| Electronics > Beginners |
| Opto attenuator, will it work? and more... |
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| FriedMule:
First, the attenuator are ment as a "do not touch audio" attenuator and should be fairly independent of the components audio quality on the diode-site of the circuit. But will it work? Second I have tried to design a simple PSU that shall be used for many light tasks in my DIY amplifier. What I do not know is how the PSU part would react with different loads. i.e. the attenuator on +10V and GND, an other load between +10V and -10V, a third load between some other combination. All with different resistance and different current needs. |
| JohnnyMalaria:
Have a look at this: http://diyaudioprojects.com/Solid/DIY-Lightspeed-Passive-Attenuator/ The important thing is that the resistance/current response between devices is sufficiently variable that you'll need to buy many devices and find two that have close responses. |
| FriedMule:
Great, it looks like the original post that inspired the description I found. So it would work? What about the PSU question? Oh, forgot to ask abaut the grounding of the audio signal? |
| T3sl4co1l:
More or less. But: 1. Can you still get vactrols? Must not be many out there. Fine for one-offs, even if you need second hand parts to do it, but keep that in mind in case anyone else wants to try it! 2. You may want a regulated supply, as the supply ripple modulates the output. 3. Dual gang log potentiometer is probably not what you want, for a couple of reasons. One, the logs don't track (they aren't opposite). Two, you need one exponential (the normal kind, actually; why they're called log, well, who knows), one log (the inverse of the usual curve!). Three, they need to go in opposite ways (give or take pinout). Fourth and final, the vactrol response itself is probably not proportional, so you ideally need something even different from that. This is more of an argument about subtle aspects of the control: how smoothly the attenuation follows the setting, and how stable the load side impedance is (which affects bandwidth, and maybe gain of the source/load as well). The overall response should be about right. But it may be sloppy enough that, for example, you find better results with a linear than log type pot, and still that it's not really all that nice. The "duh"-iest question, though: You're using potentiometers. Dual ganged. You're adjusting stereo signals (in the linked case, not in the OP case I guess?). What's wrong with putting the signal through the stupid pots in the first place? Photoresistors are unquestionably worse than pots for signal quality. Their distortion is surprisingly low (except for photoFETs, but those aren't being used here), but, not as low as a normal resistor. Tim |
| FriedMule:
Hi T3sl4co1l thanks for your reply, some of it did go way over my head but I'll try to respond:-) --- Quote from: T3sl4co1l on June 05, 2018, 06:57:02 am ---More or less. But: 1. Can you still get vactrols? Must not be many out there. Fine for one-offs, even if you need second hand parts to do it, but keep that in mind in case anyone else wants to try it! --- End quote --- It looks like DigiKey still sells them: https://www.digikey.com/product-detail/en/luna-optoelectronics/NSL-32SR2/NSL-32SR2-ND/5039808 --- Quote from: T3sl4co1l on June 05, 2018, 06:57:02 am --- 2. You may want a regulated supply, as the supply ripple modulates the output. --- End quote --- I properly may but here are we way over my head:-) I'll try to answer and hope that you bare with my "answer". The transformer are way over sized for it's needs. So I think that it will not dip voltage or current. The components I use can operate between 5 and 48 voltage (i think) so I hope that it will not effect anything to use a normal supply. --- Quote from: T3sl4co1l on June 05, 2018, 06:57:02 am --- 3. Dual gang log potentiometer is probably not what you want, for a couple of reasons. One, the logs don't track (they aren't opposite). Two, you need one exponential (the normal kind, actually; why they're called log, well, who knows), one log (the inverse of the usual curve!). Three, they need to go in opposite ways (give or take pinout). Fourth and final, the vactrol response itself is probably not proportional, so you ideally need something even different from that. This is more of an argument about subtle aspects of the control: how smoothly the attenuation follows the setting, and how stable the load side impedance is (which affects bandwidth, and maybe gain of the source/load as well). The overall response should be about right. But it may be sloppy enough that, for example, you find better results with a linear than log type pot, and still that it's not really all that nice. --- End quote --- Yes I see (I hope) It will give a lot more sense to use linier pot instead of log! :-) --- Quote from: T3sl4co1l on June 05, 2018, 06:57:02 am --- The "duh"-iest question, though: You're using potentiometers. Dual ganged. You're adjusting stereo signals (in the linked case, not in the OP case I guess?). --- End quote --- Way over my head again:-) Both pot are on the same turn-pin, hope that it answers your question? --- Quote from: T3sl4co1l on June 05, 2018, 06:57:02 am --- What's wrong with putting the signal through the stupid pots in the first place? Photoresistors are unquestionably worse than pots for signal quality. Their distortion is surprisingly low (except for photoFETs, but those aren't being used here), but, not as low as a normal resistor. Tim --- End quote --- I have several reasons for using LDR instead of pot (do not know if they are good:-) 1) I can use a pot that I like, without having to think of it's "sound". 2) The attenuation is in mid circuit, without any wires in the signal path. 3) When the pot gets old and noisy, it wont be audible in the speakers 4) The high praise it gets form audio listeners: http://www.dms-audio.com/lightspeed-attenuator and http://diyaudioprojects.com/Solid/DIY-Lightspeed-Passive-Attenuator/ 5) Funny to try to do old things in a "new" way. :-) |
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