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UV Lamp (my variant)

kaspiysk113

New Member
Assembled a lamp for lighting a photoresist of 42 (6x7) UV LEDs.
The LEDs are turned on in parallel, the quenching resistor is set in series to the matrix, 10 Ohms 10 Watts.
Power supply 9V-12V.
It works perfectly, but at 12V. it is better to light for no more than 1 minute, since the resistor is heated.
 

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picbits

Well-Known Member
Maybe use two or three LEDs at a time in series with a current limiting resistor per pair . Then parallel these banks together. Using LEDs in parallel is bad practice unless they are perfectly matched and generally on the same heatsink / die.
 

gophert

Well-Known Member
Most Helpful Member
If you'd do it with 14 groups of 3 LEDs, with 14 resistors at 150 ohms, you'd be giving up about 1 watt of heat across the whole board (70 mW per resistor).

your way, I think it will be about 8 watts on one resistor. 9 volt drop across the 10 ohm resistor when using a 12VDC supply.

do it this way (you may want to buy new UV LEDs and start over). Also, you may notice your eyes getting sore. They are "just" LEDs but the light is still UV. Wear appropriate welding goggles (not just UV sunglasses).

71C28926-9C29-445C-B309-BBEC15BEEB1B.jpeg
 
Last edited:

kaspiysk113

New Member
Maybe use two or three LEDs at a time in series with a current limiting resistor per pair . Then parallel these banks together. Using LEDs in parallel is bad practice unless they are perfectly matched and generally on the same heatsink / die.
thanks for the comment, but I'm aware of it
it was just interesting to assemble the matrix exactly like this, with a single resistor
of course, I understand that in the event of failures of the LEDs, the remaining ones will flow even more current, which can lead to the death of the entire matrix
 

kaspiysk113

New Member
If you'd do it with 14 groups of 3 LEDs, with 14 resistors at 150 ohms, you'd be giving up about 1 watt of heat across the whole board (70 mW per resistor).

your way, I think it will be about 8 watts on one resistor. 9 volt drop across the 10 ohm resistor when using a 12VDC supply.

do it this way (you may want to buy new UV LEDs and start over). Also, you may notice your eyes getting sore. They are "just" LEDs but the light is still LED. Wear appropriate welding goggles (not just UV sunglasses).

View attachment 123973
thank you for advice
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Using LEDs in parallel is bad practice unless they are perfectly matched and generally on the same heatsink / die.
While I would agree in general, it's not a problem with 'large' numbers of LED's in series parallel, as they average out - as long as you use the same make and type of LED's.

So basically, parallel is bad, series is good, paralleled series chains is 'ok'.

Considering the overheating resistor though, I'd certainly much prefer to use separate resistors per series chain in this case, as per post #3.
 

kaspiysk113

New Member
While I would agree in general, it's not a problem with 'large' numbers of LED's in series parallel, as they average out - as long as you use the same make and type of LED's.

So basically, parallel is bad, series is good, paralleled series chains is 'ok'.

Considering the overheating resistor though, I'd certainly much prefer to use separate resistors per series chain in this case, as per post #3.
thank you, this is very constructive
 

Pommie

Well-Known Member
Most Helpful Member
You're currently supplying ~1A at 12V which is 12W and you're wasting 10W in a resistor (which unsurprisingly, gets hot). If you connect 4 LEDs in series they will have a Vf of ~8.5V. Assuming 20mA per LED then you need a series resistor of (12-8.5)/0.02 = 175Ω - this will waste ~70mW - a 1/8th W resistor will do. Parallel up 11 of these and you have 44 LEDs without the waste (and heat) - that's 2 spare ones!!!

Mike.
P.S. maths done in head after a few wines so not guaranteed to be correct.
Edit, missed post #3 which says basically the same thing.
 

kaspiysk113

New Member
You're currently supplying ~1A at 12V which is 12W and you're wasting 10W in a resistor (which unsurprisingly, gets hot). If you connect 4 LEDs in series they will have a Vf of ~8.5V. Assuming 20mA per LED then you need a series resistor of (12-8.5)/0.02 = 175Ω - this will waste ~70mW - a 1/8th W resistor will do. Parallel up 11 of these and you have 44 LEDs without the waste (and heat) - that's 2 spare ones!!!

Mike.
P.S. maths done in head after a few wines so not guaranteed to be correct.
Edit, missed post #3 which says basically the same thing.
to be precise, it is about 8W. on the resistor, however, your calculation is generally accurate, thank you
 

gophert

Well-Known Member
Most Helpful Member
to be precise, it is about 8W. on the resistor, however, your calculation is generally accurate, thank you
Except UV LEDs have about 3v to 3.5Vf instead of two as pomme's math shows. So you get a max of 3 in series for a 12VDC supply with stable current control resistor.
 

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