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Wattage Rating Of Pots Used In Volume Control

I want to build I volume control for a battery powered amplifier I use for my iPod, but I'm not sure what wattage value the pots have to be. The amplifier outputs around 1 watt per channel.

The volume circuit I found was in a post on this newsgroup from 2001 by Beau Schwabe:

o------------->Right(+) | | o----o + | | | Right(+)>----||----o----\\/\\/--o 100uF 1K Pot | o-->GND 100uF 1K Pot | Left(+)>----||----o----/\\/\\--o + | | | | o----o | o------------->Left(+)

I plan on using two separate 1K pots, allowing me two adlust the left/ right channels separately, and the fact I can't get a dual pot of the right value at Dick Smith.

Reply to
Dave.H
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Looks like you are running the pots to ground the signal at one extreme of the travel? That won't fly. Hard on the pot and hard on the amplifier.

The wiper from the pot taps off part of the signal the way it is normally done, and then the amount of power involved is usually minimal - as for how much exactly, that depends on the type of output drive you have but with an Ipod I can't see it dissipating enough power in the pot to be a concern - not with a 1K or higher pot . . .

So signal goes to one end of the pot (full clockwise), ground to the other, and the wiper takes single off to the amp you are adding - along with a separate ground connection to the amp.

So wire it the right way and don't worry about it. Most pots of a 1" size are rated from 1/2 to 2 watts depending on construction - but use it the way you show and that becomes meaningless since the rating is for the entire pot element not the bit at the end that will burn out if wired incorrectly. When that is something you must do you look at the current capacity of the pot and don't exceed that.

Your 1 watt pot becomes a 1/2 watt pot at its midpoint if you use it to short the signal. Or your 1K, 1W pot can't exceed about 30 milliamps anywhere in its range.

Reply to
default

Do not use that circuit! Unless what I'm seeing is an ASCII art translation artifact, it appears to have some serious problems.

A standard volume control has the input signal going to the top of the pot (the contact that the wiper hits when the knob is fully clockwise), and the output from the wiper. The other side of the pot is grounded, the same ground for input and output signals.

If this is to go between the Ipod and an external amp, then the pot resistance can be just about anything up to say 50K.

10K is a good value. On the other hand, if this is to go on the output of a fixed-gain amp to reduce the volume, then you should probably stick with 1K. But this is generally a poor choice compared to controlling at the input.

Wattage is not an issue for input controls, and would only be an issue on the output if the amp was to drive a speaker. If that is the case, what you really want is a "fader", a low-resistance wire-wound pot made for this purpose. Again, not recommended in most cases.

You probably don't need the capacitor, depending on the Ipod output and amp input circuits, but it shouldn't hurt.

Best regards,

Bob Masta DAQARTA v3.50 Data AcQuisition And Real-Time Analysis

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Reply to
Bob Masta

Here's a link to the amplifier schematic, to make this easier. I would probably use the volume control on my computer with my headphones also, so i want to build one that will work with that and other devices.

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Reply to
Dave.H

I've done this just now with a 5K log pot I had on hand, and I have it playing now. I know what to do now, thinking of sticking with 5K so I can reuse this pot.

Reply to
Dave.H

I've done this just now with a 5K log pot I had on hand, and I have it playing now. I know what to do now, thinking of sticking with 5K so I can reuse this pot.

Reply to
Dave.H

I've done this just now with a 5K log pot I had on hand, and I have it playing now. I know what to do now, thinking of sticking with 5K so I can reuse this pot.

Reply to
Dave.H

Ghastly! That's desighed to kill ipod batteries, cause distortion, and fry pot wipers.

Just do this. Pot wattage doesn't matter.

John

Reply to
John Larkin

The only problem with the 10K pot (incorrectly called it a 5K pot previously) is that the volume barely increases until the shaft's all the way clockwise.

Reply to
Dave.H

With the voltages and resistance values involved, ANY pot will have a large enough rating ( typically it'll be around a milliwatt ).

Graham

Reply to
Eeyore

You are expected to use an "audio taper" pot for audio. The resistance is non-linear and compensates for the perceived volume to the human ear.

Reply to
default

In this application you'd need a REVERSE log pot actually.

Graham

Reply to
Eeyore

You can replace the 10K resistors to ground on the inputs with 10K log-taper pots. Just connect the amp chip inputs to the pot wipers instead of to the top of the pot.

If you can't find a log-taper pot, you may be able to "fake" one by tying a fixed resistor between the wiper and the top leg. Experiment with different values... it's been a while since I've done this, and I've forgotten which value gives the best fit to a log curve. (I think something in the 3K range.) You won't get a log taper, but at least it will be better than linear.

Best regards,

Bob Masta DAQARTA v3.50 Data AcQuisition And Real-Time Analysis

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Scope, Spectrum, Spectrogram, FREE Signal Generator Science with your sound card!

Reply to
Bob Masta

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