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can 7.2v power a blu ray?

kruser495

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i am going to be building my new laser (sf-aw210) (lm317 driver). i read that you need around 9v to power a br. will 7.2v be sufficient or do i need to buy another battery. obviously full charger will be around 7.9v.
 



set

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You could get away with 7.2v if you used a different linear driver like the one below because it has a lower voltage drop of around 0.55v compared to the 2.5v of the lm317 driver. It is slightly more complicated to build but it can be a lot more efficient.

Power LED's - simplest light with constant-current circuit

It is designed for LED's but it will almost certainly work for LD's. I have used it many times with high power LED's and it is reliable.
 
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HIMNL9

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^ also, remember that you can also reduce the dropout through the set resistor, and its value (and, consequently, the dissipated power on the resistor), adding some voltage to the base of the transistor, via a reference voltage component.

As example, that configuration needs 0.6V minimum Vref on the resistor, this means that at, for example, 1.5A, you need 0.4 ohm resistor, that dissipate 0.9W ..... if you add, as example, 0.4V on a resistor partitor on the base of the transistor, you need additional 0.2V for get the limitation, and this means that, at the same 1.5A, you need 0.13 ohm resistor, and it dissipates 0.3W.

Just as example ;)
 
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set

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I never thought about that, good idea, but it would make it a bit more complicated and I think the op wants a simple solution.

An even better way is to replace the transistor with an op-amp/comparator as this allows you to reduce the voltage drop to around 0.05v and a small pot can be easily added to adjust the current. But it is a lot more complicated as you need a reference voltage and many more components.
 

kruser495

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is there a regulator i can switch out in place of lm317? can you also hook up 2 3.7v in seiries plus a 3.0v button cell? all in series?
 

Eudaimonium

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First, you're missing something. Combining two lithium ions in series does not result in 7.2 V. If they are freshly charged, it results in 8.4 V.

Then as they are used and current is drawn out of them, they fall down to 6V, at which point they need to be recharged.

Next, you seem to be ignoring every reply these guys have offered you, not just in this thread but in all others. Out of curiosity, how old are you?

Finally, YES you can exchange the LM317 with LM1117 (half the voltage drop) but if you would have done any kind of search on your own, you'd know it already.
 

kruser495

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First, you're missing something. Combining two lithium ions in series does not result in 7.2 V. If they are freshly charged, it results in 8.4 V.

Then as they are used and current is drawn out of them, they fall down to 6V, at which point they need to be recharged.

Next, you seem to be ignoring every reply these guys have offered you, not just in this thread but in all others. Out of curiosity, how old are you?

Finally, YES you can exchange the LM317 with LM1117 (half the voltage drop) but if you would have done any kind of search on your own, you'd know it already.

i am not ignoring anyone's post, sorry if it seemed that way. i wasn't looking to build anything more complicated, just a matter of if two would not work then i guess i would order another 10440, which i did after everyone's post. i am not trying to create any tension or hostilities. just looking for some help. my apologizes for not searching enough.

thanks everyone who posted.
 

set

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is there a regulator i can switch out in place of lm317? can you also hook up 2 3.7v in seiries plus a 3.0v button cell? all in series?

Never use different batterys in series, especially with Li-ion's as they can easily explode! I would suggest you read some more about Li-ion's as they are dangerous batteries if used incorrectly.
 
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theman101

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Given that 3.7v batteries are really 4.2v when full, wouldn't it be possible to use 2 for a total of 8.4v to power a 12x in a DDL driver? Or am I misunderstanding something? Also, wouldn't 8.4v and a DDL work with a red diode as well?
 




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