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Headphone amp needs two outputs?

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grrr_arrghh

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Hi.

I saw this on my travels, and wondered why it has two outputs? It only has one input, so it can't be anything to do with stereo...

Thanks

Tim

P.S. the original site was **broken link removed**
 

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Seems to me that some of the stereo headphones have an impedance of 32 ohms which seems obscure but that's what's rattling around in my head. Some are 8 ohms too. The resistor may address something like that.
 
That's weird , where did you get the schematic?
It may be an amp for electrostatic headphones, where there's a voltage source to the elements , but monaural??? There's a DC voltage on both outs and no value for R4 and R3, the rest is just an amp. Weird I say weird!
 
It's just a standard little transistor amplifier - C4 is used for two purposes, as an output coupling capacitor, and as a bootstrap capacitor. It's quite a common technique, it's been used many times over the years. The resistor between the two 'outputs' is to maintain the DC conditions when the headphones are unplugged.
 
Really? I've never seen anything like it unless it was phantom power for ES headphones or mics. There's also a differential DC potential between the outs......like I said the circuit is a basic cap coupled amp , but I don't understand the DC element on the outs.
 
TillEulenspiegel said:
where did you get the schematic?

**broken link removed**

Nigel said:
The resistor between the two 'outputs' is to maintain the DC conditions when the headphones are unplugged.

Fair enough, but where do I connect the two outputs. One is connected to the signal on the headphones (?), but what about the other? Presumably it wouldn't be labled as 'out' on the diag if it didn't do something!

Thanks guys

Tim
 
grrr_arrghh said:
TillEulenspiegel said:
where did you get the schematic?

**broken link removed**

Nigel said:
The resistor between the two 'outputs' is to maintain the DC conditions when the headphones are unplugged.

Fair enough, but where do I connect the two outputs. One is connected to the signal on the headphones (?), but what about the other? Presumably it wouldn't be labled as 'out' on the diag if it didn't do something!

You connect the headphones to the two pins labelled 'output', one to the live side of the headphones, the other to the ground side of the headphones. With the headphones connected you no longer need the resistor between the output pins - and when these types of circuit are used in units with built in speakers the resistor isn't fitted - as the speaker is permanently connected.
 
TillEulenspiegel said:
Really? I've never seen anything like it unless it was phantom power for ES headphones or mics. There's also a differential DC potential between the outs......like I said the circuit is a basic cap coupled amp , but I don't understand the DC element on the outs.

You are quite correct, the speaker has DC current through it, but only part of the current feeding the driver - the bootstrapping helps to reduce it. The only reason for doing this is cost, it saves you the cost of the bootstrap capacitor - it's been done quite a lot commercially in the past, in radios and record players etc.
 
one to the live side of the headphones, the other to the ground side of the headphones

ahh, I was just under the impression that you would connect the ground side of the headphones to the ground in the circuit...

Thanks for everyone's help - Much appreciated.

Tim.
 
TillEulenspiegel said:
Hmmm , You learn somthing new every day.

It's suprising what you can learn repairing domestic electronics :lol:

It's not a design I've ever liked, and I wouldn't design or build an ampilfier like that - but they were no less reliable than other amplifiers, and seem to perform perfectly well.

I've particularly never liked the idea of a standing DC current through the speaker, but it's fairly low and doesn't seem to cause any problems.
 
grrr_arrghh said:
Nigel Goodwin said:
I've particularly never liked the idea of a standing DC current through the speaker

Couldn't you use DC blocking caps on the output?

Yes you could, but that's then not the same circuit! - the whole point of the design is to remove the need for seperate speaker coupling and bootstrap capacitors.
 
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