Constant current source from a single LiPo cell at 5A

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Thread Starter

newhobby

Joined Mar 26, 2014
24
Hi,
I need help identifying a constant current source circuit to power a laser diode at about 5A from a single LiPo cell.
The Vf of the diode is about 1.9V.
I'm not interested in off-the-shelf stuff. I'd like to build it from scratch.
Is requesting help for specific chip selection something I can do or it is against the rules?
Thanks,
RI
 

Thread Starter

newhobby

Joined Mar 26, 2014
24
The cell is 3.7V nominal and I'll have it limited to a range from 3.2 to 4.2V
I'd rather have a switching circuit because this is for a handheld application and dissipating the resistor on top of the diode is going to be a task on its own.
 

bertus

Joined Apr 5, 2008
22,270
Hello,

What are you trying to do with a 10 Watt handheld laser?
A few miliwatt can already be dangerous.

Please read this page:
http://www.repairfaq.org/sam/lasersaf.htm

Intensity of a 1 mW Laser versus the Sun
Here is a comparison between the maximum intensity on the retina of the Sun and the beam from a 1 mW HeNe laser. (Adapted from one of Simon Waldman's optics lectures.)
Standard Sun:
  • Maximum intensity of sunlight at ground level (directly overhead, no smog, etc.) = 1 kW/m2 or 1 mW/mm2.

  • Assuming pupil diameter is 2 mm (i.e., radius of 1 mm), the area is approximately 3 mm2. So, the power of the sunlight through the pupil = 3 mW.

  • Focal length of eye's lens = approximately 22 mm. Angular size of Sun from Earth = 0.5 degree = 9 mR. Thus, diameter of image formed = 22 mm x 9 mR = 0.2 mm and the area of image = 0.03 mm2.

  • The intensity of the Sun on the retina (Power/Area) = 3 mW/0.03 mm2 = 100 mW/mm2.
Typical 1 mW HeNe laser (or laser pointer):
  • Power (P) = 1 mW, wavelength (l) = 633 nm, radius of beam (w) = 1 mm, focal length of eye (f) = 22 mm. So, the diameter of spot = (2 x f x l)/(w x pi) = 9 x 10-3 mm and the area of spot = 6 x 10-5 mm2.

  • The power density of the HeNe laser on the retina is 1 mW/(6 x 10-5 mm2) = 16,667 mW/mm2 = 16.667 watts/mm2.
So the 1 mW laser has the potential to produce an intensity on the retina 167 times that of direct sunlight! But there are many more factors to consider in determining the real risk of damage. In addition to those noted below, the actual focal point when looking at a laser at close range will not be at the retina so the spot size will most likely be much larger than the diffraction limit of the calculation. Even if the spot from the laser beam is smaller, natural eye movements or movement of the source (e.g., some moron waving a laser pointer) will result in it hitting any given point for a shorter time than the larger spot from the Sun (which usually doesn't move very quickly).

But, at least, perhaps you'll now have a bit more respect for that little HeNe laser or laser pointer!

(From: Jim Webb (jim@glservices.org).)

The real problem behind this is that it is assumed that the power density is the significant factor in the thermal damage mechanism. The ability of the retina to dissipate heat is not dependent on the area covered, but the periphery (circumference) of the exposed area! The blood vessels are in the retina and not the sclera (the surface under the retina) - it is the blood flow that dissipates the heat and so can only act on the *edge* not the middle of the exposed area. In circumference terms, the ratio drops to 7 times. Furthermore because the larger spot is less efficient at dissipating heat, the effective power delivered by the laser beam is only about 2 times greater than that of the spot formed by the sun.
Bertus
 

bertus

Joined Apr 5, 2008
22,270
Hello,

5 Ampere X 1.9 Volts = 9.5 Watt.
Do you have a specsheet of the laser diode?
AGAIN what are you trying to do with the laser?

In my job we use that kind of laser to burn clean the source of a massa spectrometer.
It heats up the source to about 500 C.
It is NOT something to play with.

Bertus
 

crutschow

Joined Mar 14, 2008
34,280
I agree with bertus. You don't want to be playing with a hand-help 10W (or even 5W) laser diode. Either one can put an eye out instantly and/or rapidly burn a nice hole into you.
 

Thread Starter

newhobby

Joined Mar 26, 2014
24
My laser diodes typically have 50% efficiency :(
I've already built a benchtop 20W laser device, so I can assure you I know what it is and this isn't going to be for a toy.
My package is actually a triple 5W diode in three different wavelengths, so I'll have to have either 3 drivers or a way to switch which one I want to fire. That means that size is extremely important. I think there is no way I'll be able to fit 3 drivers in the pcb space I have.
They are probably not going to be fired at the same time because the battery won't be able to handle the discharge.
 

bertus

Joined Apr 5, 2008
22,270
Hello,

At what are you "firing"?
Will the beam be confined in a closed surrounding?
Even reflections can be dangerous.

Bertus
 

Thread Starter

newhobby

Joined Mar 26, 2014
24
This is not for a toy. I'm not interested on eBay stuff.
This will be a commercial product.
I'm after some guidance in selecting an IC and possibly a freelance job if necessary. I'm not interested in hearing how dangerous lasers are.
I already know how dangerous they are and I have been unsuccessful in finding a suitable IC. That's why I'm here.
 

MrChips

Joined Oct 2, 2009
30,706
A handheld 5W laser just doesn't sound right and we don't want to hear about it and possible adverse consequences.
Hence this thread is closed.
 
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