1 beginner 4 pedals

Started by ghiekorg, September 06, 2020, 05:46:55 AM

Previous topic - Next topic

11-90-an

The pics are just fine...  ;D

The voltages I meant are these ones...  ;)
https://www.diystompboxes.com/smfforum/index.php?topic=29816.0
But read what i saw below before doing this, as probably what i state below is the problem...

Quote from: ghiekorg on September 24, 2020, 02:19:52 AM
QuoteAnyway... FS1 and FS2 control the on/off of the reverb... ;) ;) ;)
Isn't it actually for an external pedal to control the oscillator?

From the schematic...

I believe that is the oscillation switch...
and

this is the footswitch that controls on/off

The JFETs Q1 and Q2 control the switching of the circuit here.
When a JFET's gate here is given 9v, it turns "on" which lets signal pass through the drain and source. When given Ground, (here 0v,) it turns "off" and signal is basically muted
(of course, that is a drastic oversimplification, but I'm sure others can explain it better than i can..)
What the footswitch controls here is either 9v to one, 0v to the other, or 0v to one, 9v to the other.

So anyway, if you don't wire FS1 and FS2, there will be no way for either of them to get 9v, so no signal would ever reach the Belton Brick, and no signal would also pass when the pedal is disengaged.

I hope I make sense...  :icon_redface:
flip flop flip flop flip

ghiekorg

#81
Thank a lot 11-90-an
I don't know if i get it right... I think there is a misunderstanding. In the description of the pedal there is
QuoteI also recommend using this version if you intend to put the Oscillate switch on a footswitch and build a dual footswitch effect in a 1590BB box horizontally
So, as i understand, there is a toggle switch on the pedal itself (ON/ON) and then there is the possibility to add another switch to switch the OSC without using the toggle (using FS1 and 2).
Overwise what would the switch be? It can't be the main switch of the pedal, there is no IN/OUT... am i wrong?

In case i am right i have some questions: :D
1- is there a way to connect FS1 and FS2 and bypass the whole "external" FS? For example connecting FS1 to ground and FS2 to 9V (green lines). So the state is always FS1=0v and FS2=9v and then i just use the toggle switch to change them.
2- in the schematic there is a R25 (red arrow, not listed in the pdf i have) and a LED (same). If i bypass the pedal this way can i just avoid both of them and the ground connection (blue arrow)?
3- can i connect FS1 and FS2 directly at the GND and 9V of the DC input jack (it's easier) then on the board itself maybe


I am so dumb, i didn't read the description carefully before...

About the voltage (sorry i got it super wrong :D) i will check them as soon as i am going home (in 8 hrs) if what i just said wouldn't work.




11-90-an

Nope. the oscillation switch is somewhere else... (encircled in red) ;)


The switching here in this pedal is called "tails switching". Basically, it doesn't cut out the signal right after you click the footswitch. Rather, the reverb fades away and isn't as abrupt. This won't be possible with connecting the in/out to the footswitch. That is why the JFETs do the switching. How they are configured makes all this possible.

Note that this isn't true bypass.
flip flop flip flop flip

ghiekorg

#83
Aaaah now i get it! Thanks a lot! That sentence form the description completelly threw me out. Now i understand :)
I also found this:


The problems i have now are:
- how do i connect the footswitch? Mine is a 3pdt and i have no dpdt, can i just leave the right column empty and follow th image above? It seems so when watching this:

- Where do IN and OUT go now? Directly from the board to the input and output jacks' tips?


11-90-an

#84
Yes. A 3pdt switch is basically 3 spdt switches all toggled at once. A dpdt switch is 2 spdt switches all toggled at once. Therefore, when using a 3pdt for 2pdt application, the only problem is that one spdt is wasted!  :icon_lol:

Yep, the board in/out should be connected directly to the in/out jacks
flip flop flip flop flip

ghiekorg

thanks a lot 11-90-an. you saved me <3

As soon as i am home i try it out :)

duck_arse

Quote from: Mark Hammer on September 23, 2020, 02:23:13 PM
It would seem from the drawing that C16 is supposed to be non-polarized, without any + or - terminal.  It is possible to use a polarized/electrolytic cap in that position, and I imagine that the negative terminal of the cap goes to the chip and the + side goes to the pad where it joins with C15/R22.

I have to reason the opposite direction. the R22 resistor connects to ground, via the output pot, so is at 0V. the opamp input pin does what the output pin tells it to, and the output pin will sit at the bias voltage, about 4V5. so, the input pin will be at 4V5, the cap//resistor junction will be at 0V.

another way to deduce direction - that electro cap C15 has orientation marked, and shows (-) end pointing towards R22 and C15. this would suggest that point being (-) or more minuser - it doesn't make sense to have the (+) leg of one cap [C15] connected to the (-) leg of a series preceeding cap.

but your multimeter will provide a very quick result we can't argue with or about.
don't make me draw another line.

ghiekorg

It worked! 😁 now i just have to not mess it up while boxing it... Thank a lot

@duck_arse: I read  your message too late and my cap negative pole is facing right, looking at the green/white scheme (so the positive is facing the FS1 and 2 holes). But it works... What could happen if it is turned wrong?

11-90-an

QuoteWhat could happen if it is turned wrong?

BOOM

That's the worst thing that could happen. The least would be it heating up and smoking.  ::)
flip flop flip flop flip

ElectricDruid

Quote from: 11-90-an on September 24, 2020, 08:21:56 PM
QuoteWhat could happen if it is turned wrong?

BOOM

That's the worst thing that could happen. The least would be it heating up and smoking.  ::)

Don't agree. :P

POP!

...is about the best I've ever got! For genuine BOOM you need something a bit more explosive than typical electronics. Heating up and smoking is much more the typical thing.;)

11-90-an

Quote from: ElectricDruid on September 24, 2020, 10:04:49 PM
Quote from: 11-90-an on September 24, 2020, 08:21:56 PM
QuoteWhat could happen if it is turned wrong?

BOOM

That's the worst thing that could happen. The least would be it heating up and smoking.  ::)

Don't agree. :P

POP!

...is about the best I've ever got! For genuine BOOM you need something a bit more explosive than typical electronics. Heating up and smoking is much more the typical thing.;)


Well, of course, depends on the size of the capacitor.
There are those giant ones that when exploded, almost seem like landmines!  :icon_mrgreen:

flip flop flip flop flip

ghiekorg

I tried it a couple of minutes and nothing popped... how can I measure it with a multimeter? I don't know how to do it and searching online I can't find it (just how to check its polarity). 😔 and if it "pops" will it just stop working and the sound will stop or could it ruin other components? Thanks everyone

11-90-an

#92
nah... ::)

if it doesn't pop then it's good.

yes, it will stop working if it pops
flip flop flip flop flip

ghiekorg

i have let it ON for about 10 minutes and nothing popped. Not even warmed up

bluebunny

Quote from: ghiekorg on September 25, 2020, 12:41:33 AM
how can I measure it with a multimeter?

Measure the voltages at each end of the installed capacitor.  Find the end that's the most positive and (re)align the cap accordingly ("+" terminal to the most positive end).  If there's no discernible difference, then just leave it as it is.
  • SUPPORTER
Ohm's Law - much like Coles Law, but with less cabbage...

ghiekorg

Thank you bluebunny.
Problem is i have no clue on how to do it.  😔
I tried to touch the ground with the black tip and then the two ends of the cap with the red one. When i touch the now positive side (now facing the FS holes and connected to R22 and C15) i get 0. When i touch the negative side (now facing C17 and C18 and connected to C17, R23 and the IC1) i get i number around 100 (multimeter on 200m) that then drops. Does it mean it's flipped now?

bluebunny

Your method is correct.  The results suggest you might be better off re-orienting the cap.  Someone more qualified may be able to tell us what effect 100mV will have.  I can tell you what 12V does - it's exactly as Tom described above.  POP!  And it makes a mess...
  • SUPPORTER
Ohm's Law - much like Coles Law, but with less cabbage...

duck_arse

Quote from: ElectricDruid on September 24, 2020, 10:04:49 PM
Quote from: 11-90-an on September 24, 2020, 08:21:56 PM
QuoteWhat could happen if it is turned wrong?

BOOM

That's the worst thing that could happen. The least would be it heating up and smoking.  ::)

Don't agree. :P

POP!

...is about the best I've ever got! For genuine BOOM you need something a bit more explosive than typical electronics. Heating up and smoking is much more the typical thing.;)

nah, it's more like fssshhhhHHHH, with the last H's getting bigger font.

now, your  "i get i number around 100 (multimeter on 200m) that then drops" suggests to me you are using the wrong meter range. you want a Volts range, 20V probably. with [power on and] black probe to ground, cause always, red probe to the cap legs in turn. one leg should show 0V, one should show around +4V5, steady, not dropping. the one that is higher is more positive than ground, and that tells you which way the cap goes in.
don't make me draw another line.

ghiekorg

Here is how i measured it. Now the values are even lower. No idea... wait, you don't mean power on in the sense of pluggin the board, right?
https://www.youtube.com/watch?v=tqTD3vVoWGg

And here is where i measured:



11-90-an

But wait... all this stuff about the capacitor... but does the reverb work? :icon_mrgreen:
flip flop flip flop flip