balance

[Professor smith]

my speakers are about 89db. How much of a voltage difference at the amp output between the left and right channel can be heard and enough to cause a shift in image?

[weltersys]

At around 4000 Hz, where the ear is most sensitive, an almost unmeasurable voltage (less than 1/100th of a volt) difference could cause an "image shift".

A tweeter or midrange hooked up opposite polarity from the other can cause a shifting image, stereo sources stay more "still" mono sources "waver".

[Professor smith]

but if 0.01 volts can cause an image shift what chance is there of attaining a match?

[weltersys]

Originally Posted by Professor smith

but if 0.01 volts can cause an image shift what chance is there of attaining a match?

A pan control can be used to match amplifier sensitivity, which does not need to be matched within .01 volts to provide an even left/right balance.

If you have an amplifier that randomly shifts in voltage amplification, it should be repaired or replaced if it bothers you.

I would doubt that an "image shift" has anything to do with amplifier voltage, but everything to do with program, speaker, and hearing frequency response deviations.

If your speaker placement and the room are not symmetrical, the reflected room sound from each will be different, and will vary with frequency, adding further to "image shift".

[sreten]

Hi,

A very bad start to answering the OP's question, and a stupid number,
1/100 of a volt, which is meaningless without defining the real level.

The numbers to use are dB, 1dB will cause a shift, 3dB very noticeably,
6dB very severely and 10dB you might as well disconnect one speaker.

1dB is about 13% difference between the channels. 10% is a reasonable
number to say it won't make much difference, though golden ear types
might say lower, say 6% or 0.5dB, if your speakers match to that level.

rgds, sreten.

[cbdb]

Yes, those are % voltage. So the .01 volts is meaningless. If your listening quietly and the signal level is only .05v, .01v will make a differance. If its 20bd louder at .50v .01v wont matter.

[Professor smith]

Originally Posted by weltersys

If you have an amplifier that randomly shifts in voltage amplification, it should be repaired or replaced if it bothers you.

What do you mean by randomly shifts in voltage?

[Professor smith]

Originally Posted by sreten

Hi,

A very bad start to answering the OP's question, and a stupid number,
1/100 of a volt, which is meaningless without defining the real level.

rgds, sreten.

volts are more meaningful to me than db because thats what i am measuring and
seeing on the multimeter.
So what does 0.1 volt translate to in db? I used a 1khz sine wave signal going into the amplifier.

[weltersys]

Originally Posted by Professor smith

volts are more meaningful to me than db because thats what i am measuring and
seeing on the multimeter.
So what does 0.1 volt translate to in db? I used a 1khz sine wave signal going into the amplifier.

Professor,

To translate voltage into dB, or what you hear, dB SPL, you have to know the speaker sensitivity, and the distance and level listened at.
Doubling voltage will result in a 6 dB level difference, as Sreten points out, that is a large level difference.

However, I would not describe a large level difference as an "image shift", you need to describe what your perception of an "image shift" is when it comes to sound .

Sound pressure - Wikipedia, the free encyclopedia

At 1K, -8 dBSPL is roughly the sound of a mosquito flying 3 m away.
A very small fraction of a volt in to an 89 dB 1 watt one meter sensitivity speaker could equal that level, as a professor you may be able to figure that out, somebody just did in the thread titled: “A Test. How much Voltage (power) do your speakers need?

[dumptruck]

Originally Posted by Professor smith

volts are more meaningful to me than db because thats what i am measuring and
seeing on the multimeter.
So what does 0.1 volt translate to in db? I used a 1khz sine wave signal going into the amplifier.

But dB is what matters, as already explained, so you need a percentage difference, as also explained. You are measuring two voltages; just calculate the percentage difference between them.