stereo earphones with 4 conductors?

I do not know but I suspect that all earphones and headphones with separate leads to each piece can be rewired to keep the leads independent.

Kal

Sure. Just cut above the "y". ;-)

Cheers! Rich

There's a 4 conductor 1/8" plug used on Nokia cell headsets, if that helps...

--
A host is a host from coast to coast.................wb8foz@nrk.com
& no one will talk to a host that\'s close........[v].(301) 56-LINUX
Unless the host (that isn\'t close).........................pob 1433
is busy, hung or dead....................................20915-1433

Is it possible to process the two signals before amplification so that one leg of each amplifier output carries the same signal? Then conventional headphone connections could be used.

Chris

In a normal bridge-connected amplifier the load is connected between the outputs of a pair of amplifiers each with a gain A, the input signal Vi is fed to one amplifier and is inverted and fed to the other as -Vi. The voltage across the load is the difference AVi-(-AVi) = 2AVi.

In the stereo case, if the input signals are Vl and Vr the output signals from the two pairs of amplifiers are 2AVl and 2AVr.

If the circuitry is changed a little so the input signal for the opposite stereo side is added to the input of each amplifier then:

for the left channel the pair of output signals would be 2A(Vl+Vr) and

2A(-Vl+Vr) giving a (difference) output signal of 2AVl as before;

for the right channel the pair of output signals would be 2A(Vl+Vr) and

2A(Vl-Vr) giving a (difference) output signal of 2AVr as before;

but one leg of each carries the same voltage so their connections could be paralleled, or a single amplifier used for left and right channels, and only three headphone connections would be required whilst retaining the low supply voltage advantage of the BTL configuration.

Where's the flaw? Any mismatching of amplifier gains would introduce crosstalk, but a little of that may not be such a bad thing for headphone listening.

Chris

Headphones with the 4 pin DIN headphone connectors usually have separate wires.

Paul

I think that the idea is excellent. One drawback is that the output power capability is reduced (the dynamic range is lower, because now the signal that does not have to saturate is A*(Vl+Vr), instead of A*Vl. So, half the voltage, and 1/4 of the power, which is a lot.

Even that, there are cases in which you have plenty of output power capability (and you could live with 1/4 of it), and this idea would help save you two (relatively) big electrolytic capacitors. I've searched for ICs implementing this idea, but haven't found any, and I'm surprised. Ok, in case of high power speakers, it makes no sense, because they never share any node, in any connector. But for ear/headphones, with so many 3-node connectors out there, I think there would be market. Maybe I'm missing something else.

Thanks!

Sure, but we need to buy earphones and include them "as they are", with our product. We can't modify each earphone, in mass production.

Best,

If it's any use, I've tried the TI/Burr-Brown DRV134's to drive headphones. Just tied the left -ve output of the balanced output to the right -ve. Works fine.

Graham H.

I've just known of approaches like the DirectDrive by Maxim (to be able to use stereo earphones with connectors with 3 nodes). Not bad, but they use switched capacitor charge pumps, and I have to avoid noisy circuitry like that.

And then you end up with not being able to use just any headphones simply because you don't want the capacitor. Headphones are the thing that's likely to break first, and you've ended up with a situation where the user will then end up not being able to use the gizmo since they have to order the headphones from some single source in New Jersey.

I have a headphone jack on the car radio that I use as a bedside radio. I put the capacitor on it so I could use a single output of the car radio to feed the headphones. I sure didn't need the extra power of balanced output, and the capacitor was hardly "large". You should have been around forty years ago, then a 10,000uf capacitor at 17v was the size of a coke can. Now, it's not any extra size.

Since this sort of thing is dealt with all the time in commercial equipment, you've made the issue more complicated than necessary. They live with the usual headphones, and thus they must live with the "nasty" output capacitor. On the other hand, my Sansa Fuze is small enough that there must be miniscule capacitors at the needed capacitance, since all of the unit complete with batteries and LCD screen is jammed in that tiny package.

Michael

OK. You did not so specify. If you are buying them in bulk, you should query the manufacturers directly since you will also have to deal with the specification of the connector. The standard connector is tip-ring-base and only accomodates 3 lines.

That said, it is possible that any earphone with a tip-ring1-ring2-base plug is, or can be, suitable.

Kal

You're right, it throws away the output power advantage of the bridge configuration. It was getting a bit late when I wrote that!

A simpler arrangement providing the same output power and using no more parts would use conventional push-pull amplifiers for left and right, and a third amplifier fed with no signal to provide a stiff half-rail node to which both loads would be returned. That doesn't have the same crosstalk issue.

Chris

No, they must not live with the output capacitors. You can avoid them with a dual supply (or with charge pumps if you run from an explicit single supply). I'd bet that's what they do in tiny MP3s, etc.

It's only that I don't want to use noisy circuits like charge pumps, and I don't have a dual supply.

Best

In which case you contact some firms about an offer to produce them the way you want them.

Which appears to be what some $10-20 radio/cassette players use for the audio output amps in their one IC. The "ground" in the earphone jack is at 1/2 the battery voltage. (The jack is mostly plastic). There's only a + and - connection at each end of the three volt supply, so it's a derived virtual ground.

Mark Zenier snipped-for-privacy@eskimo.com Googleproofaddress(account:mzenier provider:eskimo domain:com)

- snip -

Fair enough for driving high impedance 'phones. The power supply current might get a bit high though when there's any significant difference audio components when the loads on two of the amplifiers are effectively short circuits (through the internal 50 ohm resistors). Rather limited output power into low-Z 'phones though, with those 50 ohm resistors that can't be bridged out.

As the data sheet

says, the amplifiers can be operated 'single ended' by connecting one of the outputs to earth, but then why bother using balanced-pair amplifiers at all?

Chris

- snip -

Indeed, and the question is then what is the identity of that IC?

Mention has already been made of Maxim's 'DirectDrive' scheme that creates a synthetic negative rail using a charge pump

Anyone know what Apple use in their iPods and suchlike ... is there a separate audio output IC?

Chris

Looking at my notes and digging out a screwdriver and just looking showed part numbers of LAG668F, LAG665CB and LAG665F. A google search for LAG665F indicated the manufacturer is Mitsumi. (They're only good for about three hard drops, so I have plenty of corpses to autopsy).

Mark Zenier snipped-for-privacy@eskimo.com Googleproofaddress(account:mzenier provider:eskimo domain:com)