Hi, I was wondering if you use a cdS cell (cadium sulfide resistor) in a circuit, does the circuit consume any power when it's in a completely dark environment? Does the battery get wasted even though the circuit is completely in the dark? I am thinking if the cell is used with a relay to control on/off power to the circuit.
Thanks,
FL
Since the dark resistance is lower than infinite ...
Rene
-- Ing.Buero R.Tschaggelar - http://www.ibrtses.com & commercial newsgroups - http://www.talkto.net
Most cdS cells have really high dark resistance, many megohms. So depending on the voltage across it, the dark current is likely to be quite low, microamps or so.
Why not measure the cell with an ohmmeter and see (well you wont be able to see it, hmmm......) ?
The dark resistance is usually pretty high, so dark current should be very low. But you have to think about all light levels between darkness and the brightest the light can be.
A battery saving approach might be to use a small (low current CDS cell or photo transistor and a comparator (i.e. LM393) to measure the light level and make the on off decision to operate the relay.
Since there are some thick dudes in this newsgroup who are still suffering from the delusion that since you posted a question you know what you are talking about and will then waste their time giving you some answers.....
What the f*ck are you talking about?
DNA
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no, but to get completely dark you'd need to cool it to absolute zero (to exclude all infrared)
if you just mean in a dark room or closed box it will have finite resistance and,therefore, conduct a little.
the full question though depeds on ther circuit. I have a nightlight with a CDS cell, and that circuit consumes more power in the dark than in the light.
also, a typical CDs cell won't conduct well enough to operate an ordinary relay by itself.
Bye. Jasen
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