Driver build problems

BennyF · Mar 17, 2011

So i have a 1W 445nm diode and i'm trying to build a driver to power it. However I've used the post/ tutorial that was put up by ROG8811. I've done the math for his circuit and for me to get 125mA I would need a 10 ohm resistor after the potentiometer. however when i do this i get around 11mV on my dummy load at the 1ohm resistor. I'm looking to make about 800mA so i would need a 1.5 ohm resistor. when i do that i get 12.4mV across the 1 ohm resistor. across my 6 diodes i get 4.2V and at the regulator i have 1.25V im powering my circuit with a 9V. im just not sure what im doing wrong if all the voltages are right except the one across the resistor. I've rebuilt the circuit almost 10 times and im almost 100% sure its correct. I need help. :yabbem:

lasersbee · Mar 18, 2011

If you are powering your driver with a 9Volt battery...
I would say your problem is there..
A 9volt battery can't reliably supply enough current...
(depending on the type of 9V bat)
When you test your circuit with a test load.... check
the voltage at the input of your driver..

Jerry

BennyF · Mar 18, 2011

i have exactly 9V on the input.

BennyF · Mar 19, 2011

i went and got some new batteries 3.6V 2200mAh plugged them into my circuit and still get about 10.5 volts. what am i doing wrong?

123splat · Mar 20, 2011

Benny,
How about a schematic. Much easier to figure out what is happening with a picture.

Toke · Mar 20, 2011

BennyF said:
i went and got some new batteries 3.6V 2200mAh plugged them into my circuit and still get about 10.5 volts. what am i doing wrong?

One thing you might be doing wrong is supplying your driver with less than the 7-8 Volt required for that driver/diode combination.

Apart from that there are a huge number of errors possible, it is hard to see what you are doing from here.

BennyF · Mar 20, 2011

please excuse the yellow boxes its harder to read then i thought just use the schematic and you'll understand.

Toke · Mar 20, 2011

It looks like you have the current set with 100.5 Ohm, that will give you 1.25/100.5 = 12.5mA, and app. 12.5 mV across the resistor on your testload.

No mystery here.

ETA: Resistors in series are simply added up. Resistors in parallel are inverted to conductivity (S), added together, and the result then inverted back to resistance.

e.g. 5 * 10Ohm resistors in parallel are inverted to 5 * 0.1S and then inverted back to 2Ohm.

BennyF · Mar 20, 2011

i realize that and i know the math but the 100 ohms should be giving my LM317 the 1.25 V if i take it out and use the 1.5 Ohms that i calculated for (i know i have .5Ohms on the schematic) it only gives me around 50mV. The voltage should give me around 800mV or so. This is what i'm not getting. the LM317T is new to me i haven't really ever used it so i don't know how it works. If you have any suggestions i am really open to them. please help me make this better. I refuse to hook my diode to this circuit till i get the correct readings. is my math wrong? for 800mV i need 1.6 Ohms. but now that i took the 100ohm resistor out i dont have the 1.25V for the LM317T.

Toke · Mar 20, 2011

??????
Why are you still using the 100Ohm resistor?

The idea here is that the current for the diode goes through your sense resistance and give a voltage difference of 1.25V
That does mean something like 1.5 Ohm for 800mA, the two 1 Ohm in parallel gives 0.5 Ohm, adding a 100 Ohm in series gives 100.5 Ohm.

I recommend you browse through Wiki under Ohm's law, the U=R*I thing.

Toke · Mar 20, 2011

??????
Why are you still using the 100Ohm resistor?

The idea here is that the current for the diode goes through your sense resistance and give a voltage difference of 1.25V
That does mean something like 1.5 Ohm for 800mA, the two 1 Ohm in parallel gives 0.5 Ohm, adding a 100 Ohm in series gives 100.5 Ohm.

I recommend you browse through Wiki under Ohm's law, the U=R*I thing.

alf638 · Mar 20, 2011

Yes, I agree. Just try it without the 100 ohm and see if it works.

Toke · Mar 20, 2011

NO
Not simply without the 100 Ohm, with a 1 Ohm instead of it.
So two 1 Ohm in parallel and one 1 Ohm in series with it.

The only downside is that 800mA through a 1 Ohm resistor will give 0.8V voltage drop for an effect of (P=U*I) 0.8A * 0.8V = 0.64W.
Your resistors look like 0.25W ones and will burn off pretty quickly.

BennyF · Mar 20, 2011

Like i said if i take the 100 ohms out and just use 1.5 ohms i get around 50mV. What I'm saying though is that with just 1.5 ohm's i don't get the 1.25V from the LM317. and i end up with somewhere around 4.5V across my 6 Diodes on my dummy load.

Toke · Mar 20, 2011

6 diodes will give app. 4.5V no matter what the current through them are, that is kind of the point with using them as test load, that they emulate a LD.

So, 1.5 Ohm gives you 50mV on the test load resistor. ???
Are the batteries freshly charged?
The 7.2V is kind of marginal for this.

BennyF · Mar 20, 2011

they aren't really fresh anymore since i've been trying to figure this out, so ill have to charge them.

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