I'm confused on which way to wire a potentiometer. I understand the wiper arrow, but there is another arrow (left-right) that I think shows the direction or zero or something. Below is a gain pot, linear taper, for a guitar signal. Is the wiper going the right direction? At max gain it should be a gain of 50/4.7. But if I wire the wiper backwards, the pot will probably work backwards.

 

When pins 1 & 2 are shorted together the impedance of the pot is 50KΩ. When pins 2 & 3 are shorted together it is 0KΩ

 

I'm confused on which way to wire a potentiometer. I understand the wiper arrow, but there is another arrow (left-right) that I think shows the direction or zero or something. Below is a gain pot, linear taper, for a guitar signal. Is the wiper going the right direction? At max gain it should be a gain of 50/4.7. But if I wire the wiper backwards, the pot will probably work backwards.
View attachment 306348

You'll need to take a look at the physical orientation of the pins, relative to the pot rotation, to mounting it correctly.
Otherwise, it may operate the the reverse fashion than you expect.

 

Yeah. and to make matters more confusing, many guitar pedals have the pots mounted on the bottom of the PCB, compared to all the other components. So that reverses it too. I'll definitely have to think it through.

 

What is the DWG program you used to draw it up?

 

What is the DWG program you used to draw it up?

Looks like Eagle.

 

if I wire the wiper backwards, the pot will probably work backwards.

It seems backwards to me. Clockwise rotation will decrease gain instead of increasing it.

At max gain it should be a gain of 50/4.7.

This is also backwards. You have an inverting amplifier.

 

Here is that portion of the circuit in spice. As R increases, the output of the circuit goes up, from 0-26 dB.

 

The arrow on the pot icon usually shows the direction the wiper moves for clock-wise rotation of the pot shaft.

 

As R increases, the output of the circuit goes up, from 0-26 dB.

Okay. Don't particularly like being treated like a mushroom.

 

Okay. Don't particularly like being treated like a mushroom.

No idea what this means.

 

So I think dl324 is right, I had it wired backwards. With pot at 0 (ccw) terminal 2 is shorted to 3 and provides 0 resistance. Gain = 0/4.7. Should be like this:



But since the pot is mounted on the bottom of the board, I need to mirror the pot in the PCB layout like this (writing is backwards, terminals flipped).



Am I thinking about it right?

 

A volume control as an input voltage divider is used frequently. A gain control is used one time or only a few times.

 

So I think dl324 is right, I had it wired backwards. With pot at 0 (ccw) terminal 2 is shorted to 3 and provides 0 resistance. Gain = 0/4.7. Should be like this:

View attachment 306429

But since the pot is mounted on the bottom of the board, I need to mirror the pot in the PCB layout like this (writing is backwards, terminals flipped).

View attachment 306430

Am I thinking about it right?

Yes. But I wouldn't rely on the schematic symbol to determine correct physical orientation.
Confirm the orientation with the actual physical part.
In addition, be aware that when a part is "mirrored", most PCB programs move the part's silkscreen to the bottom of the PCB board. You may or may not want that.

 

A volume control as an input voltage divider is used frequently. A gain control is used one time or only a few times.

That's not necessarily true with a guitar effect pedal.

 

You unscrew the cover to adjust the gain trimpot with the screwdriver every 10 years when you replace the guitar.

 

You unscrew the cover to adjust the gain trimpot with the screwdriver every 10 years when you replace the guitar.

no not quite. All outboard guitar preamps have a gain pot that you adjust for different guitars, etc.

 

Similar situation, notch filter. When the pot is at 100k, frequency = 55 Hz. But when the pot is a 0, frequency = 780 Hz. So again, I think this is wired backwards, right? I want to mirror it in Eagle so pin 3 is on the left and 1 is on the right?!?

As it is shown, the knob would rotate backwards, 55 Hz would be fully CW and 780 would be fully CCW. ?

 

no not quite. All outboard guitar preamps have a gain pot that you adjust for different guitars, etc.

Forgive AG he lives in Canada, everything is better there.

 

With pot at 0 (ccw) terminal 2 is shorted to 3 and provides 0 resistance.

No.

With the pot at 0 (ccw), terminal 2 is shorted to 1.

A schematic symbol doesn't know anything about physical orientation, top side / bottom side, etc. It describes the function of the component. The standard for trimpots is that the wiper (pin 2) moves across the resistance element in a clockwise direction from pin 1 to pin 3. Thus, at full CCW pin 2 is shorted to pin 1 inside the part, regardless of external connections, and the resistance from the wiper to pin 3 is the maximum value of the resistance element. The small arrow on the decal is pointing from pin 1 to pin 3 because, by convention, that is the direction of clockwise rotation.

How you connect the pins determines if clockwise rotation increases or decreases the resistances seen by the rest of the circuit. In the #12 schematic, clockwise rotation increases the negative feedback loop series leg resistance, which increases the gain of the stage.

ak