One of my concerns is that when I desoldered the two capacitors of the red LEDs and the ground cable that goes to the headphone jack, the track was ruined.
I gently scraped down to the copper and restored everything.
Checked and rechecked, there is continuity.
I wouldn't want there to be a third intermediate layer.
I gently scraped down to the copper and restored everything.
Checked and rechecked, there is continuity.
I wouldn't want there to be a third intermediate layer.
Fair enough but can we do a quick test to verify for certain.
With the amp OFF and headphones plugged in you should hear a small click in the headphones if you measure between ground and left out and also between ground and right out using the diode range on the meter. That will just prove the phones are connected to the board.
There are still other easy tests we can do. Do you have any resistors such as a 220k or 470k? anything in that sort of high value?
If you connect the resistor between the opamp side of C1 (the input cap) to the positive rail you should see the main output go to to exactly the same voltage as you measure on the opamp input the resistor is applied to.
No no no
that is the worst thing you can do. This has to be something super simple.
This is what I mean with the resistor test. You only need dab the resistor on to see, no need to solder.
Whatever voltage you apply at the opamp input should be seen at the output. Do NOT connect headphones doing this test.
Here are two similar R values and one is connected to the positive rail in one test and to the negative rail in the other. The output should follow the input exactly. Don't use lower than around 220k.
Whatever voltage you apply at the opamp input should be seen at the output. Do NOT connect headphones doing this test.
Here are two similar R values and one is connected to the positive rail in one test and to the negative rail in the other. The output should follow the input exactly. Don't use lower than around 220k.
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So:
Resistor between operational amplifier input and positive rail.
Then I measure voltage on the op amp output and input.
It has to be the same.
Amp on?
Do you have another headphone amp which can be used as a signal tracer? If so, then apply some music, or a steady tone from Audacity, to the whammy input jack. You should be able to trace from the input jack, volume control, input amp of the whammy, The root cause is probably one of input stage, output stage, or power supply. Use the tracing circuit for the input stage, not the output.
You can buy signal tracing devices on amazon but why not DIY.
You can buy signal tracing devices on amazon but why not DIY.
Amp on. The amp overall has a DC gain of '1' and so all we are doing is confirming the amp is operating correctly.
The opamp input voltage and the main headphone amp output voltage should be identical (the opamp input will not be the same as the opamp input, that is normal).
Make sure no headphones connected for this.
Hi Mooly.
I did the click test.
It can be heard distinctly on the right and left.
I found a suitable resistor for the second test but I'm struggling to understand (I'm not familiar enough).
I connect input opamp by 470kohm to +15v.
Where do I put the multimeter leads?
Between opamp input to ground and then on headphones output?
I did the click test.
It can be heard distinctly on the right and left.
I found a suitable resistor for the second test but I'm struggling to understand (I'm not familiar enough).
I connect input opamp by 470kohm to +15v.
Where do I put the multimeter leads?
Between opamp input to ground and then on headphones output?
If you did that test using the connection points on the board then that proves good continuity to the 'phones
You measure from ground (so black meter lead to ground) to the opamp input pin you have connected the resistor to. Whatever voltage you read there you should also see the same voltage at the headphone output. Do not connect the 'phones though for this test and I would also say leave the RCA inputs floating (not connected to any external equipment.
If you look at the circuit you have your resistor in series with the 100k on the board. So ohms law gives us a voltage of 2.63 volts if you have 15 volt supplies.
Worked out as I=V/R so we have 15/(100,000+470000) giving 0.0263 milliamps. V=I*R and so there is 0.0263E-3 * 100,000 which is 2.63 volts. If you connect the voltage to the negative rail the result would be -2.63 volts.
Hi Fubar3, thanks.
Yes, I have another headphone amplifier but I need to re-solder a few things.
Now I'll try the test recommended by Mooly.
Then we'll move on...
Resistor between input and rail +-15V
On both channels the same.
Input opamp to ground 2.67V
Output (jack) 2,67V
Yes, I have another headphone amplifier but I need to re-solder a few things.
Now I'll try the test recommended by Mooly.
Then we'll move on...
Done.
Resistor between input and rail +-15V
On both channels the same.
Input opamp to ground 2.67V
Output (jack) 2,67V
OK
So that shows the amp is working. If that voltage were audio you should get sound.
If you want you can try the same test but this time we use the RCA input to apply the voltage.
Look at the circuit:
You would need to link out C1 and C5 so that DC is passed.
Now you apply the test voltage to the RCA inputs. This time you can use a 9 volt battery as a source if you want or you can use a resistor as before.
The voltage will now be under control of the volume pot and be variable from zero upwards.
If you use a resistor you can use a lower value if you want because the resistor is now in series with the volume pot and that will reduce the voltage available. How low depends on the value of pot but nothing bad will happen because it is protected by R39 and R40.
So that shows the amp is working. If that voltage were audio you should get sound.
If you want you can try the same test but this time we use the RCA input to apply the voltage.
Look at the circuit:
You would need to link out C1 and C5 so that DC is passed.
Now you apply the test voltage to the RCA inputs. This time you can use a 9 volt battery as a source if you want or you can use a resistor as before.
The voltage will now be under control of the volume pot and be variable from zero upwards.
If you use a resistor you can use a lower value if you want because the resistor is now in series with the volume pot and that will reduce the voltage available. How low depends on the value of pot but nothing bad will happen because it is protected by R39 and R40.
If you use a 9 v battery you would see the voltage at the wiper of the pot alter from 0 through to 9 volts as you rotate the control.
With those caps linked that wiper voltage should appear at the opamp input.
Its a mystery to me as well
You'll find the problem. Apply the voltage to the input and just trace it through. It has to missing at some point.
Good morning.
I connected everything.
9V battery.
RCA input with R39/R40 2200ohm
Bypassed capacitors.
Opamp input to main via 220kohm.
50k potentiometer
Output to the jack I find from OV to 6.75V.
Measuring the channel that is currently not on battery, I find 0V to1.9Volt.
I hope I have explained well.
Thank you for your patience.
I connected everything.
9V battery.
RCA input with R39/R40 2200ohm
Bypassed capacitors.
Opamp input to main via 220kohm.
50k potentiometer
Output to the jack I find from OV to 6.75V.
Measuring the channel that is currently not on battery, I find 0V to1.9Volt.
I hope I have explained well.
Thank you for your patience.
The 9 volt battery goes to the input RCA.
Caps bypassed.
We don't use a resistor now if we are using the battery. The battery is the only input voltage needed.
With a 50k pot you should see the wiper voltage alterable between 0 volts and up to a little below below 9 volt.
Like this. The main output voltage will be a little bit below the wiper voltage because of R2 and R3 (which form a divider) but the output voltage should be the same as you see on pin 3 of the opamp.
Caps bypassed.
We don't use a resistor now if we are using the battery. The battery is the only input voltage needed.
With a 50k pot you should see the wiper voltage alterable between 0 volts and up to a little below below 9 volt.
Like this. The main output voltage will be a little bit below the wiper voltage because of R2 and R3 (which form a divider) but the output voltage should be the same as you see on pin 3 of the opamp.