Thanks for explaining that Fletcher.

Well here's an update to what I've been doing. I've taken a picture of the PCB and labelled the components with spoken of mostly. I have checked they all correspond to the correct paths on the schematic.
I will label every part and continue testing some more afterwards.

I think Folderol's point on the resistors is interesting and I feel they are used to correspond to signal voltage, thus the other 3 light up at a certain voltage level.
Also as Folderol said, that I may not be IC 4 completely because one LED should light when the pedal is switched on and the others when there is a sufficient level from the incoming signal:

PCB Labelled

Yes but it is still controlled by the IC. You say you should have a high on pin 7, you don't. It's either not getting there or it's being shorted. There is nothing except the IC itself which could short it (excluding blobs of solder etc). You also had 0v on pin 6 which is wrong (Folderol pointed that out). I suspect that part of the IC op amp has gone short to deck. With the unit off measure the resistance between pin 7 and pin 11, if it's short the IC needs to come out. This wont be that hard to do

Quote fletcher:

---

Yes but it is still controlled by the IC. You say you should have a high on pin 7, you don't. It's either not getting there or it's being shorted. There is nothing except the IC itself which could short it (excluding blobs of solder etc). You also had 0v on pin 6 which is wrong (Folderol pointed that out). I suspect that part of the IC op amp has gone short to deck. With the unit off measure the resistance between pin 7 and pin 11, if it's short the IC needs to come out. This wont be that hard to do

---

Yes you're right it's highly likely it's the IC. There is no resistance between 7 and 11 but are all the non-negative pins grounded internally through pin 11?

There's only resistance between terminal 5 and 6 and by the picture below 2+3, 9+10 12+13 have no resistance which means they're shorting also right?



Looks like the last owner gave it real good bashing.

Thanks

Edited by Music Manic (24/08/12 03:36 AM)

Maybe. As Folderol said it is quite rare. Usually they fail if they get a "bang" from a high voltage part of a circuit failing. This unit has no high power stages so you would think it would last. Maybe it stopped working very early in the units life, and because the device still works without it the previous owner didn't care. This is more common, if an IC is faulty it is going to fail as soon as the unit is used. That's why electronic devices that have a problem usually fail within their warrenty.

Don't forget to fit the IC holder.

Edited by fletcher (24/08/12 10:05 AM)

'smee again

That's a beautifully clear pic of the PCB, and I'd say the soldering looks pretty clean. It would be interesting if you could get an equally clear shot of the component side.

All those pins of the IC apparently shorting is extremely rare and would certainly account for all the screwy readings you were getting. I'm inclined to agree with Fletcher that either it failed very early due to a manufacturing defect, or possibly never worked correctly at all.

My method of removal is to use a pair of cutters like these to cut the legs tight against the body of the chip, I then (again like Fletcher) heat the legs individually at the solder joint, but use a pair of fine medical tweezers to pull the leg out.

It's then just a matter of careful use of a solder sucker and/or wick to clean up before putting a chip holder in.

--------------------
It wasn't me!
(Well, actually, it probably was)

Quote Folderol:

---

'smee again

That's a beautifully clear pic of the PCB, and I'd say the soldering looks pretty clean. It would be interesting if you could get an equally clear shot of the component side.

All those pins of the IC apparently shorting is extremely rare and would certainly account for all the screwy readings you were getting. I'm inclined to agree with Fletcher that either it failed very early due to a manufacturing defect, or possibly never worked correctly at all.

My method of removal is to use a pair of cutters like these to cut the legs tight against the body of the chip, I then (again like Fletcher) heat the legs individually at the solder joint, but use a pair of fine medical tweezers to pull the leg out.

It's then just a matter of careful use of a solder sucker and/or wick to clean up before putting a chip holder in.

---

Back again too!

While waiting for my parts my new digimeter arrived.
Anyway I have some news for you, after a good slog today with the free time I had.There are some funny reading with regards to the amp pins 5,6 and 7:

Result of Pins of IC 4:
2 to 3 = 123k
5 to 6 = 50k
9 to 10 = 119k
12 to 13 = 120.5k
5-6 very low impedance, which means a higher voltage output right?


Ok the impedance results I got from Pin 11 to:

3 = 50k
5 = 50k
10 = 50k
12 = 50k
seems ok!

Pin 11 to:

2 = 73k
6 = 47 ohms
9 = 69k
13 = 70.5k
PIN 6 rather low isn't it?

to Pins:

1 = 906
7 = 0
8 = 906
14 = 906
I had to put the red test lead to PIN 11 to get a reading. Does that mean the output of the amps is a negative charge? But either way 7 IS dead


Ok I have checked all the resistors and the schematic is correct regarding their values but my tests came up with some odd results:

R32 (specified at 430k) gave a result of only 83.4k
R35 (specified at 82k) gave a result of only 69k. Is that ok? Will it tolerate?
R36 (specified at 47 ohms) gave a result of only 66k. Again, is that ok?
R16 (specified at 11k) gave a result of only 9k

R32 is the one I'm worried about. If the resistance is now that low does it mean it's been over powered and zapped thus leaving a too large a voltage/current to enter that part around those resistors where the LEDs are?

I couldn't check the 2.2M resistors because I have a 2000k limit.

Thanks

Quote Folderol:

---

'smee again

That's a beautifully clear pic of the PCB, and I'd say the soldering looks pretty clean. It would be interesting if you could get an equally clear shot of the component side.

---

Yes that's macro mode sussed. Sunlight seems to work best. Here you go Misters:


Components side picture 1


Components side Picture 2

Quote Folderol:

---

'smee again




My method of removal is to use a pair of cutters like these to cut the legs tight against the body of the chip, I then (again like Fletcher) heat the legs individually at the solder joint, but use a pair of fine medical tweezers to pull the leg out.

It's then just a matter of careful use of a solder sucker and/or wick to clean up before putting a chip holder in.

---

Yes I have those cutters already Folderol. Will the pedal still function without this IC?
If you get a chance could you explain the signal flow from the guitar signal and how it relates to the circuit. I know that the schematics current is based on the opposite of true electron flow.


I think I'll try to make a clone of this baby, now that I know it so well, but I'd leave out the fancy LEDs this time.

Thanks to you and Fletcher for your perseverance. You don't know how much I've learnt from this one thread.

Another question

V1 AND V3 are labelled 1MB pots. Does that mean they are 1 MegaOhm?
V2 is 100k
V4 is 2k right?

Also the switch is a 6 pin type. Does the grounding track change when it is on and off?

Let me know if I'm going on, but this stuff is interesting.

Thanks

Hi again,

Would it be possible to work out, by the schematic, what type of transistors Q1 and Q2 are?

I've looked for them everywhere but I can't seem to find them. Here's the label code (I've posted in another topic):

1 - Code 2206 9J
2 - C245B OBA

Thanks

Well now, those latest pictures reveal a very interesting story. The chip that does the 'expanding' is IC5, and originally I expected it to be a voltage controlled amplifier. It isn't. I recognised the part number instantly and it is actually a dual opto-coupler! These devices are most used in logic switching isolation circuits, and this is a very novel application

Getting down to basics.

Reading resistors in circuit will give very unreliable results (unless you have special kit) as everything around them will affect the results. In the case of the 430k resistor, you are probably reading the other resistors in the chain, along with the power supply resistance!

You've actually read R36 wrongly. It goes between Pin 6 of IC4 and ground, and when you were on the IC itself you read 47 ohms

Q1 can be just about any small signal PNP transistor e.g. BC327. It is wired as an emitter follower (the emitter follows the base) and drives the internal LEDs in IC5, as this is a bit much for the lower half of IC3 to manage.

Similarly Q2 can be almost any small signal NPN transistor e.g. BC337 (if you're not going to bother with IC4 then it's not needed).

In your unit both of these are working fine.

Leaving IC4 out won't affect the rest of the circuit - well, it'll draw slightly less current.

You're quite correct about the pots. Dunno why they stuck a B there.

The switch is two electrically isolated switches that just happen to be in the same housing, operated by the same lever.

--------------------
It wasn't me!
(Well, actually, it probably was)

Quote Folderol:

---

Well now, those latest pictures reveal a very interesting story. The chip that does the 'expanding' is IC5, and originally I expected it to be a voltage controlled amplifier. It isn't. I recognised the part number instantly and it is actually a dual opto-coupler! These devices are most used in logic switching isolation circuits, and this is a very novel application

---

IC 5 deals with the gain. It looks like ICs 1 and 2 are feeding it also. It give a typical Billy Gibbons style distortion which is very different from any guitar amps gain.
I'm not quite sure what you mean with what it does. I'm trying to work out the signal flow and it looks like things are going in and out everywhere. Are all the capacitors resistors etc used just to create the right voltage for the pots and amps?

Quote Folderol:

---

Getting down to basics.

Reading resistors in circuit will give very unreliable results (unless you have special kit) as everything around them will affect the results. In the case of the 430k resistor, you are probably reading the other resistors in the chain, along with the power supply resistance!

You've actually read R36 wrongly. It goes between Pin 6 of IC4 and ground, and when you were on the IC itself you read 47 ohms

---

Wow! that's news to me. I thought if I placed the probes on the wire just at the head of the resistor nothing would interfere with the reading.
Yes I meant 47 ohms

Quote Folderol:

---

Q1 can be just about any small signal PNP transistor e.g. BC327. It is wired as an emitter follower (the emitter follows the base) and drives the internal LEDs in IC5, as this is a bit much for the lower half of IC3 to manage.

Similarly Q2 can be almost any small signal NPN transistor e.g. BC337 (if you're not going to bother with IC4 then it's not needed).

In your unit both of these are working fine.

Leaving IC4 out won't affect the rest of the circuit - well, it'll draw slightly less current.

You're quite correct about the pots. Dunno why they stuck a B there.

The switch is two electrically isolated switches that just happen to be in the same housing, operated by the same lever.

---

I've been searching all day to find out the transistors and you can answer all that just by looking at a picture. You're a genius.

If I reconstruct the circuit without the LED stuff how would I go about worrying what I can leave out? Would I just link output 1 of IC 1 to the switch?

Edited schematic

Ok I've rubbed out the LED part from the schematic and I'm trying to work out the direction of the signal flow and the current. After readinf your explanation in the other topic about how a guitar pickup sends its voltage to a circuit, the capacitor C1 is where the guitar signal starts right? But what is the track coming in from the left of the page in the resistor R18 just above the zener diode? Is that the positive terminal of the power supply/battery?
Do we then have negative terminal of the power supply/battery attached to the grounding pin of the jack?

Thanks

Edited by Music Manic (26/08/12 01:33 AM)

Quote Folderol:

---

Q1 can be just about any small signal PNP transistor e.g. BC327. It is wired as an emitter follower (the emitter follows the base) and drives the internal LEDs in IC5, as this is a bit much for the lower half of IC3 to manage.

---

Does the transistor affect the sound of the signal? If so are there alternate types, such as germanium ones, I can add to change the quality?

Thanks

MM I think you are trying to run before you can walk. If you want help to fix the pedal that's cool, just replace the IC and done. If you want to understand how the pedal works, well that's a huge ask.

Electronics is a huge subject, even just the basics. There is no simple answer to anything you want to know without first having a lot of background knowledge. It's like trying to understand tritone substitution before you can play a chord. It's not that myself or someone else is unwilling to help you, it's just not possible without writing a book. If I was you I would get some electronics books and read up first, asking questions here if you need something explained. Of course if you need tips fixing something that's different, it is possible to repair something without understanding it. I am getting the impression here though that your on a mission to learn this stuff properly (commendable) and need the big picture. However it is a lot to learn and takes time if you want deeper understanding. It's quite easy to build things from kits, repair the odd dry joint etc. Quite hard (without the background knowledge) to understand the reasons behind the design etc. If you don't do some studying every question you ask will have an answer that just poses you more questions, and it will confuse you.

Folderol!

After reading you post on the guitar signal flow I can, now clearly see the path of the audio signal and how it gets processed.

C1 - capacitor - receives the positive (tip of guitar lead) voltage and the negative is grounded through a line of components.

C12 - capacitor - has its negative terminal attached to the ground of the output jack and its positive to the output of the level pot.


Brilliant

Thanks

P.S. Why does the capacitor C23 link to C1? I see its positive terminal is linked to the tip of the lead.

Edited by Music Manic (27/08/12 02:40 PM)

....So while C1 charges C23 discharges right?

Quote Music Manic:

---

....So while C1 charges C23 discharges right?

---

Umm. No.
C1 feeds the 'ordinary' signal path.
C23 feeds the buffer amplifer for the rectifier, and eventually, via IC4 the LEDs
At normal audio frequencies neither of these will charge or discharge significantly. They charge once at switch-on to the required DC voltage and stay there.

Without going too much into detail, there are other low value caps in that circuit that behave differently, and if fact do some frequency shaping and stabilisation, but it's best to ignore them for now.

You really need to read up on operation amplifiers (OP-AMPs) - just the basics so you know what to expect and will have a much better understanding of how they are used.

To avoid confusion, as most of the chips are pairs of op amps I'm calling the upper row 'a' and the lower row 'b'

Briefly, the signal line is:
IC1a inverting amplifier with variable gain - that's what the pot above it does.
IC2a non-inverting amplifier with variable gain (controlled by IC5 along with the pot)
IC5 behaves like a voltage controlled resistor
IC3a inverting amplifier with varibable gain - yet another pot!

IC2a (along with IC5) is the expander proper, and will also produce a small amount of distortion, but the main distortion element is the two back-to-back diodes D1,D2 fed by R8,C8. This is a classic design.

Control line:
IC2b unity gain non-inverting buffer
IC3b fixed gain non-inverting amplifier-driver
IC1b unity gain non-inverting buffer
IC4 4x comparator

IC3b works with Q1 to drive IC5s LEDs. It works backwards, in that the the more signal there is, the less drive there is to the LEDs (notice there is a rectifier D3). This makes IC5s apparent resistance higher, and that in turn gives less negative feedback, and finally more gain in IC2a.

Finally, in your rub-out, you don't need to do anything to the 'spare' line from the switch, just rub that out too. Also, IC1b won't be doing anything, so you don't need any of the components feeding it.

P.S.
I aint no genius, but have been working with electronics (literally man & boy) since the late 1950s. In that time I've cooked a few valves, burned out a number of transformers and popped countless transistors and chips. I am fortunate to have a very good memory, so rarely make the same mistake twice There are component configurations that you come across so often that you could eventually draw them blindfolded.

--------------------
It wasn't me!
(Well, actually, it probably was)

Quote Folderol:

---

Umm. No.
C1 feeds the 'ordinary' signal path.
C23 feeds the buffer amplifer for the rectifier, and eventually, via IC4 the LEDs
At normal audio frequencies neither of these will charge or discharge significantly. They charge once at switch-on to the required DC voltage and stay there.

---

Ah ok. As fletcher said I'm running before I can walk. I assumed that because C1 has its negative terminal on the guitar tip's rail where C23 has its positive there.

Quote Folderol:

---

Without going too much into detail, there are other low value caps in that circuit that behave differently, and if fact do some frequency shaping and stabilisation, but it's best to ignore them for now.

You really need to read up on operation amplifiers (OP-AMPs) - just the basics so you know what to expect and will have a much better understanding of how they are used.

---

I have a book or so on that so I'll read up and give you a deserved break.

Quote Folderol:

---

To avoid confusion, as most of the chips are pairs of op amps I'm calling the upper row 'a' and the lower row 'b'

Briefly, the signal line is:
IC1a inverting amplifier with variable gain - that's what the pot above it does.
IC2a non-inverting amplifier with variable gain (controlled by IC5 along with the pot)
IC5 behaves like a voltage controlled resistor
IC3a inverting amplifier with varibable gain - yet another pot!

IC2a (along with IC5) is the expander proper, and will also produce a small amount of distortion, but the main distortion element is the two back-to-back diodes D1,D2 fed by R8,C8. This is a classic design.

Control line:
IC2b unity gain non-inverting buffer
IC3b fixed gain non-inverting amplifier-driver
IC1b unity gain non-inverting buffer
IC4 4x comparator

IC3b works with Q1 to drive IC5s LEDs. It works backwards, in that the the more signal there is, the less drive there is to the LEDs (notice there is a rectifier D3). This makes IC5s apparent resistance higher, and that in turn gives less negative feedback, and finally more gain in IC2a.

Finally, in your rub-out, you don't need to do anything to the 'spare' line from the switch, just rub that out too. Also, IC1b won't be doing anything, so you don't need any of the components feeding it.

---

This is getting clearer by the day, but you're right, I need to know more about amps and how they work.

Quote Folderol:

---

P.S.
I aint no genius, but have been working with electronics (literally man & boy) since the late 1950s. In that time I've cooked a few valves, burned out a number of transformers and popped countless transistors and chips. I am fortunate to have a very good memory, so rarely make the same mistake twice There are component configurations that you come across so often that you could eventually draw them blindfolded.

---

Well, you're making a walk in the Labyrinth seem like a stroll in the park. People that can teach like that are one in a million, believe me.

Thanks again and I'll be back after I've sussed op-amps and fixed this pedal (I'll upload pictures when done).

Just received parts and all WORKING now:



Chip Removed:


Solder job - I know bad but was my first go:


New Chip in IC holder:


Operational LED:


All LEDs working with guitar signal:



Weeeeee!

Guys thanks you so much, it was an enjoyable experience in finding the fault and finding out what's going on as well at the same time.
Folderol I am going to read up on op amps as well now and I will go through your last post so that I now how this circuit works in its entirety.

Fletcher thanks for your input and the IC tip was really helpful. If the chip went wrong then it would be easy to replace.

Time to read up some more so I don't become a PITA.

Thanks

nice one!

IC holders are good practice, in the future you may find something is causing them to blow and you miss it, at least you only have to plug another new one in (you probably would only see this in circuits of higher power). More useful in fault finding, you can remove the IC and get useful voltage readings which can help you, as the IC would alter those voltages as it works.

good luck with your mission to learn

Very pleased to see this working.

I must warn you however that this sort of thing can become very addictive. Once you've sorted out one bit of kit you want another, and another...

Your soldering is not at all bad for a first attempt (I've seen a lot worse form people calling themselves experts). It looks as if you applied the solder to the iron, and the iron to the joint. You should actually apply both the iron and the solder to the join at the same time so your iron and solder form a 'V' with the point of the V at the joint. It just takes practice. Oh, and do make sure your iron is hot enough. The solder should flow smoothly and quickly, but without the flux burning off.

--------------------
It wasn't me!
(Well, actually, it probably was)

Quote Folderol:

---

Very pleased to see this working.

I must warn you however that this sort of thing can become very addictive. Once you've sorted out one bit of kit you want another, and another...

---

It's already happening.

Quote Folderol:

---

Your soldering is not at all bad for a first attempt (I've seen a lot worse form people calling themselves experts). It looks as if you applied the solder to the iron, and the iron to the joint. You should actually apply both the iron and the solder to the join at the same time so your iron and solder form a 'V' with the point of the V at the joint. It just takes practice. Oh, and do make sure your iron is hot enough. The solder should flow smoothly and quickly, but without the flux burning off.

---

Yep! right again. I did the V on one of them and it was instant.

Time to search for some broken pedal....


Thanks again, I owe you a beer (and a half for Fletcher,haha!). You're a gent!

Hi again,

I've got quite a few of the parts for this board and am going to try and re-create parts of it
soon.
Could you tell me what type of diodes D1,D2,D3,D5, and D6 are. Do some of them act as rectifiers?

Thanks

Quote Music Manic:

---

Could you tell me what type of diodes D1,D2,D3,D5, and D6 are. Do some of them act as rectifiers?

---

Hi MM!

The most likely is 1N4148 or equivalent silicon diodes, which you can buy for a few pence. Diodes always act as rectifiers (that's their job )

D1/D2 are clipping the incoming waveform.
D3 seems to be a half-wave rectifier that in conjunction with C13 turns the incoming amplified AC voltage into a DC envelope.

D6 removes the negative part of the waveform so only the positive half gets measured by the four-way comparator for the LED level readout.

Hope this helps!


Martin

--------------------
YewTreeMagic

Quote Martin Walker:

---

Hi MM!

The most likely is 1N4148 or equivalent silicon diodes, which you can buy for a few pence. Diodes always act as rectifiers (that's their job )

D1/D2 are acting as clippers.
D3 seems to be a half-wave rectifier that in conjunction with C13 turns the incoming amplified AC voltage into a DC envelope.

Hope this helps!


Martin

---

Hi Martin

So specific resistance doesn't matter?

D1 and D2 are stated as 1588. Is that a particular model?

P.S. The Zener Diode - D4 regulates the voltage doesn't it?

Thanks.

Ah, the 1588 refers to the venerable 1S1588 diode, but as I suspected just replacing them with a 1N4148 will probably make little difference.
Specific resistance isn't an issue – the forward voltage is what counts here, and that's likely to be around 0.6 volts for both. If it IS any different then the tone of the clipping will change a bit that's all.

Yes, the zener is creating a stable 9.1v reference voltage from your battery supply, and then dividing that by two (using R16 and R17) to produce a reference for the other opamps, to place their non-inverting inputs half way between the power supply and ground.


Martin

--------------------
YewTreeMagic

Quote Martin Walker:

---

Ah, the 1588 refers to the venerable 1S1588 diode, but as I suspected just replacing them with a 1N4148 will probably make little difference.
Specific resistance isn't an issue – the forward voltage is what counts here, and that's likely to be around 0.6 volts for both. If it IS any different then the tone of the clipping will change a bit that's all.

---

That's interesting to know. It's all about knowing how it works so I can experiment.

Quote Martin Walker:

---

Yes, the zener is creating a stable 9.1v reference voltage from your battery supply, and then dividing that by two (using R16 and R17) to produce a reference for the other opamps, to place their non-inverting inputs half way between the power supply and ground.

Martin

---

Great info. So simple when someone shows you.

Thanks

Glad that helped


Martin

--------------------
YewTreeMagic

the zener works only with the power supply, regulating it. As the battery is 9v, well it wont do much as it's a 9.2v zener, except maybe shorten the life of a slightly lively new battery. If you were only going to use batteries you might be better off without it. Of course it would then be safer to remove the power supply socket. There again I suspect the supplied supply was unregulated, if it was replaced with a regulated one the zener would be redundant.

Quote fletcher:

---

the zener works only with the power supply, regulating it. As the battery is 9v, well it wont do much as it's a 9.2v zener, except maybe shorten the life of a slightly lively new battery. If you were only going to use batteries you might be better off without it. Of course it would then be safer to remove the power supply socket. There again I suspect the supplied supply was unregulated, if it was replaced with a regulated one the zener would be redundant.

---

Yes I'm beginning to break down parts of the system and will re-create the schematic for you to look at. I want to singularise each stage and make a circuit out of what will work as a minimum, then add and subtract and change. I am working out the capacitors. I think one in the schematic is labelled wrongly, so I'll have to go through all of them.

Thanks

Ok I've just got the 1N4148 diodes. I'm not sure which D3,D5 and D6 are. Are they the same?

The small dark blue capacitors are Tantalum and the light blue are Ceramic. I have the correct value types, but would it matter if they have different voltage ratings if I can't find the same for all of them?

After that I can get the breadboard out and start playing......

Thanks

Yes, the same diodes should be fine. However I'm a bit puzzled about D5. It is inline with the supply to IC3, which would drop the voltage by about 0.6V. Maybe I'm getting old or sommat, but I can't see any benefit from that in this circuit

All the caps can be rated at anything over 9V (10V and 16V are common figures). Some could probably be rated at 6V, but that might actually be more expensive, and why take any risk?

--------------------
It wasn't me!
(Well, actually, it probably was)

Quote Folderol:

---

Yes, the same diodes should be fine. However I'm a bit puzzled about D5. It is inline with the supply to IC3, which would drop the voltage by about 0.6V. Maybe I'm getting old or sommat, but I can't see any benefit from that in this circuit

All the caps can be rated at anything over 9V (10V and 16V are common figures). Some could probably be rated at 6V, but that might actually be more expensive, and why take any risk?

---

Thanks Folderol.

V3 is the volume pot at the end of the chain and that is working off IC3. It has a 1M ohm resistor going into IC3 too. Is it anything to do with the output current?

Quote Music Manic:

---

V3 is the volume pot at the end of the chain and that is working off IC3. It has a 1M ohm resistor going into IC3 too. Is it anything to do with the output current?

---

In a word, No!
In more words
This is an inverting amplifier which tends to act as a voltage see-saw, so as the input goes up the output goes down and vice-versa. Without going too deep into OpAmp theory, the ratio of the input and output voltages is the same as the ratio of the input and feedback resistances.

When the pot (the feedback resistor) is at maximum the resistor ratio is 1:1 so if there is 100mV in there will be 100mV out, but going in the opposite direction.

If the pot was at half resistance, then the ratio would be 1:0.5 so you'd only get 50mV out, and if the pot was at zero...

--------------------
It wasn't me!
(Well, actually, it probably was)

Edited by Folderol (07/09/12 09:22 PM)

I've just bought the small dark blue capacitors, which are the tantalum ones, which atre polarised. The light blues are causing a bit of a problem. Would any normal type with the same rating do? These are the non-polarised ones on the schematic and you can see them clearly in the .gif:

Bixonic

There is a 20pF capacitor across V4 - the volume knob.

The capacitors I'm not sure about are:

1 - 223K(0.022uf)
2 - 102K(0.001uf)
3 - 101J(100pF)
4 - 104k(0.1uf)

These are the closest looking ones I've found:
Capacitors - Non polarised

As long as they have those figures stamped on does it matter. Some have some massive voltage ratings.

I may be going on but could you tell me how the capacitors operate? Are some there just for the circuit's operation and do the polarised ones get "triggered" by the incoming audio signal.

Thanks again for your invaluable help.

Edited by Music Manic (08/09/12 03:47 AM)

Hi again,

Looking at the schematic, even when the battery source or mains is connected the circuit will still be open until a lead is inserted into the input right?
Is this how pedals protect the battery from draining? If I leave an input lead inserted though will this drain the battery quicker?

Quote Music Manic:

---

Hi again,

Looking at the schematic, even when the battery source or mains is connected the circuit will still be open until a lead is inserted into the input right?
Is this how pedals protect the battery from draining? If I leave an input lead inserted though will this drain the battery quicker?

---

Correct on all counts

Incidentally, saying a capacitor is polarised simply means it has to be connected the right way round. The - marking must be negative.

Don't worry about the voltage rating on the other caps - it is actually hard to get less than a 50V rating on them!

Most of the caps are either for coupling (as we discussed before) or for de-coupling (C14? & C15). That is, they effectively short AC signals to ground to stop them interfering where they are not wanted.

Some of the caps around IC2a are for frequency shaping, and the one across V3 is there to keep the amp stable if the pot is set for high gain.

--------------------
It wasn't me!
(Well, actually, it probably was)

Quote Folderol:

---

Quote Music Manic:

---

Hi again,

Looking at the schematic, even when the battery source or mains is connected the circuit will still be open until a lead is inserted into the input right?
Is this how pedals protect the battery from draining? If I leave an input lead inserted though will this drain the battery quicker?

---

Correct on all counts

---

I'm stunned, I've actually got something right!

Quote Folderol:

---

Incidentally, saying a capacitor is polarised simply means it has to be connected the right way round. The - marking must be negative.

Don't worry about the voltage rating on the other caps - it is actually hard to get less than a 50V rating on them!

---

Ok! there are so many different types of materials. Would any particular type affect the sound or cause noise like resistors? They obviously have their different benefits.

Quote Folderol:

---

Most of the caps are either for coupling (as we discussed before) or for de-coupling (C14? & C15). That is, they effectively short AC signals to ground to stop them interfering where they are not wanted.

Some of the caps around IC2a are for frequency shaping, and the one across V3 is there to keep the amp stable if the pot is set for high gain.

---

Yes I'm getting to terms with their behaviour within different contexts.

I'm catching up on Op-amps and know there is a simplicity to their function of acting like a valve. Setting them up is a different matter where it comes to biasing etc.

Any pointers welcome.

P.S. I'm really surprised you haven't told me where to go yet!

Thanks

to help give Folderol a break, you asked about how capacitors work.

If you brought two pieces of wire with a p.d. across them close together the electrons on the more negative side can "feel" the pull of the more positive side. They can't get across the gap and sort of pile up. How many electrons can "pile up" is very small as it's related to the cross-sectional area of the wire, which being a wire is not much:) However if the two wires were connected to some big metal plates the area would be much bigger and the "capacitance" for electrons to pile up would also be much bigger. This is in fact what a capacitor is. Two big metal sheets would be a bit impractical so they are wrapped up into a cyclinder and sealed up. The capacitance is measured in Farads and is related to the surface area (as we already said), bigger area = bigger capacitance. It is also related to the space between the sheets, further apart = less capacitance. Finally it also is dependant on the dilectric which is what's between the plates. On large electolytic caps the dilectric is maintained by the voltage across the plates. When not used for awhile it can break down and cause the plates to short. Although they do self-repair to an extent once a voltage is restored, older caps might dry out and lose this ability.

So now you can see why DC is blocked - it can't cross the gap. AC can though as the changing voltage causes the electrons to run in and out of the plates, thus passing on the signal. The other factor to understand is it takes time for the capacitor to "fill up" or "empty". This is why a capacitive circuit has properties which are frequency dependant.