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Topic: How to TTL a laser? (Read 36377 times) previous topic - next topic

123Splat


In that circuit i see a constant 5v regulator and a constant current of 300ma.....

So what are you saying?

In the circuit, as drawn, I see a 7805 linear voltage regulator being fed by a 1.5V cell.  But assuming that the poster meant that the source voltage was, infact 7V or above, as required by the regulator, the regulator will supply 5V out, as long as the supply can keep input at, or over, 7V; at a MAXIMUM of 1A draw, as long the supply can supply 1A or more.  The output is only regulated to 5V at a max of 1A.  if the input current (source battery depleation, for example) drops below 1A, so does the output current.  One Amp in, through 10 Ohms (disregarding losses in the transistor) is 500mA, not 300mA. What I am saying is read the datasheet.

MarkT,
"But a series resistor is one perfectly good way to get constant current drive..." for a first year E.T. student, or maybe more accurately said, Until the resistor heats up and changes value (remember what lousey temp coefficients resistors have?).

I do a lot of playing around LASER diodes at various wavelengths, and the first rule (after use your goggles) is ALWAYS use a constant driver driver circuit.  Then you worry about things like heatsink requirements...
You CAN depend on a resistor for your current control. if you want to (they call it 'Kip Kay'ing'), but you are gonna loose more diodes to over-current-burn-outs than you will to ANY other cause.

You do it however you want. Listen, or not, read the datasheet, or not  I did my part. See ya!


Grumpy_Mike

There is a lot of rubbish in this thread but fortunately 123Splat is speaking sense.
Quote
You do it however you want. Listen, or not, read the datasheet, or not  I did my part.

Thanks.

If you want more info about driving lasers see:-
http://www.repairfaq.org/sam/laserdps.htm

kerimil

Thx for your input - will look into it. Though I think it's a more profound problem for those who run everything at max rated current.  It's naive to assume that there is some magic threshold below which everything is fine and then 2mA above it you've got a burnt diode. Not that I am saying that you aren't right - quite the opposite I think current regulating is such a big deal because of running everything at max rated current

cjdelphi



In that circuit i see a constant 5v regulator and a constant current of 300ma.....

So what are you saying?

In the circuit, as drawn, I see a 7805 linear voltage regulator being fed by a 1.5V cell.  But assuming that the poster meant that the source voltage was, infact 7V or above, as required by the regulator, the regulator will supply 5V out, as long as the supply can keep input at, or over, 7V; at a MAXIMUM of 1A draw, as long the supply can supply 1A or more.  The output is only regulated to 5V at a max of 1A.  if the input current (source battery depleation, for example) drops below 1A, so does the output current.  One Amp in, through 10 Ohms (disregarding losses in the transistor) is 500mA, not 300mA. What I am saying is read the datasheet.


MarkT,
"But a series resistor is one perfectly good way to get constant current drive..." for a first year E.T. student, or maybe more accurately said, Until the resistor heats up and changes value (remember what lousey temp coefficients resistors have?).

I do a lot of playing around LASER diodes at various wavelengths, and the first rule (after use your goggles) is ALWAYS use a constant driver driver circuit.  Then you worry about things like heatsink requirements...
You CAN depend on a resistor for your current control. if you want to (they call it 'Kip Kay'ing'), but you are gonna loose more diodes to over-current-burn-outs than you will to ANY other cause.

You do it however you want. Listen, or not, read the datasheet, or not  I did my part. See ya!




I was taking into acount the forwarf voltage drop of a typical 2v diode.

kerimil

Awesome now I am lost... forward voltage drop has to be taken into consideration, right ? so I should choose resistor value based on:

(supply voltage - forward voltage drop of the LD) / LD planned current

...precisely what I wrote earlier in the thread.

What's more, while temperature has effect on value of resistors I found that the temperature coefficient of most resistors is 0.0050% per 1 deg C. So in other words if the temperature increases by 100 deg C the change in resistance will be 0.5% so way below tolerance of most resistors.

so... who is right ??

Grumpy_Mike

Laser diodes are a lot more complex than LEDs. The forward voltage drop changes with temperature, that is why a resistor will not cut it. Read the link I posted.

MarkT

It does if you have enough voltage drop (12V supply to blue laser for instance),
although its crude and wasteful of power, its good enough as a constant current
source - you make sure you aren't pushing the diode too hard of course...

Ideally a DC-DC converter + constant current driver would be employed, indeed,
but thats extra cost and complexity and the budget went on the laser diode and a 12V
supply was already there...
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

runaway_pancake

Why not link to some data regarding this "laser module" of yours?
To some it's about a loose "laser diode", I guess, while others figure it's a prepped ready-to-go "module" (like for "pointers") or something.

"Who is like unto the beast? who is able to make war with him?"
When all else fails, check your wiring!

kerimil

Now here is another question related to this topic. It might be a bit stupid but anyway...
Can I use a multimeter to check polarity of a laser diode ? (using resistance setting)

Can I check it's voltage drop using a "Diode check" function that some multimeters have ?
If, yes I assume I might as well test it by measuring voltage across it when more current is passed through it - as described here ->

Grumpy_Mike

Quote
Can I use a multimeter to check polarity of a laser diode

Probbley, I do this to test LEDs especially surface mount ones.

The diagram is the way to measure the volts drop.

kerimil

Thx. Yeah the pic shows an alternative method because diode check function supposedly provides voltage drop value of the diode, but because there's not much current flowing it is a bit off.

Grumpy_Mike

When I test an LED with the resistance or continuity function of a meter the LED glows very dimly when the red meter wire is on the anode and the black is on the cathode.

123Splat

If you are talking LASER diode in a TO type can, looking at the rear (bottom), you will notice that there are three notches in the rim (Base) of the can, two V shaped and one Rectangular.  If you hold the can with the square or rectangular notch to the top (12 o'clock position) and the V shaped notches at 9 o'clock and 3 o'clock, you will see three or four pins: pin #1 at 9o'clock, #2 at 12 o'clock (attached to the case), #3 at 3 o'clock, and, if there is one, #4 at 6 o'clock.

USUALLY, for I.R. diodes (also what is used in 'Green DPSS modules), the case (and pin#2) are positive. For reds,blues and most violets, the case is negative.  Pin #1 will usually be negative for I.R. and POSITIVE for the rest.  Observe current and static electricty precautions. Don't look into the light.......

123Splat

Also, not usually a very good idea to use the multimeter in Ohms setting to check LASERs. Could cause current damage.

polymorph

A bare LASER diode typically has a built-in photodiode for power feedback. So make sure you are actually measuring the LASER diode itself.

However, if it has the controller circuit board on it, you can't just check it with a continuity meter.
Steve Greenfield AE7HD
Drawing Schematics: tinyurl.com/23mo9pf - tinyurl.com/o97ysyx - https://tinyurl.com/Technote8
Multitasking: forum.arduino.cc/index.php?topic=223286.0
gammon.com.au/blink - gammon.com.au/serial - gammon.com.au/interrupts

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