Need help with infrared flashlight, please take a look

[dr pepper]

Yes that looks like a professional solution.
Note the 1000ma version needs a minimum of 6v supply, so you'd be better powering this from a 7.2v battery pack, the good thing about using a switching reg like this is that its much more effecient, ie your batteries will last longer than if you used just a simple resistor, and the brightness of the led will be constant.
You could put a 1k ohm pot acrross 0v and 7.2v supply, and connect the wiper to the analogue dimming i/p on the board, then you would have a brightness control.
To be honest I'd forget the icop board, it looks like an old inefficient design, and those electrolytics have probably gone high impedance.
I'm near certain that your device was nothing like the image you posted, I think it was nothing more than a glorified torch/detector probably only capable of detecting movement.

 

[rumpfy]

Hi Proto,
Considering all the stuff you have researched and the postings till now, my recommendation is to go with what Dr. pepper says. I note that you seem to be more into photography than electronics, and you seem to have a need to get a result.
The SFH is a 1 amp diode and the driver you picked will drive that diode to its full brightness capability. However, dr. pepper notes in post 37 that the diode needs a heat sink. Using the driver to drive the diode at full power, the maximum power dissipation in the diode would be 3.4 watt. This heat will need to be dissipated through the mounting lugs of the diode; presumably through BOTH connecting lugs. The thermal resistance of the total heatsink area will have to be not more than 23 degrees C per watt. If you need more help with this aspect of your project, then get back to us. You dont want to cook your new diode so you'll need to keep it cool. There is some published data on the thermal resistance of printed circuit boards and your problem is not difficult. From memory, a piece of normal PCB material of about 1000 mm square has enough thermal conductivity. So two pieces about 15 mm x 30 mm will possibly do the job.

 

[protocol217]

ok, i have one more question: i already have a driver that gives 800 mA so if my led can work on that amperage it wouldn't be heated to maximum.
so the question is can my led work on 800 mA and if yes, what would be the % percentage of the glow ?

and if it can not work on 800 mA i will buy the 1A one that will drive the led to 100 %

 

[rumpfy]

Presumably the driver you have will limit the output current to 800 mAmp. This combination is OK cos the diode is more capable than the driver. ie the driver cant really overdrive the diode. Nevertheless, you have a meter to check the actual operating current thru the diode.
The Osram data sheet for the SFH on page 5 shows several graphs. One graph is the 'Total Radiant Flux' and this shows radiant flux at a diode current of 1 amp as a base and then plots the 'RELATIVE' light output at other currents. At 1 amp (ie 10 to the power '0') the relative radiant flux is 1; ie (10 to the power '0'). At a current of one tenth of an amp (10 to the power -1) the RELATIVE power is one tenth of the power at 1 amp. So the diode is quite linear in its light output versus diode current.
At 800 mAmp you will have 0,8 of the light output compared to that at 1 amp.
There is another graph on page 5 which is of interest. This is the 'Permissible Pulse Handling Capability'. This graph shows how to drive the diode in a pulse mode. The square wave in the top RH corner shows D=tp/T so if tp equals T then the diode current is ON all the time and the graph line for D=1 shows a horizontal line at If = 1 amp. If you wanted to run the diode in a pulsed mode to give three times the peak light output compared to the DC mode, then looking at the curve for D=0.33 shows that the peak current varies with the pulse ON time tp. For tp = 10 to the power -5, (0.00001 seconds) the peak current is about 2.3 amp. If the ON time is longer at 0.01 second, then the peak current is only 1.5 amp peak.
For the first case at tp = 10 * -5, the repetition frequency is 30 kilo hertz, and in the second case the repetition frequency is 3 kHz. The first case will give a peak light of 2.3 times the 1 amp DC value, and in the second case the light output will be only 1 1/2 times the 1 amp DC value.
One thing I neglected to mention was that with the ICOP thing, you already had the mechanical arrangement. With what you want to do now, you have a bit of mechanical stuff to do and mostly this involves the heat sink design. There was a good design brief from (I think) texas Instruments. I thought I had it, but cant quickly retrieve it. It was about using solderable types of IGBT transistors on PCB material.
hope this helps.

 

[dr pepper]

Try it, I bet it works.
So long as you have a reasonable chunk of ally in contact with the led it should be cool enough, try a piece of 1 1/2 x 1 1/2 x 1/4 ally angle on the bench, see how warm it gets.
Dont forget to look away from the led.

 

[Menticol]

hello Protocol, on this video you will see how to use your new multimeter, hope it may help

 

[protocol217]

OK, today my OSRAM - SFH4235 - IR EMITTER diode arrived after 30 days. and i bought this driver for it:

it has two trimmers one is for voltage, one is for current, the man i bought the driver from set it to output 3 volts and 800 mAmp output current, when i attach the diode to driver it gloves but when i put my multimeter between one of the outputs to measure current that the diode drowse i get 140 mAmps and rising slowly ( slowly i mean 1 mAmp in 5-10 seconds ).
is that good or should i get a reading like 800 mAmps.

i installed the diode in a white led flashlight housing with magnifying glass for focus. the diode is glowing like my build in diode in my night vision, maybe a bit darker but very little. i thought that it would be much brighter. is 140 mAmps on my multimeter mean that it is working with just 15 %.
and should i buy an ampermeter to measure directly the output current from the driver.

 

[dr pepper]

The voltage o/p with the diode disconnected wants to be higher, the circuit cannot regulate otherwise.
Dont adjust the current if your sure its 800ma, however you need to increase the volatge to over 3v with no load so theres headroom so the module can have enough spare voltage to regulate.
Be very carefull not to blow the led, using a resistor as a load might be a better idea, 3.3 ohm at 5 watts is probably a good dummy load, you can probably find one on a scrap board or series parallel some to get a similar value.

 

[protocol217]

i increased a voltage to 3,4 volts but its not glowing better, its the same, i think a diode like this should glow better than my build in nigh vision ir diode.

can you explain more how that works with a resistor, what battery should i use and where to put the resistor..

 

[rumpfy]

Will try to help:
1. What exactly did you buy; OR is there an instruction leaflet for it. Reason I sak this is because it is normal to have only a current limit for a diode and not both voltage and current. So lets see what the unit you bought actually does. There must be a procedure for setting the volt and amp control trimmers.
2. If the SFH is a IR led, how can you judge the light output by 'looking at it'. If its IR you cant really see it, can you?
3. Your said that, 'without the multimeter the current was 800 mA, and with the multi meter, the current was 140 mA'. Is this exactly what you said? It is possible that this is true, but if it is true, then it suggests that the multimeter has had an affect on the current because the small voltage drop due to the multimeter has reduced the diode voltage enough to give a large current change. This could happen, but it would suggest that the PCB unit is delivering a constant output voltage and that the current limit circuitry is not set to deliver a 'constant current'. The voltage drop due to the connecting of the multimeter could be around 0.2 volt. According to the SFH data sheet, at a current of say 500 mA, if you reduced the diode voltage by 0.2 volt, this could change the diode current to say 50 mA. So you need to be aware that changing the series resistance in the diode circuit can change the operating conditions under some conditions of setting the PCB unit. Using a different ammeter wont really change the situation; just stick with your multimeter.
4. Have you mounted the diode properly on a good heat sink? is there a photo?
5. What is the purpose of the LED on the pcb?

 

[dr pepper]

I cant improve on rumpfy's comments, your driver will probably be ok we need to nkow more on what you have as requested.

 

[protocol217]

i managed to get the diode going about 90 % it is much brighter now. the man that i bought the driver from helped me. the diode is now attached to the aluminum heat sink with some glue and thermal paste.

i can see the ir light with my , i manged to focus the light to be a square of light. i mounted the driver and the diode with heat sink into my old icop housing, the driver is tho big for anything else, now i am using it with my night vision binoculars, works great when total darkness. it is ruining of two 9 v batteries connected in parallel.

Thanks for all the patient and the effort you all put in my project.
i will post the photo of a finished ir flashlight.

 

[rumpfy]

Well done proto,
Can I finally suggest you look at the data sheet for the diode. On page 7 of the Osram SFH 4235 data sheet, there is a drawing of the 'Recommended Solder Pad Design'. Can you look at this and try to see how they conduct the heat out of the diode. One path is through the connecting leads but there seems to be another path from directly under the centre of the diode. There is reference to 'Thermal Enhanced PCB' which is just 'PCB with thick copper material'. There is another comment; 'Anode and Heatsink are electrically connected'.
Just go over this part of the spec and understand it. I am not sure EXACTLY what they mean by 'anode' and heatsink', but the drawing shows '3 solder points'. Itys not important to do the mounting exactly as they say but it is important to understand where the heat flow is occurring and to get YOUR heatsink in thermal contact with the required parts of the diode.
AND, just check the LED current.

 

[dr pepper]

Excellent, I'm pleased too, you'll have much more power from this than the original icop jobby.
90% is also a good idea, a little headroom isnt a bad thing.
Do you have any pics of the gadget in action?
Your driver will pull every last joule form the batteries which is a good thing you'll get your moneys worth, take care though when your batteries are totally depleted they might leak.