Help with DOD250-type OD

Started by Wadey86, April 23, 2018, 01:36:27 AM

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Wadey86

Hi all,

Attempting my first stompbox, and my first circuit in a few years tbh, and having a difficulty with this.

I'm building one of these:
http://www.diyguitarpedals.com.au/shop/boms/Vulcan_DOD250_schematic.pdf
with one modification - I have replaced the clipping diodes (D2 and D3-D4) with a switch each selecting between two LEDs (on-off-on) for the clipping.  (it's a bit of an experiment for funsies)

I have built the circuit with the following results when plugged in -
Firstly, there is no signal coming from guitar to amp.  at all.
If I dime the volume and gain pots i get a small amount of hiss
Connecting/Disconnecting the battery gives a small flash from the selected LED(s)

I have tried it with the RC4558, and a TL072.  The TL072 is currently installed.
V1 = 6.25
V2 = 5.9
V3 = 4.25
V4 = 0
V5 = 6.3
V6 = 6.4
V7 = 6.4
V8 = 12.8

Can anyone advise me what the next step in debugging this would be?

Thank you

bluebunny

Welcome to the forum, Wadey.

Quote from: Wadey86 on April 23, 2018, 01:36:27 AM
Firstly, there is no signal coming from guitar to amp.  at all.

Do you mean when it's in bypass?  Or ever?  It may be time for, a) an audio probe (to find out what's happening to your guitar signal), and b) pictures of what you've built.  BTW, I assume you're powering this from a nominal 12V supply?  In which case your readings look reasonably ball-park.
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Wadey86

Thanks.

ATM I haven't hooked a switch in, just the board, so yes that's when it's engaged.

Using a fresh 9v battery to power it.

I'll have to look up audio probes, and ill try to post some photos asap, thanks for the tip and for your thoughts on the readings.

Rob Strand

The voltages look good so maybe it's the wiring.    Maybe check the foot switch wiring especially the input.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

antonis

12.8V on pin 8 from 9V battery (without any voltage converter) should be considered peculiar, shouldn't it..??  :icon_wink:
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Rob Strand

Quote12.8V on pin 8 from 9V battery
I was thinking a plug pack but he does say battery.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Wadey86

Thanks Rob, but I haven't put a switch in yet, I have that to wire up but just wanted to test out the circuit before I go boxing it up, so I don't have to pull it all out again...

I'm getting a reading of 11.8 across the 9V battery straight up, so might be that my multimeter is off, or is that normal from a packet-fresh battery?

I'll post a photo when I can

Slowpoke101

You may need to put a new battery into your multimeter.
The accuracy of some multimeters can be very poor when their battery is nearly dead.

A fresh 9V battery can read a bit higher than 9V when not loaded by a circuit but very rarely above 10V.
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..

Rob Strand

QuoteI'm getting a reading of 11.8 across the 9V battery straight up, so might be that my multimeter is off, or is that normal from a packet-fresh battery?
I think 10.6V is the highest I've ever seen.   I go with Slowpoke101's suggestion.

Have you got any 5V regulators?  they should be accurate to about +/-4%.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Kipper4

Side note

How come that R4 doesnt go to the output of U1 2 (buffered Vref) Otherwise why not just tie down the spare op amp.

Also instead of 500k gain pot to gnd? Wouldnt it be better also at the output of U1 2

There has to be a reason. It's not like Paul.
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ElectricDruid

Quote from: Kipper4 on April 23, 2018, 06:57:33 AM
Side note

How come that R4 doesnt go to the output of U1 2 (buffered Vref) Otherwise why not just tie down the spare op amp.

Yes, that's weird. It looks to me like a error by whoever traced the circuit. Otherwise why not use the extra op-amp to buffer the Vref? You might as well.

Quote
Also instead of 500k gain pot to gnd? Wouldnt it be better also at the output of U1 2

There has to be a reason. It's not like Paul.

The odd thing about that for me is the huge values. Bigger resistors produce more noise, especially in circuits with plenty of gain. The advantage is the current consumption goes down, so I'd assume this was designed for long battery life, and we'll tolerate a bit of background hiss.



reddesert

Quote from: ElectricDruid on April 23, 2018, 02:45:24 PM
Quote from: Kipper4 on April 23, 2018, 06:57:33 AM
Side note

How come that R4 doesnt go to the output of U1 2 (buffered Vref) Otherwise why not just tie down the spare op amp.

Yes, that's weird. It looks to me like a error by whoever traced the circuit. Otherwise why not use the extra op-amp to buffer the Vref? You might as well.

Quote
Also instead of 500k gain pot to gnd? Wouldnt it be better also at the output of U1 2

There has to be a reason. It's not like Paul.

The odd thing about that for me is the huge values. Bigger resistors produce more noise, especially in circuits with plenty of gain. The advantage is the current consumption goes down, so I'd assume this was designed for long battery life, and we'll tolerate a bit of background hiss.

The 1M feedback resistor and 500K gain pot + 4.7K resistor to ground are faithful to the classic DOD 250 circuit https://sites.google.com/site/snmavronis/NeoClassicFX/741overdrive/research.  That used a single 741 op-amp - buffering the bias is an addition.

Wadey86







Ok so from what I have gathered the opamp is more or less doing what it should, so tge most likely scenario is I've made a rookie error in my off board wiring... So attached are photos, what I have wired is (clockwise on the board)
- +9v - obvious, the red wire from the battery snap
- volume pot - facing the base of the pot, pins at 6 o'clock pins number 3-2-1
- clipping diodes - colour wires go parallel to + side of 2 LEDs for each diode slot, and from the flat edge side of each LED to a on-off-on switch (one for each diode slot), to a common black wire (per slot) to the - marked side of the diode slot
- output jack - white to the tip, sleeve to Ground marked on the board
- ground - to black wire of the battery snap
- input jack - white to the tip, sleeve to ground
- gain pot - as volume pot

Have I made a rookie mistake here? Any help/suggestions appreciated...

ElectricDruid

Quote from: reddesert on April 23, 2018, 03:44:09 PM
The 1M feedback resistor and 500K gain pot + 4.7K resistor to ground are faithful to the classic DOD 250 circuit https://sites.google.com/site/snmavronis/NeoClassicFX/741overdrive/research.  That used a single 741 op-amp - buffering the bias is an addition.

Fair enough. I used to design distortion pedals with an eye to getting current drawn below 3mA. I was fifteen. TL062 or LF444 were my favourite chips and a 1M gain pot wasn't something to be afraid of.

Moving it to a dual op-amp and then not using the other half does seem a bit odd though, doesn't it?! I mean, it's not like the TL061/71/81 wouldn't have done the job.

T.


thermionix

I can't quite follow what you've done with the LED clippers, but possibly some mistake there.  Try it simple first, just two anti-parallel diodes (LEDs) with no switches, see if you get sound.

Rob Strand

QuoteHave I made a rookie mistake here? Any help/suggestions appreciated...
Well, I gave it a look over and I can't seen anything obvious.

The first thing that comes to mind is you have a short. Check for solder bridges and wires bent over.
It will be at the input, output or around the diodes.   Best to check shorts by eye but you can use your DMM continuity buzzer - test with the unit powered off.

The only other thing I could think of is D4 on the PCB is actually D2 on the schematic and the wire shunt in location D4 is shorting the output.  (unlikely but worth a sanity check)
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

ElectricDruid

Ok, sorry for derailing the thread. I'm going to be more helpful this time, I promise! ;)

If there's no signal coming through, and assuming the op-amp voltages are ok (big assumption - this we need to check; >11V  on a 9V battery is not realistic.), the most likely cause is something shorting it to ground. That could be an actual short, but if you've installed switches for the diodes on the output, I'd be inclined to suspect that one/both of those is casing a short to ground when it's not supposed to? Do you get no signal in all positions of the switches? If you *do* get a signal in one position, I would expect it to be heavily distorted, but it's still a big step forward. If not, desolder both switches - does the signal come back? If you *still* don't get anything, that means the problem must be earlier in the circuit. That narrows it down quite a lot since there isn't that much ahead of the diodes in the audio path of the circuit.

HTH,
Tom

Wadey86

Thanks all.

Rob - I'll have another look for shorts/bridges. Had one or two but thought I'd got them, I'll check again though with fresh eyes when I get home from work.

ElectricDruid - thanks, that helps. The switches are on-off-on, so the middle position should cut the - side of the LEDs, so that should be like removing them entirely, right? Either way, I get no signal at any position of both switches.

Thermionix's suggestion I think is the next step - to just put one LED in each slot and see if I get anything there. If bit then like Rob suggested the issue is most likely early in the circuit then.

Rob Strand

#18
Quoteto just put one LED in each slot and see if I get anything there.
The circuit should pass signal with no diodes loaded at all.
You might not want to pull the wires out as they might be hard to get back in.
You could just desolder the wires from the switches but just desoldering the centre pin on the switch will pretty much remove everything to do with the diodes.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

PRR

> Bigger resistors produce more noise, especially in circuits with plenty of gain.
> The advantage is the current consumption goes down,


The 470K is "bypassed" by the source. Has almost no effect on hiss. Would have less effect if it was larger. The 10K series has a small effect. The main hiss is the TL072(?) JFET hiss. None of it is "large" in guitar world.

Total current consumption is 90% the TL072's internal bias, and 9% the 22K divider. When CLIPPING there's a tenth mA in R7 to the LEDs. The bias and NFB values have no effect on current consumption.

Anyway why worry? His "9V" battery makes 12V! (I do agree that his meter is on the edge of crapping-out. It has an internal, say, 1.00V reference. As the meter main battery sags this falls-off to 0.9V, 0.8V... By comparison, the external voltage reads high. Until the meter battery falls so low the display and brain punk-out.)
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