cdw3423
Newly Enlightened
I saw a thread on the topic of LED Camera flash but it was over 3 years ago and didn't seem to reveal much so I did some reading and thinking and this is the result.
First how much light do we need and how fast do we need it.
When I say how much light I mean light intensity on the subject for a given time period.
Since in photography less light just means you need it for a longer time.
Here is what I know, or think I know, about camera flashes.
Typically full output means the flash is on for about 1/1000 of a second.
Obviously if we want to make a camera flash with LEDs it needs to also fire for a similarly short time.
Camera flashes use a xenon arc lamp as their light source.
Wikipedia says that xenon arc lamps put out between 30 to 50 lumens/watt.
That is for continuous running xenon arc lamps, if the number is any different for a strobe, I would guess would be lower.
Studio lights are rated in Watt Seconds. Not Watts/Sec like some people have stated.
I can't seem to find a similar rating for typical portable camera flashes. However I did read somewhere that they are between 30 to 80 watt seconds.
So what is a watt second. Well since a watt is joules/sec at watt second would just cancel out the seconds and leave us with a joule.
Since I happen to have a flash (Nikon SB-26) taken apart for repair, I decided to just look at the capacitor and see how much energy it can hold. It is a 350V 1,400 microfarad capacitor. According to this web site, http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html that capacitor can store 86 joules of energy when fully charged to 350V.
Now for some math. (quick note I round of the numbers I have in here but use the non rounded numbers for further calculations, so if you pick them up in the middle, you will get different numbers.)
If my flash is 80 Watt Seconds or Joules (it can't use all of the 86 joules of energy in the capacitor we'll say that 6 is lost, probably a lot more than that is lost) and it uses that energy in 1/1000 of a second (.001sec), then the xenon arc lamp must use 80joules/.001 seconds which = 80,000 joules/sec which is watts. Seriously 80,000 watts!? That's a lot. I am going to assume that is correct. However given the way the system works, I am guessing that when you discharge 80,000 watts into that xenon lamp from that capacitor so fast there is a lot of inefficiency. Since I have no idea how much, I am going to proceed as though it does transfer most of the energy into the lamp that then creates 30-50 lumens/watt for a very short time. So lets' just take the average of 40 lumens/watt * 80,000 watts = 3.2 million lumens.
Ok now for what I know about the Cree XP-G S2.
Driven at 1 amp it puts out 367.1 lumens.
Assuming we drive them at 1 amp to get 3.2 million lumens, we need about 8,700 Cree XP-G S2 LEDs. Not at all practical. Lets take some liberties in our specifications. A camera flash delivers it's light in 1/1000 of a second. Lets say it is acceptable to take 1/100 of a second to deliver the same light. Then we only need 870 LEDs. Still ridiculous.
What would reality have to be to make using the Cree XP-G S2 for a flash?
since the LED's are only going to be on for a very short time, 1/100 of a second at most. I'm guessing that we can pack them together pretty tight. Say one every 6 mm. I think, although fairly expensive, you could justify using 20 LED's in 2 rows of 10, so an array 60mm by 12mm of 20 LED's Driven at 1 amp we will get about 7,300 lumens. Since we are going to have it on 10 times as long as the xenon arc lamp, it would have to put out 73,00 lumens. Since we already determined we are dumping 80,000 watts into that bulb that would be an efficiency of about 0.92 lumens per watt. Although I doubt the flash is putting out anywhere near 40 lumens/watt when you consider the whole system efficiency (not just the lamp efficiency), I bet it is doing better than .92 lumens per watt.
Sounds like the LED would need put out probably 10 times more light than the Cree XP-G S2 does to be practical for a camera flash with similar light out put as my Nikon SB-26.
However that said I think an experiment might be worth while and tell us more about the "actual" light output of the flash.
Setup:
Good digital camera with manual settings.
Good camera flash with manual settings.
Rig up a reflector for 1 to 3 LED's of known lumen output, so they provide a similar light pattern to the camera flash.
A room with no other light sources.
Set the camera to use a slow ISO setting, fully closed aperture, (so we need a lot of light to get our exposure correct) flash on manual at whatever setting will provide a good exposure on your subject. After the right setting is found for your exposure, record the information
Mount your LEDs in the same location the flash was. Turn them on and adjust your shutter speed/ aperture/ ISO setting till you achieve the same exposure with the LED's. It should be simple math to determine how many led's flashing for 1/100 of a second would be required to get the same light output as your flash did.
For example. Let's say you had your camera set at an ISO of 100, f16 aperture and the flash fired for a full 1/1000 of a second (have to find the specs for your flash or some way to tell how long it fires.)
Then suppose to get the same exposure with your LED's (Let's say it was 3 of the Cree XP-G S2.
If you had to set the film speed to 800 (8 times 100) The aperture to f4 (for those who don't know f4 means the aperture opening is 1/4 of the lens focal length, and f16 would be 1/16th the lens focal length and since it is a round opening you have to square the size of the opening to get the relative area of each so (1/4)squard/(1/16)squared = 16 so now we have 16 * 8 times more exposure or 128 times more exposure and then say the shutter speed was 1/10 of a second which would be 100 times more than 1/1000 so now we are up to 8 * 16 * 100 or 12,800 times more exposure. So it took 12,800 times more exposure in my hypothetical scenario. Since our flash with the same light intensity will only be on for 10 times as long we can now calculate that we need 1,280 times more light than we got from our 3 LED's.
Of course that was all made up numbers just to show how the calculations would work. However I suspect if someone were to actually do this experiment I would bet you would find you need far less than that to make an led flash that would provide the same amount of light in 1/100 of a sec that a good camera flash provides in 1/1000 of a second and you would probably get several times longer battery life. But unless the number of LED's needed is close to 20, it seems very impractical to me.
I do plan to do this experiment but with the things I have to do and the time it will likely take me to come up with a reflector that will do what I want, it will probably be some time before I get around to it. But that's ok since I am pretty sure we are going to need LED's that put out a lot more light before this will be practical.
Chris W
First how much light do we need and how fast do we need it.
When I say how much light I mean light intensity on the subject for a given time period.
Since in photography less light just means you need it for a longer time.
Here is what I know, or think I know, about camera flashes.
Typically full output means the flash is on for about 1/1000 of a second.
Obviously if we want to make a camera flash with LEDs it needs to also fire for a similarly short time.
Camera flashes use a xenon arc lamp as their light source.
Wikipedia says that xenon arc lamps put out between 30 to 50 lumens/watt.
That is for continuous running xenon arc lamps, if the number is any different for a strobe, I would guess would be lower.
Studio lights are rated in Watt Seconds. Not Watts/Sec like some people have stated.
I can't seem to find a similar rating for typical portable camera flashes. However I did read somewhere that they are between 30 to 80 watt seconds.
So what is a watt second. Well since a watt is joules/sec at watt second would just cancel out the seconds and leave us with a joule.
Since I happen to have a flash (Nikon SB-26) taken apart for repair, I decided to just look at the capacitor and see how much energy it can hold. It is a 350V 1,400 microfarad capacitor. According to this web site, http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html that capacitor can store 86 joules of energy when fully charged to 350V.
Now for some math. (quick note I round of the numbers I have in here but use the non rounded numbers for further calculations, so if you pick them up in the middle, you will get different numbers.)
If my flash is 80 Watt Seconds or Joules (it can't use all of the 86 joules of energy in the capacitor we'll say that 6 is lost, probably a lot more than that is lost) and it uses that energy in 1/1000 of a second (.001sec), then the xenon arc lamp must use 80joules/.001 seconds which = 80,000 joules/sec which is watts. Seriously 80,000 watts!? That's a lot. I am going to assume that is correct. However given the way the system works, I am guessing that when you discharge 80,000 watts into that xenon lamp from that capacitor so fast there is a lot of inefficiency. Since I have no idea how much, I am going to proceed as though it does transfer most of the energy into the lamp that then creates 30-50 lumens/watt for a very short time. So lets' just take the average of 40 lumens/watt * 80,000 watts = 3.2 million lumens.
Ok now for what I know about the Cree XP-G S2.
Driven at 1 amp it puts out 367.1 lumens.
Assuming we drive them at 1 amp to get 3.2 million lumens, we need about 8,700 Cree XP-G S2 LEDs. Not at all practical. Lets take some liberties in our specifications. A camera flash delivers it's light in 1/1000 of a second. Lets say it is acceptable to take 1/100 of a second to deliver the same light. Then we only need 870 LEDs. Still ridiculous.
What would reality have to be to make using the Cree XP-G S2 for a flash?
since the LED's are only going to be on for a very short time, 1/100 of a second at most. I'm guessing that we can pack them together pretty tight. Say one every 6 mm. I think, although fairly expensive, you could justify using 20 LED's in 2 rows of 10, so an array 60mm by 12mm of 20 LED's Driven at 1 amp we will get about 7,300 lumens. Since we are going to have it on 10 times as long as the xenon arc lamp, it would have to put out 73,00 lumens. Since we already determined we are dumping 80,000 watts into that bulb that would be an efficiency of about 0.92 lumens per watt. Although I doubt the flash is putting out anywhere near 40 lumens/watt when you consider the whole system efficiency (not just the lamp efficiency), I bet it is doing better than .92 lumens per watt.
Sounds like the LED would need put out probably 10 times more light than the Cree XP-G S2 does to be practical for a camera flash with similar light out put as my Nikon SB-26.
However that said I think an experiment might be worth while and tell us more about the "actual" light output of the flash.
Setup:
Good digital camera with manual settings.
Good camera flash with manual settings.
Rig up a reflector for 1 to 3 LED's of known lumen output, so they provide a similar light pattern to the camera flash.
A room with no other light sources.
Set the camera to use a slow ISO setting, fully closed aperture, (so we need a lot of light to get our exposure correct) flash on manual at whatever setting will provide a good exposure on your subject. After the right setting is found for your exposure, record the information
Mount your LEDs in the same location the flash was. Turn them on and adjust your shutter speed/ aperture/ ISO setting till you achieve the same exposure with the LED's. It should be simple math to determine how many led's flashing for 1/100 of a second would be required to get the same light output as your flash did.
For example. Let's say you had your camera set at an ISO of 100, f16 aperture and the flash fired for a full 1/1000 of a second (have to find the specs for your flash or some way to tell how long it fires.)
Then suppose to get the same exposure with your LED's (Let's say it was 3 of the Cree XP-G S2.
If you had to set the film speed to 800 (8 times 100) The aperture to f4 (for those who don't know f4 means the aperture opening is 1/4 of the lens focal length, and f16 would be 1/16th the lens focal length and since it is a round opening you have to square the size of the opening to get the relative area of each so (1/4)squard/(1/16)squared = 16 so now we have 16 * 8 times more exposure or 128 times more exposure and then say the shutter speed was 1/10 of a second which would be 100 times more than 1/1000 so now we are up to 8 * 16 * 100 or 12,800 times more exposure. So it took 12,800 times more exposure in my hypothetical scenario. Since our flash with the same light intensity will only be on for 10 times as long we can now calculate that we need 1,280 times more light than we got from our 3 LED's.
Of course that was all made up numbers just to show how the calculations would work. However I suspect if someone were to actually do this experiment I would bet you would find you need far less than that to make an led flash that would provide the same amount of light in 1/100 of a sec that a good camera flash provides in 1/1000 of a second and you would probably get several times longer battery life. But unless the number of LED's needed is close to 20, it seems very impractical to me.
I do plan to do this experiment but with the things I have to do and the time it will likely take me to come up with a reflector that will do what I want, it will probably be some time before I get around to it. But that's ok since I am pretty sure we are going to need LED's that put out a lot more light before this will be practical.
Chris W
