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: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
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:
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.
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.
http://www.inphora.com/LED_intensity.pdf
This study talks about the phenomena but still doesnt give probably what you are looking for.
This study talks about the phenomena but still doesnt give probably what you are looking for.
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.
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.
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.
If you control the LED with a stabilized current source, even if the voltage drop across the diode changes, the current through the diode will remain constant since the active control circuitry of the current source will compensate to keep current constant. I wonder if LED light output changes if you keep current constant regardless of voltage drop? Hmmm.
Just put a DMM set at high Z, across the cell only, (no need to power the led), and you will get a stable Z reading, then hold the body with just your fingers and you will see the Z change quite a bit.
This is why I have always said to have all 4 units together and then pot them in hard snow ski base wax.
And you can use one of these for matching your LDR's
Cheers George
This is why I have always said to have all 4 units together and then pot them in hard snow ski base wax.
And you can use one of these for matching your LDR's
Cheers George
Attachments
Last edited:
I dont recall doing this but George's finding means that the resistive material changes with temp and we already know the LED does that as well. So double edge sword.
Yikes!
Now you know why I strongly advise against putting the Lightspeed inside an amp, and to pot them in hard wax as well.
Cheers George
LDR L+R imbalance problem
Hi,
A friend of mine built an attenuator based on LDR design.
I noticed that L+R channels are imbalanced.
when the pot knob is turned up, it starts from R channel +5dBFS louder then L channel (minimum),
but when knob turning up, the imbalance gets smaller and smaller, until around -20dbFS (around middle) it's almost perfect, then Left channel (when turning up) became louder and louder - up to -5dBFs, and it's almost perfect when in full 'open' position.
I was told it is just a problem with LDR and it possibly can't be resolved, but I remember reading some audio purist using LDR att. and I don't think they wouldn't notice a 5dB difference between channels, so I presume it's possibly to have it ideal (0.1dB - I can live with, 0.2dB - would be maximum)
Could you please help me - what could cause this sort of problem ?
How to resolve it ?
Is it simply swapping a pair of LDRs or sth else ?
I really appreciate your help, here - thank you in advance
If LDR design is not perfect and it's very difficult to find pair of LDR with 0.1dB difference - could you please advise me, what type of passive att. would be good ? - in a 'human - 99%' price range
Hi,
A friend of mine built an attenuator based on LDR design.
I noticed that L+R channels are imbalanced.
when the pot knob is turned up, it starts from R channel +5dBFS louder then L channel (minimum),
but when knob turning up, the imbalance gets smaller and smaller, until around -20dbFS (around middle) it's almost perfect, then Left channel (when turning up) became louder and louder - up to -5dBFs, and it's almost perfect when in full 'open' position.
I was told it is just a problem with LDR and it possibly can't be resolved, but I remember reading some audio purist using LDR att. and I don't think they wouldn't notice a 5dB difference between channels, so I presume it's possibly to have it ideal (0.1dB - I can live with, 0.2dB - would be maximum)
Could you please help me - what could cause this sort of problem ?
How to resolve it ?
Is it simply swapping a pair of LDRs or sth else ?
I really appreciate your help, here - thank you in advance
If LDR design is not perfect and it's very difficult to find pair of LDR with 0.1dB difference - could you please advise me, what type of passive att. would be good ? - in a 'human - 99%' price range
do you claim, that buying a matched pair LDRs off a reliable seller (here's one) would sort it out completely ?
I am thinking about either giving up on this project or using other option (ALPS pot), I am not a real diy'er, so please explain;
thank you