Can you please measure the DC on both in and out when the pedal is bypassed? On the outer side of the in/out cap.

Quote from: Baran Ismen on February 22, 2024, 02:36:24 AMOne thing that I could not still understand (the logic) is the term of impedance.

Can someone explain to me what impedance is, and how does it affect like explaining to a monkey? I mean, literally the most basic way possible with real-life examples even 

Seems my pick-ups have an output impedance of 1K. How good is it? That means the output current has just a little "resistance" to go out from the guitar? 

To make it super simple, think of impedance as of a resistance, represented by a resistor. On the input, that resistor is between the input and ground. On the output, that resistor is in series. In reality, circuits are full of frequency dependent parts, and various invisible parasite parts that may be also frequency dependent. When those are also taken into account, and they must be, the resistance becomes impedance, which is a complex term that covers them all.

Now, when you place a small resistor between the input and ground, that resistor will be in parallel to the input resistance (represented by the resistor mentioned above). When you have two resistors in parallel, the result is always less than the smaller of them. The small resulting resistor then loads the input signal and sucks the life out of it.

The term Impedance covers resistance (represented by the resistor mentioned above), and also caps and coils, because they are there, and you have to take them into account. Caps have their resistances, resistors have capacities, and things get very messy. So as you see, in reality not all the frequencies will be dragged down / let through the same way, because the frequency dependent parts affect them differently.

Output impedance can be simplified a similar way, but this time as a resistor in series with the output. When you place something behind it and that something loads the output (by its own small input impedance, remember it is between the input and ground), the signal will drop on the output impedance. Like when you use a weak DC voltage source to feed a hungry pedal - the voltage will drop and the pedal won't work.

That's why we typically want a big input impedance (so it won't load the input signal) and a small output impedance (so it won't drop the output signal when loaded by something behind it).

Try to understand the way a humble resistive voltage divider works, arkadaşım..

Then replace resistors with impedances (capacitive and inductive reactances - no reactance for pure resistive materials)
e.g. impedance of a string of series R, L & C is considered R + Z_C + Z_L, where Z_C = 1/(2π*f*C) and Z_L = 2π*f*L with f = frequency of interest..

Then you'll understand why 1V input of a 100nF/100k HPF results into 500mV output (50% attenuation) at 15.9Hz..

P.S.
No need for phase lead/lag, complex numbers and transfer functions involvement at the moment..

Quote from: Lino22 on February 22, 2024, 03:26:22 AMCan you please measure the DC on both in and out when the pedal is bypassed? On the outer side of the in/out cap.

You mean the DC voltage or the current?

1- Shall I keep the pedal powered when I do this?
2- Shall I remove the resistors I've soldered?
3- I doubt I can find the input and output caps, I only see 5 or 6 caps that are thru-hole, and most of them are around the power input. There's one at the voltage regulator and there's another one close to an IC whose writings are deleted.

Quote from: antonis on February 22, 2024, 06:26:28 AMTry to understand the way a humble resistive voltage divider works, arkadaşım..

Then replace resistors with impedances (capacitive and inductive reactances - no reactance for pure resistive materials)
e.g. impedance of a string of series R, L & C is considered R + Z_C + Z_L, where Z_C = 1/(2π*f*C) and Z_L = 2π*f*L with f = frequency of interest..

Then you'll understand why 1V input of a 100nF/100k HPF results into 500mV output (50% attenuation) at 15.9Hz..

P.S.
No need for phase lead/lag, complex numbers and transfer functions involvement at the moment..

That still includes some maths and science, file mou    Trust me, in theoretical or even basic knowledge, I'm a bit donkey in electronics 

Quote from: Baran Ismen on February 22, 2024, 07:34:16 AM

Quote from: Lino22 on February 22, 2024, 03:26:22 AMCan you please measure the DC on both in and out when the pedal is bypassed? On the outer side of the in/out cap.

You mean the DC voltage or the current?

1- Shall I keep the pedal powered when I do this?
2- Shall I remove the resistors I've soldered?
3- I doubt I can find the input and output caps, I only see 5 or 6 caps that are thru-hole, and most of them are around the power input. There's one at the voltage regulator and there's another one close to an IC whose writings are deleted.

I mean DC voltage.
Yes, keep the pedal powered, leave both in and out jacks in, make sure the circuit is bypassed by the 3PDT.
Yes, please remove the resistor.
Don't look for the caps, measure it on the switch.

Where exactly did you put the resistor?

Quote from: Lino22 on February 22, 2024, 07:54:35 AM

Quote from: Baran Ismen on February 22, 2024, 07:34:16 AM

Quote from: Lino22 on February 22, 2024, 03:26:22 AMCan you please measure the DC on both in and out when the pedal is bypassed? On the outer side of the in/out cap.

You mean the DC voltage or the current?

1- Shall I keep the pedal powered when I do this?
2- Shall I remove the resistors I've soldered?
3- I doubt I can find the input and output caps, I only see 5 or 6 caps that are thru-hole, and most of them are around the power input. There's one at the voltage regulator and there's another one close to an IC whose writings are deleted.

I mean DC voltage.
Yes, keep the pedal powered, leave both in and out jacks in, make sure the circuit is bypassed by the 3PDT.
Yes, please remove the resistor.
Don't look for the caps, measure it on the switch.

Where exactly did you put the resistor?

I soldered them on the switch, you can check it from the video I shared last night. There it is.

https://www.youtube.com/watch?v=pzV6-AYBkaQ&t=1s

Are you sure you know which lug is which on the switch? Can you please rewrite it the way you see it?

Does the LED stay ON in bypass mode..??

Quote from: Lino22 on February 22, 2024, 08:06:08 AMAre you sure you know which lug is which on the switch? Can you please rewrite it the way you see it

I'm not sure, lets go over this picture.



The one in the middle-bottom is ground, that's I know.
On bypass, middle one and ground is shorted,
On activation, middle one and top-middle both have 3.3 Volts.

The ones that I've soldered the resistors (outer ones) are in and out but whether they're PCB or Jack, I'm not sure :/ I mean, I checked them with beep mode of the DMM, but how can I be sure which one is in and which one is out? As they're connected to each other, wouldn't it confuse me?

Quote from: antonis on February 22, 2024, 08:08:09 AMDoes the LED stay ON in bypass mode..??

Normally no, but when I short the middle or upper middle leg with the ground, it does, AND it also lets the circuit run in the background and the abrupt delay on engage disappears. Fixes something, but breaks another lol..

I think its best to make another video tonight

This?





It looks like the pulldowns are correctly placed.

Quote from: Lino22 on February 22, 2024, 08:25:19 AMThis?





It looks like the pulldowns are correctly placed.

Most likely yes. One thing in addition (I suppose) is that probably the input of the circuit is also grounded on bypass. Because when I engage the pedal, effect "starts" from the scratch, and there's around 10-15ms of delay. I mean, I could not find another answer to explain this.

When I short the middle and top-middle legs of the switch, this is no longer a case, no delay, working like a charm, no delay, decay goes on, but LED stays on permanently.

QuoteMost likely yes. One thing in addition (I suppose) is that probably the input of the circuit is also grounded on bypass. Because when I engage the pedal, effect "starts" from the scratch, and there's around 10-15ms of delay. I mean, I could not find another answer to explain this.

I don't think it happens because of the input being grounded. Is it grounded? Can you please measure it?

QuoteWhen I short the middle and top-middle legs of the switch, this is no longer a case, no delay, working like a charm, no delay, decay goes on, but LED stays on permanently.

Sure when you connect them you will connect the LED minus to the ground permanently.

Try to find out what is connected to the top middle lug and the top right one. This is an unusual way of connecting 3PDT, and those two unused lugs may have a reason which is not apparent.

6 Reasons Guitar Pedals Pop When Switched

http://www.muzique.com/lab/pop.htm

Best regards, Jack

There you go, prepared a long video..

https://youtu.be/RiucQMcivC0

Quote from: amz-fx on February 22, 2024, 09:44:03 AM6 Reasons Guitar Pedals Pop When Switched

http://www.muzique.com/lab/pop.htm

Best regards, Jack

I indeed read this several times, mister

Quote from: Baran Ismen on February 22, 2024, 07:40:32 AMThat still includes some maths and science, file mou    Trust me, in theoretical or even basic knowledge, I'm a bit donkey in electronics 

Better now..?? 

Quote from: antonis on February 22, 2024, 03:23:00 PM

Quote from: Baran Ismen on February 22, 2024, 07:40:32 AMThat still includes some maths and science, file mou    Trust me, in theoretical or even basic knowledge, I'm a bit donkey in electronics 

Better now..?? 

In this regard I should use 1K resistor as the output impedance of my guitar seems 1K?   

1k might be the impedance of your guitar pickups (coils) at a particular frequency..
(usually measured with 1Vppk at 1kHz..)
https://www.syscompdesign.com/wp-content/uploads/2018/09/guitar-pickups.pdf

That only stands for Volume pot FCW and considering no loss through Tone pot.. 

Quote from: Baran Ismen on February 23, 2024, 03:39:43 AMIn this regard I should use 1K resistor as the output impedance of my guitar seems 1K?

That's impedance matching, used when you need best power transfer (e.g. between tube output transformer and speakers). In small signal audio we use bridging, for best voltage transfer.

I vote for matching..!!

(bridging requires a lot of maths and science..)