Help with resistors and capacitors

hey in closed post Gazoo posted


"The resistor is put in series with the diode. I would recommend a 2 ohm .5 watt resistor if you are going to be using a minimag with two cells. The electrolytic capacitor goes across the diode. The recommend value is a 47uf 16 volt capacitor. The capacitor will be marked with the negative side. The negative side of the capacitor gets soldered to the negative side of the diode, and the positive side of the capacitor to the positive side of the diode. "

(in place of a driver)
 i went to my local radio shac and got a 47uf 35V as the closest i could find. for a resister i got a 2.2 ohm .5 watt as the closest i could find how would that affect my 16X dvd laser diode. im not sure if i should hook it up yet. im planing on hookin up to 2 AA's
will this change anything

You are fine with those components, but keep in mind this is the bare minimum needed to protect your laser diode. The now famous circuit Daedal provided is the very best for the money.

The resistor won't be much of a problem, they normally have a few percent of tolerance anyway. The 35V cap won't be quite as effective at its job, but its not a crucial part anyway. Overall, it looks like you're good to go. In the future, shop online! Much better prices and selection than the rat shack.

the 47uf 35V cap is just fine.

as for the resistor; 2.2 is too low.

putting 2 2.2ohm resistors in parallel would result in 1.1 ohms @ 1 watt.... which is faaar too low.

putting 2 2.2 ohm resistors in series is only 4.4 ohms @ .5 watt.

this is still far too low.

your safest combination is putting 2 10 ohm resistors in parrallel. 5 ohms for a torch is about as low as you'd want to go. plus, you should get proper heatsinking..

5ohms for a lab laser, would be fine, still though, you need good cooling.

conclusion:

return the 2.2ohm resistors and buy 10 ohm resistors.

In actual testing with two batteries, I have found a 5 ohm resistor to be to high. I will try it again later this evening. I haven't hooked up any of my diodes from this second batch I received yet.

Ok...I just tested it out with the GB diode using a 2.2 ohm resistor. With Nimh batteries (2.57 volts) the current draw was 100ma's. With brand new Alkaline batteries (3 volts) the current draw was closer to 170ma's (almost perfect). This was using the 10 amp scale on my meter. Of course using E2 Lithiums the current would be much higher...but I don't have any to test it with.

Using a 5 ohm resistor with Alkaline's the current draw was 110ma's. 110ma's is a very comfortable current but it will not burn well and I know everyone wants to be able to light matches and pop balloons and so on. Thus a 2.2 ohm resistor will accomplish what everyone wants to do, and still maintain a semi-comfortable current using 2 Alkaline batteries.

Gazoo --

Good post for the noobs who don't want to make a current source.
You posted a lot of info here.

Mike

lol yea thx

Gazoo said:

Ok...I just tested it out with the GB diode using a 2.2 ohm resistor. With Nimh batteries (2.57 volts) the current draw was 100ma's. With brand new Alkaline batteries (3 volts) the current draw was closer to 170ma's (almost perfect). This was using the 10 amp scale on my meter. Of course using E2 Lithiums the current would be much higher...but I don't have any to test it with.

Using a 5 ohm resistor with Alkaline's the current draw was 110ma's. 110ma's is a very comfortable current but it will not burn well and I know everyone wants to be able to light matches and pop balloons and so on. Thus a 2.2 ohm resistor will accomplish what everyone wants to do, and still maintain a semi-comfortable current using 2 Alkaline batteries.

Click to expand...

i was doing ok with ohms law then i confused my self :-/ but
volts =amps x ohms or i x r
so with a 5 ohms x 110ma=5x(150/1000)(?)=.55
iam really confused atm :'(

volts=e
amps=e/r
3/5ohms .6 or 600ma?
but up there u siad 5 ohms= about 110 ma thats were i got confused and also would 3.3ohms be about 150ma or 3 volts/ 3.3 ohms=.90909... or 909ma i x r says 909ma x 3.3ohms = .909 x 3.3=2.9997e or volts

You have to account for the voltage drop across the diode which is going to be appx. 2.5 volts.

ah i figured there was a voltage drop like on leds and i new i had to be forgetting something lol

also that owuld explain were my .55 came from lol