Your last Veroboard project!
A UV exposure box is an extremely useful piece of kit. It can be used to make proper PCB's. It can also be used to make other things such as intricate photo etched parts (a subject for another instructable). The trouble is they can be a little pricey for the hobby enthusiast especially if you want the double side type.
This instructable outlines the construction of a double sided UV exposure box using the recent generation of high brightness UV LEDs.
Why use LEDs?
LEDs are far more energy efficient than either incandescent or fluorescent lamps offering between 5-10 times more efficiency making them cheaper to run and kinder to the environment. They also (unlike fluorescent tubes) do not contain mercury. LEDs have a far greater life span than the other types of lamp measured in decades rather than months. The frequencies being emitted are also in a tighter band making UV LEDs safer than the traditional UV tubes. There's also just something cool about LEDs, I can't put my finger on it, but ever since I was a kid I've found them to be one of the more fascinating electronic components.
Is there a disadvantage to using LEDs?
Not really, however the UV exposure box I have detailed here is a little less powerful than the commercially available ones. This means that your exposure times will be around 2 ~ 3 minutes as opposed to 30 ~ 40 seconds, but come on, do you really need your PCB's to be produced that quickly? Anyway sometimes having a slightly slower exposure time can be an advantage allowing you a little more control.
This UV Exposure box will consist of 2 UV panels; each having 84 LEDs a total of 168 LEDs. Each panel will draw about 700mA at 12v. This makes each panel 8.4watts a total of 16.8 watts for the whole thing.
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Signing UpStep 1: Materials
I bought two 100psc packs from eBay.
If you find something better then ensure that they are;
- At least 2000mcd in brightness
- Have a peak wavelength of less than 400nm.
- A viewing angle of at least 20 degrees.
You will also need 2x 160mm x 100mm pieces of Veroboard and 56x 75R resistors.
Another important choice is the PSU. I used a plug in, 12 volt 24 watt switch mode power supply. Switch mode power supply's are far more energy efficient than most other types and are also very stable.
All the other parts and materials are easy to find. Some I bought, some I salvaged. This is where you own judgement and taste comes in. In the end it's up to you how closely you follow my design. I've included all the CAD drawings and schematics as metafiles so they're easier to read when you print them out.
CAD files.zip423 KB




























































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I think using a non-toxic spray method would be a lot quicker and cheaper in the end.
Which will give better and faster results; total watts, total mcd or both?
Instead of painting the wood white, I covered it with metallic duck tape which gives me extra brightness.
I also tested some pcb that usually take 120s in a retail exposure box with UV lamps and they took the same 120s in my box with the LED.
Thanks for the idea and the design, it is excellent!
Most efficient way is to use make strings of few LEDs in series with resistor (and you can connect many such strings in parallel)
R=(V-n*Vf)/If
P=R*(If)^2
where
R= resistance (Ohms)
V= power supply voltage (Volts)
Vf= LED forward voltage drop (usually about 2.2-3.8V for most LED, read specs)
If= LED forward current (usually 20mA, read datasheet)
n=number of LEDs in series (in one string)
P = resistor power (Watts, minimum rating)
note: you should aim to get the voltage difference to be about 2V:
(V-n*Vd) ~ 2V
reason:
if the difference is greater, the more energy is wasted into heat by resistor (you need more powerful resistor and power supply).
if the difference is too small you loose current regulation because LEDs are non-linear device.
calculation example, assuming you are using
V=12V
Vf=3.2V
If=0.020A (that is 20mA)
n=3
R=(12-3*3.2)/0.020=120 Ohm
P=120*0.02^2=0.048W (you can use 1/8W or 1/4W without problem)
note how P is small. that is because we have (12-3*3.2)=2.4V
using large resistors makes no sense because to get current to match spec, you would need high voltage (this is not just dangerous but also means that most of the energy is wasted in heat so those resistors would have to be for high power).
for example using same scenario
Vf=3.2V
If=0.020A (that is 20mA)
n=3
R=47k
we would need resistor to have power rating of at least
P=47000*0.020^2=18.8W
(and that was just power three LEDs)
Also DC power supply would need to provide much higher voltage which is
V=n*Vf + R*If
V=3*3.2 + 47000*0.020
V=9.6 + 940
V=949.6V
That is way too much just for poweering few LEDs.
So that was introduction for current limiting resistors.
For bleeding resistors we calculate it from time constant.
t=RC
where
t = time (seconds)
C=capacitor in power supply
R= resistor (note this is combined resistance of load and bleed resistor)
R=t/C=Rb*Rc/(Rb+Rc)
Rc=V/(m*If) this is load resistance (for m LED strings drawing current If)
Rb=bleed resistor
Say you have simple 12V PSU with bridge rectifier and 10000uF capacitor and 6 LED strings drawing 20mA
C=0.010 F
m=6
Rc=12/(6*0.020)=100 Ohm
Without bleeding resistor t=RC=0.010 * 100= 1 seconds
which is shourt enough
adding bleed resstor in parallel, we can bleed the capacitor faster (total resistance is smaller and t=RC is smaller).
Say we use Rb=18 Ohm
R=18*100(18+100)=15.25 Ohm
t=RC=15.25 * 0.010 = 0.1525 sec
so turn off time was reduced by about 6x.
note that value of bleed resistor was quite small. usinglarge value such as 10k or 47k would not make any difference.
also keep in mind power disipated by bleed resistor P=12^2/18=8W
(which is more than used power for LED circuit in this example).
good luck
Just enter your number of leds and the wizard will help you to decide the resistors, the arrange of leds and the PSU that you will need.
My 5 cents, I hope it helps.
Cool instructable by the way, really cool stuff.
I'll assume you're using the same kind of LEDs and resistors that are specified then for every veroboard you've got 84 LEDs in bundles of three connected in parallel. Thus for every bundle you've got a forward current of:
I_f = (12V - 3x3.4V)/75ohm = 1.8/75 =24mA
So one veroboard, with 84 LEDs, will draw a current of
I_vero = (84/3)x24mA = 672mA
So if I've understood you correctly you wish to have 8 veroboards of UV LEDs then after you've connected them in parallel the total power drawn from the source would become:
P_total = 12Vx(8x672mA) = 12Vx5.375mA = 64.5W
So you'd need a 12V 65+W power supply, though to be on the safe side I'd personally use a 75W. Though as I said, you might be using LEDs with a different forward voltage so here's a formula where V_f is the forward voltage of your diodes:
P_total = 2304(4-V_f)/25 W
Just out of curiosity, what kind of setup needs that big an exposure surface?
Regards,
Dean
Great instructable!
Don
http://www.bivar.com/product-details.asp?PartNo=UV5TZ-400-30
and
http://www.bivar.com/product-details.asp?PartNo=UV5TZ-395-30
Hope this can help :D
Thanks
They work pretty well for me!
You should make your own test pattern with 30sec exposure increment because I might not be using the same photo resist PCB clad board. Each brand is different.
Good luck