Homebrew_PCBs

http://www.crystalaser.com/ has QUV 355-25 Crystal Laser of 25mW at 349 or
351 or 355nm, just inside the photoresist window.
I don't know how much it is, for the moment, but the head is not so bulky.
One can mount it on a CNC Machine easily.
Of course, some simple optics is necessary, because the beam is 0.9mm diameter.

Calculations:
Photoresist needs maximum 100mJ/cmp, that means 4 seconds/cmp with 25mW laser.
That means 240 seconds for 100x60mm. Four minutes of XY Plotter work.
If my calculations are correct, that is not too much.
Cristian

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Jon Elson is the autority, he's done one and he's been using one for
years. http://pico-systems.com/photoplot.html

I tried to go that route and I am half way through (all hardware is
completed) with no desire to integrate it at this time. Reason? If I
say that one more time a certain Alien will ask again about the
current moon phase and I don't want that to happen.

Mike

--- In Homebrew_PCBs@yahoogroups.com, Cristian <bip@f...> wrote:
> http://www.crystalaser.com/ has QUV 355-25 Crystal Laser of 25mW
at 349 or
> 351 or 355nm, just inside the photoresist window.
> I don't know how much it is, for the moment, but the head is not so
bulky.
> One can mount it on a CNC Machine easily.
> Of course, some simple optics is necessary, because the beam is
0.9mm diameter.
>
> Calculations:
> Photoresist needs maximum 100mJ/cmp, that means 4 seconds/cmp with
25mW laser.
> That means 240 seconds for 100x60mm. Four minutes of XY Plotter
work.
> If my calculations are correct, that is not too much.
> Cristian
>
> ----------
>
>
> ---
> Outgoing mail is certified Virus Free.
> Checked by AVG anti-virus system (http://www.grisoft.com).
> Version: 6.0.733 / Virus Database: 487 - Release Date: 02/08/04
>
>
> [Non-text portions of this message have been removed]

Sorry, wrong direction:
It was NOT about a photo plotter.
My idea was to use a plotter with an 'UV pen' to expose directly the
Photosensitive PCB!!
No film is to be in between.
Cristian

>Jon Elson is the autority, he's done one and he's been using one for
>years. http://pico-systems.com/photoplot.html
>
>I tried to go that route and I am half way through (all hardware is
>completed)
>Mike

----------


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Hi Christian
I am looking at your calcs from another angle here trying to figure all this out.... I'd like to know more about your Calculation if you'd care to share it with us in detail.

If there is any logic in my thoughts here, so 2 UV Flouro Lamps @ say 20Watts each, thats 40 Watts total and I spose that a %age of this is scattered and would not be directed at the PCB, so how much actual wattage is being used to develope the PR ??
What I am getting at here is that there was mention that UV LED's would not be strong enough that maybe so in the raw state, If the UV LED was concentrated using a Collimating Lense for a LD to get that small concentrated spot, maybe that could /would equal the equivalent of the UV Flouro wattage, the point is that these UV Lasers or UV LED Arrays at 1 Watt are $$ and regular High Output UV LED's are cheap if they could do the job

some neat stuff here http://www.roithner-laser.com/index.htm

Thomas

----- Original Message -----
From: Cristian
To: Homebrew_PCBs@yahoogroups.com
Sent: Wednesday, October 06, 2004 7:13 PM
Subject: [Homebrew_PCBs] UV laser exposure


http://www.crystalaser.com/ has QUV 355-25 Crystal Laser of 25mW at 349 or
351 or 355nm, just inside the photoresist window.
I don't know how much it is, for the moment, but the head is not so bulky.
One can mount it on a CNC Machine easily.
Of course, some simple optics is necessary, because the beam is 0.9mm diameter.

Calculations:
Photoresist needs maximum 100mJ/cmp, that means 4 seconds/cmp with 25mW laser.
That means 240 seconds for 100x60mm. Four minutes of XY Plotter work.
If my calculations are correct, that is not too much.
Cristian

----------




[Non-text portions of this message have been removed]

Hi Thomas,
My posting was meant to open a discussion in order to avoid making films or
other transparencies.
Of course, multiple UV LEDs will do the job, with suitable summing optics.
Unfortunately I live in Romania, where is very hard to find UV leds or
lasers. Optics, cheap, to allow experimenting, NO way.
E-bay is there, but no PalPay for Romania, so.....
The medium monthly salary here is $150, so everything is expensive for us.
We have to go straight to the target, no funds for experimenting.
What remains for me is to initiate a discussion hoping someone there, with
more possibilities....
Cristian

Hi Christian

> Calculations:
> Photoresist needs maximum 100mJ/sqcm, that means 4 seconds/sqcm with
> 25mW laser.

mW/sqcm =mJ

> That means 240 seconds for 100x60mm. Four minutes of XY Plotter work.
> If my calculations are correct, that is not too much.
> Cristian

----------


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[Non-text portions of this message have been removed]

I'm writing a reasonably detailed explaination on the use of photoresist for the traintools group. The polymerization reaction generated by UV light in photoresist is more complex than you would imagine. There are many ways to expose photoresist, but you have to do it right or resolution goes to pot. When I get it done - a week or two from now - I'll post it to the photo section of this group as well.

----- Original Message -----
From: Thomas
To: Homebrew_PCBs@yahoogroups.com
Sent: Wednesday, October 06, 2004 12:20 PM
Subject: [Homebrew_PCBs] UV laser exposure


Hi Christian
I am looking at your calcs from another angle here trying to figure all this out.... I'd like to know more about your Calculation if you'd care to share it with us in detail.

If there is any logic in my thoughts here, so 2 UV Flouro Lamps @ say 20Watts each, thats 40 Watts total and I spose that a %age of this is scattered and would not be directed at the PCB, so how much actual wattage is being used to develope the PR ??
What I am getting at here is that there was mention that UV LED's would not be strong enough that maybe so in the raw state, If the UV LED was concentrated using a Collimating Lense for a LD to get that small concentrated spot, maybe that could /would equal the equivalent of the UV Flouro wattage, the point is that these UV Lasers or UV LED Arrays at 1 Watt are $$ and regular High Output UV LED's are cheap if they could do the job

some neat stuff here http://www.roithner-laser.com/index.htm

Thomas


----- Original Message -----
From: Cristian
To: Homebrew_PCBs@yahoogroups.com
Sent: Wednesday, October 06, 2004 7:13 PM
Subject: [Homebrew_PCBs] UV laser exposure


http://www.crystalaser.com/ has QUV 355-25 Crystal Laser of 25mW at 349 or
351 or 355nm, just inside the photoresist window.
I don't know how much it is, for the moment, but the head is not so bulky.
One can mount it on a CNC Machine easily.
Of course, some simple optics is necessary, because the beam is 0.9mm diameter.

Calculations:
Photoresist needs maximum 100mJ/cmp, that means 4 seconds/cmp with 25mW laser.
That means 240 seconds for 100x60mm. Four minutes of XY Plotter work.
If my calculations are correct, that is not too much.
Cristian

----------




[Non-text portions of this message have been removed]



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[Non-text portions of this message have been removed]

Cristian wrote:

> http://www.crystalaser.com/ has QUV 355-25 Crystal Laser of 25mW at 349 or
> 351 or 355nm, just inside the photoresist window.
> I don't know how much it is, for the moment, but the head is not so bulky.
> One can mount it on a CNC Machine easily.
> Of course, some simple optics is necessary, because the beam is 0.9mm diameter.
>
> Calculations:
> Photoresist needs maximum 100mJ/cmp, that means 4 seconds/cmp with 25mW laser.
> That means 240 seconds for 100x60mm. Four minutes of XY Plotter work.
> If my calculations are correct, that is not too much.
> Cristian

Where did you get 4 seconds from ? Shouldn't the time to expose 1cm^2 be
equal to: 100mJ/cm^2 / 25mW * (1cm / (0.09cm * 0.09cm)) = 493 secons =
8 minutes. A typical 100 x 60mm single sided PCB with 50% track usage
will take at least 4.1 hours !

Christian & Adam,
Where I'm having an issue understanding is where the time comes from.
It looks like the 100mJ/cm^2 is really 100 mW*Seconds/cm^2 as 1 Joule =
1 Watt Second.
To get 100mJ/cm^2, you need 100 mWSeconds/cm^2, or (25 mW * 4 Seconds)/cm^2.
With 25mW from the LED, concentrate it onto 1 cm^2, & it takes 4 seconds
to expose.
Since the area of exposure is a 0.9mm diameter (?) circle, the area is:
pi * r^2 = 3.14159*(0.45mm)^2 = 0.64 mm^2 = 0.0064 cm^2.
The power required to expose this area = 0.0064 cm^2 * 100 mW Seconds/cm^2
= 0.64 mW*Seconds
The time required = 0.64 mW * seconds / 25 mW = 0.0256 seconds.
Since the area is 0.64 mm^2, you can "pretend" that is the spot is a
square 0.8mm on each side. What that buys you is the ability to
calculate a nominal velocity
for the spot to expose the board. if the spot is moving at a constant
speed, to expose a point,
it must be within the spot for 0.0256 seconds,
V = 0.8mm / 0.0256 seconds = 31.25 mm/second
Now comes the magic:
I assumed different things in different parts of this equation for a reason.
I assumed the spot was circular to determine the area. This is probably
right.
I assumed the spot was a rectangle to calculate the speed. This is
patently wrong.
But the 0.8 mm sides provides us with a better working velocity than
using the 0.9mm diameter.
This is because using the 0.9mm diameter means only the exact center of
the beam provides a
full 0.0256 second exposure to the board. Further, the energy
distribution of the LED is
probably gaussian (I don't know that, but when you don't know, gaussian
is a good guess).
That just means the center of the beam is 'brighter' then the edges, and
to expose the edges,
you have to expose them for longer.
By varying the velocity of the 'print head', you vary the exposed trace
width. I am sure that
if I were smarter (or more stubborn) I could tell you what velocity
would give you what trace
width, but I expect there are too many variables to know it well: what
is the actual energy curve,
what is the thickness of the photo resist, how 'active' is this
particular batch, etc. My gut reaction
is to try a series of traces in all 8 cardinal directions. Vary the
velocity from about 1.5 cm/sec
to 3.5 cm per second, develop it and check the trace width in each
direction. While I like the
idea, I am a little worried that you will be generating inconsistent
widths, but as long as it works,
then it's 'wicked good'.
Good luck, Christian
Richard
PS, sorry, I can't help it, but you two together have a seriously
religious naming convention ;)

> From: Adam Seychell <a_seychell@...>
>Subject: Re: UV laser exposure
>
><snip>
>
>
>Where did you get 4 seconds from ? Shouldn't the time to expose 1cm2 be
>equal to: 100mJ/cm2 / 25mW * (1cm / (0.09cm * 0.09cm)) = 493 secons =
>8 minutes. A typical 100 x 60mm single sided PCB with 50% track usage
>will take at least 4.1 hours !
>
>

[Non-text portions of this message have been removed]

>Richard
>PS, sorry, I can't help it, but you two together have a seriously
>religious naming convention ;)

What it really meant?
**************************************************
Thanks for confirming my calculations.
I'll try to buy the CrystaLaser, if they'll respond and will not be a
fortune for me.
A laser is convenient as it's ray are collimated.
The CNC milling machine is already on my desk.

Also www.nichia.co.jp has under development an UV LED of 100mW radiant power.
Hope cheaper than laser, but with more complex optics to collimate.

A collimate ray is better than a classic conic ray with one focal point,
because you don't care so much about the distance between the head and the PCB.

Any idea about collimating lens calculations there?
Cristian


----------


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[Non-text portions of this message have been removed]

RMustakos wrote:

> Christian & Adam,
> Where I'm having an issue understanding is where the time comes from.
> It looks like the 100mJ/cm^2 is really 100 mW*Seconds/cm^2 as 1 Joule =
> 1 Watt Second.
> To get 100mJ/cm^2, you need 100 mWSeconds/cm^2, or (25 mW * 4 Seconds)/cm^2.
> With 25mW from the LED, concentrate it onto 1 cm^2, & it takes 4 seconds
> to expose.
> Since the area of exposure is a 0.9mm diameter (?) circle, the area is:
> pi * r^2 = 3.14159*(0.45mm)^2 = 0.64 mm^2 = 0.0064 cm^2.
> The power required to expose this area = 0.0064 cm^2 * 100 mW Seconds/cm^2
> = 0.64 mW*Seconds
> The time required = 0.64 mW * seconds / 25 mW = 0.0256 seconds.
> Since the area is 0.64 mm^2, you can "pretend" that is the spot is a
> square 0.8mm on each side. What that buys you is the ability to
> calculate a nominal velocity
> for the spot to expose the board. if the spot is moving at a constant
> speed, to expose a point,
> it must be within the spot for 0.0256 seconds,
> V = 0.8mm / 0.0256 seconds = 31.25 mm/second
> Now comes the magic:
> I assumed different things in different parts of this equation for a reason.
> I assumed the spot was circular to determine the area. This is probably
> right.
> I assumed the spot was a rectangle to calculate the speed. This is
> patently wrong.
> But the 0.8 mm sides provides us with a better working velocity than
> using the 0.9mm diameter.
> This is because using the 0.9mm diameter means only the exact center of
> the beam provides a
> full 0.0256 second exposure to the board. Further, the energy
> distribution of the LED is
> probably gaussian (I don't know that, but when you don't know, gaussian
> is a good guess).
> That just means the center of the beam is 'brighter' then the edges, and
> to expose the edges,
> you have to expose them for longer.
> By varying the velocity of the 'print head', you vary the exposed trace
> width. I am sure that
> if I were smarter (or more stubborn) I could tell you what velocity
> would give you what trace
> width, but I expect there are too many variables to know it well: what
> is the actual energy curve,
> what is the thickness of the photo resist, how 'active' is this
> particular batch, etc. My gut reaction
> is to try a series of traces in all 8 cardinal directions. Vary the
> velocity from about 1.5 cm/sec
> to 3.5 cm per second, develop it and check the trace width in each
> direction. While I like the
> idea, I am a little worried that you will be generating inconsistent
> widths, but as long as it works,
> then it's 'wicked good'.
> Good luck, Christian
> Richard
> PS, sorry, I can't help it, but you two together have a seriously
> religious naming convention ;)
>

Well, to be pedantic we should be strictly following scientific
notation, with SI units and use standard errors were applicable :)

Your 4 seconds sounds completely reasonable. I must of had too much
coffee in the system. I agree its impossible to get accurate
calculations using such simple assumptions, but I guess we only need
ball park numbers to know if what Christian is trying to achieve is a
waste of time.

So a typical 10 x 6 cm single sided PCB with 50% track usage will take
on the order of a few minutes using a 25mW 355nm light source.
At 10 watts input power, thats 0.25% efficiency ! Now I see why mercury
discharge lamps are the choice for photopolymerization.

Adam

I'm looking forward to it!

But please, post it in Files, not Photos. The Photos section only
allows everyone to see a 300x400 max size, only the moderators and the
uploader can see the original higher res images.

There are no such limits in the Files section.

Steve Greenfield
Listowner

--- In Homebrew_PCBs@yahoogroups.com, "Earl T. Hackett, Jr."
<hacketet@c...> wrote:

> I'm writing a reasonably detailed explaination on the use of

photoresist for the traintools group. The polymerization reaction
generated by UV light in photoresist is more complex than you would
imagine. There are many ways to expose photoresist, but you have to
do it right or resolution goes to pot. When I get it done - a week or
two from now - I'll post it to the photo section of this group as well.

>