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Old 25th October 2011, 08:24 PM   #4281
benb is offline benb  United States
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Quote:
Originally Posted by portreathbeach View Post
Thanks Uriah.

I don't necessarily want to work out the dB difference, but I was curious as to how it is worked out. I found a couple of equations for dB power and dB voltage. The dB voltage equation is : dB = 20 x log10 (v1 / v2)

I should hopefully get ordering my parts this week
Where R1 is the top resistor and R2 is the bottom resistor of a resistive divider, the voltage ratio generated by the divider is R2 / (R1 + R2), so the dB attenuation is:

dB = 20 x log10 (R2 / (R1 + R2))

Hope this helps.

ETA: This assumes a high impedance load, otherwise the load has to be included: the load resistance would be in parallel with R2, and that parallel combination would be the new value for R2 in that equation.

Last edited by benb; 25th October 2011 at 08:26 PM.
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Old 30th October 2011, 09:24 PM   #4282
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Thanks benb
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Old 5th November 2011, 04:54 PM   #4283
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So the LDR changes resistance as temperature varies.

Can anyone tell me the reason for the change? Is it the LED changing brightness due to change in temperature, or does the resistive material change resistance when the temperature changes? Or something else happening?

Thanks.
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Old 5th November 2011, 05:16 PM   #4284
udailey is offline udailey  United States
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Measure voltage and current across your LED then change the temp of the LDR and you will know. I think its the resistive material.
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Old 5th November 2011, 07:36 PM   #4285
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Originally Posted by udailey View Post
Measure voltage and current across your LED then change the temp of the LDR and you will know. I think its the resistive material.
I don't see how this will tell me if it's the LED or the resistive material that is responsible for the resistance change. I was hoping that the company had published a technical paper, or something.
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Old 5th November 2011, 08:11 PM   #4286
udailey is offline udailey  United States
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Because if you measure voltage and current and then change the temp and measure again you will know if voltage or current is changing with temp and this will tell you if its the LED.
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Old 5th November 2011, 08:34 PM   #4287
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Originally Posted by udailey View Post
Because if you measure voltage and current and then change the temp and measure again you will know if voltage or current is changing with temp and this will tell you if its the LED.
Perhaps I'm picking nits; my thought (given I know nothing of the technology of diodes) that perhaps an LED could become more or less efficient in converting a given amount of current to light at a different temperature. So my question would still be unanswered.

Update: Your comment caused me to actually do some research, and I found this on Wikipedia:

" Like other lighting devices, LED performance is temperature dependent. Most manufacturers’ published ratings of LEDs are for an operating temperature of 25 °C. LEDs used outdoors, such as traffic signals or in-pavement signal lights, and that are utilized in climates where the temperature within the luminaire gets very hot, could result in low signal intensities or even failure.[41]

"LED light output actually rises at colder temperatures (leveling off depending on type at around −30C ... "

But that doesn't answer the question because it doesn't say how much it changes as temperature changes and it doesn't address the specific LED used in the LDR device, and it of course doesn't address the resistive material at all which may, or may not, add its own temperature variation to the mix. Sure wish the company would publish a factual paper on this issue.
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Old 5th November 2011, 09:11 PM   #4288
udailey is offline udailey  United States
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http://www.inphora.com/LED_intensity.pdf
This study talks about the phenomena but still doesnt give probably what you are looking for.
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Old 5th November 2011, 09:17 PM   #4289
udailey is offline udailey  United States
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Wikipedia under just DIODE rather than LED says this:

Temperature measurements
A diode can be used as a temperature measuring device, since the forward voltage drop across the diode depends on temperature, as in a Silicon bandgap temperature sensor. From the Shockley ideal diode equation given above, it appears the voltage has a positive temperature coefficient (at a constant current) but depends on doping concentration and operating temperature (Sze 2007). The temperature coefficient can be negative as in typical thermistors or positive for temperature sense diodes down to about 20 kelvins. Typically, silicon diodes have approximately −2 mV/˚C temperature coefficient at room temperature.
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Old 5th November 2011, 09:22 PM   #4290
udailey is offline udailey  United States
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If you could figure out the tempco of the diode you could counter it with a parallel thermistor. I tried this with moderate success a few summers ago but didnt see a reason to continue. L and R LDRs, if matched, are going to act the same way in the same temp so we should not detect a change in the sound as temp changes, unless of course its inside a Class A or near a tube which might be extreme enough to change audibly.
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