[robzy]

So I was thinking about the problem of large output caps, and the stress they put on headphones during start up.

What if, instead of a power switch, you used a potentiometer with a built in switch?

Have the switch wired to power like normal, but wire up the pot between the output cap and the output themselves.

That way, to turn on the amp, you turn the potentiometer all the way on. After the first click, power is turned on and the pot eats up the big voltage that the headphones would otherwise see, and when the pot is turned all the way up it's shorted out doing no harm.

Of course, it's not quite as elegant as a relay based system, but it's quite a bit simpler/cheaper.

Turning on your amp by turning a potentiometer all the way to the right might feel a bit strange, but when you think about it, it's not mindblowingly confusing

So, would this work? And - has it been done before?

Rob.

 

[S3TUP]

Wouldn't work. The pop comes from the amplifier itself no matter what is connected to it, when it turns on. And your pot would just disconnect the input of the amp, so you'll hear the pop as you always did.
Try it with the volume set low and you'll still hear the pop.

The quick and dirty way to make the anti-pop thing is using relay driven thru resistor and with capacitor connected to the coil. The relay will connect/disconnect the output of the amp with delay, so the output will be switched on after some time you power on the amp.
You'll need to play with the resistor and capacitor values to get the thing work.

I did it on some little headphone amp and it works pretty well.

A better way is to put some transistor in between to turn the relay on, but it gets more complicated. Tou can put some 555 timer too, or opamp, or microcontroller, or... but the relay stays there in schematics.


The more "mechanic" way is to use 3-position switch, in which the first position is off, second is power, and third is power+output relay "on" signal.

3-position switches are used widely in tube amplifiers where you have to turn on the filament power and heat the thing up before turning the whole amplifier, to prolong tube's life.



But the built-in knob switch is a nice way keeping front panel clean of additional controls. Too bad there is no quality ALPS doing that...

 

[robzy]

Quote:

Originally Posted by S3TUP /img/forum/go_quote.gif Wouldn't work. The pop comes from the amplifier itself no matter what is connected to it, when it turns on. And your pot would just disconnect the input of the amp, so you'll hear the pop as you always did. Try it with the volume set low and you'll still hear the pop.

I think you misunderstood what I meant... or rather, I wasn't clear in what I meant.

The pot will be connected on the output of the amp, directly before (in series with) the headphones.

This is so that when the amp is first turned on, there is the equiv. of a 50k resistor in series with the headphones - eating up all the excess voltage.

Rob.

 

[S3TUP]

well, than it's ok,
Do you plan to control the volume with the same pot?

 

[tomb]

Quote:

Originally Posted by robzy /img/forum/go_quote.gif I think you misunderstood what I meant... or rather, I wasn't clear in what I meant. The pot will be connected on the output of the amp, directly before (in series with) the headphones. This is so that when the amp is first turned on, there is the equiv. of a 50k resistor in series with the headphones - eating up all the excess voltage. Rob.

I understand what you mean, but no offense - I think you misunderstand the real issue you're trying to protect your headphones from.

It's not the output caps that's causing the stress. It's the DC voltage that's present before they charge up. A charged cap protects your headphones against that DC. IOW, if you could have the cap charge instantly, you would - that way your headphones would be protected immediately.

A headphone relay-delay is there in more sophisticated amps to give time for the output caps to charge. They cutout the circuit to the headphone until such time that the output caps have fully charged.

Also, the only thing that will charge the caps is current to ground. So, for your method to have a chance, you'd need to put the resistance between cap terminal and ground (note that any amp with output coupling caps already has resistors from the caps to ground). Otherwise, if the pot was in series with the headphone, the caps would never charge and your headphones would still be subjected to the voltage when you turned the pot up to zero resistance - until the caps had time to charge (through the headphone). Your headphones would again be exposed to the voltage until the caps charged.

If the pot is placed from the caps to ground, the caps would charge gradually according to the resistance of the pot - just as you propose. However, this would mean that the headphone is actually in parallel with the pot resistance and would again be exposed to the voltage by virtue of it's much lower resistance pulling more of the current compared to the higher resistance pot.

No, it won't work.

 

[linuxworks]

a turn-on delay circuit is really the best way to deal with designs that 'pop' or have turn on/off transients.

 

[Juaquin]

Small signal relays are cheap (plus a couple resistors/caps) - no more than a good pot with switch.

 

[robzy]

Quote:

Originally Posted by tomb /img/forum/go_quote.gif Also, the only thing that will charge the caps is current to ground. So, for your method to have a chance, you'd need to put the resistance between cap terminal and ground (note that any amp with output coupling caps already has resistors from the caps to ground).

Ah, gotcha, this makes sense.

Rob.

 

[robzy]

Quote:

Originally Posted by tomb /img/forum/go_quote.gif Also, the only thing that will charge the caps is current to ground.

Wait a second, if that is true, then what's the logic behind unplugging headphones for start up? The caps still won't be charged and will create a nasty spike when you plug the headphones in, right?

Quote:

Originally Posted by S3TUP /img/forum/go_quote.gif well, than it's ok, Do you plan to control the volume with the same pot?

Nope, not at all. During normal amp function it's turned all the way up, thus shorted.

Quote:

Originally Posted by Juaquin /img/forum/go_quote.gif Small signal relays are cheap (plus a couple resistors/caps) - no more than a good pot with switch.

But in this case you don't need a good pot Any old cheap one will do.

Rob.