I have a setup where an amplifier with 2 channels drives four 8 ohms 80 watts speakers on each channel. The amplifier has a single volume knob which controls volume of both channels at once. What I would like to do is add a volume control for both channels separately between amplifier and speaker. This as I understand is not the right way because the potentiometer required will have to handle 80 watts of power, and thats why a volume knob is used at the input of amplifier instead of output. Unfortunately i have to use the setup i described. I have looked into T-pad and L-pad attenuaters/resistor divider.

I would like to know whether I can get some good ones in Europe? I am presently living in Austria, and I really cannot find anyone who deals with it in Europe. Also does anyone have experience/suggestions with controlling volumes of 2 channels separately?

  • 3
    \$\begingroup\$ Is there any way you could intercept the audio signals going into the amplifier? \$\endgroup\$ Commented Aug 27, 2018 at 14:29
  • \$\begingroup\$ @calcium3000: I dont think i can. Its a Yamaha amplifier and i dont want to open it up to void the warranty. \$\endgroup\$
    – RAN
    Commented Aug 28, 2018 at 13:28
  • \$\begingroup\$ I don't mean open up the amp -- I mean put a potentiometer between the signal source (mic preamp, CD player, etc.) and the amplifier. That way you can use a little guy for volume control. \$\endgroup\$ Commented Aug 28, 2018 at 13:32
  • \$\begingroup\$ @calcium3000: But then it will control volume of both the channels. Like i mentioned in my question, i would like to control volume of both channels separately! \$\endgroup\$
    – RAN
    Commented Aug 28, 2018 at 13:46
  • 1
    \$\begingroup\$ No, unless you're sending a mono signal into the amp you'll have two channels -- left and right. They may be on the same cable (e.g. with a TRS-style plug) or separate (e.g. paired RCA plugs). \$\endgroup\$ Commented Aug 28, 2018 at 13:57

3 Answers 3


There are a couple of ways of doing what you want.

The most obvious method is to use a commercially-available "L-Pad". I've seen these available with power ratings up to about 100 Watts. The higher-power versions can be expensive.

Note that the internal wire-wound resistor elements have an appropriate audio taper. The larger versions have resistance elements that change wire size as the resistance gets lower and the power being handled gets higher.

Another method is to use a tapped transformer. These are available for both 70V and 25V distributed loudspeaker systems.

What is not commonly known about these transformers is that although they are intended for use in a constant-voltage loudspeaker distribution system, they can be used directly between a low-impedance amplifier and a low-impedance speaker. It is simply a matter of choosing the right transformer.

Look for a tapped transformer that will work with both 70V and 25V systems with a power rating of at least 100W. Depending upon how low you want the frequency response to go, they can be relatively inexpensive. The least expensive versions of these only go down to about 200Hz and are intended for ceiling speakers. The better ones go much lower and are used in large venues such as sporting facilities (with much better speakers).

  • \$\begingroup\$ Raid: Thank you for your comment. Will this part do the trick? conrad.at/de/… \$\endgroup\$
    – RAN
    Commented Aug 28, 2018 at 13:45
  • \$\begingroup\$ I believe that it will work very well for what you want. Moreover, it is a stereo control and will control both channels of your system if you wish. \$\endgroup\$ Commented Aug 28, 2018 at 21:13
  • \$\begingroup\$ Ok, but should i be concerned about resistances and power ratings as Andyaka pointed out in the answer above? \$\endgroup\$
    – RAN
    Commented Aug 29, 2018 at 4:37
  • \$\begingroup\$ The power rating of the L-Pad that you linked to is 200 Watts. That means that it will handle a power amp and loudspeaker combination of up to 200 Watts. Recall that I mentioned above that the resistance winding within the device changes resistance per unit of rotation and wire size changes as appropriate to handle the power required. \$\endgroup\$ Commented Aug 29, 2018 at 16:01
  • \$\begingroup\$ Think of a L-Pad as two very-special purpose rheostats coupled to a single shaft (one channel). They are designed specifically to attenuate speakers. In the case of the device that you linked to, up to 200 Watts (per section). \$\endgroup\$ Commented Aug 29, 2018 at 16:03

Let me try and put you off using a potentiometer at the output in series with the speaker. Let's say (for argument) that you'd like to control the speaker level to about one-tenth of normal full-level. That would require adding 80 ohms in series with an 8 ohm speaker thus the voltage level on the speaker is about 10%. OK so far. You would use an 80 ohm pot (more accurately a rheostat).

Now if you turned the pot up so that there is the same voltage across both speaker and pot each would dissipate equally the same power. If the speaker power is 50 watts then the pot would also dissipate 50 watts but, it would dissipate it on about one-tenth of the pot track.

So, the pot you would need would have to be one that is rated at 500 watts. It's going to be the size of a small dinner plate and cost beyond $200. In other words, you are probably better off just buying another amplifier and using its volume control to give independent channel amplification.

  • \$\begingroup\$ 8 ohms 80 watts implies voltage of 25 of volts approximately across speaker. Now if i use a rheostat/pot between amp and speaker of 80 ohms then 2.5 volts will appear across speaker which 0.7 watts and across the pot will 6.25 watts. If i turn the pot all the way to minimum resistance, then all 80 watts will be across the speakers and and minimum dissipation across the pot, right? \$\endgroup\$
    – RAN
    Commented Aug 28, 2018 at 13:43
  • \$\begingroup\$ And when you have the pot set to 8 ohms, it will dissipate 40 watts and the speaker will dissipate 40 watts. The travel on the pot is about 10% hence the pot rating would need to be 400 watts. You only analyzed the two end cases and didn't consider the worst case (pot impedance = speaker impedance). \$\endgroup\$
    – Andy aka
    Commented Aug 28, 2018 at 14:39
  • \$\begingroup\$ perfect, understood thank you, and that is why they use L-Pad? \$\endgroup\$
    – RAN
    Commented Aug 28, 2018 at 15:21
  • \$\begingroup\$ @RAN an L-Pad has the "benefit" of maintaining a constant load on the amplifier and this of course is even more wasteful but some amplifiers need this. Other than that they are made from rheostats and are pretty much what I'm trying to get you to avoid. \$\endgroup\$
    – Andy aka
    Commented Aug 30, 2018 at 8:30
  • \$\begingroup\$ Yes I understand now. Hence to avoid overheating, i have decided to bypass the L-pad completely with a switch when no attenuation is needed, and keep it by default to half the value when attenuation is needed. I hope i am not missing anything. \$\endgroup\$
    – RAN
    Commented Aug 30, 2018 at 16:23

Look for a Rheostat. These are high power potentiometers.


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