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Carlp 14th January 2010 02:30 AM

LDR attenuator questions
I'm playing with some photo-resistors (LDRs) I have on hand in order to try building an LDR-based volume pot/attenuator (probably starting with diyparadise - all about volume control...). But in looking into the photo resistors I have, I find that the lowest resistance under fairly intense LED light is about 1.5k ohm (I've seen some optocouplers that supposedly go to 60 ohm, but I want to play with a few DIY ideas).

Given the relatively high resistance compared to other Pots, it seems to me these are likely to throw away a bunch of signal. Then again, it's rare indeed to put the attenuator/vol control all the way "up" so maybe there's ultimately no signal waste. Anyone care to help me think about this?

Also, how do I assess the input impedance when I use these in a given design? If the design calls for a 100k pot, do I need to find a photo resistor with a "dark" resistance of 100k ohm? That seems unlikely based on my research so far.

Finally, I have two photo resistors that have 3 leads, but every other one I've seen has only 2 leads (as I'd expect). Any ideas what the 3rd lead would be for? These were stashed away in my junk box and were probably purchased in the 1980s.

Thanks a ton.

georgehifi 14th January 2010 04:39 AM

Hi Carl, first is 1.5Kohm is way too high for this purpose, the one I use is the NSL32SR2S these are typicaly as low 40ohms, even at 40ohms there is quiet sound comming through the speakers at minimum level on an average system with around 90db speakers. With the ones you have you may not get down further than normal listening level.

Cheers George

Carlp 14th January 2010 05:35 AM

Thanks, George. I've looked at those and that's where I saw the 60 ohm figure, but 40's even better. So I'm still trying to get a grip on high efficiency speakers, but at 95-99 dB SPL speakers, would you say the Lightspeed attenuator would still shut down the sound level sufficiently?


georgehifi 14th January 2010 05:49 AM

Yes my friend has 100db JBL's and he get's down quite enough for background music.

Cheers George

bear 15th January 2010 01:15 PM

people seem to like them - Georgehifi makes them - they do seem to produce their own harmonics, afaik.

There are some schemes that optimize the use of LDRs - I mentioned them in passing in an earlier thread. They were used commercially in the famous Lumitron pots in mix boards.

The usual scheme is to add a switch at the end of the "throw" of the control used to adjust the volume that somehow causes a relay or switch to permit an "infinite" attenuation (completely off).

Of the many ways of controlling level, this is one.


Carlp 15th January 2010 07:53 PM

Thanks, Bear. That's a great suggestion and I'll probably play with it. I'm mostly trying to figure out how to make these work with an amp or pre that is designed for a specific pot (e.g. thus far my completed amps are 10k and 100k) since the "dark" resistance is often in the M ohms. But I've decided to just play with them and see what I find out. Should be illuminating! :D


bear 15th January 2010 09:40 PM

as long as the driving impedance is below the load impedance, you have no worries.

otoh if you put the LDRs into the front end of your amp that changes things slightly since the input Z for most actual devices is rather high. Of course that varies a bit depending on the specifics of the input stage, but it's a safe bet that it will be high enough.

myself, I'd opt for other means...


udailey 17th January 2010 05:06 PM

I have not yet tried sticking the LDRs directly on the inputs of my chip amp. Yeah yeah a chip amp is not that great but its what I am using. So, if I did this how do I figure what the impedance of the LDRs should be and do I remove all other resistors in the signal that normally go in series and shunt to the inverting input?
Try using more than one LDR in parallel for your shunt LDRs. Then you can take that min volume down farther and farther the more you parallel.

Carlp 17th January 2010 05:36 PM


Thanks for that suggestion. I was starting to think about just that when I saw your reply. BTW, I'm still struggling to figure out how to determine the input impedance for any given component, particularly amps/preamps.


udailey 17th January 2010 05:53 PM

I wonder about this to. What I have seen several times is a suggestion to multiply the source output by 10 and that would be your recommended pot value and multiply that again by 10 and that is the input amp value. However, I have not heard this to be particularly true as my cdp I think puts out 600R (this I am most unsure of) and my LDR attenuator is at anywhere from 6k to 25k as I listen to different impedances to see what I like best and my amp is at 47k. So I just picka value and listen and I have not had any audible oscillations or other bad side effects.
I have had the LDR attenuator up to 250k in the same set up and it does not sound good. Sounds like I put a bad cap in the signal. For me the lower the better as it gets brighter/crisper with lower and lower impedances. Maybe this is stressing my cdp when I go really low. Maybe Bear or George can help us understand this.

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