you ! . . .SHOULD IT THEN HAPPEN TO BE THE UNITS M5218 I.C. . . instead.

USE THIS PROCEDURE TO PULL THAT M5218 I. C., OFF THE BOARD . . . . .
( if you are not being adequately adept in electro-mechanical-skillset coordination of totally floating its pin 8 of ANY solder presence, thereby leaving its pin 8 FLOATING of any electrical circuit connectivity. )

HOWTODOITTOIT . . . . . ( waiting to see if the related circuitry's E-cap was bad . . . . thus not needing to do this)


Re REFERENCING . . . . . .
https://i.ibb.co/Tcp6rwJ/Casio-CT638-Diagnostic-Op-Cit.png

73's de Edd . . . . .


View attachment 51357

.

We have some numbers here for you Sir. I tested the capacitor to the right of m5218 with the diode tester on my multi meter. it charged up to 1.18 V and did not go any furthur.... (see picture below)

So than I desoldered it and tested it off the board and it went up to just before 3v and than it read OL. (see picture below)

I went to read the voltage of PIN6 of CA5668 and it read 1.7 V

so than I desoldered pin 8 of m5218 and I confirmed it was floating by doing a continuity test with the positive hole of that capacitor I just took off, and I confirmed that it was not touching the holes walls...

So than I measured voltage at pin 6 of CA5668 and it read 1.7 V .

no music noted.

Let me know what you think of if i've missed anything.

I did resolder the 2 remaining resistors. and currently there are 3 capacitors off the board. and pin 8 of M5218 is floating.

Should I resolder anything back just yet?

one of the pictures ive attached has a blue arrow going to where I took the capacitor off. and a red arrow going to the floating pin 8 of m5218.

thank you

 

[LJ89]

So it sounds like 73's de Edd has a great plan. His highest suspicion is the electrolytic capacitor just under M5218 (or to the right of this IC if looking at Edd's diagram from Post #74). Test that first in-circuit (again with power off, at least one battery removed and shorting the + and - lead to discharge the capacitor). Do the continuity test and see if this capacitor charges quickly (i.e. should be some number reading on meter before it quickly goes to OL). If it doesn't do this, remove the capacitor and do continuity test off-board. Then, if capacitor seems OK, desolder pin #8 of M5218 (which is where this IC gets it's 5ish volt power) to disconnect ("float") this pin from the PCB (you need to get all the solder out of the hole and ensure the pin is in the middle of the hole not touching any sides of the hole). After desoldering this pin, the best way to ensure it is isolated ("floating") from the board trace is to check for any resistance from pin #8 to where that electrolytic capacitor's positive (+) lead was before your removed it. Make sure you resolder the resistor leads you had desoldered before (i.e. the 47Ω and 22Ω resistors which feed AVDD voltage to M5218 and LC7881, respectively). Now, put your battery (ies) back in, power on your keyboard and see if you are reading 5Vish volts DC at pin #6 of CA5668. If you are, check pin #10 of LC7881 too (which should also be 5Vish volts DC). If you get these proper voltages, then most likely M5218 is damaged and needs to be replaced. Let us know what happens with these tests or if you have any questions.

thankyou for the instructions, I have posed my results, let me know what you think. The voltage of my LC7881 is 0 V after floating pin8 of M5218

 

[vbplayer]

Excellent job of testing things. I would not resolder the capacitors yet (will explain below).

Here are some observations:
a) you removed the correct capacitor (next to M5218),
b) you desoldered pin #8 on M5218 well (although I see little flakes of solder in places around the area from your picture),
c) you have a Klein Tools MM600 multimeter (which has a capacitance reading capability),
d) obviously, 1.7V wasn't the DC measurement we were hoping for on pin #6 of CA5668 (this pretty much matches what you were reading in Post #82), and
e) I am confused by your last statement in Post #90 that "The voltage of my LC7881 is 0 V after floating pin8 of M5218". That makes no sense (unless there is some kind of short in that circuitry). Hhhhmmmm.

Here's what I would do at this point. But, if you would prefer to let 73's de Edd chime in first, I completely understand.

My suggestions:
1) Clean your PCB a bit in the areas where you have desoldered and resoldered components (i.e. 3 desoldered capacitors, 2 resoldered resistors, and pin #8 of M5218). If you have some rubbing alcohol, put a little on a clean rag (or Q-tips) and rub around the worked areas on the green side of the PCB. Don't scrub, just rub to remove any solder specks and flux residue being careful not to catch the rag or Q-tip threads on the protruding wire leads and pins. Then, let this dry (should be quick) and inspect very closely for solder flecks still on the PCB. To me, from your close-up picture with your red and blue arrows, it looks like there could be some potential solder bridges between some M5218 pins which might be bad. If you need to, use a toothpick (or something that won't scratch the PCB) to gently get rid of these solder flecks.
2) Retake the DC voltage measurement at pin #6 of CA5668 to see if it has changed from 1.7V. If it hasn't changed to around +5VDC, take some measurements (or all 12 except #4) I asked you to take before when going through the traces from CA5668 pin #6. Other than the initial 22Ω resistor just after pin #6 of CA5668, there shouldn't be any influence on the AVDD voltage until the trace gets to the circuitry associated with LC7881.
3) At this point, we need to first, ensure the CA5668 IC isn't damaged. Then, if the IC is OK, we need to isolate the the problem around LC7881 (and beyond). I recommend desoldering the wire jumper I discussed earlier in Post #78. You don't need to remove the whole jumper, but just desolder one end and carefully bend the end (don't lift trace) so it is "floating" in the hole (just like pin #8 of M5218). Do a continuity test to ensure this wire jumper end is "floating" (i.e. not contacting the trace). This will completely remove all LC7881 circuitry (and beyond) from the picture. After "floating" one end of the jumper, check pin #6 of CA5668 again (hopefully it will be around +5VDC). If it isn't around +5VDC, then either CA5668 is damaged, or there is still some path to ground (a short) somewhere. If it is around +5VDC, then the problem is around LC7881. I'll stop here to see where we need to go next based on your findings.
4) Since your multimeter has a capacitance reading capability, I would check the capacitance of each of the 3 electrolytic capacitors you have removed (since they are off the PCB). Again, you can ONLY accurately check capacitance when they are removed from the PBC (i.e. off-board, not in-circuit). Remember, before you test capacitance (as I have stated before), make sure you discharge the capacitor by shorting across it's leads (if you don't do this, you could damage your multimeter). Set your multimeter to the capacitance setting (i.e. -||- symbol). The actual capacitance symbol has the right vertical line arched a little, but I cannot represent that well with a "C" or "c" (look at your User's Guide for symbol on p. 6 and operation on p. 15). Let us know what the 3 capacitors measure.

That's it for now.

 

[LJ89]

Excellent job of testing things. I would not resolder the capacitors yet (will explain below).

Here are some observations:
a) you removed the correct capacitor (next to M5218),
b) you desoldered pin #8 on M5218 well (although I see little flakes of solder in places around the area from your picture),
c) you have a Klein Tools MM600 multimeter (which has a capacitance reading capability),
d) obviously, 1.7V wasn't the DC measurement we were hoping for on pin #6 of CA5668 (this pretty much matches what you were reading in Post #82), and
e) I am confused by your last statement in Post #90 that "The voltage of my LC7881 is 0 V after floating pin8 of M5218". That makes no sense (unless there is some kind of short in that circuitry). Hhhhmmmm.

Here's what I would do at this point. But, if you would prefer to let 73's de Edd chime in first, I completely understand.

My suggestions:
1) Clean your PCB a bit in the areas where you have desoldered and resoldered components (i.e. 3 desoldered capacitors, 2 resoldered resistors, and pin #8 of M5218). If you have some rubbing alcohol, put a little on a clean rag (or Q-tips) and rub around the worked areas on the green side of the PCB. Don't scrub, just rub to remove any solder specks and flux residue being careful not to catch the rag or Q-tip threads on the protruding wire leads and pins. Then, let this dry (should be quick) and inspect very closely for solder flecks still on the PCB. To me, from your close-up picture with your red and blue arrows, it looks like there could be some potential solder bridges between some M5218 pins which might be bad. If you need to, use a toothpick (or something that won't scratch the PCB) to gently get rid of these solder flecks.
2) Retake the DC voltage measurement at pin #6 of CA5668 to see if it has changed from 1.7V. If it hasn't changed to around +5VDC, take some measurements (or all 12 except #4) I asked you to take before when going through the traces from CA5668 pin #6. Other than the initial 22Ω resistor just after pin #6 of CA5668, there shouldn't be any influence on the AVDD voltage until the trace gets to the circuitry associated with LC7881.
3) At this point, we need to first, ensure the CA5668 IC isn't damaged. Then, if the IC is OK, we need to isolate the the problem around LC7881 (and beyond). I recommend desoldering the wire jumper I discussed earlier in Post #78. You don't need to remove the whole jumper, but just desolder one end and carefully bend the end (don't lift trace) so it is "floating" in the hole (just like pin #8 of M5218). Do a continuity test to ensure this wire jumper end is "floating" (i.e. not contacting the trace). This will completely remove all LC7881 circuitry (and beyond) from the picture. After "floating" one end of the jumper, check pin #6 of CA5668 again (hopefully it will be around +5VDC). If it isn't around +5VDC, then either CA5668 is damaged, or there is still some path to ground (a short) somewhere. If it is around +5VDC, then the problem is around LC7881. I'll stop here to see where we need to go next based on your findings.
4) Since your multimeter has a capacitance reading capability, I would check the capacitance of each of the 3 electrolytic capacitors you have removed (since they are off the PCB). Again, you can ONLY accurately check capacitance when they are removed from the PBC (i.e. off-board, not in-circuit). Remember, before you test capacitance (as I have stated before), make sure you discharge the capacitor by shorting across it's leads (if you don't do this, you could damage your multimeter). Set your multimeter to the capacitance setting (i.e. -||- symbol). The actual capacitance symbol has the right vertical line arched a little, but I cannot represent that well with a "C" or "c" (look at your User's Guide for symbol on p. 6 and operation on p. 15). Let us know what the 3 capacitors measure.

That's it for now.

Thankyou very much for the reply. I will get to this asap, I will try and do this tomorrow!

Thanks

 

[LJ89]

Excellent job of testing things. I would not resolder the capacitors yet (will explain below).

Here are some observations:
a) you removed the correct capacitor (next to M5218),
b) you desoldered pin #8 on M5218 well (although I see little flakes of solder in places around the area from your picture),
c) you have a Klein Tools MM600 multimeter (which has a capacitance reading capability),
d) obviously, 1.7V wasn't the DC measurement we were hoping for on pin #6 of CA5668 (this pretty much matches what you were reading in Post #82), and
e) I am confused by your last statement in Post #90 that "The voltage of my LC7881 is 0 V after floating pin8 of M5218". That makes no sense (unless there is some kind of short in that circuitry). Hhhhmmmm.

Here's what I would do at this point. But, if you would prefer to let 73's de Edd chime in first, I completely understand.

My suggestions:
1) Clean your PCB a bit in the areas where you have desoldered and resoldered components (i.e. 3 desoldered capacitors, 2 resoldered resistors, and pin #8 of M5218). If you have some rubbing alcohol, put a little on a clean rag (or Q-tips) and rub around the worked areas on the green side of the PCB. Don't scrub, just rub to remove any solder specks and flux residue being careful not to catch the rag or Q-tip threads on the protruding wire leads and pins. Then, let this dry (should be quick) and inspect very closely for solder flecks still on the PCB. To me, from your close-up picture with your red and blue arrows, it looks like there could be some potential solder bridges between some M5218 pins which might be bad. If you need to, use a toothpick (or something that won't scratch the PCB) to gently get rid of these solder flecks.
2) Retake the DC voltage measurement at pin #6 of CA5668 to see if it has changed from 1.7V. If it hasn't changed to around +5VDC, take some measurements (or all 12 except #4) I asked you to take before when going through the traces from CA5668 pin #6. Other than the initial 22Ω resistor just after pin #6 of CA5668, there shouldn't be any influence on the AVDD voltage until the trace gets to the circuitry associated with LC7881.
3) At this point, we need to first, ensure the CA5668 IC isn't damaged. Then, if the IC is OK, we need to isolate the the problem around LC7881 (and beyond). I recommend desoldering the wire jumper I discussed earlier in Post #78. You don't need to remove the whole jumper, but just desolder one end and carefully bend the end (don't lift trace) so it is "floating" in the hole (just like pin #8 of M5218). Do a continuity test to ensure this wire jumper end is "floating" (i.e. not contacting the trace). This will completely remove all LC7881 circuitry (and beyond) from the picture. After "floating" one end of the jumper, check pin #6 of CA5668 again (hopefully it will be around +5VDC). If it isn't around +5VDC, then either CA5668 is damaged, or there is still some path to ground (a short) somewhere. If it is around +5VDC, then the problem is around LC7881. I'll stop here to see where we need to go next based on your findings.
4) Since your multimeter has a capacitance reading capability, I would check the capacitance of each of the 3 electrolytic capacitors you have removed (since they are off the PCB). Again, you can ONLY accurately check capacitance when they are removed from the PBC (i.e. off-board, not in-circuit). Remember, before you test capacitance (as I have stated before), make sure you discharge the capacitor by shorting across it's leads (if you don't do this, you could damage your multimeter). Set your multimeter to the capacitance setting (i.e. -||- symbol). The actual capacitance symbol has the right vertical line arched a little, but I cannot represent that well with a "C" or "c" (look at your User's Guide for symbol on p. 6 and operation on p. 15). Let us know what the 3 capacitors measure.

That's it for now.

Hey,

1) I tried to clean the pcb as best i could using alcohol swabs, there was definitely some specks I lifted. I've attached some after pictures of the "clean pcb"

2) Here are my readings using battery power.

1) pin #6 of CA5668 = 1.7vdc
2) the right lead of the 22Ω resistor (which is just below CA5668), which should be the same as 1) above, = 1.7vdc
3) the left lead of the same 22Ω resistor in 2) above = 1.7vdc
4) pin #8 of M5218 (which is left bottom pin as pin #1 is marked by the dot), which should be the same as 3) above = 300mv
5) the right lead of the 22Ω resistor (which is below LC7881), which should be the same as 3) above = 1.7vdc
6) the left lead of the 22Ω resistor in 5) above = 1.7vdc
7) pin #10 of LC7881 (which is the far right bottom pin as pin #1 is marked by the dot), which should be the same as 6) above = 1.7vdc
8) the left lead (which directly below pin #2 of LC7881) of the diode (which looks like a resistor but is mostly clear),
I did both right and left lead of this diode and produced different results (not sure if important)
RIGHT = 1.7vdc
LEFT = 1.14vdc
9) the left lead of the 10Ω resistor (which is alone about 1 1/2 inches right of the right side of LC7881), which should be the same as 5) above = 1.7vdc
10) the right lead of the same 10Ω resistor in 9) above. = 1.7vdc
11) pin #10 of CA5668 with the keyboard turned OFF, = = 0vdc
12) pin #10 of CA5668 with the keyboard turned ON.= 4.97vdc

3)

So i desoldered one end of jumper wife and bent upwards , see attached picture.

I noticed that when I went to go power up the red power LED would not turn on. So to confirm it was the jumper wire that caused this I put the jumper wire back in the hole and held it to the side to contact trace and the red LED turned on. So than I removed the one end of the jumper wire out of the hole again and tested voltages.

PIN 10 of ca5668 = 4.97vdc

PIN 6 of ca5668 = 0vdc

So than I took my multimeter and put one lead on the jumper wire and one lead on ground. it read 1.8vdc


4 ) Thanks for letting me know the multimeter has a capacitance option! I had no idea lol

I took the readings and attached the pictures bellow, I want to say they were all pretty close to the uf rating on the capacitor.

Let me know what you think, thank you again.

 

[vbplayer]

OK, good job on everything. your voltage measurements, although not the expected +5VDC, look consistent. I think we are back to your CA5668 is bad. You ordered a new one as I remember (from China, which will probably take several weeks to get here; mine did).

I believe I determined that my LA5668 (same as your CA5668) is bad. I had replaced the original in the beginning (months ago) and it didn't fix my problem. In fact, the replacement LA5668 went bad too. At first, like your situation, I thought something else in the trailing circuitry caused the replacement to go bad. But, there is a chance it was a) already bad when I got it (as I did not test it before putting it onto my PCB), b) it was good, but a cheap part that failed immediately in installation, or c) I damaged it when I was installing it on my PCB. Whatever the reason, the replacement LA5668 is not working now.

[SKIP this paragraph if you DON'T want the LONG STORY]
Therefore, I was very reluctant to just slam my last LA5668 onto my PCB for fear it would get destroyed. This presented a problem for me being able to test any trailing circuitry from pin #6 since I had no +5VDC power source (and I didn't want to use pin #8 for a variety of reasons which I can explain if you would like me to). So. I came up with an idea. I searched some old, old equipment I had buried in a closet and found a 7805 IC which is a simple (but reliable) voltage regulator which puts out close to +5VDC. I made up a simple 5V power supply (with some ceramic capacitors per the 7805 datasheet and a switch to turn this power supply on/off when I wanted). I went through a similar debugging flow that you did by desoldering pin #8 of M5218 and one end of a resistor (similar to your jumper wire) to remove M5218 and the LC7881 circuitry. Then, I ran a test with my new 5V power supply and the voltage levels looked good through the pathways (i.e. ~5.25VDC). So, I soldered the end of the resistor back in (associated with the LC7881 circuity), re-ran my test and voltage levels still looked good. So, I finally soldered pin #8 of M5218 back in, re-ran my test and voltage levels still looked good.

To make a long story short, after doing everything in the above paragraph, I tried turning up the main volume just a little bit and pressed a key on the keyboard. Lo and behold, the keyboard works. I tested out every key, a few different tones and the beat bank. Seems like things are good for the moment (with my work-around 5V power supply I created in the above paragraph.

My next step is to attempt to desolder the bad replacement LA5668 and solder in my last known-good LA5668 (as I tested it off-board first). Hopefully, I won't damage it. We'll see.

But I digress. Back to your issue. Hopefully, 73's de Edd will chime in. But, if I were you, I would solder your 3 capacitors back onto the PCB since their values look good (i.e. close to 5%). Make sure you put them in the right places and the polarity (i.e. +/-) is correct on the PCB. Then, inspect that soldering and ensure no solder flacks or residue exists. I would leave pin #8 of M5218 "floating" and the one end of the jumper wire "free". That way when you get your new CA5668 and replace the old one, you can take 1 step at a time adding circuitry back in after verifying you have a good +5VDC from pin #6. In your 4th picture of the bottom of the PCB (green side), it looks like there is a little residue (appears just a little bit darker green) around one solder joint (just to the right of center) and a smear in the right to upper right area. Just use some rubbing alcohol and Q-tip (but again don't rub to hard scratching the PCB; you might even be able to just dab with alcohol on a cloth, then wipe gently).

That's it for now.

 

[LJ89]

OK, good job on everything. your voltage measurements, although not the expected +5VDC, look consistent. I think we are back to your CA5668 is bad. You ordered a new one as I remember (from China, which will probably take several weeks to get here; mine did).

I believe I determined that my LA5668 (same as your CA5668) is bad. I had replaced the original in the beginning (months ago) and it didn't fix my problem. In fact, the replacement LA5668 went bad too. At first, like your situation, I thought something else in the trailing circuitry caused the replacement to go bad. But, there is a chance it was a) already bad when I got it (as I did not test it before putting it onto my PCB), b) it was good, but a cheap part that failed immediately in installation, or c) I damaged it when I was installing it on my PCB. Whatever the reason, the replacement LA5668 is not working now.

[SKIP this paragraph if you DON'T want the LONG STORY]
Therefore, I was very reluctant to just slam my last LA5668 onto my PCB for fear it would get destroyed. This presented a problem for me being able to test any trailing circuitry from pin #6 since I had no +5VDC power source (and I didn't want to use pin #8 for a variety of reasons which I can explain if you would like me to). So. I came up with an idea. I searched some old, old equipment I had buried in a closet and found a 7805 IC which is a simple (but reliable) voltage regulator which puts out close to +5VDC. I made up a simple 5V power supply (with some ceramic capacitors per the 7805 datasheet and a switch to turn this power supply on/off when I wanted). I went through a similar debugging flow that you did by desoldering pin #8 of M5218 and one end of a resistor (similar to your jumper wire) to remove M5218 and the LC7881 circuitry. Then, I ran a test with my new 5V power supply and the voltage levels looked good through the pathways (i.e. ~5.25VDC). So, I soldered the end of the resistor back in (associated with the LC7881 circuity), re-ran my test and voltage levels still looked good. So, I finally soldered pin #8 of M5218 back in, re-ran my test and voltage levels still looked good.

To make a long story short, after doing everything in the above paragraph, I tried turning up the main volume just a little bit and pressed a key on the keyboard. Lo and behold, the keyboard works. I tested out every key, a few different tones and the beat bank. Seems like things are good for the moment (with my work-around 5V power supply I created in the above paragraph.

My next step is to attempt to desolder the bad replacement LA5668 and solder in my last known-good LA5668 (as I tested it off-board first). Hopefully, I won't damage it. We'll see.

But I digress. Back to your issue. Hopefully, 73's de Edd will chime in. But, if I were you, I would solder your 3 capacitors back onto the PCB since their values look good (i.e. close to 5%). Make sure you put them in the right places and the polarity (i.e. +/-) is correct on the PCB. Then, inspect that soldering and ensure no solder flacks or residue exists. I would leave pin #8 of M5218 "floating" and the one end of the jumper wire "free". That way when you get your new CA5668 and replace the old one, you can take 1 step at a time adding circuitry back in after verifying you have a good +5VDC from pin #6. In your 4th picture of the bottom of the PCB (green side), it looks like there is a little residue (appears just a little bit darker green) around one solder joint (just to the right of center) and a smear in the right to upper right area. Just use some rubbing alcohol and Q-tip (but again don't rub to hard scratching the PCB; you might even be able to just dab with alcohol on a cloth, then wipe gently).

That's it for now.

Thanks for the response!

I recieved my replacement CA5668 in the mail yesterday (aliexpress)

Is there a way I can bench test it before I install it? That way I know it is in working order before I install ?

Thanks

 

[LJ89]

OK, good job on everything. your voltage measurements, although not the expected +5VDC, look consistent. I think we are back to your CA5668 is bad. You ordered a new one as I remember (from China, which will probably take several weeks to get here; mine did).

I believe I determined that my LA5668 (same as your CA5668) is bad. I had replaced the original in the beginning (months ago) and it didn't fix my problem. In fact, the replacement LA5668 went bad too. At first, like your situation, I thought something else in the trailing circuitry caused the replacement to go bad. But, there is a chance it was a) already bad when I got it (as I did not test it before putting it onto my PCB), b) it was good, but a cheap part that failed immediately in installation, or c) I damaged it when I was installing it on my PCB. Whatever the reason, the replacement LA5668 is not working now.

[SKIP this paragraph if you DON'T want the LONG STORY]
Therefore, I was very reluctant to just slam my last LA5668 onto my PCB for fear it would get destroyed. This presented a problem for me being able to test any trailing circuitry from pin #6 since I had no +5VDC power source (and I didn't want to use pin #8 for a variety of reasons which I can explain if you would like me to). So. I came up with an idea. I searched some old, old equipment I had buried in a closet and found a 7805 IC which is a simple (but reliable) voltage regulator which puts out close to +5VDC. I made up a simple 5V power supply (with some ceramic capacitors per the 7805 datasheet and a switch to turn this power supply on/off when I wanted). I went through a similar debugging flow that you did by desoldering pin #8 of M5218 and one end of a resistor (similar to your jumper wire) to remove M5218 and the LC7881 circuitry. Then, I ran a test with my new 5V power supply and the voltage levels looked good through the pathways (i.e. ~5.25VDC). So, I soldered the end of the resistor back in (associated with the LC7881 circuity), re-ran my test and voltage levels still looked good. So, I finally soldered pin #8 of M5218 back in, re-ran my test and voltage levels still looked good.

To make a long story short, after doing everything in the above paragraph, I tried turning up the main volume just a little bit and pressed a key on the keyboard. Lo and behold, the keyboard works. I tested out every key, a few different tones and the beat bank. Seems like things are good for the moment (with my work-around 5V power supply I created in the above paragraph.

My next step is to attempt to desolder the bad replacement LA5668 and solder in my last known-good LA5668 (as I tested it off-board first). Hopefully, I won't damage it. We'll see.

But I digress. Back to your issue. Hopefully, 73's de Edd will chime in. But, if I were you, I would solder your 3 capacitors back onto the PCB since their values look good (i.e. close to 5%). Make sure you put them in the right places and the polarity (i.e. +/-) is correct on the PCB. Then, inspect that soldering and ensure no solder flacks or residue exists. I would leave pin #8 of M5218 "floating" and the one end of the jumper wire "free". That way when you get your new CA5668 and replace the old one, you can take 1 step at a time adding circuitry back in after verifying you have a good +5VDC from pin #6. In your 4th picture of the bottom of the PCB (green side), it looks like there is a little residue (appears just a little bit darker green) around one solder joint (just to the right of center) and a smear in the right to upper right area. Just use some rubbing alcohol and Q-tip (but again don't rub to hard scratching the PCB; you might even be able to just dab with alcohol on a cloth, then wipe gently).

That's it for now.

GOOD NEWS!

So I resoldered the 3 caps.

Than I soldered the new CA5668 on the board and I measured 5v at PIN 6 of CA5668!

I didn't have any sound so than I resoldered that bare jumper wire and pin 8 of m5218.

And I pressed a key and voila! Sound .. clear and loud!!!

Is there any other measurements I should take before I put this bad boy together ?

So my other concern was that when i was adding flux to the pins with a brush and than soldering ( as I saw online) and than when the flux was heated it splattered everywhere on the board and I've been trying to clean the residue with alcohol and it's seems to make a whitish sort of staining that I just can't get rid off.

I looked up the flux that I was using and it's acid paste flux sp-30... apparently that's used for plumbing ... :/ so I'm assuming I need to get this board cleaned off before reassembly.. should I just keep using alcohol? I just don't want it to cause corrosion?

Also what solder is better with lead or without lead? With rosin core or without?

Thankyou!!!!

 

[73's de Edd]

Sir LJ89 . . . . .

Back on post # 83 I saw no voltage reading being given for for pin 8 of M5218 I. C., but with all of the two other supply loops . . . . u/Processer at the very bottom circuitry and for the . .. . .DAC chip 7881. . . . . .after reconnecting their isolation / feed resistors which then, powered up their circuitry.

Hopefully BOTH of those are now being up to an ~5 VDC level ; BUT If you don't find a full 4V Plus at pin 8 of the M5218 I. C., it could STILL have a internal problem, where it is pulling too much current from the POWER LA/CA 5668 I.C. pin 6 and thereby overtaxing its regulator section - - - - -> its premature failure.
Come back with that voltage level on pin 8.

Now if you will take DC metering ACROSS . . . .one meter probe on each side of the resistor . . . . of each of the 22 . . . . .47 . . . . . 22 . . .and . . . .10 ohm feed off resistors we can then compute how much power each section is now consuming.


73's de Edd . . . . .

Efficiency is just a highly refined and developed form of laziness.


.

 

[vbplayer]

Sorry I didn't get back to you before you soldered CA5668 on the PCB, but it sounds like it working, so that's great.

Sounds like the flux you used may have been a type that is used for cleaning the tip of your soldering iron (to help recondition the tip for "tinning" purposes), but really isn't for use on the PCB. In any case, it is what it is (and there's no point in dwelling on the past).

I am not sure what to use to clean your PCB at this point (i.e. don't want to do more damage than good). I would recommend doing some Internet searches (as that's all I could do). I remember from high school chemistry that the opposite of an 'acid' is a 'base' with 'neutral' being a pH of around 7. Maybe this will help with your searches, but make sure you are also specific as it relates to PCB cleaning. You obviously don't want to create any additional corrosion. And, you also want to be careful not to scratch the PCB to much either. Whatever you find to counteract the acid (in the flux), you will want to ensure it doesn't have a harsh reaction with the materials that make up the PCB. And, when it's all said and done, you want to ensure the PCB is dried off properly because moisture itself can be corrosive as well as not reacting well with electricity.

Here is something I found when doing quick search on "acid base neutralization", that "HCl(aq)+NaOH(aq)⇌NaCl(aq)+H2O(l)" which I think means when combining Hydrochloride acid (HCl) with Sodium Hydroxide (NaOH), the result is Sodium Cloride (NaCl; salt) and water (H2O), therefore salt water. Like I said above, "you obviously don't want to create any additional corrosion" and "you also want to be careful not to scratch the PCB", like with salt water which is corrosive and abrasive.

Do some searching and maybe 73's de Edd can weigh in too on a good solution. Let us know what you discover.

 

[vbplayer]

Based on what 73's de Edd stated in his Post #97, I am going to do what he said for additional measurements myself.

I am going to add one more measurement into my testing too, that is measuring the DC voltage at pin #10 of LC7881. I think the chances of this being lower than 5VDC is slim as this is just a Digital to Analog Converter (ADC). The M5218 IC is an Operational Amplifier which I believe tends to need more power. This may be why 73's de Edd didn't mention LC7881.

 

[LJ89]

Sir LJ89 . . . . .

Back on post # 83 I saw no voltage reading being given for for pin 8 of M5218 I. C., but with all of the two other supply loops . . . . u/Processer at the very bottom circuitry and for the . .. . .DAC chip 7881. . . . . .after reconnecting their isolation / feed resistors which then, powered up their circuitry.

Hopefully BOTH of those are now being up to an ~5 VDC level ; BUT If you don't find a full 4V Plus at pin 8 of the M5218 I. C., it could STILL have a internal problem, where it is pulling too much current from the POWER LA/CA 5668 I.C. pin 6 and thereby overtaxing its regulator section - - - - -> its premature failure.
Come back with that voltage level on pin 8.

Now if you will take DC metering ACROSS . . . .one meter probe on each side of the resistor . . . . of each of the 22 . . . . .47 . . . . . 22 . . .and . . . .10 ohm feed off resistors we can then compute how much power each section is now consuming.


73's de Edd . . . . .

Efficiency is just a highly refined and developed form of laziness.


.

Sir de Edd,

I have some measurements here for you.

M5218 PIN # 10 = 4.6v

22ohm resistor directly under CA5668 = 161mv

22ohm resistor under LC7881 = 44mv

10ohm resistor to the right of LC7881 = 19.7mv

47ohm resistor to the left of LC7881 = 164.6mv


Let me know if I took the readings properly.. I turned the keyboard on... and measured each resistor with my meter.

Thank you