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LED based safelights
When I started enlarging recently I bought a second hand Photax safelight, the bulb popped the 3rd time I switched it on . As I work in electronics and had access to a massive stock of LEDs I decided to make a safelight by wiring up a large handful of red leds. This seems to work well enough, they have a very narrow spectrum of light etc. With my recent haul of paper some of it requires a Kodak OC 'light amber' safe light, this got me thinking. kodak OC appears to be a 580nm filter and I believe I have a amber LEDs with I believe 580nm wavelengh, so any reason not to use them?
Does anyone else use or make safelights using LEDs? |
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We make hand-held safelights, "SafeTorch", in both red and amber variants. The amber ones are 589nm which is the minimum sensitivity for RA4 paper as well as being safe for b+w.
I have a very bright red LED lamp in the darkroom which appears to be very safe despite the level of illumination. |
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When Ilford was bought out there was a LED safelight on ebay which I bought. It was made by Ilford but never put on the market, or so I was told by the man I bought it from who was employed in darkroom R&D, as it was deemed far too expensive to produce. I have never had any problems with fogging with any type of paper & the light output (variable) is far better & brighter than the normal safelight. Highly recommended.
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Are you able to provide us with the type/model of the LED that you are using so that others may adopt them if they wish?
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I have some red and amber LEDs used in car rear light clusters which are very bright. Despite my best intentions, I haven't tried them as safelights yet. Steve. |
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I'll post pics of my red safelight when I'm in work next, basically its about 200 (that is a yorkshire metric handfull) square red LEDs inserted into a breadboard and you then need about a 3v power supply which could be some simple transformer and rectifier (you could wire the LEDs in series parallel and get something to work off 12V if you liked), I'm using a bench psu from work. Mine was cobbled together with stuff I had at work that no one was using. I have about 20,000 3mm amber leds too which is why I thought about the RA4 safelight as I've just inherited the RA4 paper in my haul yesterday.
Rather than Richard divulge his LEDs as he is running a business selling them I will find out what my ones are next week and let people know. I'm sure SOMEWHERE at work we must have a means of measuring the wavelength of the LEDs I have. Rapid Electronics are my prefered supplier for LEDs as they seem to be the best value, you may find suitable datasheets on their website to tell you the wavelength of the different LEDs they sell. I have also thought it would be neat to have red LED illumination in my enlarger head for focusing without having to use a red filter. that way when you hit the focus button you are guaranteed safe red light. |
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LED Lesson No. One:
If you remember Ohm's Law (assuming you knew it already) then it will be useful when sorting out how to wire up LEDs. A red LED needs about 1.7 volts. Rather than create a power supply specifically at that voltage, LEDs are wired up with a resistor in series. If you have one LED and a power supply of 5 volts then you subtract the 1.7 volts across the LED from 5 leaving 3.3 volts across the resistor. This is where Ohm's Law comes in. The value of that resistor sets the current through itself and therefore, through the LED (as they are in series). So if we want 10mA of current, we need to drag the Ohm's Law equation from the back of our brain and remember that it is: V = I x R Re-arrange it for resistance and it is R = V/I So for 10mA, the resistor needs to be 5/.01 (current in amps, not milliamps). This is 500 ohms. The closest standard resistor is 470 ohms which is close enough. If you want a bit more current e.g. 20mA then either re-calculate or halve the resistance. In this case the closest standard resistors are 220 or 270 ohms. Either will do. For LEDs in series, just add their voltages. So for three red LEDs in series, the voltage required will be 5.1 volts. Obviously a bit more is needed to drop across the resistor so a higher voltage power supply will be needed. e.g. 12 volts. For this, just treat the three LEDs as one LED, work out the voltage drop then what the rest of the voltage across the resistor will be and do the maths again. Did that help or hinder? Steve. Last edited by Steve Smith; 28th April 2011 at 03:32 PM. |
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