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FrozenGate by Avery

I know it's DANGEROUS, but...

B

Bearded_Galaxy

0
Joined
Sep 7, 2008
Messages
396
Points
18
So, I want to do jayrob's keychain build using the Lowe's flashlight and the AW 10280. I broke my FlexDrive. I know it's dangerous, but it shouldn't be too bad if I use a 10ohm resistor and a short duty cycle if I just connect my LPC diode directly to the battery, right? Or should I put in a couple of 1N4001 diodes to lower the voltage, and use less resistance (looking at ohm's law and the lowered voltage).
 





You _could_, however the flexdrive is a boost driver. If you do what you describe, as your batteries run out of power, your laser dims. With the flexdrive, the laser output is pretty much the same right up until the battery is almost dead.

It'd probably be safer just to buy a new flexdrive ..
 
If you're absolutely sure of your maths and use a good capacitor across the diode, then in theory it should be fine. I wouldn't run it like this for very long, however - use a driver for any longer-term use, this should only be for very basic testing.
 
It'll really be for short duty cylces, 10 seconds or less, for the most part. I could get the FlexDrive repaired for $12, but there's no guarantee that I won't break it again, making a diode, capacitor, and resistor the best option.
 
So here's what I did: I consulted an old, retired electrical engineer, and he suggested I just use resistors and a capacitor, skipping the silicon diodes. I'm now using a 10uf (no micro sign on iPod touch keyboard) tantalum capacitor, and from my rckstr kit: 3x 12R5, 1x 20R5, 1x 33R5. With just 1 12R5, I read 1.6V across the resistor, and my end resistance is really close to 3ohms. With that voltage, I guess I'm getting a bit more than 500mA current. Great at lighting matches fast, super bright, and it doesn't even heat up too fast. When I had accidentally shorted it to the 10280, it heated up VERY fast, and burned matches VERY quick. I think I'll be happy with this build!
 
You shorted resistors and had the diode directly across the batteries? If that diode lives, you're very lucky. The only thing limiting the current would be the internal resistance of the batteries, they supplied all the current they could. Which is in no possible way healthy for the diode or the battery. So I hope you didn't do just that.
 
I can't for the life of me understand that a member that has been on the
Forum as long as I have has not yet understood what Dummy Test Loads
are for....:whistle:

As to using a battery with a Current Limiting resistor... you know what
the concensus is here on the Forum....

If you do want to go that route it is your choice... you've already stressed
your LD as far as I'm concerned by shorting the Current limiting Resistor
and dumping the full Battery Amperage potential across an unprotected
LD... IMO
Good luck with your build..

Jerry
 
to loosely quote Tech Junkie, just climb up the utility pole, and connect to the transformer, that should work real nice.
 
Bluefan said:
You shorted resistors and had the diode directly across the batteries? If that diode lives, you're very lucky. The only thing limiting the current would be the internal resistance of the batteries, they supplied all the current they could. Which is in no possible way healthy for the diode or the battery. So I hope you didn't do just that.
Click to expand...

So far the diode is fine. It was only directly connected to the battery (and on) for a total of like 15-20 seconds, and no more than 3 seconds at a time. Also, it's not just the internal resistance of the batteries, it's the LD's internal resistance too. If it was only the battery's resistance, it probably would not have survived a full second, as lithiums can provide several C amperes (sp?), and the batteries I have probably would've given it well over 1A, maybe as much as 2A, and the diode wouldn't have survived that, at least not for very long.
 
Bearded_Galaxy said:
So far the diode is fine. It was only directly connected to the battery (and on) for a total of like 15-20 seconds, and no more than 3 seconds at a time. Also, it's not just the internal resistance of the batteries, it's the LD's internal resistance too. If it was only the battery's resistance, it probably would not have survived a full second, as lithiums can provide several C amperes (sp?), and the batteries I have probably would've given it well over 1A, maybe as much as 2A, and the diode wouldn't have survived that, at least not for very long.
Click to expand...

It hardly matters at all about how long you keep it on. The thing that's not protecting it is voltage spikes, very little limiting on current... It's a bad idea to power it straight off batteries, in general. Do what you like, but why do you think every single person here uses a driver? You want something that will stably limit the current. Short duty cycles mainly affect the increase in temperature. Current is the amount of charge per second - it doesn't matter if you had that current for 1sec or 40sec... the object took that much charge within that 1sec timeframe... and it can easily end up being too much for it to handle. Personally, it's a better investment to spend $30 getting a good fix for it rather than having a much lower quality laser, which can potentially break at any time, and will not perform up to it's potential.

Well, if I'm wrong anywhere, someone please correct me. I'm only an AP Physics C student in high-school, and have not yet pursued electrical engineering in college yet.
 
Near as I can tell, you're not wrong anywhere, you're just telling me things I've known for a while. And it can matter how long, because PHR at 100mA will last noticeably longer than a PHR at 150mA, although I've obtained reasonable life from 150mA on my PHR diodes. It only doesn't matter how long if the diode itself can't handle the amount of current at all. Like trying to run a PHR at 500mA: if it even shines at all before dying, it will not last more than probably a fraction of a second. In which case one second or 10 seconds doesn't matter. Think about it like this: I can walk for hours, or jog for a little while and then be SUPER tired, because I exerted myself. The faster I try to run, the less time it's OK for me to do that. If I sprint for only, say, 30 feet, then it's no big deal. But if I try to sprint for a full 100 meters, then I'm going to be really out of breath at the end there. If I was tied to the back of a truck, and it accelerated all the way up to 50mph, I just plain couldn't run that fast. So time is completely relevant, when the current is limited enough that the diode itself can handle it. Also, if you ask any veteran on here, they'll tell you that the duty cycle is a very important consideration when running LDs at high currents.

Also, I have discovered a *slight* advantage to not using a FlexDrive: as the battery power drops, the diode stays a nice bright red, even when the current gets pretty low. I had a BRIGHT red laser pointer, even below 3.2V on a 3.7V lithium cell.
 
Bearded_Galaxy said:
Also, I have discovered a *slight* advantage to not using a FlexDrive: as the battery power drops, the diode stays a nice bright red, even when the current gets pretty low. I had a BRIGHT red laser pointer, even below 3.2V on a 3.7V lithium cell.
Click to expand...

As current through a Laser Diode drops... so does the beam output
power.... The human eye is not a good judge of Laser Power by
brightness..
Just check any plotted Laser graph on the forum for ANY Laser
Diode...

Jerry
 
bearded galaxy: the internal differential resistance of the laser diode is incredibly low. It's a diode, these have an exponential relation between voltage and current. On your average diode, the current doubles with each 0.025V rise in voltage. The operating point of your LD may be around 2V. The battery is 3.7V. The limit will be the internal resistance of the battery and the resistance of the wires. Wires probably only have a few milliohm resistance. The only thing limiting the current is the internal resistance of the batteries, which is probably strongly related to how full it's charged and the temperature. And it's a far too high current

Running in duty cycles this long only prevent long term thermal effects, nothing more of the many possibilities. Another thing waiting to happen: some batteries may build up charge like a capacitor, making them capable of delivering a pulse far above the short circuit current, frying a diode even faster. Drivers have a reason, stability and reliability.
 
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