Need help understanding a power circuit (caps and diodes)
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I'm trying to repair an old electric piano (Baldwin Piano Pro EP101) and I've gotten advice to check out the filter caps in the power circuit. I've metered out the voltages and I found power where I didn't expect it. Please see image:
Power from transformer coming in the top. Pins "I" and "J" are ground on the bottom. I've mirrored the board so the traces match up with the components on the top side.
- Why do I find voltage on the (-) side of the black capacitor?
- Why is the (+) of the black capacitor going to ground?
- I think both of these capacitors are in series, but why is there ground in the middle of the two caps?
- Does it mean I have power coming in through the "M" pin at that bottom that shouldn't be? Or maybe one of the diodes at D9 and D10 (2nd and 3rd from the left) are bad and letting the power go the wrong way?
Am I on the right track? Should I just start pulling parts and testing them out of circuit? If you're interested in the overall problem, see the short youtube video here: Baldwin Piano Pro - Very loud noises
Edit: Thanks for the feedback.Larger picture and my attempt of a diagram. Surprised to see AC and DC volts when I measured. Not sure what that is about.
simulate this circuit – Schematic created using CircuitLab
power capacitor diodes
asked Mar 30 at 18:39
lopazopylopazopy
436
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lopazopy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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show 1 more comment
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I'm trying to repair an old electric piano (Baldwin Piano Pro EP101) and I've gotten advice to check out the filter caps in the power circuit. I've metered out the voltages and I found power where I didn't expect it. Please see image:
Power from transformer coming in the top. Pins "I" and "J" are ground on the bottom. I've mirrored the board so the traces match up with the components on the top side.
- Why do I find voltage on the (-) side of the black capacitor?
- Why is the (+) of the black capacitor going to ground?
- I think both of these capacitors are in series, but why is there ground in the middle of the two caps?
- Does it mean I have power coming in through the "M" pin at that bottom that shouldn't be? Or maybe one of the diodes at D9 and D10 (2nd and 3rd from the left) are bad and letting the power go the wrong way?
Am I on the right track? Should I just start pulling parts and testing them out of circuit? If you're interested in the overall problem, see the short youtube video here: Baldwin Piano Pro - Very loud noises
Edit: Thanks for the feedback.Larger picture and my attempt of a diagram. Surprised to see AC and DC volts when I measured. Not sure what that is about.
simulate this circuit – Schematic created using CircuitLab
power capacitor diodes
asked Mar 30 at 18:39
lopazopylopazopy
436
New contributor
lopazopy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
measure the voltage across the black capacitor ..... what do you get? .....Why is the (+) of the black capacitor going to ground?because the negative terminal of the black capacitor is connected to a voltage that is more negative than ground
$endgroup$
– jsotola
Mar 30 at 19:14
2
$begingroup$
I'm with @Transistor , Some details of that transformer are missing. Either those red & green transformer wires are connected somehow to the yellow transformer wires (perhaps inside the transformer), or there are other wires coming from the transformer not shown. You seem certain that I,J are ground...could there be a transformer connection to this point?
$endgroup$
– glen_geek
Mar 30 at 19:15
$begingroup$
The overall problem looks like keyboard trouble.
$endgroup$
– AltAir
Mar 30 at 19:34
$begingroup$
Measure Vdc across every part and Vac across the Caps. You should expect +Vdc across each cap and Vac<5%Vdc. Suspect any with 0V
$endgroup$
– Sunnyskyguy EE75
Mar 30 at 20:07
$begingroup$
@glen_geek You're right. There is a wire coming from the transformer that I labeled as 0 V. I think I am wrong to call that Ground.
$endgroup$
– lopazopy
Mar 30 at 22:22
|
show 1 more comment
$begingroup$
I'm trying to repair an old electric piano (Baldwin Piano Pro EP101) and I've gotten advice to check out the filter caps in the power circuit. I've metered out the voltages and I found power where I didn't expect it. Please see image:
Power from transformer coming in the top. Pins "I" and "J" are ground on the bottom. I've mirrored the board so the traces match up with the components on the top side.
- Why do I find voltage on the (-) side of the black capacitor?
- Why is the (+) of the black capacitor going to ground?
- I think both of these capacitors are in series, but why is there ground in the middle of the two caps?
- Does it mean I have power coming in through the "M" pin at that bottom that shouldn't be? Or maybe one of the diodes at D9 and D10 (2nd and 3rd from the left) are bad and letting the power go the wrong way?
Am I on the right track? Should I just start pulling parts and testing them out of circuit? If you're interested in the overall problem, see the short youtube video here: Baldwin Piano Pro - Very loud noises
Edit: Thanks for the feedback.Larger picture and my attempt of a diagram. Surprised to see AC and DC volts when I measured. Not sure what that is about.
simulate this circuit – Schematic created using CircuitLab
power capacitor diodes
asked Mar 30 at 18:39
lopazopylopazopy
436
New contributor
lopazopy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm trying to repair an old electric piano (Baldwin Piano Pro EP101) and I've gotten advice to check out the filter caps in the power circuit. I've metered out the voltages and I found power where I didn't expect it. Please see image:
Power from transformer coming in the top. Pins "I" and "J" are ground on the bottom. I've mirrored the board so the traces match up with the components on the top side.
- Why do I find voltage on the (-) side of the black capacitor?
- Why is the (+) of the black capacitor going to ground?
- I think both of these capacitors are in series, but why is there ground in the middle of the two caps?
- Does it mean I have power coming in through the "M" pin at that bottom that shouldn't be? Or maybe one of the diodes at D9 and D10 (2nd and 3rd from the left) are bad and letting the power go the wrong way?
Am I on the right track? Should I just start pulling parts and testing them out of circuit? If you're interested in the overall problem, see the short youtube video here: Baldwin Piano Pro - Very loud noises
Edit: Thanks for the feedback.Larger picture and my attempt of a diagram. Surprised to see AC and DC volts when I measured. Not sure what that is about.
simulate this circuit – Schematic created using CircuitLab
power capacitor diodes
power capacitor diodes
asked Mar 30 at 18:39
lopazopylopazopy
436
New contributor
lopazopy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 30 at 18:39
lopazopylopazopy
436
New contributor
lopazopy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 30 at 21:40
lopazopy
asked Mar 30 at 18:39
lopazopylopazopy
436
New contributor
lopazopy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 30 at 18:39
lopazopylopazopy
436
asked Mar 30 at 18:39
lopazopylopazopy
436
436
New contributor
lopazopy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
lopazopy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
lopazopy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
measure the voltage across the black capacitor ..... what do you get? .....Why is the (+) of the black capacitor going to ground?because the negative terminal of the black capacitor is connected to a voltage that is more negative than ground
$endgroup$
– jsotola
Mar 30 at 19:14
2
$begingroup$
I'm with @Transistor , Some details of that transformer are missing. Either those red & green transformer wires are connected somehow to the yellow transformer wires (perhaps inside the transformer), or there are other wires coming from the transformer not shown. You seem certain that I,J are ground...could there be a transformer connection to this point?
$endgroup$
– glen_geek
Mar 30 at 19:15
$begingroup$
The overall problem looks like keyboard trouble.
$endgroup$
– AltAir
Mar 30 at 19:34
$begingroup$
Measure Vdc across every part and Vac across the Caps. You should expect +Vdc across each cap and Vac<5%Vdc. Suspect any with 0V
$endgroup$
– Sunnyskyguy EE75
Mar 30 at 20:07
$begingroup$
@glen_geek You're right. There is a wire coming from the transformer that I labeled as 0 V. I think I am wrong to call that Ground.
$endgroup$
– lopazopy
Mar 30 at 22:22
|
show 1 more comment
1
$begingroup$
measure the voltage across the black capacitor ..... what do you get? .....Why is the (+) of the black capacitor going to ground?because the negative terminal of the black capacitor is connected to a voltage that is more negative than ground
$endgroup$
– jsotola
Mar 30 at 19:14
2
$begingroup$
I'm with @Transistor , Some details of that transformer are missing. Either those red & green transformer wires are connected somehow to the yellow transformer wires (perhaps inside the transformer), or there are other wires coming from the transformer not shown. You seem certain that I,J are ground...could there be a transformer connection to this point?
$endgroup$
– glen_geek
Mar 30 at 19:15
$begingroup$
The overall problem looks like keyboard trouble.
$endgroup$
– AltAir
Mar 30 at 19:34
$begingroup$
Measure Vdc across every part and Vac across the Caps. You should expect +Vdc across each cap and Vac<5%Vdc. Suspect any with 0V
$endgroup$
– Sunnyskyguy EE75
Mar 30 at 20:07
$begingroup$
@glen_geek You're right. There is a wire coming from the transformer that I labeled as 0 V. I think I am wrong to call that Ground.
$endgroup$
– lopazopy
Mar 30 at 22:22
1
1
$begingroup$
measure the voltage across the black capacitor ..... what do you get? .....
Why is the (+) of the black capacitor going to ground? because the negative terminal of the black capacitor is connected to a voltage that is more negative than ground$endgroup$
– jsotola
Mar 30 at 19:14
$begingroup$
measure the voltage across the black capacitor ..... what do you get? .....
Why is the (+) of the black capacitor going to ground? because the negative terminal of the black capacitor is connected to a voltage that is more negative than ground$endgroup$
– jsotola
Mar 30 at 19:14
2
2
$begingroup$
I'm with @Transistor , Some details of that transformer are missing. Either those red & green transformer wires are connected somehow to the yellow transformer wires (perhaps inside the transformer), or there are other wires coming from the transformer not shown. You seem certain that I,J are ground...could there be a transformer connection to this point?
$endgroup$
– glen_geek
Mar 30 at 19:15
$begingroup$
I'm with @Transistor , Some details of that transformer are missing. Either those red & green transformer wires are connected somehow to the yellow transformer wires (perhaps inside the transformer), or there are other wires coming from the transformer not shown. You seem certain that I,J are ground...could there be a transformer connection to this point?
$endgroup$
– glen_geek
Mar 30 at 19:15
$begingroup$
The overall problem looks like keyboard trouble.
$endgroup$
– AltAir
Mar 30 at 19:34
$begingroup$
The overall problem looks like keyboard trouble.
$endgroup$
– AltAir
Mar 30 at 19:34
$begingroup$
Measure Vdc across every part and Vac across the Caps. You should expect +Vdc across each cap and Vac<5%Vdc. Suspect any with 0V
$endgroup$
– Sunnyskyguy EE75
Mar 30 at 20:07
$begingroup$
Measure Vdc across every part and Vac across the Caps. You should expect +Vdc across each cap and Vac<5%Vdc. Suspect any with 0V
$endgroup$
– Sunnyskyguy EE75
Mar 30 at 20:07
$begingroup$
@glen_geek You're right. There is a wire coming from the transformer that I labeled as 0 V. I think I am wrong to call that Ground.
$endgroup$
– lopazopy
Mar 30 at 22:22
$begingroup$
@glen_geek You're right. There is a wire coming from the transformer that I labeled as 0 V. I think I am wrong to call that Ground.
$endgroup$
– lopazopy
Mar 30 at 22:22
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
I can only guess what those clever Japanese designers had in mind with these diodes.
Large E-caps must have a Dissipation Factor of << 1% where DF=ESR/Xc(120Hz) with 120Hz ripple current thus with the same current means Vac/Vdc =DF= <1%
Thus C1 is bad , C2 seems OK C3,C4,C5 are all bad.
I would replace all 5 Caps and consider replacing any other E-caps found on other circuit boards from Digikey or Mouser. Consult with tech support for equivalent or better in similar sizes.
simulate this circuit – Schematic created using CircuitLab
answered Mar 31 at 1:21
Sunnyskyguy EE75Sunnyskyguy EE75
70.6k226103
$endgroup$
add a comment |
$begingroup$
+1 for mirroring the underside of the board. Best practice is to draw the schematic and mark up the measured voltages. You can add one in using the CircuitLab button on the editor toolbar. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. I suspect that there's a transformer centre-tap connected somewhere other than the top of the board so try to draw that too.
Why do I find voltage on the (-) side of the black capacitor?
The circuits must require both positive and negative supplies with respect to ground. This is common in audio circuits.
Why is the (+) of the black capacitor going to ground?
So that correct polarity is maintained.
I think both of these capacitors are in series, but why is there ground in the middle of the two caps?
Time for a schematic.
simulate this circuit – Schematic created using CircuitLab
Figure 1. (a) A single-rail supply. (b) A split-rail supply giving both positive and voltage power outputs.
Does it mean I have power coming in through the "M" pin at that bottom that shouldn't be?
Or maybe one of the diodes at D9 and D10 (2nd and 3rd from the left) are bad and letting the power go the wrong way?
No. All is well in that regard.
Post a schematic as best you can and we'll update the answer.
answered Mar 30 at 19:16
TransistorTransistor
88.2k785189
$endgroup$
$begingroup$
Sorry but your schematic here looks nothing like the board as all inputs go only to Anodes while an AC bridge input connects to one Anode and one Cathode on each input. But the caps are correct and there is no obvious 0V reference except may I,J which are paired.
$endgroup$
– Sunnyskyguy EE75
Mar 30 at 20:11
1
$begingroup$
My Figure 1 isn't meant to be a board schematic. All the inputs go to anodes only. That's why I requested more details.
$endgroup$
– Transistor
Mar 30 at 20:21
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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oldest
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$begingroup$
I can only guess what those clever Japanese designers had in mind with these diodes.
Large E-caps must have a Dissipation Factor of << 1% where DF=ESR/Xc(120Hz) with 120Hz ripple current thus with the same current means Vac/Vdc =DF= <1%
Thus C1 is bad , C2 seems OK C3,C4,C5 are all bad.
I would replace all 5 Caps and consider replacing any other E-caps found on other circuit boards from Digikey or Mouser. Consult with tech support for equivalent or better in similar sizes.
simulate this circuit – Schematic created using CircuitLab
answered Mar 31 at 1:21
Sunnyskyguy EE75Sunnyskyguy EE75
70.6k226103
$endgroup$
add a comment |
$begingroup$
I can only guess what those clever Japanese designers had in mind with these diodes.
Large E-caps must have a Dissipation Factor of << 1% where DF=ESR/Xc(120Hz) with 120Hz ripple current thus with the same current means Vac/Vdc =DF= <1%
Thus C1 is bad , C2 seems OK C3,C4,C5 are all bad.
I would replace all 5 Caps and consider replacing any other E-caps found on other circuit boards from Digikey or Mouser. Consult with tech support for equivalent or better in similar sizes.
simulate this circuit – Schematic created using CircuitLab
answered Mar 31 at 1:21
Sunnyskyguy EE75Sunnyskyguy EE75
70.6k226103
$endgroup$
add a comment |
$begingroup$
I can only guess what those clever Japanese designers had in mind with these diodes.
Large E-caps must have a Dissipation Factor of << 1% where DF=ESR/Xc(120Hz) with 120Hz ripple current thus with the same current means Vac/Vdc =DF= <1%
Thus C1 is bad , C2 seems OK C3,C4,C5 are all bad.
I would replace all 5 Caps and consider replacing any other E-caps found on other circuit boards from Digikey or Mouser. Consult with tech support for equivalent or better in similar sizes.
simulate this circuit – Schematic created using CircuitLab
answered Mar 31 at 1:21
Sunnyskyguy EE75Sunnyskyguy EE75
70.6k226103
$endgroup$
I can only guess what those clever Japanese designers had in mind with these diodes.
Large E-caps must have a Dissipation Factor of << 1% where DF=ESR/Xc(120Hz) with 120Hz ripple current thus with the same current means Vac/Vdc =DF= <1%
Thus C1 is bad , C2 seems OK C3,C4,C5 are all bad.
I would replace all 5 Caps and consider replacing any other E-caps found on other circuit boards from Digikey or Mouser. Consult with tech support for equivalent or better in similar sizes.
simulate this circuit – Schematic created using CircuitLab
answered Mar 31 at 1:21
Sunnyskyguy EE75Sunnyskyguy EE75
70.6k226103
edited Mar 31 at 1:39
answered Mar 31 at 1:21
Sunnyskyguy EE75Sunnyskyguy EE75
70.6k226103
answered Mar 31 at 1:21
Sunnyskyguy EE75Sunnyskyguy EE75
70.6k226103
answered Mar 31 at 1:21
Sunnyskyguy EE75Sunnyskyguy EE75
70.6k226103
70.6k226103
add a comment |
add a comment |
$begingroup$
+1 for mirroring the underside of the board. Best practice is to draw the schematic and mark up the measured voltages. You can add one in using the CircuitLab button on the editor toolbar. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. I suspect that there's a transformer centre-tap connected somewhere other than the top of the board so try to draw that too.
Why do I find voltage on the (-) side of the black capacitor?
The circuits must require both positive and negative supplies with respect to ground. This is common in audio circuits.
Why is the (+) of the black capacitor going to ground?
So that correct polarity is maintained.
I think both of these capacitors are in series, but why is there ground in the middle of the two caps?
Time for a schematic.
simulate this circuit – Schematic created using CircuitLab
Figure 1. (a) A single-rail supply. (b) A split-rail supply giving both positive and voltage power outputs.
Does it mean I have power coming in through the "M" pin at that bottom that shouldn't be?
Or maybe one of the diodes at D9 and D10 (2nd and 3rd from the left) are bad and letting the power go the wrong way?
No. All is well in that regard.
Post a schematic as best you can and we'll update the answer.
answered Mar 30 at 19:16
TransistorTransistor
88.2k785189
$endgroup$
$begingroup$
Sorry but your schematic here looks nothing like the board as all inputs go only to Anodes while an AC bridge input connects to one Anode and one Cathode on each input. But the caps are correct and there is no obvious 0V reference except may I,J which are paired.
$endgroup$
– Sunnyskyguy EE75
Mar 30 at 20:11
1
$begingroup$
My Figure 1 isn't meant to be a board schematic. All the inputs go to anodes only. That's why I requested more details.
$endgroup$
– Transistor
Mar 30 at 20:21
add a comment |
$begingroup$
+1 for mirroring the underside of the board. Best practice is to draw the schematic and mark up the measured voltages. You can add one in using the CircuitLab button on the editor toolbar. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. I suspect that there's a transformer centre-tap connected somewhere other than the top of the board so try to draw that too.
Why do I find voltage on the (-) side of the black capacitor?
The circuits must require both positive and negative supplies with respect to ground. This is common in audio circuits.
Why is the (+) of the black capacitor going to ground?
So that correct polarity is maintained.
I think both of these capacitors are in series, but why is there ground in the middle of the two caps?
Time for a schematic.
simulate this circuit – Schematic created using CircuitLab
Figure 1. (a) A single-rail supply. (b) A split-rail supply giving both positive and voltage power outputs.
Does it mean I have power coming in through the "M" pin at that bottom that shouldn't be?
Or maybe one of the diodes at D9 and D10 (2nd and 3rd from the left) are bad and letting the power go the wrong way?
No. All is well in that regard.
Post a schematic as best you can and we'll update the answer.
answered Mar 30 at 19:16
TransistorTransistor
88.2k785189
$endgroup$
$begingroup$
Sorry but your schematic here looks nothing like the board as all inputs go only to Anodes while an AC bridge input connects to one Anode and one Cathode on each input. But the caps are correct and there is no obvious 0V reference except may I,J which are paired.
$endgroup$
– Sunnyskyguy EE75
Mar 30 at 20:11
1
$begingroup$
My Figure 1 isn't meant to be a board schematic. All the inputs go to anodes only. That's why I requested more details.
$endgroup$
– Transistor
Mar 30 at 20:21
add a comment |
$begingroup$
+1 for mirroring the underside of the board. Best practice is to draw the schematic and mark up the measured voltages. You can add one in using the CircuitLab button on the editor toolbar. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. I suspect that there's a transformer centre-tap connected somewhere other than the top of the board so try to draw that too.
Why do I find voltage on the (-) side of the black capacitor?
The circuits must require both positive and negative supplies with respect to ground. This is common in audio circuits.
Why is the (+) of the black capacitor going to ground?
So that correct polarity is maintained.
I think both of these capacitors are in series, but why is there ground in the middle of the two caps?
Time for a schematic.
simulate this circuit – Schematic created using CircuitLab
Figure 1. (a) A single-rail supply. (b) A split-rail supply giving both positive and voltage power outputs.
Does it mean I have power coming in through the "M" pin at that bottom that shouldn't be?
Or maybe one of the diodes at D9 and D10 (2nd and 3rd from the left) are bad and letting the power go the wrong way?
No. All is well in that regard.
Post a schematic as best you can and we'll update the answer.
answered Mar 30 at 19:16
TransistorTransistor
88.2k785189
$endgroup$
+1 for mirroring the underside of the board. Best practice is to draw the schematic and mark up the measured voltages. You can add one in using the CircuitLab button on the editor toolbar. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. I suspect that there's a transformer centre-tap connected somewhere other than the top of the board so try to draw that too.
Why do I find voltage on the (-) side of the black capacitor?
The circuits must require both positive and negative supplies with respect to ground. This is common in audio circuits.
Why is the (+) of the black capacitor going to ground?
So that correct polarity is maintained.
I think both of these capacitors are in series, but why is there ground in the middle of the two caps?
Time for a schematic.
simulate this circuit – Schematic created using CircuitLab
Figure 1. (a) A single-rail supply. (b) A split-rail supply giving both positive and voltage power outputs.
Does it mean I have power coming in through the "M" pin at that bottom that shouldn't be?
Or maybe one of the diodes at D9 and D10 (2nd and 3rd from the left) are bad and letting the power go the wrong way?
No. All is well in that regard.
Post a schematic as best you can and we'll update the answer.
answered Mar 30 at 19:16
TransistorTransistor
88.2k785189
answered Mar 30 at 19:16
TransistorTransistor
88.2k785189
answered Mar 30 at 19:16
TransistorTransistor
88.2k785189
answered Mar 30 at 19:16
TransistorTransistor
88.2k785189
88.2k785189
$begingroup$
Sorry but your schematic here looks nothing like the board as all inputs go only to Anodes while an AC bridge input connects to one Anode and one Cathode on each input. But the caps are correct and there is no obvious 0V reference except may I,J which are paired.
$endgroup$
– Sunnyskyguy EE75
Mar 30 at 20:11
1
$begingroup$
My Figure 1 isn't meant to be a board schematic. All the inputs go to anodes only. That's why I requested more details.
$endgroup$
– Transistor
Mar 30 at 20:21
add a comment |
$begingroup$
Sorry but your schematic here looks nothing like the board as all inputs go only to Anodes while an AC bridge input connects to one Anode and one Cathode on each input. But the caps are correct and there is no obvious 0V reference except may I,J which are paired.
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– Sunnyskyguy EE75
Mar 30 at 20:11
1
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My Figure 1 isn't meant to be a board schematic. All the inputs go to anodes only. That's why I requested more details.
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– Transistor
Mar 30 at 20:21
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Sorry but your schematic here looks nothing like the board as all inputs go only to Anodes while an AC bridge input connects to one Anode and one Cathode on each input. But the caps are correct and there is no obvious 0V reference except may I,J which are paired.
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– Sunnyskyguy EE75
Mar 30 at 20:11
$begingroup$
Sorry but your schematic here looks nothing like the board as all inputs go only to Anodes while an AC bridge input connects to one Anode and one Cathode on each input. But the caps are correct and there is no obvious 0V reference except may I,J which are paired.
$endgroup$
– Sunnyskyguy EE75
Mar 30 at 20:11
1
1
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My Figure 1 isn't meant to be a board schematic. All the inputs go to anodes only. That's why I requested more details.
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– Transistor
Mar 30 at 20:21
$begingroup$
My Figure 1 isn't meant to be a board schematic. All the inputs go to anodes only. That's why I requested more details.
$endgroup$
– Transistor
Mar 30 at 20:21
add a comment |
lopazopy is a new contributor. Be nice, and check out our Code of Conduct.
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1
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measure the voltage across the black capacitor ..... what do you get? .....
Why is the (+) of the black capacitor going to ground?because the negative terminal of the black capacitor is connected to a voltage that is more negative than ground$endgroup$
– jsotola
Mar 30 at 19:14
2
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I'm with @Transistor , Some details of that transformer are missing. Either those red & green transformer wires are connected somehow to the yellow transformer wires (perhaps inside the transformer), or there are other wires coming from the transformer not shown. You seem certain that I,J are ground...could there be a transformer connection to this point?
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– glen_geek
Mar 30 at 19:15
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The overall problem looks like keyboard trouble.
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– AltAir
Mar 30 at 19:34
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Measure Vdc across every part and Vac across the Caps. You should expect +Vdc across each cap and Vac<5%Vdc. Suspect any with 0V
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– Sunnyskyguy EE75
Mar 30 at 20:07
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@glen_geek You're right. There is a wire coming from the transformer that I labeled as 0 V. I think I am wrong to call that Ground.
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– lopazopy
Mar 30 at 22:22