Replace a pot with a ¿resistor?

[Yib]

My speaker's volume knob is totally broken and now I have to replace it with something that let me use the speakers without a pot and hence, without letting me control the volume (I'm ok with this, I will just control it in windows).

What I thought about is just short circuit the two pots but I don't think this is a good idea. I've outlines a simple circuit to show you how it's wired:

So, thinking it twice I thought it would be better to replace the pot with a resistor an my questions are:

• Each pot has three legs, between which should I wire the resistors? Hot and middle onw, leaving the ground one alone?
• How many ohms the resistor should have?

Note: If I connect it letting the pot circuit open I will hear the sound but very very low, being able to hear it just if I pull the speaker up to my ear.

PS. I'm sorry if my english ain't so good, it's not my native language. If you want me to clarify anything just tell me.

Thank you. Regards.

 

[BobK]

A pot can be replaced not by one resistor, but by two. You would attach both to the middle pin of the pot and the other two ends to the two other pins. You could size to two resistors to select any volume level you wanted. The sum of the two resistors should be close the resistance of the pot. For instance, if replacing a 10K pot, you could use two 4.7K resistors to get an intermediate volume.

Alternately, if setting the volume all the way up is an option, you could just wire the middle pin to the one that is not grounded.

Bob

 

[Yib]

So, since it has two pots (I guess its because this way it has more "volume points"?) should I attach two resistors in each pot the way you said? having each resistor the half impedance each pot has?

Thank you.

 

[BobK]

The two pots are for the 2 channels, I presume. Yes, two resistors would always work, hower equal resistance might not be the best. Try temporarily substituting a working pot and measure the two resistors (center pin to each out pin) at the volume you want to set it at.

Bob

 

[Yib]

That is not possible because the two pots are totally torn apart and I dont have a polimeter too. It doesnt have to be the perfect volume, I can control it from windows, but I don't know what resistors should I try. A range like from 0.33-0.66 (in a 0 to 1 volume scale) should be fine so maybe someone could take a guess.

PS. The pot is labelled as A50K

Thank you.

 

[davenn]

yup well that was an essential bit of info

A50k = 50k Ohms ( 50,000 Ohms)

cant remember if the A is for a linear or logrithmic pot
some one else is sure to chime in on that


Dave

 

[Yib]

Wow, that's a lot of ohms for someone who studies electrical engineering (and has little idea of electronics)

 

[(*steve*)]

Wow, that's a lot of ohms for someone who studies electrical engineering (and has little idea of electronics)

Yeah, we like our resistors big and our currents small.

 

[Yib]

So the sound in a speaker is based in the current signals received by the computer/whatever and the volume pot is a multiplier of that current? More impedance means less current (hence less volumen) and vice versa? I guess the pot is put in series with the speaker and changing the impedance (with the pot) will change the current that flows to the speaker? Or is it put in parallel, but it wouldn't make sense that letting the potshort circuited would set the volume all the way up.

In this case, wouldn't you be able to take a wild guess and tell me what ~ohms should the resistors (4 of them) have?

Thank you, it's nice to learn new things.

 

[(*steve*)]

The pots are almost certainly used to vary the level of the signal into the amplifier.

It would be grossly inefficient (not to mention the effect on damping) to place a variable resistor in series with a speaker.

 

[Yib]

They are in parallel then? But then, why the volume lowers when impedance tends to infinite (open circuit)?

 

[BobK]

The pos is a voltage divider. If turned one way fully, the entire signal going into the amplifier is alowed through. If turned fully the other way, none of the signal goes through. Any position in between will allow some part of the the voltage through.

What you need to do is replace the pot with two resistors that will give you the maximum volume you want.

Or, as I said earlier, if the max volume is okay, just put a wire from the center to the other terminal that is not connected to ground. This effectively puts the volume control at max.

Bob

 

[EinarA]

To answer Dave's question: A stands for audio taper. The pot is split into linear sections with the lower half being typically 10 percent of the total ( but can be 5 to 20 ), and the upper half 90 %. B is used for linear. Various other letters indicates other tapers, check with Mfg for details.
Try Bob's idea to jumper in to out and see what you get. If for headphones don't do this with them on your head.