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Old 11-13-2007, 04:52 AM #1
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Default Once again another topic regarding amk's DRIVER!

Okay, so after I fried my LD, I went on to bigger and better things.

I found out that I was using a 4004 diode, I think its supposed to be 4001??
Also, Im using a 10ohm resistor, not a 3.9, cuz radio shack dosnt have anyhting less than 10 ohms. WHAT WILL HAPPEN TO ME??
I checked the current output (or at least I think I did) and I got .12 when it was on the 10A mode, which I presume means 120mA, is this what I want, or not? BTW this is with the load.

and one last thing, how do I use a multimeter to see what the polarities are on a LD? do I use the ohms or something? thanks

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Old 11-13-2007, 09:05 AM #2
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Default Re: Once again another topic regarding amk's DRIVE

a 4004 is the same thing, so thats ok
with the resistors, if you use a 10ohm resistor, you will only be able to get 120ma of current. what you CAN do is put some resistors in parallel if you can't get a small enough value. for example, 3 x 10ohm resistors in parallel will get you 3.33ohms which will get you a max current of 360ma.
google 'resistor laws parallel' if you want to find out more about that.

if your multimeter has a diode function you can use that. or set it to measure resistance. you should probably see the diode glow dimly if you have it the right way too.
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Old 11-13-2007, 11:50 PM #3
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Default Re: Once again another topic regarding amk's DRIVE

Quote:
Originally Posted by woop
a 4004 is the same thing, so thats ok
with the resistors, if you use a 10ohm resistor, you will only be able to get 120ma of current. what you CAN do is put some resistors in parallel if you can't get a small enough value. for example, 3 x 10ohm resistors in parallel will get you 3.33ohms which will get you a max current of 360ma.
google 'resistor laws parallel' if you want to find out more about that.

if your multimeter has a diode function you can use that. or set it to measure resistance. you should probably see the diode glow dimly if you have it the right way too.
Okay so I got this link: http://www.physchem.co.za/Current%20.../Resistors.htm
and about halfway through it has the resistors in parallel thing. So basically i have
| (10ohm R) |
Lead from LM chip--- |-------| (10ohm R) |-----|---- Potentiometer
| (10 ohm R)|

WOW that took me a while to type ;D
So the leads from one eod of the resistors would be soldered together and to the LM, the other three would be soldered together and to the pot...

Im off to solder, just tell me if I did it right.
THanks

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Old 11-14-2007, 12:12 AM #4
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Default Re: Once again another topic regarding amk's DRIVE

Okay back now with results:

I hooked up the resistors in parallel, just to my luck I had 3 extra 10 ohms lying around.
after hooking up a load, and testing, I got:
380mA
Which is better than 120 for sure, yet one question:
Will 380 kill a SenKat diode (16x Sony)?
Ill try and get some pics of my driver/enclosure onto this thread

AND

about the multimeter, it has an hFE thing, with six holes to stick leads into, the six are seperated onto two groups by ink lines, three of the holes are labled NPN and the other group is PNP. I was wondering if this is for the diode, my uncle says its to test transistors. Please help,
THanks,

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Old 11-14-2007, 12:17 AM #5
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Default Re: Once again another topic regarding amk's DRIVE

Your uncle is correct..

380ma's is too much...but your circuit is working properly. Remove one of the resistors and you should end up with 250ma's...this should be OK with the GB diode..I have been running mine with 250ma's.

To figure the resistor you want you take 1.25 and divide it by the milliamps. 1.25/.250=5ohms..
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Old 11-14-2007, 12:22 AM #6
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Default Re: Once again another topic regarding amk's DRIVE

Quote:
Originally Posted by Gazoo
Your uncle is correct..

380ma's is too much...but your circuit is working properly. Remove one of the resistors and you should end up with 250ma's...this should be OK with the GB diode..I have been running mine with 250ma's.

To figure the resistor you want you take 1.25 and divide it by the milliamps. 1.25/.250=5ohms..
okay thanks, so I should take one of the 10 ohms off. Will 250mA's be high current to the point where it will be powerful? (there was a thread where the author said hooking a 4ohm gives it lots of power making it a good burner) Will I get more power if I hook up a 4 ohm? thanks, im off to de solder one of the resistors.

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Old 11-14-2007, 12:33 AM #7
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Default Re: Once again another topic regarding amk's DRIVE

The diode begins to burn very well with only 160ma's..so the answer is yes...250ma' will give you ~150mw's out..plenty to burn.. Also make sure all your connections are good and that you have your meter connected BEFORE you apply power to the diode..
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Old 11-14-2007, 12:58 AM #8
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Default Re: Once again another topic regarding amk's DRIVE

Quote:
Originally Posted by Gazoo
The diode begins to burn very well with only 160ma's..so the answer is yes...250ma' will give you ~150mw's out..plenty to burn.. Also make sure all your connections are good and that you have your meter connected BEFORE you apply power to the diode..

okay I fixed it up and I can get a max of 260 out of my circuit, i turn the pot a bit and get 250.
I thought about it, and anyhting over 250 or so will shorten diode life, right? Whats the mW approx I can get out of 300mA? or will I get an LED with 300mA lol?

Thanks for the advice, okay Ill hook up my meter and set it to 250 before connecting the diode.

One thing, when I hook the meter on, and insert the red handle metal point thingy that is hooked up to my cricuit to the meter (the black is already hooked up) my load stops working and I get an amp reading. My load stays off untill I disconnect my red thingy from the meter. Is this natural? thanks.

*update* one more thing will my diode be safe at 260mA? will I get any more mW from it to the point where it would be worth it? thanks. *update*

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Old 11-14-2007, 01:06 AM #9
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Default Re: Once again another topic regarding amk's DRIVE

Do you have your meter connected in series with the circuit? It should be...

Your diode should be safe at 260ma's... but every diode has its own personality so it is hard to determine when it will turn into an LED. No doubt it will last longer with a lower current. Also make sure it doesn't get too warm as this will also effect its life.
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Old 11-14-2007, 01:57 AM #10
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Default Re: Once again another topic regarding amk's DRIVE

Quote:
Originally Posted by Gazoo
Do you have your meter connected in series with the circuit? It should be...
idonno here it is
|+ on load
+ lead from circuit----------|
|+ on meter

|- on load
- lead from circuit-----------|
|- on meter


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Old 11-14-2007, 02:08 AM #11
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Default Re: Once again another topic regarding amk's DRIVE

You are connecting it in parallel with the load which is wrong. You need to connect it in series with the load. The diagram on the following web page will help:

http://www.scienceshareware.com/how-...with-a-dmm.htm
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Old 11-14-2007, 02:53 AM #12
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Default Re: Once again another topic regarding amk's DRIVE

I would not recommend doing what you are about to do. *Do not measure the current by putting the DMM in series with the load. *

Will it work and give accurate results - YES.

Is it likely that a test lead will slip off the circuit in test and cause a momentary open circuit - YES. *
Ask youself what happens in this situation. *When the lead slips off the current in the circuit goes to 0. The LM317 will attempt to maintain the designed current - it does this by increasing the Vout. *The Vout will almost instantly rise to the V(batt) - V(LM317dropout). *When the lead touches again and the circuit is restored, the Laser Diode will see this voltage. *(another good reason to not put in too much capacitance on the load side of the LM317).

OK, that was the DO NOT. *Here's the DO:

You have 2 other good options to measure the current with a LM317 set as a current source.

1. Measure the current from the battery side of the LM317. *It will be the same as the LD current due to Kirchoff's Law. If the lead slips - who cares - the power source becomes disconnected so no damage to the LD.

or

2. Measure the voltage drop across the fixed resistors and calculate the current. *For example: let's say you have 2x 10ohm resistors in parallel for an effective resistance of 5ohms. *If you measure a 1.25V drop across them, the current must be 0.25A (250mA). *(I= V/R = 1.25/5)
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Old 11-14-2007, 02:54 AM #13
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Default Re: Once again another topic regarding amk's DRIVE

Quote:
Originally Posted by Gazoo
You are connecting it in parallel with the load which is wrong. You need to connect it in series with the load. The diagram on the following web page will help:

http://www.scienceshareware.com/how-...with-a-dmm.htm
WHen in parallel, load or no load, it shows anywhere from 50mA to 270mA depending on what my pot is set to.

When in series with a load, I get 20mA, which is totally wrong....

help plz

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Old 11-14-2007, 02:57 AM #14
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Default Re: Once again another topic regarding amk's DRIVE

Quote:
Originally Posted by chimo
OK, that was the DO NOT. *Here's the DO:

You have 2 other good options to measure the current with a LM317 set as a current source.

1. Measure the current from the battery side of the LM317. *It will be the same as the LD current due to Kirchoff's Law. *If the lead slips - who cares - the power source becomes disconnected so no damage to the LD.

or

2. Measure the voltage drop across the fixed resistors and calculate the current. *For example: let's say you have 2x 10ohm resistors in parallel for an effective resistance of 5ohms. *If you measure a 1.25V drop across them, the current must be 0.25A (250mA). *(I= V/R = 1.25/5)
for #1, does that mean just hook up DMM to output leads and measure current?

how do I measure the voltage drop for option 2? plz help

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Old 11-14-2007, 03:14 AM #15
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Default Re: Once again another topic regarding amk's DRIVE

Quote:
Originally Posted by amkdeath
[quote author=chimo link=1194933143/0#11 date=1195012423]
OK, that was the DO NOT. *Here's the DO:

You have 2 other good options to measure the current with a LM317 set as a current source.

1. Measure the current from the battery side of the LM317. *It will be the same as the LD current due to Kirchoff's Law. *If the lead slips - who cares - the power source becomes disconnected so no damage to the LD.

or

2. Measure the voltage drop across the fixed resistors and calculate the current. *For example: let's say you have 2x 10ohm resistors in parallel for an effective resistance of 5ohms. *If you measure a 1.25V drop across them, the current must be 0.25A (250mA). *(I= V/R = 1.25/5)
for #1, does that mean just hook up DMM to output leads and measure current?

how do I measure the voltage drop for option 2? plz help

amk[/quote]

#1 measure current on THE BATTERY SIDE
#2 measure voltage

I would recommend that you google about measurement techiques using a DMM and do some reading. The process may take you longer but it will end up costing less in fried LDs. Good luck.
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Old 11-14-2007, 05:24 AM #16
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Default Re: Once again another topic regarding amk's DRIVE

Maybe I missed it but what are you using for a load?
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