Passive LED-based headphone protection mod?

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OlympicPhil

Well-known member
Joined
Sep 13, 2004
Messages
70
Can anyone help?

I've been set the task of modding 150-or-so pairs of Beyer headphones with passive LED-based driver protection circuits.

I've scoured Google but can't find anything that would fit the bill.

My supervisor assures me it's possible using only an LED and a resistor.

Anyone done this or anything similar?

Thanks in advance, any help gratefully received.

Sarcasm or witty comments also welcome.


Phil.


P.S. Some engineer jokes I found on another message board:

--------------------

To the optimist, the glass is half full.

To the pessimist, the glass is half empty.

To the engineer, the glass is twice as big as it needs to be.

--------------------

A priest, a doctor and an engineer were waiting one morning for a
particularly slow group of golfers. The engineer fumed, "What's with
those guys? We've been waiting 15 minutes!" The doctor chimed in, "I
don't know, but I've never seen such inept golfers!" The priest said,
"Here comes the greens keeper, let's ask him." He said, "Hello, George!
What's wrong with the group ahead of us? They're rather slow aren't
they?" The greens keeper replied, "Oh, yes. That's a group of blind
fire fighters. They lost their sight saving our clubhouse from a fire
last year, so we always let them play for free anytime." The group fell
silent for a moment. The priest said, "That's so sad. I think I will
say a special prayer for them tonight." The doctor said, "Good idea.
I'm going to contact my ophthalmologist colleague and see if there's
anything he can do for them." The engineer said, "Why can't they play
at night?"

--------------------

An engineer was crossing a road one day, when a frog called out to him
and said, "If you kiss me, I'll turn into a beautiful princess." He
bent over, picked up the frog and put it in his pocket. The frog spoke
up again and said, "If you kiss me and turn me back into a beautiful
princess, I will stay with you for one week." The engineer took the
frog out of his pocket, smiled at it and returned it to his pocket. The
frog then cried out, "If you kiss me and turn me back into a princess,
I'll stay with you for one week and do ANYTHING you want." Again the
engineer took the frog out, smiled at it and put it back into his
pocket. Finally, exasperated the frog asked, "What is the matter with
you? I've told you that I'm a beautiful princess and that I'll stay
with you for one week and do anything you want. Why won't you kiss me?"
The engineer said, "Look, I'm an engineer. I don't have time for a
girlfriend, but a talking frog, now that's cool."
 
ledmodcloseupip4.jpg


ledsworkingav7.jpg



Sorted.

Apparently all I needed was a resistor and an LED per channel.

Who'd'a thunk it? :-D
 
It's like this:


+ve input to headphones left _________________> to headphone driver +
....................................................................|
................................................................resistor
....................................................................|
...............................................................v LED v
....................................................................|
ground _________________________________> to headphone driver -


then the same for the right-hand channel.
 
What is the purpose of this mod - actually protecting the headphones, or just letting a bystander know when they're driven too hard?

The first isn't achieved with this circuit, the second doesn't seem important really..?

I've seen some Sennheiser's protected by two back-to-back 1.3W-zeners (15V?)inside - but that does not prevent them from blowing from time to time. But is better than your scheme for sure..

Jakob E.
 
[quote author="sodderboy"]The pic is fuzzy, but is that an 820R? I am game for any new (for me) tricks, but I am sceptical that this is any protection against some programmer/songwriters I know. . .
Mike[/quote]

They're 1.8k resistors, and the driver inserts are 400R.

Just about to test the system with a "before and after" approach.

I'll post my results later today.
 
[quote author="gyraf"]What is the purpose of this mod - actually protecting the headphones, or just letting a bystander know when they're driven too hard?

The first isn't achieved with this circuit, the second doesn't seem important really..?

I've seen some Sennheiser's protected by two back-to-back 1.3W-zeners (15V?)inside - but that does not prevent them from blowing. But is better than your scheme for sure..

Jakob E.[/quote]

The LEDs will be enclosed so they can't be seen, and they're intended as protection against blowing the drivers, not as indicators.
 
No... it can't work as described.

If you have a low source impedance, merely strapping things in parallel (specially when they have comparably HIGH impedances) cannot do anything to help. Like putting a 9V battery onto a 3V light bulb, it doesn't matter WHAT you put in parallel (near-short-circuits excluded of course) you can't save the bulb from being overdriven.

This can be an indicator, nothing more.

Keith
 
Nope, you're all right... after proper testing it's made absolutely no difference at all.

:mad:

It's pretty though ;-)

Any ideas?
 
[quote author="gyraf"]
Any ideas?

The Sennheisser scheme - two back-to-back Zeners across each driver? Select voltage to taste..

Jakob E.[/quote]

Would that be with the zeners facing away from each other, in parallel with the driver like the resistor/LED package in my first attempt?

edit: I guess you probably mean with the zeners in parallel with each other, one facing in each direction.
 
Two zeners in cross-parallel will act like two regular diodes, and clip at ~0.6V. They need to be in reverse-series. You still need a series resistive element for any short/shunt protection to work, otherwise voltage won't decrease, and thus dissipated power will remain unchanged.

Keith
 
[quote author="SSLtech"]Two zeners in cross-parallel will act like two regular diodes, and clip at ~0.6V. They need to be in reverse-series. You still need a series resistive element for any short/shunt protection to work, otherwise voltage won't decrease, and thus dissipated power will remain unchanged.

Keith[/quote]

In reverse series, won't it prevent signal from passing at all?
 
I'm a bit sketchy on how zeners work... do they act like regular diodes in the negative direction, i.e. no throughput allowed, but allow voltage up to a certain threshold in the positive direction, or are they "circuit-invisible" in the positive direction and only allow voltage up to the threshold in the negative direction?
 
In the forward direction (greater potential on anode), they have a 0.6V drop like a regular silicon diode. In the reverse direction (greater potential on cathode), they do not conduct until the "zener voltage" is reached.

If you connect two back-to-back zeners together across a signal line (between "hot" and "cold") they will not conduct in either polarity until the zener voltage + 0.6V has been reached. So, a pair of 5V zeners would clip positive and negative signal peaks of 5.6V.
 
[quote author="NewYorkDave"]In the forward direction (greater potential on anode), they have a 0.6V drop like a regular silicon diode. In the reverse direction (greater potential on cathode), they do not conduct until the "zener voltage" is reached.

If you connect two back-to-back zeners together across a signal line (between "hot" and "cold") they will not conduct in either polarity until the zener voltage + 0.6V has been reached. So, a pair of 5V zeners would clip positive and negative signal peaks of 5.6V.[/quote]

Yes. There is a thread debating the merits of various schemes for signal clipping from about a year ago iirc. Since zeners may not be too well-matched, and diodes like 1N4148s from the same production batch can be very close in forward drops, I advocated putting a single zener in the bridge as mentioned by NYDave earlier in this thread. It's five parts per channel plus a series resistor (unless that R is already in your 'phone feed) but at least clips symmetrically. The clip point will be about two 4148 drops plus the zener voltage. It will also be a little softer due to the two-series-diode characteristic in series with the zener, which may or may not be desirable.
 
[quote author="OlympicPhil"][quote author="SSLtech"]Two zeners in cross-parallel will act like two regular diodes, and clip at ~0.6V. They need to be in reverse-series. You still need a series resistive element for any short/shunt protection to work, otherwise voltage won't decrease, and thus dissipated power will remain unchanged.

Keith[/quote]

In reverse series, won't it prevent signal from passing at all?[/quote]

Whoah. We've taken a left turn here...

In reverse series with each other, but still in parallel with the load. -My fault perhaps for not being fully clear.

And no. Assume 0.6V forward drop and 5.0V reverse drop (four-eggs-arm-pull), two in series will drop (i.e. limit the total voltage across the series-pair) to 5.6V in either direction.

If they're in series with the load, the load will see nothing until the signal exceeds the voltage... a sort of humungous crossover distortion. If they're in parallel with the load, they'll "clip" the load by shunting excess voltage off as heat. (to the limit of their power handling capacity, and assisted by series resistance between them and the cource. The less series resistance, the more readily they'll pop.)

You must have a linear resistance in series. Only the shunt leg must be allowed to be nonlinear (semiconductor) in a passive setup. -And it really needs to be near symmetrical... At the very least it MUST act on both poles of the signal.

Keith
 
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