my 1st effort at building a driver - help

[rockers]

Hello all, hope this is the correct place to post,

After reading this thread - http://laserpointerforums.com/f51/i-want-build-laser-thread-52972.html

I thought I would have a go at making the driver from this - Laser driver - It can be done

Got all the componets and a pcb, the only problem i had my local store (maplins) didnt have the correct pot i wanted, one close to the spec was this - Cermet Preset Potentiometers : Potentiometers : Maplin

Got it all together as per the diagram, used a normal white led to test it worked, which it didnt but as i move somthing it did light up.
bassically shorting it across regulator the + point and the middle one made it work, using a digital multi meter across the led i can adjust using the pot from 50ma up to 800ma.

So does the end result sound like how it should work ? i thought buying the 100R pot would give me 0-100ma range ? or could they have given me the wrong one ? as only came in a clear plastic bag.

Dont under stand how this isnt working without having to short it, could the regulator be not working there for shorting the + in to the middle point continued the circuit ?

Thanks

Rocky

 

[Asherz]

Using an LED to emulate the current across a Laser diode just doesn't work, the current your reading across the LED will be completly different to what you would be measuring across a laser diode.

To measure the current output of your driver accuratly you need to use a test load, normally consisting of 4-6 1N4001 diodes (how many depends on which colour laser diode you want to set your drivers current for) in series with a 1ohm resistor, you then set your DMM to read volts and measure across your 1ohm resistor.

If you use ohm's law I=V/R, as you are measuring across a 1ohm resistor, the reading you get in volts, will be the same as the current, thus you can set the current before you put your laser diode in to be powered by the driver. (essentially what you are doing with the LED, but with the correct test load.)

 

[rockers]

Hello, thanks for the reply

I did wonder if a LED would give me the readings i need, will be hopefully using a red LD so will get a test load circuit made up for that.
Just need to know why I'm having to short the regulator for the circuit to work ?

Thanks

Rocky

 

[rockers]

Hello again, right have replaced the regualtor with a new one and the circuit is complete.
Have also made a test load now consisting of 4 1n4001 diodes and a 1 ohm resistor as per my 2nd link, set it all up and measured across the resistor with DMM set to volts, and i can adjust between .01 and .1 so would i right in guessing that is 1 and 100 mah i can adjust up between ?


Didnt seem half as hard todo once i had read those links a few times, thanks to who ever made them.

next need a laser diode and lens
Cheers

 

[Wolfman29]

No no no no. A 100 Ohm pot does NOT mean you can set between 0 and 100mA in current. Depending on how your circuit is set up, it could mean various things. Essentially, all a potentiometer is is a variable resistor, letting you use varying resistances to set current.

However, the current you get follows this rule, generally speaking:

I(current in Amps) = 1.25/R.

So if you are only using a 100 Ohm pot as your sole resistor, then at max you will be getting something like 1.25mA, and at max, you will get a couple of amps (inherent current in the wiring, potentiometer, etc. will prevent you from getting "infinite" current).

 

[rockers]

OH dear, so back to being confused again, what i have made is laid out the same as in the 1st link i posted the same componets aswell.

Cheers

 

[lasersbee]

If you used this exact drawing and parts and
voltage input...

then the Minimum current your Driver will put
out is 1.25/105 (5 +100) Ohms = ~11mA

And the Maximum current your Driver will put
out is 1.25/5 Ohms = ~250mA

There are always tolerances in passive parts
hence the ~...


Jerry

 

[Wolfman29]

So let me explain how the LM317 DDL driver works.

Linear regulators work by assuring that there is a constant voltage between the output and the adjust pins. Generally, that constant voltage, also known as the reference voltage, is 1.25V.

So what *we* do is take advantage of that constant voltage and apply Ohm's Law: we put a resistor between the output and the adjust pins so that the current that flows from the output pin, through the resistor, and then to the adjust pin is what we want it to be.

Therefore, the equation is I=1.25/R.

We then stick our output lead on the adjust pin so that the current can then go to our diode from there.

It gets a bit more complex if you start using parallel and serial resistors across the adjust pin, but the basic idea is the same - just stick your resistors between output and adjust, measure the resistance, and then, using that equation will give you the current your driver will output.

 

[rockers]

Hello, cheers for the help guys.
Ok so the figure 1.25 is what i use to work out the current my circuit should be putting out min and max ?
Equation you have - I=1.25/R I - the mA ? and R the resistance ? or ohm?

Tbh I cant see anything wrong with the circuit i have created, other than the pot I'm using is there a right and wrong way ? as the middle pin had to bend towards the others a little so i could use it in the way the circuit diagram states.

Cheers

 

[Wolfman29]

So essentially, you have to work out the minimum resistance the maximum resistance in your circuit (if you are using a pot).

There are numerous parallel and serial calculators out there on the web, so google those.

Regarding the equation, I is in amps, R is in resistance, which is the same thing as Ohms (Ohms is the unit we use for resistance).

For the pot, you should only have to use two leads (the single one on one side and one of the other ones (both are not needed).

Mind taking a picture of your set up? And what resistors are you using? Just the pot? Or are you using resistors in series/parallel with the pot, as well, like on rog's site? Please elaborate on your set-up.

 

[rockers]

Yes, I am using the same components in the same setup that are used on rog's site the only thing different is the pot I'm using this one Cermet Preset Potentiometers : Potentiometers : Maplin
100ohm one.
Will get some pictures tomorrow, will have to excuse the soldering

Cheers

 

[rockers]

Yes, I am using the same components in the same setup that are used on rog's site the only thing different is the pot I'm using this one Cermet Preset Potentiometers : Potentiometers : Maplin
100ohm one.
Will get some pictures tomorrow, will have to excuse the soldering

Cheers

 

[Wolfman29]

In that case, Lasersbee is right - the range is 1.25/(5+100) = 11.9mA to 1.25/5 = 250mA.

Turning your pot all the way one way will result in 11.9mA, and the other way is 250mA. Regarding the right pot - really, the only thing a different pot will restrict is the lower limit. The higher the maximum resistance on the lower limit, the lower the minimum output current.

 

[rockers]

When Laserbee has those figures, where does the 5 come from ? 100 being the 100ohm - max resistance on the pot ?

On a little of topic how would get more than 250mA ? as thats no resistance from the pot so where would the current be changed from ?

Will get some pictures of what i have so far and also of my test load circuit.

Sorry for all the questions, just trying to get my head round things.

Cheers

 

[Wolfman29]

The 5 comes from the two 10 Ohm resistors in parallel. Their equivalent resistance is 5 ohms. And yes, you are right about the 100.

In order to get more than 250mA, you would need to decrease the resistances of those set resistors, i.e. replace them with 5 Ohm resistors or something.

And it's not a problem.

 

[rockers]

ok so finally got some pitcures. not brilliant tho,
1st picture, battery + bottom right, battery - bottom middle/left, there is a break in the pcb i have put in to stop them shorting
there the pot is, the middle pin in between the other two, there is also a break there between the 1st pin and 2nd two pins.


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and 2nd is the load test circuit.


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hope somthing is under stood, i need to set my Dmm to volts to check the current ?

Cheers