Is my pedal FUBAR? Debugging Report Posted

[brokenbones]

1.What does it do, not do, and sound like? - The pedal is non-functioning. I believe it was input with a different polarity adapter causing the permanent outage. Since i'm kind of new at this i'd like to know if this thing can be fixed.

Is it strange for all the voltages to read so close to one another??!

My electrolytic caps all measure the same voltages for + and -. I assume this means they're fried?

***Also, I tried the audio probe method but to no avail. I'd start on the input section and Q1 and C1 both were dead. I even tried swapping out them out for new components and even a new R1 but nothing worked. Once I got no signal from R2,R3, and R4 something weird happened to my Macbook headphone jack. I was running audio out into the pedal from an online signal generator. That was the sign I needed to stop and decided to post my findings on here. Like I said, i'm fairly new to this game. I've got a couple pedal builds under my belt and now i'm trying to tackle my "stack" of old, broken pedals. I'm sure this pedal is dunzo but i'd like to learn and know for sure what could have happened and MAYBE if it can be fixed.

2.Name of the circuit = Boss DS-1, stock, late 90's
3.Source of the circuit (URL of schematic or project) = http://www.diystompboxes.com/DIYFiles/up/Build_Your_Own_DS-1_Distortion.pdf
4.Any modifications to the circuit? N
5.Any parts substitutions? Possibly D1. I think I swapped out a new 1N4004 a few years back.
6.Positive ground to negative ground conversion? N
7.out of circuit battery voltage? => 9.6V
Now insert the battery into the clip. If your effect is wired so that a plug must be in the input or output jack to turn the battery power on, insert one end of a cord into that jack. Connect the negative/black meter lead to signal ground by clipping the negative/black lead to the outer sleeve of the input or output jack, whichever does not have a plug in it. With the negative lead on signal ground, measure the following:
Voltage at the circuit board end of the red battery lead = 9.56V
Voltage at the circuit board end of the black battery lead = -.03V via input jack ground to circuit board (schematic yellow wire/Hole 14)

Q1 
C = 9.25
B = 8.96
E = 9.37

Q2
C= 9.53
B= 9.56
E= 9.52

Q3
C = 9.56
B = 8.68
E = 9.55

Q4
C= 9.54
B= 9.51
E= 9.51

Q5
C = 9.54
B = 9.51
E = 9.51

Q6
S = 9.54
G = .11
D = 9.54

Q7
S = 9.54
G = .14
D = 9.54

Q8
S = 9.54
G = .14
D = 9.54

IC1
P1 9.14
P2 9.14
P3 9.42
P4 9.54
P5 9.14
P6 9.32
P7 9.42
P8 9.42


D1
A (anode, the non-band end) = 9.55
K (cathode, the banded end) = 9.56

D2 5.6V Zener
A = 7.04
K = 9.56

D3 Omitted

D4
A = 9.55
K = 9.52

D5
A = 9.52
K = 9.55

D6
A = .11
K = 8.66

D7
A = .11
K = 8.66

D8
A = 9.55
K = 9.42

[GibsonGM]

Sorry man, it happens!  What I would do is as follows:

Replace D1 with an appropriate diode, the 400x series is fine.  Re-connect proper power supply.  Study the board when you power it up to be sure nothing is getting HOT!  Pull power rapidly if so...I think some Q's could be popped here (Q=transistor)

- what voltage do you then get at the top of D1 after this? Should be same as power supply.  What's the voltage between R24 and R25?  Should be about 4.5V   

What you've posted suggests to me that the diode (to start with...) could be porked.   You can do this fast if you know how to test a diode....you should have a high resistance to ground from the diode cathode to ground.   If we know the power supply is ok, we'll move on. 

If it 'starts to straighten out' when you change the diode, please re-do your other measurements so we can see if ICs and/or Q's are affected...the bias voltages are way out, so I need to know if the ref. at the power supply is ok first...this is a place to start, anyway!

[R.G.]

What he said.

Good one, Mike.

[brokenbones]

D1 has failed before and been replaced. It worked for a few months afterwards. I have one Diode left and i'm debating to try this again. The PCB is burnt around one anode/cathode of D1. I'm sure this part of the failure.

[GibsonGM]

What he said.

Good one, Mike.

Thanks, R.G.! 

Bones - yeah, that diode is there for protection.  It doesn't conduct normally, it is reversed in the circuit and that prevents it from 'turning on'.    If you reverse polarity, tho, (ahem...lol), it will conduct and kinda 'make a loop' right there, and hopefully save your circuit.    The diode is KEY...if it has failed as a short, it will be a short circuit when you apply power and continue causing problems.  So, I'd suggest you start there - at least remove it for now, but then you MUST NEVER EVER reverse that polarity as there will be nothing protecting your circuit!

You know how to test a diode?  Use your meter, set on "ohms".   With black probe on the band end (cathode), red probe on the other (anode), you should get a low resistance reading - the diode is turned on and conducts thru your meter.    If you reverse the probes (red on band end), you would see if very high reading as this is now "backwards" (reverse biased) and only microscopic current can flow.     Odds are good you read very low in both directions, or "error" (burned, open, no connection).

The burn marks tell me that it needs to come out so it's in this equation...that 'it has failed before' means the power supply was reversed  before, too.    Can't go further til it's taken out and replaced, or you remove it and carefully apply proper voltage, and get us some measurements... 

Order some more diodes - you'll need them, ha ha...1N4001 and 1N914 types...and that's just for starters!!!  Addictive hobby, but the parts are cheap....

[R.G.]

What he said, plus a radical alternative.

If the reverse protection diode actually shows burns around it on the PCB, there has been some serious current flowing in there. Given that the diode was replaced, and it failed again implies to me that it's been damaged, and only limped along for a little while.

If you want this to be fixed, you can replace the one most-damaged part, like what has already been done. If you want it to be reliable in the future, you have to assume that there may be a lot of damage.

I first commented that the electronics in effects pedals might as well be free, compared to the box, the switches, jacks, controls and so on. That's still true. If I were repairing this, I would make up a list of every single semiconductor and electrolytic cap and order new ones for them all.  Resistors and non-electro caps are usually either dead or not, but semiconductors and electro caps can have a lingering death. So I would replace every polarized part on the PCB. The parts are almost certain not to be expensive. Electro caps are about US$0.10, transistors are about the same, opamps are US$0.25 and up.

It sounds radical, but it does cut off the likelihood that in a week, another part will fail; then a month after that, another one.

I would get the parts in, clip off the old ones on the component side and then pull out the stubs of the leads from the holes and poke the holes open with a wooden toothpick, and put in new ones, carefully observing the polarity of the original part.

It's a way to sidestep some continuing misery.

[GibsonGM]

Really, what R.G. says is the very best way to approach a problem like this in such a relatively simple circuit.   We could help you waste a lot of time spend time going thru all the components, trying to interpret voltage readings that might be skewed by a blown component elsewhere...err, lots of 9V's in there, LOL...I was heading in this direction, but those 9's are suggesting some pretty hefty problems...

In the end, it's just your time and a few dollars, and you'll get a LOT of great experience, hands-on with the soldering iron!   Up for the challenge?   

[antonis]

Just to add something on the above well said.. 

Place a 100R (or less) 1/4 W resistor in series with the protecting diode.
In case of reverse polarity, it saves battery life or lower PS "strain" and it works as a fuse at lower current than diode..

[GibsonGM]

Exactly.  The reversed polarity causes the diode to conduct, but it will do so at a HIGH current, as its "on" resistance is very low.  The 100 ohm resistor in series, so you have "+" >> 100R >> reversed diode to ground >> rest of circuit, gives the diode a load.  And like Antonis said, it is easier on whatever is powering the device.

It won't restrain ("current limit") the circuit in any way, as 9v/100ohms = 90mA, FAR more than most any 9V circuit would ever draw. 

[R.G.]

As a further aside, reverse shunt protection diodes like this don't usually conduct enough current to die, except in two circumstances.

1. A **big** power supply connected to the pedal, in reverse; this provides enough current so the ~1V forward drop of the diode generates enough current to get very hot. Nominal 1A diodes as are very common will only generate on the order of 1W (1V * 1A) at the diode max current, and only a couple of watts with a 2A supply connected backwards. 1-2 W is in general not enough to visibly burn PCB stock. Well, maybe phenolic.

So the current is often bigger than 1-2A. Some of the boy-genius setups with a computer power supply that can make 12V at 20-30A might do something like this unless there is a current limiter somewhere to keep the pedal from getting the power supply's full and undiluted attention.

2. More common would be connection to a 9V *ac* adapter as used by some digital pedals. These things can supply a lot of current, several amps into diode, and are guaranteed to be the wrong polarity every 1/60 of a second. AC power connected to pedals is almost certain to fry the diode.

If the diode fries shorted, then the power supply goes on to overheat and possibly kill itself. If the diode fries open, the power supply is then free to spend a lot of special time frying all the parts on the board.

[thermionix]

How burnt is the pcb?