Help with Casio keyboard not working

[LJ89]

Sir LJ89 . . . . .

OUR CONTINUANCE . . . . .

Now you are finding fault ! . . . . .
with that output pin 6 of the LA / CA5668, being now only at a 1.248 VDC level is being suggestive of it feeding out to a circuit and then encountering a short /or excessive loading on that branched off line, thereby loading down and pulling the +5V AVDD supply on down to that lower voatage level, instead of being the expected ~ 5VDC.

I magged up and then digitally enhanced one of your top board photos to give us the supplied reference and a relevant piecemeal researched schematic that agrees with your boards LC 7881C D/A converter , LM5218 dual op amp, used for filtering out of digital artifacts and trash , and then the final Dual Audio Output Power Amp of the LA4127.

My closest mark up reference is being at the CA5668 regulator chip and its RED rectangle designation marking and in its RED square foreground seems to show the last half of . . . ?'2 ohms printed on the PCB.

SET UP . . . . .

In a NO POWER /BATTS disconnected test, use metering in its low ohms function.
Then you measure from from IC pin 6 to what looks to be its associative resistor connection . . . .of one of the leads of that ( RED-RED-BLACK band, color code marking) 22 ohms resistor. Then you confirm that the touching of one of its leads results on an almost zero reading / like shorting the meters two test probe leads together. OR . . . . .you could just visually trace the direct foil path to the terminated 22 ohm resistor lead.
Then you go to the OTHER resistor lead and just see if it is being connected to the references small RED square marked up side of the smaller blue cased E-cap.
If so, then you ohmically test across that E-caps two leads to see if it is being dead shorted .
If so . . . . . . .then . . . . .as per the video, lightning actually CAN strike twice on Casio players of this design, using a 6.3 VDC rated E-cap with 5 + volts subjecture on it.
You can then unsolder and lift the defective cap from the board. Or if being a nonsolder-er-er . . . .as of yet!
Just a back and forth /side to side full bending of the cap case should let metal fatigue break loose its two wire leads after 363 full flexures.
Then a retest of the unit . . . .sans the bad E- capacitor . . .with batteries installed should hopefully have the unit making beautiful music for mama-san again.
Then you worry about getting / ordering a like capacitive value of E-cap at on upwards of a 16VDC rating, thereby providing a specs voltage cushion for the future..

If that is not the problem, then we move on to my top center VIOLET square markup as well as the right top sides YELLOW square markup.
BECAUSE like situations with 22 ohm isolative resistors and their following bypass caps also exist at those locations..
And there is being a final u/p reset IC located at the GREEN square markup, its also being associated with that +5V AVDD supply line .

AN INITIAL ASSIST . . . . .

Using another photo, I will now start you on the initial flow path of the pin 6 of the LA5668 and its +5V AVDD supply source.
Pin 6 only has one additional foil connection to its side which likely will be for a bare wire jumper on the other side of the board and it then ends up at the pin 10 area and that foil associated with the end of this jumper, it then only shares its foil with what must be THE 22 ohm resistor, then the other side of that resistor has its foil pad which only has two other adjoining pad connections, one of which must be the pad connection to the + wire of THE suspect E-cap and the other pad, must be associated with carrying on the path of the +5V AVDD supply ever on to its other connection areas of the system.
Now, I can't flop the board over . . . nor meter out . . . to confirm those routings , but I highly suspect that I am being correct.


Le REFERENCE . . . . . .

https://i.ibb.co/FYSTcKX/Casio-CT-638-Audio-Flow-Circuitry-II.png



73's de Edd . . . . . .

View attachment 51296

.

Providing a quick update.

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.

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

I will test the other components that you highlighted in your schematic. Thank you

 

[LJ89]

I think it might be a good idea to test 6V220μF capacitor out-of-circuit (i.e. desolder, pull off PCB, and test by itself). I may try this too and will let you know my results.

In Post #57, you mentioned "The output of pin 6 of CA5668 is 1.248vdc." Very interesting as I read 2.14V DC here (as I stated in Post #46). Sounds like the same symptom.

Would that be testing it with the diode option on the multimeter?

 

[LJ89]

I think it might be a good idea to test 6V220μF capacitor out-of-circuit (i.e. desolder, pull off PCB, and test by itself). I may try this too and will let you know my results.

In Post #57, you mentioned "The output of pin 6 of CA5668 is 1.248vdc." Very interesting as I read 2.14V DC here (as I stated in Post #46). Sounds like the same symptom.

So I tested that capacitor out of circuit with the multimeter on diode setting. It charges up to just to approx. 2.9V and than it goes OL. So I tried again, same thing..

Can I swap it with another capacitor from the board to see if that does anything?

 

[LJ89]

So I tested that capacitor out of circuit with the multimeter on diode setting. It charges up to just to approx. 2.9V and than it goes OL. So I tried again, same thing..

Can I swap it with another capacitor from the board to see if that does anything?

Hmm I removed another capacitor with the same specs 6V220μF ( it was located underneath the Sanyo LC7881 IC. It did the same thing changed up to 2.9V and went OL..

So I'm not sure if I've found the fault. If I understand correctly should these capacitors charge to 6V? And stay at 6v? They are not supposed to go OL ?

 

[vbplayer]

Sorry I wasn't available for a while to answer your questions in Post #62 & #63. But, since you went ahead and tried some things, I will try to just answer your questions in Post #64. I hope someone else out there, like 73's de Edd, can confirm that I am answering your questions correctly.

I don't think you have found any fault yet.

The short answer to 1st two questions "If I understand correctly should these capacitors charge to 6V? and And stay at 6v?" is 'no'. To elaborate, the '6V' in 6V220μF means this capacitor is rated up to a maximum of 6V. In fact, I think it's actually rated up to 6.3V if you look at the specs (they just don't have room on the PCB to print this detail). Therefore, if the voltage exceeds this maximum, the capacitor will be damaged. Normally, the voltage at this point in the circuit would be around 5V, therefore this capacitor is here. If you trace from your wall wart connector on the PCB, you will find a 16V470μF capacitor going to ground. At this point in the circuit, you want something around 9V here, therefore you have a 16V rated capacitor. If you put a 6V rated capacitor in this part of the circuit, you would fry it.

You stated that the 6V220μF capacitors both charged up to 2.9V, then your meter went to OL. I think this is exactly what should have happened. Here's why. First, I don't think you are measuring voltage (V). I believe the values you are seeing on your meter as the capacitor is charging are increases in resistance which eventually (and usually very quickly) ends up at OL (which represents infinite resistance). So, I think these capacitors are OK. Again, without a capacitance meter, you cannot know for sure if a capacitor is within its tolerance (i.e. a 220μF capacitor rated at 10% tolerance should read 220μF ± 22μF, or 198μF - 242μF).

 

[LJ89]

Sorry I wasn't available for a while to answer your questions in Post #62 & #63. But, since you went ahead and tried some things, I will try to just answer your questions in Post #64. I hope someone else out there, like 73's de Edd, can confirm that I am answering your questions correctly.

I don't think you have found any fault yet.

The short answer to 1st two questions "If I understand correctly should these capacitors charge to 6V? and And stay at 6v?" is 'no'. To elaborate, the '6V' in 6V220μF means this capacitor is rated up to a maximum of 6V. In fact, I think it's actually rated up to 6.3V if you look at the specs (they just don't have room on the PCB to print this detail). Therefore, if the voltage exceeds this maximum, the capacitor will be damaged. Normally, the voltage at this point in the circuit would be around 5V, therefore this capacitor is here. If you trace from your wall wart connector on the PCB, you will find a 16V470μF capacitor going to ground. At this point in the circuit, you want something around 9V here, therefore you have a 16V rated capacitor. If you put a 6V rated capacitor in this part of the circuit, you would fry it.

You stated that the 6V220μF capacitors both charged up to 2.9V, then your meter went to OL. I think this is exactly what should have happened. Here's why. First, I don't think you are measuring voltage (V). I believe the values you are seeing on your meter as the capacitor is charging are increases in resistance which eventually (and usually very quickly) ends up at OL (which represents infinite resistance). So, I think these capacitors are OK. Again, without a capacitance meter, you cannot know for sure if a capacitor is within its tolerance (i.e. a 220μF capacitor rated at 10% tolerance should read 220μF ± 22μF, or 198μF - 242μF).

Thank you very much for the response. That makes sense. I will resolder those capacitors back!

 

[vbplayer]

LJ89 ...... have you resoldered those capacitors yet? If not, don't yet and let me know.

 

[LJ89]

LJ89 ...... have you resoldered those capacitors yet? If not, don't yet and let me know.

I have not resoldered them yet.

 

[vbplayer]

Can you check voltage in a number of places to see if not having these capacitors makes a difference?

If so, set your multimeter to DCV and probably like 20V. Connect your negative lead of your multimeter to ground somewhere (on the ground plane would be fine). Then, check voltage (on the component side ensuring you only touch what I say to avoid shorting any pins) with the positive lead of your multimeter at:
1) pin #6 of CA5668,
2) the right lead of the 22Ω resistor (which is just below CA5668), which should be the same as 1) above,
3) the left lead of the same 22Ω resistor in 2) above,
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,
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,
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,
8) the left lead (which directly below pin #2 of LC7881) of the diode (which looks like a resistor but is mostly clear),
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,
10) the right lead of the same 10Ω resistor in 9) above.

Sorry this took so long for me to respond. I wanted to ensure I traced everything correctly on your PCB. Thanks.

 

[LJ89]

Can you check voltage in a number of places to see if not having these capacitors makes a difference?

If so, set your multimeter to DCV and probably like 20V. Connect your negative lead of your multimeter to ground somewhere (on the ground plane would be fine). Then, check voltage (on the component side ensuring you only touch what I say to avoid shorting any pins) with the positive lead of your multimeter at:
1) pin #6 of CA5668,
2) the right lead of the 22Ω resistor (which is just below CA5668), which should be the same as 1) above,
3) the left lead of the same 22Ω resistor in 2) above,
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,
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,
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,
8) the left lead (which directly below pin #2 of LC7881) of the diode (which looks like a resistor but is mostly clear),
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,
10) the right lead of the same 10Ω resistor in 9) above.

Sorry this took so long for me to respond. I wanted to ensure I traced everything correctly on your PCB. Thanks.

Ok sounds good, I have to go to work in a hour I will get those measurements tomorrow

Thankyou very much for the response!

 

[LJ89]

Can you check voltage in a number of places to see if not having these capacitors makes a difference?

If so, set your multimeter to DCV and probably like 20V. Connect your negative lead of your multimeter to ground somewhere (on the ground plane would be fine). Then, check voltage (on the component side ensuring you only touch what I say to avoid shorting any pins) with the positive lead of your multimeter at:
1) pin #6 of CA5668,
2) the right lead of the 22Ω resistor (which is just below CA5668), which should be the same as 1) above,
3) the left lead of the same 22Ω resistor in 2) above,
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,
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,
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,
8) the left lead (which directly below pin #2 of LC7881) of the diode (which looks like a resistor but is mostly clear),
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,
10) the right lead of the same 10Ω resistor in 9) above.

Sorry this took so long for me to respond. I wanted to ensure I traced everything correctly on your PCB. Thanks.

Hey,

Here are the numbers you have requested ( I brought it to work lol)

1) 1.24v
2) 1.24v
3) 1.24v
4) 0.52v
5) 0.55v
6) 0.55v
7) 1.3v
8) 0.79v
9) 1.24v
10) 1.24v

Let me know if you see anything!

thanks

 

[LJ89]

Sir LJ89 . . . . .

OUR CONTINUANCE . . . . .

Now you are finding fault ! . . . . .
with that output pin 6 of the LA / CA5668, being now only at a 1.248 VDC level is being suggestive of it feeding out to a circuit and then encountering a short /or excessive loading on that branched off line, thereby loading down and pulling the +5V AVDD supply on down to that lower voatage level, instead of being the expected ~ 5VDC.

I magged up and then digitally enhanced one of your top board photos to give us the supplied reference and a relevant piecemeal researched schematic that agrees with your boards LC 7881C D/A converter , LM5218 dual op amp, used for filtering out of digital artifacts and trash , and then the final Dual Audio Output Power Amp of the LA4127.

My closest mark up reference is being at the CA5668 regulator chip and its RED rectangle designation marking and in its RED square foreground seems to show the last half of . . . ?'2 ohms printed on the PCB.

SET UP . . . . .

In a NO POWER /BATTS disconnected test, use metering in its low ohms function.
Then you measure from from IC pin 6 to what looks to be its associative resistor connection . . . .of one of the leads of that ( RED-RED-BLACK band, color code marking) 22 ohms resistor. Then you confirm that the touching of one of its leads results on an almost zero reading / like shorting the meters two test probe leads together. OR . . . . .you could just visually trace the direct foil path to the terminated 22 ohm resistor lead.
Then you go to the OTHER resistor lead and just see if it is being connected to the references small RED square marked up side of the smaller blue cased E-cap.
If so, then you ohmically test across that E-caps two leads to see if it is being dead shorted .
If so . . . . . . .then . . . . .as per the video, lightning actually CAN strike twice on Casio players of this design, using a 6.3 VDC rated E-cap with 5 + volts subjecture on it.
You can then unsolder and lift the defective cap from the board. Or if being a nonsolder-er-er . . . .as of yet!
Just a back and forth /side to side full bending of the cap case should let metal fatigue break loose its two wire leads after 363 full flexures.
Then a retest of the unit . . . .sans the bad E- capacitor . . .with batteries installed should hopefully have the unit making beautiful music for mama-san again.
Then you worry about getting / ordering a like capacitive value of E-cap at on upwards of a 16VDC rating, thereby providing a specs voltage cushion for the future..

If that is not the problem, then we move on to my top center VIOLET square markup as well as the right top sides YELLOW square markup.
BECAUSE like situations with 22 ohm isolative resistors and their following bypass caps also exist at those locations..
And there is being a final u/p reset IC located at the GREEN square markup, its also being associated with that +5V AVDD supply line .

AN INITIAL ASSIST . . . . .

Using another photo, I will now start you on the initial flow path of the pin 6 of the LA5668 and its +5V AVDD supply source.
Pin 6 only has one additional foil connection to its side which likely will be for a bare wire jumper on the other side of the board and it then ends up at the pin 10 area and that foil associated with the end of this jumper, it then only shares its foil with what must be THE 22 ohm resistor, then the other side of that resistor has its foil pad which only has two other adjoining pad connections, one of which must be the pad connection to the + wire of THE suspect E-cap and the other pad, must be associated with carrying on the path of the +5V AVDD supply ever on to its other connection areas of the system.
Now, I can't flop the board over . . . nor meter out . . . to confirm those routings , but I highly suspect that I am being correct.


Le REFERENCE . . . . . .

https://i.ibb.co/FYSTcKX/Casio-CT-638-Audio-Flow-Circuitry-II.png



73's de Edd . . . . . .

View attachment 51296

.

I did your violet mark up (22ohm resistor)

with one lead on pin 6 of ca5668 and the other on the left pin of the violet 22ohm resistor it read 44ohm

with the lead this time on the RIGHT pin of that resistor it read 22ohms.


I was not sure what i was supposed to do with the yellow markup. I jsut assumed that I was to place one lead on pin 6 of ca5668 and the other lead on one of the pins of that capacitor just bellow your yellow square. SO i did that and it read OL and I did it again for the other pin of that capacitor and it read OL aswell.

and I was not sure what to do with the green square markup...

let me know if I have missed / screwed up anything!

thankyou

 

[vbplayer]

Thanks LJ89. Right now, I just glanced at the numbers and I am a bit confused. But, I need to look through these numbers in more detail and go back to your PCB traces to ensure I didn't miss something.

I realized I didn't tell you to take these measurements with the keyboard turned on (which is what I want). If it was OFF for these measurements, can you please take them again with keyboard turned ON. Sorry about that.

Can you also take two other measurement for me:
11) pin #10 of CA5668 with the keyboard turned OFF,
12) pin #10 of CA5668 with the keyboard turned ON.

 

[73's de Edd]

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 . . . . .




.

 

[LJ89]

Thanks LJ89. Right now, I just glanced at the numbers and I am a bit confused. But, I need to look through these numbers in more detail and go back to your PCB traces to ensure I didn't miss something.

I realized I didn't tell you to take these measurements with the keyboard turned on (which is what I want). If it was OFF for these measurements, can you please take them again with keyboard turned ON. Sorry about that.

Can you also take two other measurement for me:
11) pin #10 of CA5668 with the keyboard turned OFF,
12) pin #10 of CA5668 with the keyboard turned ON.

I did the first 10 measurements with the keyboard turned ON.

11) 0v ( it was initially 0.9v (right after I turned the keyboard off) but it trickled down to 0v within 30 seconds. (I'm assuming this is the capacitors discharging?)

12) 4.97v

 

[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 ?

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 transitin, to then being of a RED 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 M5218 I.C..
Now, back to the RED supply line, since it then takes very round about routing to ALSO then 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 branched to the left to feed pin 4, 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 filtering and bypass decoupling by a BLUE E-cap, with its + being connected into 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 at 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. . . .thats a guuud !

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

A 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 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 lever the wire up to being straight up and it then no longer having a folded over cinch action. Then stop and get a LARGE sewing needle /pointed scribe / or / probably your very smallest jewelers screwdriver.
That will go to the component side of the board and insert under the resistor lead that was just straightened out on its foil side.. Then you reheat the solder blob and very slowly and easily lever that lead up and out until it just clears the board.
Move on to the other supply resistors in taking one resistor end out of circuit.
When all are floating, power up and Meter for DC voltage at the point where the RED . . . . . AVDD 5+ supply connects to the 22 ohm resistor (that resistor has its other end connected to POWER LA/CA 5668 I.C pin 6 via a jumper wire link).
Then see if that voltage is now up to and around the specified 5V, if so drop power and the FIRST circuit that you want to re establish hooking up with its floating resistor lead, is being that bottom 10 ohm resistor to the u/ Processor and . . .PRAY . . . that the then to be retested AVDD 5+ supply is still up to +5 VDC normal level.
Otherwise, the only other hhing on that line is being that circuits BLUE E-cap.
Then you add in the other 2 supply resistors one at a time and then be fully expecting one of them to reinstate the voltage pull down.
It should then be the leaky ? or shorted ? BLUE E-cap associated with that sole circuit or the IC itself is actually being defective inside.
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.

Just pin 8 on the cheapest M5218 is used . . . and its BLUE E-cap.. ( Let one of them be it ! )

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


.

Thankyou very much for the very detailed response Dr. de Edd. I will be getting to this asap, I will try my best to get it done before the end of the day.

Thankyou

 

[LJ89]

Doctor 73's de Edd

Thankyou very much for the very detailed response. I will be getting to this asap, I will try my best to get it done before the end of the day.

Thankyou

 

[vbplayer]

Thanks for the measurements, LJ89. Some of it makes sense to me and some doesn't:

1) 1.24v - OK; should be around 5V, but this is where we started noticing a problem.
2) 1.24v - consistent; at least the same as 1) above.
3) 1.24v - consistent; at least the same as 2) above.
4) 0.52v - I felt this should be the same as 3) above, but it isn't; I did miss the 47Ω resistor 73's de Edd pointed out in his excellent diagram (Post #74). But, apparently there is a drop across this resistor. This makes the M5218 IC highly suspect.
5) 0.55v - this confuses me completely as this should be a directly connected pathway from 3) above, therefore 1.24V. Just a guess, but unless there is some kind of short somewhere, I think perhaps this wasn't measured correctly.
6) 0.55v - interesting that this is the same voltage as 5) above. Again, this might not have been measured correctly.
7) 1.3v - this makes some sense to me; there should be a direct pathway between this pin #10 (on LC7881) and the left side of the 22Ω resistor in 6) above. Another indicator that perhaps 5) & 6) above where measured at the wrong places.
8) 0.79v - not sure what this should be (LM7881 pin #2, VrefH1); Datasheet says this should be min of VDD - 0.5V up to max of VDD. In our current state (with VDD at 1.24V), this should be between 0.74V - 1.24V which it is.
9) 1.24v - consistent; I expected this to be the same as 5) above which should be the same as 3) above. It isn't the same as 5) (which confuses me), but is the same as 3) above.
10) 1.24v - I have no idea what this should be or what the purpose of this is (related to some signals on the μProcessor as 73's de Edd pointed out in his diagram).
11) 0v - good, this is correct when the keyboard is turned OFF.
12) 4.97v - good, this is correct when the keyboard is turned ON; this pin #10 (APO) is used by microprocessor to activate the 5V (AVDD) power on CA5668 (pin #6).

One thing that occurred to me as I was reading Post #74 by 73's de Edd is maybe there is a way to narrow down our search for the problem area related to the AVDD +5 supply (maybe a divide and conquer method). 73's de Edd mentioned removing some of the resistors to test them off of the PCB. He also mentioned how these resistors (and ceramic capacitors, along with diodes) have been "stapled in" requiring caution when removing so as to not damage PCB pathways. He also mentioned that 'innumerable bare jumper wires have been "stitched" to the boards'. This got me thinking about the jumper wires. Is there a way that removal of some key jumper wires could help us better isolate and diagnose operation of some circuits on the PCB. I think there is is one jumper that could perhaps help you separate the circuitry associated with M5218 and the circuitry associated with LM7881. I have attached a marked-up version of 73's de Edd's diagram with a black arrow showing the jumper. Removing this jumper (CAREFULLY because of bent-over leads as 73's de Edd pointed out) could in essence disconnect the LM7881 circuitry from the AVDD +5 supply. In this way, voltage measurements could be taken to determine if this circuitry is causing the problem or not. If the AVDD voltage does return to around +5V (from 1.24V), then most likely the problem is in the LM7881 circuitry. If the AVDD voltage doesn't return to around +5V (from 1.24V), then most likely the problem is in the M5218 circuitry (but there still could be a problem in the LM7881 circuitry if there are 2 failures which is typically unlikely).

BEFORE REMOVING THIS JUMPER WIRE, I would like to get confirmation from 73's de Edd that running this test will not cause any further damaged to the PCB circuitry. I would think that providing no supply voltage to the LM7881 circuitry (by removing this jumper) would be essentially the same as what is currently being supplied (1.24V). Once we have confirmation, then we can proceed. Otherwise, I would respect 73's de Edd's advance and work this from whatever direction he decides.

NOTE: Right now, I am having problems attaching the marked-up version of 73's de Edd's diagram with a black arrow. Don't know what's going on with my uploading capability. Therefore, I will describe where this jumper is in 73's de Edd's diagram. It is the jumper wire to the right of the M5218 IC (just to the right of the electrolytic capacitor; cannot see the value of the capacitor though, maybe 6V10μF or 6V100μF).

 

[LJ89]

Thanks for the measurements, LJ89. Some of it makes sense to me and some doesn't:

1) 1.24v - OK; should be around 5V, but this is where we started noticing a problem.
2) 1.24v - consistent; at least the same as 1) above.
3) 1.24v - consistent; at least the same as 2) above.
4) 0.52v - I felt this should be the same as 3) above, but it isn't; I did miss the 47Ω resistor 73's de Edd pointed out in his excellent diagram (Post #74). But, apparently there is a drop across this resistor. This makes the M5218 IC highly suspect.
5) 0.55v - this confuses me completely as this should be a directly connected pathway from 3) above, therefore 1.24V. Just a guess, but unless there is some kind of short somewhere, I think perhaps this wasn't measured correctly.
6) 0.55v - interesting that this is the same voltage as 5) above. Again, this might not have been measured correctly.
7) 1.3v - this makes some sense to me; there should be a direct pathway between this pin #10 (on LC7881) and the left side of the 22Ω resistor in 6) above. Another indicator that perhaps 5) & 6) above where measured at the wrong places.
8) 0.79v - not sure what this should be (LM7881 pin #2, VrefH1); Datasheet says this should be min of VDD - 0.5V up to max of VDD. In our current state (with VDD at 1.24V), this should be between 0.74V - 1.24V which it is.
9) 1.24v - consistent; I expected this to be the same as 5) above which should be the same as 3) above. It isn't the same as 5) (which confuses me), but is the same as 3) above.
10) 1.24v - I have no idea what this should be or what the purpose of this is (related to some signals on the μProcessor as 73's de Edd pointed out in his diagram).
11) 0v - good, this is correct when the keyboard is turned OFF.
12) 4.97v - good, this is correct when the keyboard is turned ON; this pin #10 (APO) is used by microprocessor to activate the 5V (AVDD) power on CA5668 (pin #6).

One thing that occurred to me as I was reading Post #74 by 73's de Edd is maybe there is a way to narrow down our search for the problem area related to the AVDD +5 supply (maybe a divide and conquer method). 73's de Edd mentioned removing some of the resistors to test them off of the PCB. He also mentioned how these resistors (and ceramic capacitors, along with diodes) have been "stapled in" requiring caution when removing so as to not damage PCB pathways. He also mentioned that 'innumerable bare jumper wires have been "stitched" to the boards'. This got me thinking about the jumper wires. Is there a way that removal of some key jumper wires could help us better isolate and diagnose operation of some circuits on the PCB. I think there is is one jumper that could perhaps help you separate the circuitry associated with M5218 and the circuitry associated with LM7881. I have attached a marked-up version of 73's de Edd's diagram with a black arrow showing the jumper. Removing this jumper (CAREFULLY because of bent-over leads as 73's de Edd pointed out) could in essence disconnect the LM7881 circuitry from the AVDD +5 supply. In this way, voltage measurements could be taken to determine if this circuitry is causing the problem or not. If the AVDD voltage does return to around +5V (from 1.24V), then most likely the problem is in the LM7881 circuitry. If the AVDD voltage doesn't return to around +5V (from 1.24V), then most likely the problem is in the M5218 circuitry (but there still could be a problem in the LM7881 circuitry if there are 2 failures which is typically unlikely).

BEFORE REMOVING THIS JUMPER WIRE, I would like to get confirmation from 73's de Edd that running this test will not cause any further damaged to the PCB circuitry. I would think that providing no supply voltage to the LM7881 circuitry (by removing this jumper) would be essentially the same as what is currently being supplied (1.24V). Once we have confirmation, then we can proceed. Otherwise, I would respect 73's de Edd's advance and work this from whatever direction he decides.

NOTE: Right now, I am having problems attaching the marked-up version of 73's de Edd's diagram with a black arrow. Don't know what's going on with my uploading capability. Therefore, I will describe where this jumper is in 73's de Edd's diagram. It is the jumper wire to the right of the M5218 IC (just to the right of the electrolytic capacitor; cannot see the value of the capacitor though, maybe 6V10μF or 6V100μF).

Hello vbplayer,

Yes that sounds good to me to wait for 73's de Edd for confirmation. I have not yet completed his direction as I haven't had time.

Would you like.me.to re measure some of those values that didn't seem right to you and i can confirm my lead placement in a post.

Thankyou

 

[vbplayer]

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.