I am building a simple audio circuit that will only be used for headphones. It is starting with a very low signal going into a passive volume and tone control board. The output of the volume/tone board goes into a small amp that has a 3.5 watt output. When the vol/tone board is turned to max it is too loud for headphones. I want to limit that 3.5 watts to maybe 3/4-1 watt so when the volume is turned to max it doesn't blow the headphones or my ears. What is the best and most reliable way to reduce the output of this small amp?

Thank you

  • 1
    \$\begingroup\$ share the details of amp circuit you have. or do not turn the volume to max. \$\endgroup\$
    – User323693
    Feb 7, 2017 at 6:23
  • 2
    \$\begingroup\$ (There is no "best" or "most reliable" answer. It sure would be nice, though. Just learn the "best" and "most reliable" ways to do things and completely eliminate imagination and creativity from the equation and just write a program to punch out the best answer every time. Oh well.) You might consider applying global negative feedback as a means of reducing gain. What's the "small amp?" Can you provide a link to it? \$\endgroup\$
    – jonk
    Feb 7, 2017 at 7:55
  • \$\begingroup\$ Here is the amp: ebay.com/itm/… \$\endgroup\$ Feb 8, 2017 at 6:38
  • \$\begingroup\$ Here is the tone board: ebay.com/itm/… \$\endgroup\$ Feb 8, 2017 at 6:39

3 Answers 3


I would recommend configuring the amplifier for a lower closed-loop gain. The amplifier board you linked to is built around a Diodes Inc. PAM8403. You can pull the datasheet for that part to learn how to configure it's gain.

Unfortunately, the datasheet is not very clearly written. It gives the closed-loop gain of the amplifier as:

AVD = 20*log [2*(RF/RI)]

However, "RF" is not identified on any schematic. Based on the verbage, though, I think it is safe to assume that RF is the feedback resistor inside the IC, with a fixed value of 142kohm. "RI", in the above equation, is the sum of the internal input resistor, which has a value of 18kohm, and any additional external resistor in series with the input to the chip (labeled RI on the front page application schematic).

What this means for you is that you probably want to add additional resistance in series with the input to the amplifier board. If there is already an external RI on the little board that you have, then try replacing it with a larger value until you are satisfied with the volume range. If there is no footprint for it on the PCB, you will have to find a way to kludge the resistor into the path - maybe splice it into the connection between the tone board and the amplifier board.

It is worth noting that your tone board is passive and, thus, has significant output impedance itself. In this way, it is possible that the tone control will interact with the gain of the amplifier and vice-versa, altering the control law of the volume and/or tone potentiometers. I won't offer a complete explanation here, but this is something to keep in mind if experimentation yields confusing results.


Perhaps, you are using a turn-pot to adjust your headphone volume.

So, it is a simple voltage divider circuit at the opamp feedback which determines your gain.


simulate this circuit – Schematic created using CircuitLab

I would suggest that you restrict your Gain using a fixed resistor such that you always have a set maximum gain which can't be exceeded!

Refer to the schematic above where the R2 (1K) is placed to restrict the gain to 10, in this circuit. The calculations have to be done by you though, the values in this schematic are not to scale.

By the way our ears are logarithmic and thus, for audio applications instead of using a simple linear potentiometer, we use logarthimic one. Do account for the same as well. Your objective should be to place a cap on the maximum gain attainable by the circuit! Refer to the logarthimic amplifiers and write down their equations to set the maximum gain. You can make a logarthimic amplifier by using a transistor and a resistor. If you can upload your schematic, it would be easier for us to debug it.

Happy prototyping!!

  • \$\begingroup\$ The OP's amplifier has output capable of 3.5 watts. It's very unlikely to just be an opamp. We can't really give an answer until we know what it is though. \$\endgroup\$
    – Colin
    Feb 7, 2017 at 8:45
  • \$\begingroup\$ Re. your comment on pots: the gang-to-gang matching of dual-gang lograthmic pots makes them undesirable for precision stereo audio applications. In the best designs, linear pots are used in a network with some additional fixed resistors to approximate a logarithmic law while avoiding poor channel-to-channel matching. See for example: sound.whsites.net/project01.htm \$\endgroup\$
    – user49628
    Feb 7, 2017 at 17:31

There are two ways to do this: First is to limit the input signal, perhaps by putting a resistive attenuator before the amplifier input. For example, you can insert a series resistor at the input side of the volume control. This is similar to not allowing the volume control to be turned up to full volume. Using an attenuator at the amplifier input will increase your "noise floor", but probably not enough to be noticeable.

The other way is to use a resistive attenuator at the amplifier output, before the headphone jack. A series resistor will work, but then the attenuation will be largely dependent on the headphone impedance. Many headphones are around 30 Ohms or so, but others may have an impedance of 8 Ohms. You can reduce the headphone-impedance dependency by using a voltage divider (an "L network" attenuator). Depending on the attenuator design you may have to use power resistors to dissipate the "thrown away" power. The disadvantage with the output attenuator is that you will be wasting power, but this probably won't matter in your case. A resistive output attenuator may also affect the frequency response of the headphones, probably in the low-bass region. Again, this may not be enough to matter.

Or, you can use both techniques together. I would probably choose the output attenuator if all other factors were equal.

[edit]: I just looked at impedance ranges for modern headphones. iPod-style phones are typically around 50 Ohms. Others run up to 600 Ohms, and some run well under 50 Ohms. With any of these headphones, the amplifier output power will be way under 2.5W, but this isn't the important factor. This does mean that a output reduction that a resistive output attenuator will give you is headphone-dependent. I would probably use a series resistor from the amp output (selected for the desired attenuation, but 220 Ohms will give you about a 20dB reduction), feeding a 47 Ohm resistor to ground, and connecting the headphone across the 47 Ohm resistor. You can probably use 1/4W resistors.

  • 1
    \$\begingroup\$ A third way would be to reduce the gain of the amplifier. \$\endgroup\$
    – Colin
    Feb 7, 2017 at 7:43
  • \$\begingroup\$ I find a simple resistive voltage divider works well. Choose two resistors that add up to the impedance expected by the amplifier. Connect the headphones across the lower of the two resistors. \$\endgroup\$
    – Simon B
    Feb 7, 2017 at 14:11
  • \$\begingroup\$ Simon B, most non-tube amps are quite happy to drive higher than rated output impedances. I would keep the attenuator input impedance as high as I could manage to reduce power dissipation. \$\endgroup\$ Feb 7, 2017 at 16:34
  • \$\begingroup\$ As you have observed, a resistive attenuator at the input comes at the cost of noise, and a resistive attenuator at the output introduces a dependence on the headphone impedance. I would think that changing the feedback network on the buffer amplifier would be the most appealing solution - hence the downvote. Indeed it is hard to comment until we see the circuit... \$\endgroup\$
    – user49628
    Feb 7, 2017 at 17:35
  • \$\begingroup\$ I forgot to mention an even bigger problem with the output resistive divider. The impedance of a pair of headphones varies over frequency, so a divider formed by a resistor and the headphones will produce dramatic frequency response errors. Check this out: google.com/… \$\endgroup\$
    – user49628
    Feb 7, 2017 at 17:44

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