Current/driver/test load question

[SwearBear]

Hey, people. I'm new to the whole laser community and as you can see, this is my first post on the forum. So I was wondering if someone could answer me a couple of questions about my first "build".
I'm trying to set current on an AixiZ driver for a 803t? 150mw Blu-ray diode. I made a testload from salvaged diodes (3x1N4004, 2x1N4001, 1x1N4007, all of which are almost the same according to datasheets I've come across) and an old 11.2 ohm resistor, as registered on my DMM (although it's marked as 1 ohm - brown, black, black stripes). I'm actually measuring current, not voltage. I'm using analog ampere meter. With the test load in series, it shows about 97-98 mA. Then I tried to take a measurement without a test load and it shows 100 mA. Now, if the resistor is indeed ~10 ohm, then shouldn't the reading with load be 10 times smaller? I also gotta mention that my DMM is acting funny (showing random 0.1 - 0.3 volts/amps when probes aren't even touching anything), so it may be out of calibration, so 1ohm-marked resistor may really be a 1ohm resistor. But even so, why is the mA reading almost the same with test load as without?

Test load

One of my early mistakes - I connected the driver to the batteries with wrong polarity. Something blew. I assume it was a fuse of some sort. I saw that it had same white stripe on one side as diodes do, so I replaced it with some diode I salvaged from somethig. The whole thing does work, with the Blu-ray, but I don't want to make anything permenant out of it until I set the current right.

 

[Cyparagon]

it's marked as 1 ohm - brown, black, black stripes

No, brown black black gold is 10*10° = 10Ω±5%. The resistance of your meter leads is probably around 1Ω, and your meter is probably accurate to ±5%.
All is accounted for

if the resistor is indeed ~10 ohm, then shouldn't the reading with load be 10 times smaller?

Drivers are constant-current devices.

Something blew. I assume it was a fuse of some sort.

Drivers don't generally have fuses



The whole point of a test load is so that you don't have to use an ammeter. If you know the voltage drop across a resistor and you know the resistor value, you can calculate the current through the resistor. Google ohm's law.

 

[SwearBear]

No, brown black black gold is 10*10° = 10Ω±5%. The resistance of your meter leads is probably around 1Ω, and your meter is probably accurate to ±5%.
All is accounted for
.

Ahh, so in that case would it be black, brown, black?

The whole point of a test load is so that you don't have to use an ammeter. If you know the voltage drop across a resistor and you know the resistor value, you can calculate the current through the resistor. Google ohm's law.

Ohm's law: V=IR

Resistor value = 11.1ohm
Voltage w/o load = 6.5v
Voltage across resistor = 1.2v
Voltage drop = 5.3v

So this is what I got: I=5.3/11.1=0.4775

Something's fishy.

 

[123splat]

With your meter set to Volts or Amps and the leads not attached to anything, the leads (particularly the +) are acting as antennas and picking up stray radiated AC. Try shorting the leads together, your meter should read zero (if not, get a new meter). then set meter to Ohms to see what your leads introduce into any R measurements (subtract that number from any LOW resistance readings to get a better idea of true value).

 

[Cyparagon]

No, the resistor's voltage drop is the voltage across the resistor. 1.2 / 11.1 = .108A = 108mA.

The color code of a 1Ω is brown black gold (gold is 10^-1)

 

[chipdouglas]

if you have an android operating system on your phone go to the app store and download "electrodroid" It is free. it lets you input color of resistor bands and tells you what they are. it has many other electronics features.

 

[SwearBear]

With your meter set to Volts or Amps and the leads not attached to anything, the leads (particularly the +) are acting as antennas and picking up stray radiated AC. Try shorting the leads together, your meter should read zero (if not, get a new meter). then set meter to Ohms to see what your leads introduce into any R measurements (subtract that number from any LOW resistance readings to get a better idea of true value).

Hey, thanks that explains a lot. When I touched the leads it did read zero. Learn something new every day.

No, the resistor's voltage drop is the voltage across the resistor. 1.2 / 11.1 = .108A = 108mA.

The color code of a 1Ω is brown black gold (gold is 10^-1)

Yesss, now it makes sense. On both counts. Thanks big time, Cyparagon.

if you have an android operating system on your phone go to the app store and download "electrodroid" It is free. it lets you input color of resistor bands and tells you what they are. it has many other electronics features.

No, I'm old school. I still use one of those brick cell phones. But no, I just use this site: Resistors. Thanks anyway.


Appreciate the help, guys. All that set me back on the right track. For now, anyways.

 

[Kevlar]

if you have an android operating system on your phone go to the app store and download "electrodroid" It is free. it lets you input color of resistor bands and tells you what they are. it has many other electronics features.

Sorry for the thread jack, I just wanted to chime in and say thanks for the tip Chip!!! I have the Droid X and this app will be very useful. :beer:

 

[chipdouglas]

good luck swearbear... and kevlar. np/. there are other apps. but i found this one to be suited for laserists. it even has an lm317 referrence tool.

michael.

 

[Kevlar]

I checked it out and it just about everything a laser hobbiest could ask for!

 

[chipdouglas]

i'm gonna make a thread in the general section. check it out and add input...