Small/cheap/easy 3V power supply

I found a 3.7V DC wall wart. I will see what is its actual open circuit voltage, etc, maybe I can rig up some simple voltage divider to reduce voltage a little bit.

i

A 1N4001 will drop .7V for you.

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John B

This is VERY smart. I love your idea. I have about 100 of them, or similar diodes, if not more. Now it looks like a one hour project (time mostly spent making sure that the wire from wall wart is securely fastened to the chop saw).

Thank you!!!!!!!!!!!

i

On a sunny day (Wed, 31 Jan 2007 11:30:25 -0600) it happened Ignoramus20785 wrote in :

Cool idea :-) But 5% mains voltage change will be 5 V..... I'd personally get a small 9switching) AC/DC adapter (very small ones exist, and either combine the cables, or mount it on the tool.

Well, if you have some URL for something, let me know. For now, I think that dropping .7 volt (or whatever I need, I will see what is OCV on this supply), will do what I want. My mains voltage is pretty stable at 123-125 volts. Probably would drop somewhat when the saw is actually cutting, but by that time I should not care about the laser light.

i

On a sunny day (Wed, 31 Jan 2007 12:17:48 -0600) it happened Ignoramus20785 wrote in :

Eh, one came with my Nokia cellphone I think it is 3.7 V.

Better to use a series capacitor,

C1

----||------------ a diode k-------- +4.3V --- diode ---- 3.6 -- diode --- 2.9 V |k | mains 5V zener === |a --- | |

------------------------------------- 0V

Can you calculate C1? For 60Hz and the current you expect?

There are many problems with using a capacitor to drop line voltage, as discussed in the thread on the PIC16F84. It might be worthwhile to replace the original AAA batteries with rechargeable ones, and connect a charger full time. This will stabilize the voltage. However, NiCd and NiMH would provide only 2.4 to 2.6 volts, which may not be enough for the laser. A Lithium battery with a charger might be better.

Paul

Paul, I will check with a scope what kind of waveform I get out of this power supply, that will shed some light on what I can do with it.

i

On a sunny day (Wed, 31 Jan 2007 14:23:17 -0500) it happened "Paul E. Schoen" wrote in :

Don't you think that is overkill? I have several applications that simply use a cap... I bought some remote conrolled light recently, it too has (I open everything up) the cap method. As do my PIR alarms... etc.. Just calculate the current, use 1/jwc to calculate the cap, that is it. Make sure you pass AC and not DC, caps do not work on DC :-0 I have used this method since the eighties in last century ;-)

I assume he meant dropping the 3.7V to 3V.

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John Devereux

He was not proposing dropping the mains voltage with diodes, he was proposing using a nominally 3.7V output from a Wall-Wart thing and dropping

0.7V with a diode. That sounds much more sensible.

I don't know how sensitive the laser module is to the supply voltage. If it is not critical (as one would hope since it is battery powered) then the above solution should be OK, otherwise it would be better to use a 3V LDO regulator, but these are not all that common/cheap compared to a LM317. The problem with the LM317 is that it needs about 6V input to put out 3V reliably. If he has a 6V wall-wart and a LM317 then that would be ideal.

Whatever solution is chosen, it might be good to put plenty of edcoupling capacitance right across the battery contacts of the laser module, as I have read that they are easily damaged by spikes, static etc.

Chris

Ignoramus20785 wrote in news:YbWdndTcL54hKF3YnZ2dnUVZ snipped-for-privacy@giganews.com:

Use Energizer E2 AAA lithium batteries,they work in low temps. They also have a 10 yr shelf life. Or you could use a single CR123 lithium cell; rated at 3V,1300 maH

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Jim Yanik
jyanik
at
kua.net

I have used it as well, but I had a problem when I added an 18 ohm series resistor to limit line surges. Even with normal current of about 60 mA (60 mW), a 1/2 W resistor showed burn marks and melted a nearby IC socket. The capacitor appears as essentially a short circuit to high frequencies and high dV/dT spikes, as may be common when used with a power tool. The zener might be damaged with no current limiting. An 18 ohm resistor might allow

100 amps at 1800 volts, but that is better than the ESR of the capacitor, which is probably 0.5 ohms or less. A series inductor may be even better, except for the possibility of resonance.

Paul

On a sunny day (Wed, 31 Jan 2007 22:32:51 +0000) it happened John Devereux wrote in :

Must have been confuesd by 'have a hundred'. :-)

On a sunny day (Thu, 1 Feb 2007 02:58:46 -0500) it happened "Paul E. Schoen" wrote in :

OK I see your point. maybe here in the Netherlands the mains is much cleaner (I scoped it many times, no huge short peaks here). In the remote light switch I bought they have (looking) they use 100 Ohm

2 W (wirwound??) and 330 nF (at 230V 50Hz).

So 330 nF = an impedance of 1 / (6.28 x 50 x 33.10^-8) = 100 000 000 / 10363 =

9650. the current would then be 230 / 9650 = about 24 mA. 25 mA in 100 Ohm series is about 6 mW, so no heat in normal use.

I have always used a series R, and also a series fuse (this thing has what looks like a fusable PC track..... A 1kV short spike will see 100 Ohm, and power will be dissipated in the series R. I forgot to look for the zener, but it looks like all 200mW type stuff in there. it switches a relay in fact.

I have also used this system by using a series R, cap, bridge, zener to 12 V, and then mains isolation with a simple self oscillating one transistor ferrite core converter 1:1, with only a few turns primary and secundary wound with normal wire :-)

Yes, that is my plan.

I think that it ought not be too sensible, being battery powered.

i

Determine the current I your laser uses. Then do this:

------- ----- | Wall +|---Vin|LM317|Vout---+ | Wart | ----- | | 6V DC | Adj [R] R = 1.25/I | | | | | | +----------+---[Laser]---+ | | | | -|----------------------------------+ -------

You can measure the current your laser draws from the batteries. The LM317 will limit the current to that value in the above configuration. The input supply needs to be a few volts higher than the 3V you need, thus a 6V wall wart. However, you can use 6 volts or higher, so if you have a

12V wall wart on hand, you can use that. It makes the LM317 hotter - it will dissipate (Vin-Vout)*I watts. But even the small LM317 TO39 package will handle over 1/2 watt, so you should be ok, depending on the current your laser needs and the voltage of the wall wart.

Ed