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How to determine what size Audio Potentiometer is needed? Answered

Second attempt to get this question to publish.....sigh...

In a small project with approximately (1/4 watt) .25 watt audio output into an 8 ohm speaker, what size Audio (I assume non-linear for audio) Potentiometer should I purchase?

I tried a 10k but either I got a Linear (from what I've read) or 10k is too large. When adjusting to volume, I get nothing for most of the turn and then I get all of the volume. I can adjust to the volume I need, but it's very oversenitive.

I would like it to gradually adjust the volume.



3rd attempt to answer this, hope it finally posts...

Pot(entiometer)s in audio amps usually run from 5k ohms to 500k ohms, depending on where and their intended use. Inputs are the high values and are ususally low amperage (1/4 watt is a very safe value), normally, while outputs can be extremely low values and are in the range of 4 to 10 x's the target load (closest standard value, in this case 25-100 ohms) and usually a hefty wattage rating (somewher near 1/10 to 1/2 the power output), in comparison. They are generally audio taper (logarithmic) when an external user interface is needed. Both linear and logarithmic pots will work with the only difference being the way that your ear will perceive the change made while twisting/sliding the pot.

That said, I am guessing from your question that you are building a LM386 (based) amp. There are 3 possible places for a pot in the circuit: at the input, at the output, and within the gain loop. Since you are looking for volume control, a 10k ohm pot on the input or a 25 ohm pot at the output. Look at the datasheet or Runoffgroove.com for the correct way to wire them.


Since we now know what you are dealing with and how you would prefer to work this...

#1. The first PDF you reference says that you can drive it single ended with an accompanying electrolytic capacitor (as is customary in that configuration) with a reduction in power (1/4 the bridged rating). See if that puts you in the ballpark first.

#2. Since it is a solid state output, higher impedance (resistance in this case) won't hurt it. To make it more of a smooth transition, try no more than a 1k ohm pot (500 ohm would probably be better) as a variable resistance - looking down from the top: + to the left tab, center and left tab joined, right tab to + on speaker, - on speaker to - on ISD chip. This will attenuate the output, not completely kill it, unless you find the resistor value that prevents the speaker from getting enough power to work. You don't want to put the pot across the + & - lines. The resulting short could possibly kill the amp section in the chip.

#3. You could use the + speaker line as Steveastrouk suggests. Don't forget a coupling cap.

#4. Try a higher ohm rating speaker. I've seen some in the 100 ohm range. That ought to pull the volume down all by itself.


You can use an LPad in your setup. A rheostat would work also. Problem with these is cost. A quick check of major sites (Jameco, Mouser, Digikey and Parts Express) show prices over $20 US - many over $30 - which is probably 2 to 3 times (or more) what the whole circuit cost. That much for volume control? Kinda makes you sick. For about $3-5, you could add a LM386 amp section, with volume control.

That is why I suggested a different type of attenuation. I was able to find a 25 ohm, 5 watt pot at Mouser for $3.61 apeice, but I believe there is a minimum order amount. If you make enough of these or need other things, then it is worth it. It  is linear, but at least it is in the range you are looking for with a power rating far above the ISD's power output.

The only problem I have with putting a pot across the speaker outputs in a voltage divider fashion is that the normal output is differential and should not be treated the same as a single ended output, that is that the - (negative) output is not the same as ground. Therefore any resistance you put across it will see "ground potential" at 1/2 the resistance. Above that will be positive, below that will be negative. So at 1/2 the pot across the + and - pins should act as no volume at all because that is where the + and - signals meet and should cancel out. Below that would mean that you are then sending the same signal to the both sides of the speaker, just differing levels. If the output was single ended, you would have full travel on the pot, not 1/2 as with differential.

I would like to know if Steveastrouk and Gmoon concur or am I missing something myself.

All in all, I think it would be less costly to add the output stage in a similar fashion to Steveastrouk's original post, the one referenced by Gmoon or my own above, or even the low powered old style amp by Brunoip would work - as long as you use the input volume control setup mentioned in earlier posts. Just a note: if you use these additional amp stages, you'll probably want to add external wall wart power as an option. All amps draw a small amount of quiescent current even when no sound is being produced, so there will always be a small drain on the batteries.

Hope I haven't muddied the electronic waters for you too much.


I think you have a nice summary there. The output of these little buggers is trickier to handle than you would think



6 years ago

If I'm reading you correctly, then the links provided by Quercus austrina should be very helpful.

You cannot use a regular pot (volume control) on the output of a power amp, even a .25 watt amp--most POTs are 1/4 watt at best (and probably not for a smaller section of the rotation).

Use a 25 ohm rheostat as an attenuator, like the Little Gem amp. It has a high enough wattage rating to survive.

Why can't a larger POT like a 1K or 10K work here? Because the amp is driving an 8 ohm load. Adding additional impedance between the amp and the speaker reduces the output volume quickly--as little as 2-to-5 ohms should be noticeable (best to setup as a voltage divider).

There's an old adage that the output impedance should be 1/10 of the input impedance of the next stage, or there will be signal loss. So the output of the chip is probably less than 1 ohm. Clearly adding a 1K POT is a HUGE increase in output impedance, and no signal will pass through (until you turn it down to it's minimum resistance).

An L-Pad would work fine too, as you noted. It just might be overkill for a .25 watt amp...

I'm not totally with you on the "differential" part of your last reply. (you did, and should, receive the "best answer").

Attenuators are used on just about every class amp I've seen, not just class A. That includes class AB tube amps (push-pull). I've also seen them on solid state car amps, too.

The speaker (or the attenuator) doesn't care about a ground reference--unless there's a NFB loop on the output, then it's necessary.

There's a point where the outputs are equal, sure. But that's true in any topology. It's the difference in potential between the two that's important.

Don't get me wrong--I don't claim to be an expert on solid state amps, especially the small ones. There may be short cuts and safe guards they use that make certain things impractical.

Ah, this is why I asked for your view on this as well as Seveastrouk's view - to see any angles I may have missed.

True, it is the potential between the pos and neg tabs of the speaker, not the immediate value between the amp outputs, that drive the coil or other "motor" of the driver plane. Therefore, a voltage divider setup should work. I was thinking in terms of waves, not potential. But along those lines, increasing amplitude waveforms (the exact same one) sent to opposite sides of the "speaker" would be 180 degrees out of phase and therefore have more of a cancelling effect than just voltage division. So at the low end of the divider, there would be more fall-off in the differential, or bridged, amp. That doesn't happen with single ended amps that feed direcly to ground after the speaker.

And there's another problem. To get a Lpad, rheostat or pot with enough power rating (even at the 1/4 watt output) at a low enough ohm value would generally be prohibitive in cost (if you could even find one that low powered AND that low of a value). Buying parts for a small amp is much less than these options and ultimately more versatile.


Sure, it's an interesting question...

In "my world" (not an EE world ;-), the vast majority of attenuators are used on amplifiers in the 30-100 watt power range. Those amps are almost exclusively push-pull (smaller, cheaper transformers, and some noise cancellation), and as such they qualify as differential amps--they all include phase inverters, and most use the classic long-tail pair, operate part of the time in Class B, etc.

It certainly is valid to compare the effect of an small-signal volume control vs. an attenuator, of course. Since power output vs. perceived volume isn't linear anyway, the "taper" of an attenuator surely wouldn't be linear. Perceived volume isn't linear anywhere in the amp, anyhoo.

Since audio L-pads are fairly common, I'd expect the manufacturers have a handle on that aspect, though.

Personally, I think most guitar players don't expect their attenuators to function seamlessly as another volume control. There's usually a sweet spot; a compromise between loudness and "feel."

L-pads are certainly better than a simple rheostat, as they present the same load to the amplifier regardless of the setting. In an audio sense that can make a difference.

Despite what I said above, a rheostat as voltage divider might be a problem when the load impedance get's too low (like around zero!). So adding another resistor to limit the low end is probably wise. Although the 386 in the "Little Gem" doesn't seem to have that problem. Depends on the safeguards built into the amp. And I doubt that a rheostat in a simple current-limiter configuration would have a very linear response at all...

There are plenty of fixed or switchable attenuators, and not all are classic L-pads. And an L-pad doesn't need to be adjustable...

Forgive my bone-headedness--a rheostat wired as a voltage divider would still "present" the rheostat's resistance as a load, not zero... So it wouldn't be a problem for the amp.

The ISD 1820 doesn't like not directly driving a speaker - it has a built in differential output, which should really be treated by taking it into a differential amp, rather than a single ended amp. You can make a 386 do that.

Take the output from the ISD chip into a resistor and from the resistor into a 5K log pot, take the wiper from that into a LM386, built for minimum gain. The resistor on the pot will let you scale the output lower if you need to, so you need to experiment.

I must really be missing something....???

From these PDF's it shows direct connection to 8 ohm speakers.

Please explain if you can please.

I have used ISD2540 in kits before. This ISD1820 Voice Recorder Module is a completed module ready to used for projects. I assume it would about the same output volume wise. I have just ordered one and have not received it yet for testing. The ISD2540 provided plenty if not more volume needed connected direct to 8ohm speakers.


Thanks for eveyones input. It's greatly appreciated!

I'm I not seeing this correctly.
Should I really use this module like a pre-amp to a LM386 with the volume control between this module and the LM386???

Or will a simple L-PAD between the Output and the 8hom speaker accomplish the effect I'm lookin for?

You have no independent volume control with a ISD chip, your "output volume" is recorded in your source, so you have to do something externally. An L-pad would work here.


6 years ago

You need an Audio Taper Pot.
I would need to see the circuit to advise you on a value.
In general a volume pot is best placed on the input of an amplifier.


The output is actually from a ISD1820 Voice Recorder Module.
I have added a LM386 amplifier in the past, but found it to be way too loud for my application.

You need an amplifier, pot's alone don't work like that.