Twin Peaks Tremolo Debugging

Started by Bielorusse, January 16, 2016, 08:01:18 AM

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Bielorusse

Hi everyone,

I finished this build yesterday, but I wasn't getting anything when I tested it. I will just take a break and think about something else for a while, but I figured I would post here in case someone sees what's wrong :).

It seems like I have a problem with the power supply, but I'm not quite sure. Below is a copy-paste of the "debugging - what to do when it doesn't work" text with my answers.

But first, some pictures! (I'm sorry it's such an awful "spaghetti dish" build, it looked better in the box. Ask me if you ever think another better picture might be useful.)

Build from under: (it's not on the picture, but I slip a baggie between the board and the pots to avoid contact)
http://image.noelshack.com/fichiers/2016/02/1452949029-p1030265.jpg

Board from under:
http://image.noelshack.com/fichiers/2016/02/1452949025-p1030266.jpg

Build from above:
http://image.noelshack.com/fichiers/2016/02/1452949026-p1030270.jpg




1.What does it do, not do, and sound like?

It works bypassed, but when the effect is engaged, the led's not turning on and there's no sound


2.Name of the circuit =

Twin Peaks Tremolo from David Rolo


3.Source of the circuit (URL of schematic or project) =

http://tagboardeffects.blogspot.fr/2015/10/twin-peaks-tap-tempo-tremolo.html

http://www.madbeanpedals.com/forum/index.php?topic=17253.0


4.Any modifications to the circuit? Y or N

No


5.Any parts substitutions? If yes, list them.

Pretty much followed the instructions closely. Instead of using a 2P4T switch, I used a 3P4T because I had one available on hand.


6.Positive ground to negative ground conversion? Y or N

Nope


7. What is the out of circuit battery voltage? => 8.3v

Voltage at the circuit board end of the red battery lead = 0.77v
Voltage at the circuit board end of the black battery lead = 0v


First 78L05 (the upper one)
I = 0.77
G = 0
O = 0

Second 78L05
I= 0.77
G= 0
O= 0

D1
A = 0
K = 0

IC1 (5532)
P1= 0
P2= 0
P3= 0
P4= 0
P5= 0
P6= 0
P7= 0
P8= 0

IC2 (TL074)
P1= 0.3
P2= 0.3
P3= 0.7
P4= 0.8
P5= 0.7
P6= 0.7
P7= 0.7
P8= 0.3
P9= 0.3
P10= 0.35
P11= 0
P12= 0.7
P13= 0.3
P14= 0

D2
A = 0.8
K = 0


IC3 (TAPLFO)
P1= 0
P2= 0
P3= 0
P4= 0
P5= 0
P6= 0
P7= 0
P8= 0
P9= 0
P10= 0
P11= 0
P12= 0
P13= 0
P14= 0





Thanks in advance for the help !



chuckd666

Well, you're only getting 0.77V at the input of what is supposed to be 9V in. There's the your problem! (Or at least the start of it) Check your battery connector and especially on that 9V jack, it looks a tad messy from the photo.

duck_arse

#2
is the battery getting hot? it seems you either have a short across the supply, or the supply is not wired to where you think it is. power off, meter to ohms, (plu a plug into the pwer switch jack), and then probe from the batter snap pins to the board, through all that other mess.

tell us what you find. (I found chuck beat me, but it's too late to stop me now.)

also, what is the battery voltage now, out of the snap?

[edit :] D2 is backwards!
You hold the small basket while I strain the gnat.

Bielorusse

Hi guys, thank you for your answers

Well duck_arse you found the source of the problem with the power supply! I reversed D1 and D2 (both were backwards...) and now the power is running.
But the pedal is still malfunctioning: I hear a loud ticking that goes away when the tone knob is fully clockwise, and the signal is very low and distorted.

Here are the voltages that I get now that there is power:

First 78L05 (the upper one)
I = 5.6
G = 1.7
O = 4

Second 78L05
I= 5.4
G= 0
O= 4

D1
A = 0
K = 1.6

IC1 (5532)
P1= 1.5
P2= 1.5
P3= 1.4
P4= 0
P5= 3.4
P6= 1.5/2
P7= 0.8/1
P8= 1

IC2 (TL074)
P1= 4
P2= 3.7
P3= 4.4
P4= 3.9
P5= 3.9
P6= 3.8
P7= 3.8
P8= 3.7
P9= 3.7
P10= 1.9
P11= 0
P12= 3.8
P13= 3.6
P14= 3.6

D2
A = 6.4
K = 0


IC3 (TAPLFO)
P1= 4.5
P2= 0
P3= 2.2
P4= 4.7
P5= 1/3
P6= 0
P7= 1/3
P8= 0
P9= 2.3
P10= 1.9
P11= 3.4
P12= 4.9
P13= 3
P14= 4.9

garcho

#4
QuoteFirst 78L05 (the upper one)
I = 5.6
G = 1.7
O = 4

a 78L05 is a DC voltage regulator. you give it something a few volts over 5V, and it gives you 5V on the money, up to a certain amount of current. you're only giving it 5.6V at the 'I' (input), why? it should be getting 9V, or whatever your power supply or battery is putting out. look for trouble there. the G is for ground, which should always be 0V, so another problem there. the O is output, it should be 5V on the button. it's not, because of the other two issues.

QuoteI= 5.4
G= 0
O= 4
same goes for this, sans ground issues.

QuoteD1
A = 0
K = 1.6
hmmm...
Quote
IC1 (5532)
P1= 1.5
P2= 1.5
P3= 1.4
P4= 0
P5= 3.4
P6= 1.5/2
P7= 0.8/1
P8= 1

OK, before going into any further detail, you have a major power supply problem. whatever you did before was only part of the solution, or maybe not even. keep looking there. follow the positive terminal on the battery and measure voltage relative to the negative terminal each step of the way along the entire circuit.

you're doing a good job with the troubleshooting post etiquette, bravo
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bluebunny

Were you powering this from a battery in your original post?  And if so, are you still using the same battery?  If also true, then your battery is toast.  Treat yourself to a fresh one.
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Ohm's Law - much like Coles Law, but with less cabbage...

duck_arse

ah-hah! rule number one - the battery is now flat!
You hold the small basket while I strain the gnat.

Bielorusse

Hi everyone,

I tested the build again, this time using a 9v DC adapter to avoid the battery problem.

First of all, I'm getting 11.5v between the pins of the DC jack. Which surprises me, shouldn't I get 9v?






Voltage from + pin of DC jack - pin of DC jack
to - pin of DC jack
11.5
0
to ground on the board
11.5
0
to power on the board
0
11.5

So I assumed that the problem is not coming from the wiring of the power supply, right?

Then I get these voltages on the board:

First 78L05 (the upper one)
I = 11.2
G = 1.7
O = 6.6

Second 78L05
I= 6.8
G= 0
O= 4.8

D1
A = 0
K = 1.7

IC1 (5532)
P1= 2
P2= 3.3
P3= 3.2
P4= 0
P5= 6.6
P6= 5
P7= 3.2
P8= 3.2

IC2 (TL074)
P1= 10
P2= 9.6
P3= 10
P4= 10
P5= 10.1
P6= 0
P7= 10.3
P8= 10.3
P9= 10
P10= 5
P11= 0
P12= 10.3
P13= 10
P14= 10.2

D2
A = 11.5
K = 0


IC3 (TAPLFO)
P1= 4.8
P2= 0.7
P3= 2.2
P4= 4.7
P5= 0.5
P6= 0
P7= 4 osc
P8= 0
P9= 2.4
P10= 1.8
P11= 3.3
P12= 4.7
P13= 2.4
P14= 4.7

I still get 1.7v between both sides of D1. But the diode shouldn't show any resistance, and there shouldn't be any voltage between both sides, should it?

It is this D1 that connects the G pin of the first regulator (let's call it R1) to the ground. So because of D1 I am still getting 1.7v at R1's G pin, and not 0v.

Would I resolve this problem by working on D1? The thing is, I already changed the diode, since I had put it backwards in the first place... I could still try another type of diode. I've been using 3mm red LED so far, in accordance with the instructions.


I also noticed that R2'input is still only getting 6.8v. But it is separated from the power by a 33K resistor. So I shouldn't be bothered with this 6.8v at R2's input, right? (4.7v tension & 33K resistance -> 0.2mA intensity. Would you call that normal here?)

Thank you all for your answers by the way :)

garcho

first of all, the G on the 78L05 voltage regulators' pinout stands for "ground" which means 0V, the common return for the circuit. they should be directly connected to 0V ground (negative pin of power jack), not via any diode. start with that and retake your voltages. basically, most of all the other ones are messed up too, so might as well start with the power supply. one thing at a time, until no more variables are left.
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"...and weird on top!"

bluebunny

Quote from: Bielorusse on January 23, 2016, 12:26:42 PM
I tested the build again, this time using a 9v DC adapter to avoid the battery problem.

First of all, I'm getting 11.5v between the pins of the DC jack. Which surprises me, shouldn't I get 9v?

If your 9V DC adaptor isn't regulated, then 11.5V is quite feasible until it gets loaded down by your circuit.  Any helpful labelling on the PSU, or photos?

Quote
I still get 1.7v between both sides of D1. But the diode shouldn't show any resistance, and there shouldn't be any voltage between both sides, should it?

It is this D1 that connects the G pin of the first regulator (let's call it R1) to the ground. So because of D1 I am still getting 1.7v at R1's G pin, and not 0v.

Would I resolve this problem by working on D1? The thing is, I already changed the diode, since I had put it backwards in the first place... I could still try another type of diode. I've been using 3mm red LED so far, in accordance with the instructions.

This is correct.  Diodes drop voltage, depending on their type.  This LED drops 1.7V.  Putting it between the ground terminal and 0V has the effect of evelating the output of the regulator (which should be a 78L06, btw!) by 1.7V, giving about 7.7V (or 6.7V in your case, with the wrong regulator!).

Quote
I also noticed that R2'input is still only getting 6.8v. But it is separated from the power by a 33K resistor. So I shouldn't be bothered with this 6.8v at R2's input, right? (4.7v tension & 33K resistance -> 0.2mA intensity. Would you call that normal here?)

Not sure what R2 is, but you might want to check it's not one of the ones marked "33R".  That's 33 ohms, not 33 thousand ohms.

BTW, it looks like you're counting your IC pins incorrectly, which may be confusing us: find pin 1 and count upwards counter-clockwise.
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Ohm's Law - much like Coles Law, but with less cabbage...