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What audio potentiometer to use? Answered

I am building small Bluetooth speakers for Christmas presents. I am using a 5v 2*3w Bluetooth Amplifier with a PAM8403 chip. I am wanting to add a dual potentiometer to control the output of the speaker, not the power going into the amplifier. I am going to use a logarithmic potentiometer but I cannot figure out the resistance I am needing. I have found A5k and A10k resistors, would these work for the project? Is there an equation or guide that would help me out?

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Jack A Lopez

27 days ago

Well, I think the usual way to figure these things out, is to read the data sheet. I found one here,
https://www.alldatasheet.com/view.jsp?Searchword=P...

Although, the data sheet can be confusing, and it just might be easier to search for existing tutorials, or recipes, like this one:

https://www.instructables.com/id/PAM8403-6W-STEREO...

You'll notice in Step 7 of that 'ible, the author is showing us how to wire a 10K audio taper pot (I am guessing that is the same as log taper?) pot to the input side of a PAM8403.

By the way, my interpretation of the prose in the data sheet, in particular the section labeled, "Maximum Gain" is this IC actually wants a fixed resistor, Ri, in series with each input, and the value of this fixed resistor should probably be something in the range from 20K to 100K.

I mean if you read that part, it suggests the gain of each amplifier is set by the ratio Rf/Ri, but the confusing thing is there does not seem to be any place to connect Rf. Moreover the example application circuit has a resistor Ri in it, but no sign of a resistor Rf.

It might be the case that Rf is internal, deep in the IC somewhere, and it is fixed to Rf = 142K.

So the only way to change the gain, is by changing Ri. For example, Ri =20 K would give a gain of about 7 = 142/20. Ri =100 K would give a gain of about 1.4 = 142/100.

(Also Ri =0 would give a gain of infinity, yet nothing explodes, in those designs out there which simply omit using a fixed resistor here.) ;-)

Regarding potentiometers, or voltage dividers made from fixed resistors, the usual trick is to just make sure the output resistance of the divider is smaller (by maybe a factor of 10) than the input resistance (or impedance) of whatever follows it.

For example, the maximum output resistance from a 10K pot is 2.5K, or 1/4 the total resistance, or the parallel resistance of two 5K sides together. So maybe the resistance of the next stage should be about 25K to avoid loading the pot, or voltage divider, too much.

The tricky thing is, the next stage, might be some funny thing you do not know what it is, like the input to some amplifier IC, with indecipherable documentation. Then you just hope the input impedance of that thing is high enough. If it is not, you'll probably notice it, as the output being too low.

Although I think this IC is going to be pretty forgiving. I mean I see tutorials out there, like the one linked above, of people just hooking up a variety of different things to the inputs, and then reporting that this works.

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Jack A LopezDownunder35m

Reply 26 days ago

Wow! Nice find! That is a good price too, especially considering you get the IC, the dual pot, a bunch of other little SMT parts, and a board too.

It is kind of reminiscent of that line from the movie Jaws. I think it was something like "...for that price you get the head, the tail, the whole damned thing!" Ha! ;-)

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Downunder35mJack A Lopez

Reply 26 days ago

These prices are what is killing the tinkers everywhere.
You just can't source the parts, etch a PCB and solder it together for these prices anymore.
Most of the people that were like me once just say "Why bother if I can buy something similar enough." :(

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Jack A LopezJack A Lopez

Reply 27 days ago

I just wanted to clarify what I wrote about the "output resistance" of a voltage divider.

What I mean by that is, the series resistance of a Thevenin equivalent of that voltage divider.

https://en.wikipedia.org/wiki/Th%C3%A9venin%27s_th...

For example, a typical voltage divider has V0 as its input voltage, and is made of two resistors R1 and R2 connected in series.

Vth = V0*(R2/R1+R2)), is the Thevenin equivalent source, i.e. the divided voltage.


Rth = parallel(R1,R2) = (R1*R2)/(R1+R2) is the Thevenin equivalent series resistance.

This has a maxium value when R1 = R2 = 0.5(R1+R2)

When that is true, Rth= 0.25*(R1+R2)

So the maxiumum output resistance from a 10K pot, is only 2.5K, which means I did the math wrong in my previous reply.

I'll try going back to edit that and fix it, and we will see if anyone notices.

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Downunder35m

27 days ago

How doyou want to control the output if I may ask?
On the output the amp expects to see either a 4 or 8 Ohm load, adding a potentiometer between amp and speaker won't work.

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MitchellD30Downunder35m

Reply 27 days ago

I was wanting to use a pot on the output of the amp; I have seen videos and pictures of it being done. The main problem right now is that the volume is controlled by the device connected by Bluetooth, so when I turn on the amp the speakers are at the highest volume. I was wanting to add a pot to control the volume so it could be lower when I turn it on.