When i use the 3 mm divider as show in the figure below,

The amplitue of the sound decreases if we used two head phones on the output....
...but i believe that they are parallel and at the same voltage and the current consumptions should not have changed on the head phones....i think only the power dissipation on the source would increase....

could anyone please help me disign some kind of amplifier or level shifter in between....

Intuitive ideas are very welcome....let us all work this out...I will also share my ideas in meanwhile....

nice weekends people!!

 

Well now instead of having the resistance of one headphone per channel you have two resistances in parallel. So if one headphone speaker is 32 ohms, now the source amplifier has to drive a 16 ohm load per channel and this splits the current in two, reducing the volume.

The way this circuit might look would be to have the left channel go to the input of two op amps, A and B, and op amp A would go out to headphone 1's left speaker and B would go to #2's left speaker; and then to have the right channel go to op amps C and D which would both output to the right speakers of each headphone. You can get a quad op amp which would contain all these on a single chip. If you did something like this, the source would only be outputting its voltage upon two very high impedance op amp inputs, much, much higher than the headphone's 32 ohm impedance, and the op amps would do the job of amplifying the signals; and in this way, since each of the four speakers have their own amplifier, unplugging one of the headphone sets wouldn't change the volume for the other.

Here's a similar circuit I found that would handle _one_ channel in, two channels out, just as an idea. You might need a circuit that has a little more output power though as op amps don't deliver much current on their own.

http://www.forrestwhitesides.com/node/95

 

The amp has some max current or some output resistance, something limits the max current it can produce. As patrick says two headphones splits the current.

Power can be expressed as:

I^2 * R, so if each side gets half the current the power becomes:

(I/2)^2 * R = I^2 / (2^2) * R = 1/4 * I^2 * R

So the power is 25% of what a single headphone would have.

 

Well now instead of having the resistance of one headphone per channel you have two resistances in parallel. So if one headphone speaker is 32 ohms, now the source amplifier has to drive a 16 ohm load per channel and this splits the current in two, reducing the volume.

The way this circuit might look would be to have the left channel go to the input of two op amps, A and B, and op amp A would go out to headphone 1's left speaker and B would go to #2's left speaker; and then to have the right channel go to op amps C and D which would both output to the right speakers of each headphone. You can get a quad op amp which would contain all these on a single chip. If you did something like this, the source would only be outputting its voltage upon two very high impedance op amp inputs, much, much higher than the headphone's 32 ohm impedance, and the op amps would do the job of amplifying the signals; and in this way, since each of the four speakers have their own amplifier, unplugging one of the headphone sets wouldn't change the volume for the other.

Here's a similar circuit I found that would handle _one_ channel in, two channels out, just as an idea. You might need a circuit that has a little more output power though as op amps don't deliver much current on their own.

http://www.forrestwhitesides.com/node/95

Hi,
First thankyou very much for your suggestiion...but i quite dont agree with divison of current....actually the current through each speaker remains the same but the amplifier has to source double the current ...i have made the similar representation as below: waiting for your response..

 

Hmm, I think I miscalculamated You're right, the amp has to double its current which is going to drop its output _voltage_ which will in turn reduce output current. If the amp has an output impedance of, say, 10 and your load is currently 32 ohms and you later reduce that to 16 (by splitting the signal) the voltage will drop. So your voltage is, first, [(Vsource * 32) / (10 + 32)] and, later, [(Vsource * 16) / (10 + 16)]. Then take the _new_ voltage and divide it by 32 to get the new current.

Have you found any circuits you are considering for this?

 

Hmm, I think I miscalculamated You're right, the amp has to double its current which is going to drop its output _voltage_ which will in turn reduce output current. If the amp has an output impedance of, say, 10 and your load is currently 32 ohms and you later reduce that to 16 (by splitting the signal) the voltage will drop. So your voltage is, first, [(Vsource * 32) / (10 + 32)] and, later, [(Vsource * 16) / (10 + 16)]. Then take the _new_ voltage and divide it by 32 to get the new current.

Have you found any circuits you are considering for this?

clever man....
thankyou very much for this tip...and i totally agree with this answer...thanks patrick....!!