How to populate a Morning Glory/Bbreaker/KoT pcb without instructions

[eh la bas ma]

I do have 9V on pin 8 of the LM833, and there is kind of a "soft wind" sound  when I turn the volume up.

The guitar signal doesn't go through though. 

I wonder if an extra capacitor could kill the signal : There is a 1uf cap usefull for the KoT (c15 on 1rst version's schematics and Cx5 on the 2nd, if i am not mistaking).  I placed Cx5 on the pcb : I'll try without it.

Also, there is the lug 1of the Volume pot that should be connected to ground, if the gain stage is built, or if a B250k is used instead of B100k for Gain. I did bend the lug 1 to go in the provided pad which is curiously smaller. I left the original pad empty.

"The Morning Glory adds a JFET as part of the output stage.This will increase the volume and change the output impedance.   If you use this mod, be sure to connect lug 1 of the Volume pot to the GND pad located next to the original pad.  (You'll need to bend the pin of the pot.)" and " VOL lug 1 normally goes to VB, but if using the B250k, lug 1 needs to go to GND."

[antonis]

It's time for some measurements, I presume.. 

Check for +9V on PCB respective pad..
Check for continuity between PCB GND and IC pin 4..
Check for +9V on IC pin 8.. 
Check for +4.5V on ALL IC pins..
(for pin 3 you might get lower measurement due to DMM impedance loading..)
Check for 6v to 3V say, on Q1 Drain..

P.S.
It should be convenient to take off diodes from NFB loop (those connected to IC pin 7) for more accurate measurements..
In fact, it should be better for those diodes to be connected AFTER you get some kind of sound (distorted or not..)

edit:Didn't read your last post..

[eh la bas ma]

Check for +9V on PCB respective pad..
Check for continuity between PCB GND and IC pin 4..
Check for +9V on IC pin 8.. 
Check for +4.5V on ALL IC pins..
(for pin 3 you might get lower measurement due to DMM impedance loading..)
Check for 6v to 3V say, on Q1 Drain..

Thank you for your help !

9V pad on pcb shows 9.45V
There's continuity between ground and pin 4 of the LM833

LM833 reads :
1 = 8.50V              8 = 9.45V
2 = 8.44V              7 = 9.32 V
3 = 8.49V              6 = 9.09V
4 = 0.001V            5 = 9.45V

No 4.5V value anywhere, I guess that's bad.

Q1's drain leg reads 6.37V

I took off 2 caps that seem unnecessary for the Morning Glory's version (C8 and Cx5 on my pcb, C15 and C16 on the 1rst version 's schematics), placed a jumper on C8 and C9 (C9 is connected to Rx3, Q1 drain pin, and Volume lug 3, located in the optional gain section, I am still trying to find it on schematics, it's not electrolytic).

I left the diodes for measurements, in the 1rst post's picture, you can see there is 8 of them.
If you insist, I will desolder them all .

[antonis]

No 4.5V value anywhere, I guess that's bad.

Not necessarily..

There should be 2 X 47k resistors connected to IC pin 3.. (or to pin 5, depending on the specific schematic used..)

Check for one of them to also be connected to GND..

[eh la bas ma]

 I placed them just above the polarity protection diode, and under R10

(r10 is a jumper connected to pin 5 an the positive leg of C10(100uf) on one side, to 9V and the positive leg of C11 (100uf) on the other side).

47k are R9 and R11 on my pcb.

R9 is connected to protection diode, pad IN, R11, Rpd resistor (i guess it's GND too), and GND on one side, and to cx5, C8, pad OUT on the other .

R11  is connected to, pin 8, positive leg of C11 (100uf), C6, R10 (jumper) on one side and R9, and GND on the other.

I don't know what to do with R10, it's connected to pin 5 and caps so maybe it was 47k...

[antonis]

R9 is connected to protection diode, pad IN, R11, Rpd resistor and GND on one side and to cx5, C8, pad OUT on the other .

Sorry but that can't stand..
(you realize you're talking for IN - OUT permanently connected via a 47k resistor, don't you..??)

[eh la bas ma]

I have a short between GND and IN sorry!
edit : I meant GND and OUT, i'm looking at it from the wrong side.

[antonis]

There shouldn't be neither IN nor OUT short to GND..
If it's some kind of PCB track, throw away the whole board..

[eh la bas ma]

On the pcb's wiring pads when the pedal is switched off, IN is connected to GND, it's no longer connected when I switch on.

I've cleaned aeras around solder joints, no more shorts but R9 is still ringing with IN and OUT.
Looking at Delyk's back side picture of the pcb , it seems ok to me, like it should do that :

https://www.delykpcb.com/product/el-rey-de-la-gloria-azul-ii-pcb/

(under the big pic you can choose and click on the  back side picture)

edit : the diode is just above the wiring pads and R9 is just above the diode
edit 2 : I can be wrong, the diode isn't connected to IN or OUT, so I'm not sure why it's ringing with R9
edit 3 : I will invert R9 and R10 just in case

[eh la bas ma]

It's working ! I have a signal with the stompbox activated !

And distortion ! Seems to work fine, the clip switch doesn't do much but the rest is fine.

I'm so gratefull to you !

Glory to Antonis !!

[antonis]

Hurray...!!! 

[eh la bas ma]

It's a miracle, I mean what a day ! I didn't really believe it could be done, this is your help and the strangest luck.

Many thanks again !

Now let's listen to this thing...

Edit : one last thing so I can clarify things a bit : "VB" and "VA", or VR means "voltage going to "A", to "B" or to "R", is that it ?

[antonis]

VA is the voltage feeding the circuit AFTER the protective diode..
(Power supply voltage minus Schottky diode forward voltage drop..)

VB(ias) or VR(eference) is the IC non-inverting inputs biasing voltage..
(half of VA in case of R9/R10 equal values..)
It also serves as AC ground due to C10..

[eh la bas ma]

VA is the voltage feeding the circuit AFTER the protective diode..
(Power supply voltage minus Schottky diode forward voltage drop..)

VB(ias) or VR(eference) is the IC non-inverting inputs biasing voltage..
(half of VA in case of R9/R10 equal values..)
It also serves as AC ground due to C10..

Mmmmh ? AC ground ?  I'm gonna reflect on that, do some research to really get it  inside my head. I still don't understand most of this circuit. I'm going to have to build more and specially read more . Thank you !

[antonis]

the clip switch doesn't do much

It depends on D8/9 and D10/11/12/13 type of diodes..
You can see that when switch is set to "HARD", there is only one diode forward voltage drop for each waveform polarity (+/-) where there is 2 X 1 diode forward voltage drop for switch set to the other position..
(it doubles clipping threshold hence less hard clipping..)
You can see that, in case of D8/9 voltage drop double of D10/11/12/13 one, switch is totally useless..

Here comes AC ground (VR) which simply is considered as GND for signal..
C10 (47μF) exhibits very low impedance (you can call it capacitive reactance or resistance..) so VR is effectivelly connected to circuit GND..

P.S.
Plz., don't ask the reason for connecting signal to VR instead of GND 'cause it will trigger  a veeeeery long argument and it's late enough here.. 

[eh la bas ma]

P.S.
Plz., don't ask the reason for connecting signal to VR instead of GND 'cause it will trigger  a veeeeery long argument and it's late enough here.. 

But...  but you just aroused my curiosity ? i am going to do a search on this forum.

Edit: Virtual ground? a fancy trick done to the op-amp ? there's many posts about it at first glance

[Rob Strand]

Edit: Virtual ground? a fancy trick done to the op-amp ? there's many posts about it at first glance

Virtual ground is a different thing.  It's something which comes up with inverting amplifiers.

To a first approximation the VR vs Ground thing is very simple.  You need two idea:

- Any voltage source looks like ground to AC signals.

- A capacitor prevents DC current flowing but allows AC currents to flow.

The difference between VR and Ground is VR has a DC voltage and Ground does not.  For AC signal, because of the first point, VR and Ground looks the same.

When the signal connecting to VR has a series capacitor that blocks any DC, because of the second point.   That means you can connect that point to either VR or Ground and  it will not affect the DC, and it will behave the same for AC.  So the two appear to be identical.

Optional:  Where they aren't the same is VR isn't a rock-solid voltage source it's made up of resistors and caps.  They means to some degree it's not ideal.   If the cap on VR is too small it makes this problem worse.    You might see some stuff about about VR being noisy.  It can be noisy but connecting those optional VR vs GND connection to ground doesn't get around the noise because there's other  connection points between VR and the opamp that lets the noise in anyway.    Don't worry if it doesn't make sense!

[eh la bas ma]

Almost every stompboxes are fed with DC power supply. Where does the AC current come from ? Is it "made" inside the circuit ?

P.S.
Plz., don't ask the reason for connecting signal to VR instead of GND 'cause it will trigger  a veeeeery long argument and it's late enough here.. 

Is "VR instead of GND" a specialty on Bluesbreaker-ish circuits, or many other effects are based on this, like tubescreamers for exemple ? (i am not really asking the reason here, I don't want to start any argument... I think).

[antonis]

Almost every stompboxes are fed with DC power supply. Where does the AC current come from ?

!! Signal..!!!! Signal..!!!! Signal..!!
Guitar signal is an almost perfect(*) sinewave and, as Fourier has proved, you can form any kind of waveform by superposing various sinewaves..

(*) Nothing more falseness than the above..

Is "VR instead of GND" a specialty on Bluesbreaker-ish circuits, or many other effects are based on this, like tubescreamers for exemple ?

You're confused a bit..
GND is GNS is GND, period...!!
You may consider VR as a subset of GND set..
(is works for AC as GND works for both AC and DC..)

Sometimes it's more convenient (depending on particular circuit and layout) to use VR as "ground" instead of GND..

P.S.
It seems to me you didn't read/study well what Rob told you above..

[duck_arse]

Almost every stompboxes are fed with DC power supply. Where does the AC current come from ? Is it "made" inside the circuit ?

the DC sets up the static operating conditions inside the circuit. the AC is the signal which you jamb in one end, mangle in the middle, and extract from the other end. the DC stays inside the box, the AC goes in and comes out.