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trying to find the best host/ batteries

brisco

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OK. So I bought all the parts and am not putting it together when I figured out that I need 2 3volt cr123a batteries. They are on the way to me but I found the dimensions of them and the host I have uses 3 AAA batteries side by side in this tri battery holder. Now I am stuck. I figured out that the cr123a are to long for my host. Any suggestions?
 

rkcstr

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you can use the lithium 10440 AAA-sized batteries, two of them and a dummy battery to fill the third slot would do. Or, you can do something like I did and make a tube adapter to shrink the inside diameter and use 2 of the rechargeable CR2 batteries, which will give longer battery life than the 10440s.
 

rog8811

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If you go the 10440 route you can build your drive circuit into the unused part of the battery holder Just put the capacitor and reverse diode in the Aixiz module.... unless of course you are using rckstr's mini drive :)

Regards rog8811
 

brisco

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yes. thanks. I am using the mini drive inside the Aixiz module. Wont 3.6volts X 2 be to much for the unit? I thought the chip on the driver took 2.5 and gave the rest to the LD. My LD is a phazer, I forget when they run at.
 

rkcstr

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brisco said:
yes. thanks. I am using the mini drive inside the Aixiz module.  Wont 3.6volts X 2 be to much for the unit? I thought the chip on the driver took 2.5 and gave the rest to the LD. My LD is a phazer, I forget when they run at.
Nope, you'll be fine. The chip requires AT LEAST 2.25V over the output, so going over is okay, but I wouldn't recommend going FAR over as the chip will generate more heat as voltage increases. With the average combined output of about 7.2V for two batteries, you'll be fine.
 

brisco

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So 6 volts is still good but 7.2 would be great for it or am I thinking wrong. Does your driver steady the voltage at a constant output regardless of the input or I think I read it would take the 2.25v and send the rest. The reason I ask is I have ordered the cr123 already and they are on the way and the ones I got were 3 volts each and not 3.6 I also bought a flash light off of ebay for $10 that takes 2 cr123 batteries so I think I will be ok. Its a better host than the one I have.
 

chido

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The driver keeps 2.25v to itself and it sends the remaining voltage to the laser diode, the LD will take whatever voltage it needs to work and the remaining voltage will be wasted as heat. All you need to worry about is the current going to the diode. ;)
A red laser diode (what diode are you using?) takes an average of 3v (the voltage an LD takes depends on the current it's being driven at) so you need a minimum of 5.25v in average to power your red laser diode, if that voltage gets below 5.25v the laser diode won't get enough voltage to work, and if there isn't enough voltage the current will start to drop making your laser weaker and weaker by the minute. If you use two 3v CR123 batteries, they will have to drop .75v before the driver starts to drop out, if you use two 3.6v rechargeable CR123 batteries they will have to drop 1.95v before the driver starts dropping out.
So in conclusion, the 3.6v batteries will give you a significantly longer run time than the 3v batteries.
 

rkcstr

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Yeah, what chido said is correct.

Sorry if I confused you, I didn't know you bought 3.0V ones instead of the usual 3.6V.  Yes, 3.0V batteries will work fine, I actually use the 3.0V lithium CR2s to power mine at the moment.  With that I get somewhere around 10-20min. at 400mA output(it's an estimate of that I did about 3 or 4 5min. runs at 400mA before I had to recharge), so for lower outputs and larger batteries (such as your CR123A), runtime should be longer.

I'm not really sure, but I'm thinking that 3.6V would probably have about the same runtime as the 3.0V since the current draw remains the same regardeless of the voltage, meaning that at different battery voltage but same mAH rating at a constant current draw, both batteries will have the same runtime. The extra voltage is being consumed by the regulator. If this was a buck converter, then higher voltage batteries would have longer runtime since the extra energy isn't being wasted.
 

chido

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rkcstr said:
I'm not really sure, but I'm thinking that 3.6V would probably have about the same runtime as the 3.0V since the current draw remains the same regardeless of the voltage, meaning that at different battery voltage but same mAH rating at a constant current draw, both batteries will have the same runtime. The extra voltage is being consumed by the regulator. If this was a buck converter, then higher voltage batteries would have longer runtime since the extra energy isn't being wasted.
If that was the case then 7.2v batteries wouldn't be necessary to use with DDL's driver.
The whole thing isn't really that complicated, let's say you have two 400mAh battery packs, both can deliver 400mA for one hour. One is 6v and the other one is 7.2v, you hook one of your 400mA drivers and an open can diode to both of them, both diodes are drawing 3v and the full 400mA. Let's say both batteries decrease .06v every minute, the 6v battery would deliver the necessary voltage for approx. 12 minutes, after that it would make the driver drop out. On the other hand, the 7.2v battery would deliver the necessary voltage for 49 minutes before making the driver drop out. Of course the mAh is important too, if those two batteries were 200mAh batteries instead of 400mAh they would die before the driver ever had a chance to drop out. (someone correct me if I'm wrong)
 




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