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red dot finder - fried


GazK
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Aarrgghh! I just fried the RDF on my portabowl scope.

It's one of those Daisy finders powered by a 3V watch battery. So I thought I'd be clever and replace it with a 2 x AA supply to give it more stamina. All went well until I made the final connection, switched it on; it came on but then a nasty sizzling ensued, followed by a dead laser.

I checked the polarity; it's right. The voltage is the same and I soldered onto the battery connections on the RDF. So what went wrong?

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Aarrgghh! I just fried the RDF on my portabowl scope.

It's one of those Daisy finders powered by a 3V watch battery. So I thought I'd be clever and replace it with a 2 x AA supply to give it more stamina. All went well until I made the final connection, switched it on; it came on but then a nasty sizzling ensued, followed by a dead laser.

I checked the polarity; it's right. The voltage is the same and I soldered onto the battery connections on the RDF. So what went wrong?

Assuming that you got the polarity rignt then I can't see why. 2 x 1.5v AA equals the same voltage, so that should of worked.

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I'm wondering whether the watch battery was actually chucking out 3V or something a little less. I'm no mug, I've got a degree in electrical engineering - it can't have pulled any more current than it did from the watch battery so it's not that. I want to complete the mod but I don't want to fry another £20 finder!

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Did you connect the 2 X AA (3V) in place of the 3V battery, or with the 3V still in place? Is it possible the AA batteries have ended up in series with the original making 6V total?

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Firstly, I think you'll find it's an LED, not a laser.

Secondly, have you tested the circuit to find out what is fried, i.e. the LED or the switch?

Thirdly, why not just replace the LED or the switch (having ascertained that there is no remaining fault), if that is what is fried? They are pennies at Maplin.

Thought: in soldering to the battery connections, might you have inadvertently thermally shorted a series resistor? Pop a meter onto the LED to see what the forward resistance is -- not all have an internal resistor, so you may need to add a series one.

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@Lorne - no, the watch cell was gone, as I needed to get at the contacts to solder onto.

@tetenterre - I'm equally fairly sure (!) that its a laser - how is an LED (diffuse omnidirectional light source) able to project a point of light onto a screen 3cm away? Unless physics has changed since my A levels I dont think that's possible.

Your theory about melting a resistor is the only plausible one so far. Trouble is I don't have a multimeter - I know, shame on me. I might just have to buy one, I don't really know how I got to 40 years old without owning one anyway!

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The led does not project a point of light onto the plastic screen, what you see is a reflection of the led.

Peter

Oh, I see - you learn something everyday. I won't split physics hairs (for the light to reach your eye it *must* first be projected onto the screen - dammit! I said I wouldnt) but instead I'll take your advice and see how replaceable the LED is.

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I'm equally fairly sure (!) that its a laser - how is an LED (diffuse omnidirectional light source) able to project a point of light onto a screen 3cm away? Unless physics has changed since my A levels I dont think that's possible.

:) Have they really taken geometric optics out of the A level curriculum? I've heard that it's dumbed down a bit since I last taught it, but that is ridiculous! ;)

Actually, if a RDF is correctly set up, the image of the LED on the screen is projected to infinity -- which is why it doesn't move against the background stars if you move your eye slightly.

See here (OK, not a Daisy, but the original RDFs used on telescopes were Daisys). Alternatively see here for one I made (uses a lens instead of a partially reflective mirror, but optical principle is similar).

BTW, if it's laser, not LED, I'd be curious to see how a reticle finder (like a Telrad or Rigel) works! (see here!)

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OK so optics lessons aside...

I pulled it apart and it is indeed an LED. I also think I can see what went bang... the battery terminal marked "-ve" is connected to a red wire, and the +ve terminal to the black wire. And the longer LED leg is connected to the black wire. Doh! I blame Daisy. Nice to know at least that its nothing *I* did wrong.

I can't see any way to fix this since the LED is epoxied into a tiny pinhole fitting. I've got another finder on the way from ebay, and I'll carefully check the *actual* polarity before I modify it.

Thanks for all the help and corrections.

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The Coin cells actually have a much larger internal resistance than to the AA cells.. chances are the terminal voltage will be lower under load as they are really only designed to provide at most a couple of mA (according to the data sheet 0.2mA is th emax continous drain) hence their use in watches calcualtors , memory backup etc ... infact I have sometimes used the internal resitance of the cell instead of a series resitor for some LED experiments...

http://datasheet.octopart.com/CR2032-Panasonic-datasheet-48591.pdf

Billy...

Edited by Psychobilly
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The Coin cells actually have a much larger internal resistance than to the AA cells.. chances are the terminal voltage will be lower under load as they are really only designed to provide at most a couple of mA (according to the data sheet 0.2mA is th emax continous drain) hence their use in watches calcualtors , memory backup etc ... infact I have sometimes used the internal resitance of the cell instead of a series resitor for some LED experiments...

http://datasheet.octopart.com/CR2032-Panasonic-datasheet-48591.pdf

Billy...

Aah... OK that makes sense - thanks. I am bidding on a fluke multimeter at the moment - once I have that and the new finder, I will test the battery voltage under load, and then wire a suitable compensating resistor into the modified circuit. That should do it. Oh, and connecting the circuit the right way round as well!

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