potentiometers and I.C's

I'm about to attempt to make my own effects unit (fancy a challenge plus it's got to be cheaper). I've got a load of circuit diagrams but where the pots are, theres the resistor symbol with 2 lines either side, then an arrow pointing at the middle of the resistor that seems to come from where one of the lines goes to.

I can't tell which of the 3 pins is supposed to be which, and if it's displaying the component from from the top or the base.

Also, why do you put wiring to the base of a pot? I've done it when rewiring Strats but can't figure out why.

On the I.C's front, someone once told me that the legs on an IC aren't actually in numerical order. Is there a standard layout for all IC's, are they different for all of them or is this just a load of rubbish? :confused:

Help would be greatly appreciated :D

PS. I got another floyd rose. Promise I won't file it this time :rolleyes:

The two end terminals - are the ends of the resistor (maximum resistance)

You would normally connect to either, the left terminal and the middle terminal

OR the right terminal and the middle terminal - the middle terminal connection slides along the resistor as you turn the pot - altering the resistance.

Depending on which combination you use - would alter which way you turn the pot to increase or decrease the resistance -IE clockwise or anticlockwise.

The end terminal which is not used is connected to the pot body and earth to stop Hum on guitars

If you have a Semi-circular type transistor - the two pins underneath the flat side - connect to the lines on the diagram which are opposite each other.

If the transistor is rectangular the two pins on the outside( not the middle pin ) are connected to the lines on the diagram which are opposite each other.

Originally Posted by: pogoheadI'm about to attempt to make my own effects unit (fancy a challenge plus it's got to be cheaper). [/QUOTE]
[font=trebuchet ms]Good luck! Hobbyists pay a lot more for components than equipment manufacturers do, simply because individual parts cost more when bought in small quantities. The skills and experience you will acquire is well worth it.[/font]

Originally Posted by: pogoheadI've got a load of circuit diagrams but where the pots are, theres the resistor symbol with 2 lines either side, then an arrow pointing at the middle of the resistor that seems to come from where one of the lines goes to.

I can't tell which of the 3 pins is supposed to be which, and if it's displaying the component from from the top or the base.[/QUOTE]
[font=trebuchet ms]The resistor symbol depicts the fixed resistive element of the pot. The ends of this part are attached to the two outer terminals of the pot. For the most common types, anyway.

The arrow represents the 'wiper' that moves along the surface of the resistive element when the shaft is turned. If the designer was being really nice, the schematic will have CW (ClockWise) or CCW (Counter-ClockWise) written at one end of the symbol to show which end the wiper moves toward in that direction of shaft rotation, as viewed from the knob.[/font]

[QUOTE=pogohead]Also, why do you put wiring to the base of a pot? I've done it when rewiring Strats but can't figure out why.

[font=trebuchet ms]You find a lot of this because most guitar-builders are cheapskate lazy gits that want to save a few pennies, and most guitar buyers don't know any better than to let them get away with it. The back of a pot is used as a quick way to ground part of the circuit. It also causes something called a 'ground loop' because there are several different 'ground' points created with this technique. This makes a guitar noisy because it has no protection against interference from external sources. Go to >GuitarNuts< and check out their wiring diagrams and instructions called "Quieting The Beast".[/font]

[QUOTE=pogohead]On the I.C's front, someone once told me that the legs on an IC aren't actually in numerical order. Is there a standard layout for all IC's, are they different for all of them or is this just a load of rubbish? :confused:

[font=trebuchet ms]There is usually a mark at one end of a DIP (Dual Inline Package) IC. If you look at the IC from above, with the marked end at the top, Pin #1 is at the top left corner, with the rest of the pins numbered from there in counter-clockwise order. For example, a 14-pin DIP with Pin #1 at the top left corner, will have Pin #1 through #7 down the left side, with Pin #8 through #14 up the right side. The data sheet for each IC should show this.

Transistor pin-outs vary a lot, so you will have to be very sure that you get the correct data sheet for the particular transistor you have. Bipolar transistors will have terminals marked E B and C for Emitter, Base, and Collector. FET's (Field Effect Transistors) will have terminals marked D G and S for Drain, Gate, and Source. The body of a transistor is often too small to allow the pins to be marked directly, do there is usually a tab, a mark, or a flat spot used as a way to locate the pins. The data sheet will usually have a picture to show which pin is which.

FET's, and most IC's can be damaged by even a small amount of static electricity, so read up on handling techniques that help to prevent this from happening.[/font]

Thats fantastic mate, thanks a lot :cool:

can i just confirm -

if the diagram is as so (please excuse the crudness, the dots are just so the script would allow for empty spaces) -

. !
. !
. 1 -
. ! !
. ! !<-! 2
. ! ! !
. 3 - !
. ! !
. ! !
.----------------

and we can assume its clockwise as its a control on a pedal, and the pins are 1 - 3, l-r viewing from the knob end (excuse the term) then would i connect 2 and 3 together (as the arrow comes back from where point 3 goes) and connect 1 and 3 to the other components in the diagram?

Also, i'm guessing from the thing about pot bases being used for a loop, they're not actually connected to the resistive part of the pot and are only a metal shell. Is this right?

Greatly appreciate the help :)

(edit. I cant seem to get the spaces to work in the little drawing so please use your imagination :( )

[font=trebuchet ms]I don't get much from your 'drawing', but from your description I think you've got it pegged. As the pot is adjusted towards the end that is tied to the wiper, the resistance between pins 1 and 3 will reach the maximum value. As the pot is adjusted the other way, the connection between pins 2 and 3 will progressively short out more and more of the internal resistor, until the wiper reaches the pin 1 end of its travel. At that point the resistance between pin 1 and 3 will measure as zero.

The backshell of a pot is not connected to the internal resistor, but it often is connected to the threaded ferrule that mounts the pot to the guitar. Which means that the circuit 'sees' ground via any shielding applied to the control cavity, as well as through the connections on the backs of the pots. These multiple grounding paths form loops which act as antennas for any stray radio signals in the room, including noise from computers, TV sets, and appliances.[/font]

If you send me your email address I'll send a picture over that I've done on paint, just so you see what I mean, though surely if pins 2 and 3 are connected, wont they just short out?

Originally Posted by: pogohead... surely if pins 2 and 3 are connected, wont they just short out?

[font=trebuchet ms]Let's keep this out in the open where any who wants to, can chime in. If the file size is not too big, you can attach it to a post. See "Manage Attachments" in the "Additional Options" menu below the "Reply to Thread" window.

Anyway, Pin 3 is one end of the resistor, right?
And Pin 2 is the wiper?
And they're connected together?
Which will short out only the portion of the pot between the wiper position and the Pin 3 end of the resistor, leaving the rest of the resistor in the circuit.

So, moving the wiper by turning the shaft will change the resistance to any value from zero (wiper at Pin 1 end of the resistor, shorting Pin 1 to Pin 3) to the full value of the pot (wiper at the Pin 3 end of the resistor, not shorting any of the resistor), yes?[/font]

ok, can't attach so I found this link:

pot pic

imagine that the line with the arrow (the wiper?) then comes vertically down and connects with the horizontal line. This would mean the wiper and that pin connect right?

This whole thing is so confusing. I'm REALLY looking forward to making this unit :rolleyes:

[font=trebuchet ms]Yes.

And in keeping with our discussion so far, the top would be Pin 1, the arrow would be Pin 2, and the bottom would be Pin 3.

With the wiper terminal Pin 2 connected to Pin 3, an Ohm-meter connected to Pins 1 and 3 will only 'see' the portion of the resistor bewteen Pin 1 and the wiper. The rest of the resistor is shorted out by the connection between Pin 2 and Pin 3. This makes the resistance adjustable. An adjustable resistor is sometimes called a RHEOSTAT.

The word 'pot' is short for POTENTIOMETER. Voltage is sometimes called Potential. Picture Pin 2 connected to one terminal of a Voltmeter, with the other meter terminal connected to Pin 3. If a voltage is applied between Pin 1 and Pin 3, you will be able to adjust the voltage measured by the meter by moving the wiper. This is the way a Volume pot works.[/font]

Excellent. Now then, whats a variable resistor?












Kidding mate ;)

Just to add onto my problems, the circuit diagram is split into two due to space constraints. There are arrows on both labelled A and B to show where the diagrams connect to each other, but on the second diagram theres 2 more arrows at the top and bottom marked 9V and 0V (voltage) but no arrows on the first diagram for them to join with. Should they just be connected to the power supply?

Also, can you fix crackly pots without replacing them?

Thank you :)

Ok, turns out you can fix pots with a knife and some white spirit.

Still stuck on the diagram problem though :(

[font=trebuchet ms]Toss the pots. Once they get crackly, they aren't worth the powder to blow them to hell.

The diagram? BTSOOM![/font] :confused: