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DIY Laser Torch

roSSco

roSSco

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The diode and cap are in PARALELL to the LD. If you think those items have NO EFFECT on the current that gets to the LD, you are mistaken.

I'll ask again: Do you want an estimate of the current going to the LD, or do you want to know exactly how much current is going through the diode?

knimrod, while many of the things you say above are true, they do not apply to the situation at hand.
 





roSSco

roSSco

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How many times has gazoo said the 1N4001 in paralell with the diode drops .7V? It is a paralell path for current!
 
knimrod

knimrod

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Sorry, they apply absolutely and it's not an estimate.

The 1N4001 is reverse biased in this circuit and does not conduct any current.  Well... almost no current.  The worst case reverse leakage current is rated between 5 and 50 uA at 50 Volts (0.002% - 0.02% = negligible).  At ~3 Volts it's significantly less.  If it was forward biased however, the drop would be .6 volts (actually, for the 1N4001 it's 1 Volt) and conduct the majority of the current.  

The capacitor only bypasses current while charging (which is its intended function; absorbing spikes and dropouts) and therefore has no effect on current while operating.  While there may be some current leakage in the cap, it would be so small as to be negligible also.

Do the test.. Let us know. :)
 
T

troop231

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Hey, I did what knimrod said, this is the correct way right? I took the LED out and put a jumper wire between the 2 sides. and measured where the leads to the voltmeter are. I saw around 250ma. So, is the way I did it right, and the ma? Thanks! ;D

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knimrod

knimrod

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That's not exactly what I proposed but it's effectively the same thing. You are on the ground side of the circuit but in the loop nonetheless.

You've also made a dead short for a load which is not a good simulation for an LD but obviously the regulator will still control it. I still suggest a 10 ohm resistor (it will get a little hot) to simulate load. Other than that, it looks like your circuit is working.
 
T

troop231

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knimrod said:
That's not exactly what I proposed but it's effectively the same thing. You are on the ground side of the circuit but in the loop nonetheless.

You've also made a dead short for a load which is not a good simulation for an LD but obviously the regulator will still control it. I still suggest a 10 ohm resistor (it will get a little hot) to simulate load. Other than that, it looks like your circuit is working.
Click to expand...

OK, I tried the 10ohm resistor as the load, and I now measure around 170ma. Is this really good, or awful for a burning laser?(Without frying the diode) Thanks!
 
S

spyrorocks

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Well, your diode will have a really long life, lets put it that way.

230-250ma is where you get the balance of lifetime and burning power.
 
knimrod

knimrod

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You may need to start measuring some voltages.  I think that your power source isn't up to snuff and your seeing the regulator dropout now.  You will require a power source that will maintain above 6V under load.  Four AA NiMH batteries won't do it.
 
Gazoo

Gazoo

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troop231 said:
OK, I tried the 10ohm resistor as the load, and I now measure around 170ma. Is this really good, or awful for a burning laser?(Without frying the diode) Thanks!
Click to expand...

The reason is because you are only using 4 batteries so the regulator is dropping out. It needs at least 6 volts to regulate properly. I would bet if you use six batteries you will see 250ma's using the 10 ohm resistor as a load.
 
Gazoo

Gazoo

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roSSco said:
How many times has gazoo said the 1N4001 in paralell with the diode drops .7V? It is a paralell path for current!
Click to expand...

knimrod explained this perfectly. If the diode is in series with the circuit then there will be a ~.7 volt drop. In parallel there is no voltage drop. I suggest you do as he suggested...measure the current at the input and the output. Until you do I think you will not take our word that the current readings will be exactly the same.

I will admit at one time I thought there was a voltage drop across the diode in Daedal's circuit, but that was quite a while ago.... :p
 
knimrod

knimrod

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It all adds up..  You are using 4 NiMH batteries at 1.2 volts each for a total of 4.8 volts (nominal).  The LM317 will drop a minimum of about 3 volts and at 250mA, the 10 ohm load resistor will need 2.5 volts.  The total input voltage required will be about 5.5 volts to maintain regulation.  Since you have at best, 4.8 volts, subtracting the 3 volts for the regulator gives you about 1.8 volts available for the load and yielding about 180mA.

With a laser diode in the circuit, the load drop will be about 3 volts, which brings the total to 6 volts.  This is where the 6 volt minimum comes from.
 
T

troop231

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Alrighty then, I now ditched the 4 batteries and hooked up my 12 volt power supply to the inputs of the circuit. I re-measured for current and now my reading is.... 244-245mA!!! ;D so is this what I want then? Or should I try to get it to 250mA exactly? Thanks again!
 
knimrod

knimrod

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The small difference between the target current of 250mA and the 245mA of current your seeing is probably due to the tolerance of the resistors your using (5%).

245mA is plenty of current and I think you'll be very satisified with the result.
 
Gazoo

Gazoo

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Yeah..it is looking good to me too... :) One thing I will mention, if you plan on using a 12 volt power supply you will need to put a heatsink on your regulator. The 317 has built in thermal protection and with 12 volts it will kick in and the current will drop. It did on mine after it had been running for about a minute. Powering with anything less that 9 volts and the regulator will stay cool enough. So really, once again the best power source is 6 nimh batteries or two li-ion batteries.
 
T

troop231

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Just made my laser diode order from Stontek, and ordered the MXDL and some cr123s in a 5 pack on Dealextreme. Does anyone know how long it takes for Stonetek and Dealextreme to ship to the USA? Thanks!

sku_4171_6.jpg

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GB_Diode11.JPG

red_laser_shades.jpg
 
W

wooooooolazer

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Stontek will take about 1-3 weeks normaly... dealextreme i would bet 3-5 weeks, there shipping is slowwwwww, but you cant really complain seeing as the shipping is free and you cant beat there deals.

...lazer... ;D ;D ;D
 
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