That is not the biggest issue, light sensitive switch would resolve it.
The second part of my post is the issue. The calculations for resistor values are not valid for ~100mA draw and even if they were, after temperature raises they are not anymore. There is a reason for that LEDs are normally driven with constant current source, not with resistors.
But you don't even know the failure reason of your original setup, I wouldn't be surprised if it was just heat. Look at the board, do you see any evidence?
Well, when I first pull out a board, I notice this. It looks like some condensation (glue?). But I nothing feel it when touching it with my fingers, so it seems not wet anymore.
The second thing is that maybe power in my house is not stable? We have a house and a few rooms outside. The bulbs burn out very often, like 1-2 bulbs per month. But other devices are fine. And the camera has own adapter, which I think helps to hold constant voltage.
Or why we just don't accept the fact that the end must come?
What, how? Worth it?
I gave my opinion in post#76.
Your chances to rebuild that illuminator better than the failed one are next to nothing.
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Oh, no, no externar illuminator please:)
I was thinking a bit about LDR, we can use it by avoiding camera lens (lens and LDR working together). For that need to cut his two connector wires and LDR normal install in circuit like in this video
I was tested today a "dark sensor". Using LDR from other broken camera.
Started with light sensor. That sensor works inversely of what I want-it turns light ON when is covered.
Then I spend all day to figure out how to reverse direction of lighting (that illuminate when sensor is covered) trying to found corresponding transistor and resistor.
EDIT:
2.5v battery.
LED (2~2.5v)
BC547
R1K (for LED)
R100K (for LDR)
You said it is possible that picture will have noise, but if we avoid lens sensor, I hope not. Need to test.
After I am waiting LEDs... 
I spent some time figuring out how this small scheme with one LED and LDR including to my project.
This scheme...
...into this scheme.
Can somebody tell me, did I do it right?
I will remove all wires from connector and cut links to LDR on the Board, so I can free work with it.
Don't look yet R value of resistors, that will be calculated later.
I finally received a LEDs !!!
I think that 5 mm LED will fit on the board? I was afraid that will not and that will be bigger.
In meantime, I bought a bounch of 0.25W blue resistors.
What is happened that I put 5mm LED and 22+47+10R resistors. I felt smell. Touched a resistor and they were very hot. Then I remembered that I must use 1W, like is Paul says, and not 0.25W.
Please someone confirm did I put small scheme with LDR correctly in my big scheme? (post #106) to be able to continue.
Look at the BC547 datasheet, Absolute maximum ratings and compare them to your setup.
Something popping up?
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I see from the LINK that BC547 base is 50v and emmiter 45v collector current 100mA, collector cut off current 15 nA.
My 5mm LEDs are 100mA, if something have to do with it. But they don't have to run on 100mA, maybe.
Hi,
When you have the LEDs on, measure the voltage across the series resistor and do some Ohms Law to work out the current flowing and the power dissipated in the resistors.
Tom.. 
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And if you have four of them?
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Thanks TomGeorge for visiting my Thread 
@kmin Four LEDs or BC547?
Can I use one BC for 4 rows LEDs by 6 in series?
Yesterday all day I had a problem with hidden issue - loose soldered wire.
I made a test.
- 12v (3x18650)
- one row of 6 LEDs in series
- 1x BC547
- R1200M for LDR.
- R47 (1W) for LEDs
And that works as I imagined. When I cover LDR, LEDs power UP. No heating, for now.
Current
Not sure is that right place to measure (between collector BC547 and last LED), but this place gives me only 4.5mA (LEDs are 100mA).
How much current is best to run there and how?
I know that all depends on LDR and how may light he catches. But I cover it with cap, so think that drop out 100% of current.
This is first row. Three to go. 1x 5mm and 3x 3mm LEDs, which will be just 200mA.
One series string with 6 leds at 12V should draw ~100mA +-30% (depends on actual battery voltage and actual LED forward voltage). If you have 4 of them, the current draw is ~400mA which is way too high for your transistor.
Try to measure without switching circuit first to confirm your 6S led string current (with series resistor little bit higher than 47ohm for safety margin).
Also, find a 12V power supply. Lithium batteries are dangerous. Tiny mistake and you burn your house. Also 3S battery voltage can be whatever between 9V and 12,6V, which is complicating your calculations.
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Sorry no, just measure the voltage across R1 or R2, that will give you the LED current per leg.
Multiply that value by 4 to give total LED current.
Tom...
Beginner language: Measure voltage across resistor and divide it with resistor value. For example 5V across 47ohm resistor gives you 5/47= 0.106A (106mA) current flow.
I must correct myself. It appears that base is 5v! Is that problem?
Not will be four. Will be 2 rows of 100 mA and 2 rows of 20mA.
Oh.. that are bad news 
And what you suggest me, how to solve it?
-I have more those transistors ready...
-I can order 5mm 20 mA and wait again. But still will be over.
-Some other BC's?
LIST_of_transistors
It looks that BC639 & BC640 have 1.0A, but not sure for other characteristic.
TEST:
Have only R47 + R3.2 (1W) at home right now.
Found adapter from camera, 2A. Multimeter show 12.25v, which is little over (maybe my camera burned because of that!!?).
Without circuit:
Current: I got max 79mA, which looks good.
Volts: why working on 8 volts? Is that mean that LEDs are not 1.20v?
LEDs are very bright.
Within Circuit:
Current:
-when I touch ground direct from battery, is like above, under 80mA
-when I touch ground from collector (BC 547) (trough circuit), get reading like before, only 5mA. LEDs are dim..
That is bad.
When there is light, show 0mA. On dark 5mA.
Within circuit, only 0,25 volts Resistors 50ohm.
By now, you should know that you need to limit the current through your "100mA" LED because otherwise it burns. It's up to you if you limit it to 100 or 20 mA. In your circuit you do it with series resistor.
No need for 20mA leds, 20mA is just the max current it can handle continuous.
Common BC337 should be able to handle 240mA (2x100+2x20) current (if heat can be dissipated). But you have to saturate it, which I expect you don't do with your actual switch circuit.
Difficult to follow your tests. Make just one, post the circuit and results.
If you test randomly without knowing what you do, you burn your LEDs and your third multimeter. Forget your switch circuit for now and work with your led circuit until you understand it completely and get results that make sense.
What volts??
Ps. my post#104 is still valid, just for learning purposes... 
It's time for you to find out what the forward voltage of your LEDs actually is. Six of them in series is probably very close to the voltage of your 12V power supply. If that is the case then you will need to reduce the number of LEDs in your series strings.
Should be 1.2-1.7V, but one careful consistent test setup would give all information needed.
Oh, that is it. So they are practically the same, just with limit.
LED 20mA at full (20mA) lights up the same like 100mA LED at 20%.
It's strange that they cost the same.
Haha, that was funny
Right picture above. On multimeter, 8.06v across 6x LEDs in series. There are also 50ohm resistors and a 12v adapter (12.25v). If LEDs described as 1.2v, must be 7.2v.
I thought that I was too clear in my test 
If some picture not clear, just ask, I will describe it. But I will go slover, like you said.
Easiest way would be buying that BC, But what this mean?
Also guys, I have some cheap buck step-down voltage device if I need some accurate voltage.
They have fvd1.2V at certain conditions, in practice you will never have exactly that.
That you need to complete the first step before going to the second one..
Post setup with 12V PSU, 6x 5mm LEDs and series resistor. Voltage across the resistor and voltage across one led.
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