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Topic: How to TTL a laser? (Read 39516 times) previous topic - next topic


123Splat, I think the idea of an Ohmmeter damaging diodes is a holdover from the old analog meter days. In the lowest resistance setting, a higher current flowed through them, and the old low current Ge diodes and transistors could be damaged. So the recommendation was to use the next higher Ohm setting.

That has not been an issue for modern DMMs. However, I will say that I cannot speak for all the piece of cXXp cheapo DMMs out there.
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


Look at the avatar,,,, how modern do you think I am?......

Good point, but I tend to go on the safe side with LASER diodes, most of the ones I play with are about $50 a whack.  Smoke hurts a little more than on a $1 or less transistor...

Side note:  Most manufacturers stopped building in the photodiode a couple of years ago.  unless you are getting an older unit, or scavenging out of old equipment, you will no longer find pin 3 or 4 connected to the photodiode (use external optical limiting nowdays).


Safe side - good point. I'm thinking of the $5 LASER pointers.

I didn't know they were leaving out the photodiodes, now.

As for modern - the first LASER diode I owned was $27 in the late '70s, a fortune, and required something like 2 or 3A to reach LASER threshold but could only sustain a very short pulse. So an SCR would dump a charged capacitor into it. If it didn't reach the lasing threshold, almost the entire capacitor energy would go to heat and blow the diode. Absolutely forbidden to use a VOM's ohmmeter scale to test it.

I only fired it a few times, then made a mistake and didn't let the capacitor charge high enough. Crack, no more LASER.
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


First laser was 1995 for me, and my last powerful one was a 200mw greenie,  burnt bags lit matches, then after almost losing my right eye,  i tried to increase its power and killed it $150 put me off lasers for a while, switched to high powered flashlights . .

I do love uv lasers though and green glow in the dark paper, draw by light.


ohh if anyone is curious - yes, you can find polarity of a LD with a multimeter

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.......
on the ones I have the case is negative and the 9 o'clock pin is positive...
BTW they're kind of scary to be honest - cut through black paper but are totally invisible to the naked eye
luckily I've got a cheap webcam that pics it up well. I am not even in the same room when I power it on. 


Hi to all if anyone is still monitoring this page any longer,

I have just come across the same problem (i think) i have a 3W laser diode which i want to use on a cnc machine i am building the main use for the machine is as a 3D printer but i want to add either a laser engraver or laser cutter by fitting a second head to it.

at the moment I am playing about with a 1W diode and have a laser driver/power supply which states it can be controlled by a TTL input operating at 0= off 5v = on i just do not know what a TTL input is i remember from my distant past that TTL = Transistor Transistor Logic but it is so far back in my past i forgot what it is, is it just a 5 volt signal which is switched on and off at different speeds or is it 0-5 volt adjusted in between the 0 to 5 volt to get different outputs from the laser?

this is the driver/power supply i am using :-


also if anyone can answer is the power supply just a current voltage regulator or does it switch at different frequencies as i want to start using a 3W diode which i have on order but i was wondering if i could just use a constant current/voltage regulator set to the correct requirement to do the same job as i have a couple of these that i was thinking of using:-


as these appear to give the correct current and voltage is there any way i can fit something like a solid state relay to control switching it on and off?

I dont have problems with building things like 3D printers rc quadcopters etc but my knowledge of electronics and programing is sadly lacking and at my age it is getting harder and harder to learn new things even with my having lots of time due to not working due to disability caused in the army so plenty of time + little knowledge = lots of blue smoke when trying new things but it passes the time.

any help gratefully received,

regards Poppy Ann.


Oct 26, 2015, 11:04 am Last Edit: Oct 26, 2015, 11:16 am by Shanjaq
Does this look like a decent TTL Laser Driver?  It uses an op-amp to servo the current using the base-emitter threshold voltage of an NPN transistor as a reference, relative to a sense-resistor under the main transistor and diode.  Very simple, just 3 common parts and some resistors:

TTL Laser Driver

I = Vbe / R

Vbe for NPN (check datasheet, commonly between 0.66 and 0.7) is ~572 in this sim, so that makes:
114.4 Ma = 0.572 / 5


What does "out" represent ? Shouldn't the sense ressitor be tapped at the end not connected to ground instead of on the collector of that transistor ? (I guess I'm missing something)


Oct 26, 2015, 07:00 pm Last Edit: Oct 26, 2015, 08:03 pm by Shanjaq
Oh yes, I forgot to remove that scope probe labeled "Out", for which you can see the trace when running the simulator from the link (right-click the probe and select "view in scope".)  It's basically showing that the control transistor is working to balance the sensed voltage to the internal reference voltage (~0.572V in this case) created by its base-emitter junction.  You'll need to vary this according to the Vbe value/range found in the datasheet for your particular small-signal transistor.

The sense resistor is nearest to ground so it's always relative to the reference voltage, even when the power transistor is in cutoff.  This places it on the Emitter of an NPN, which is technically high-side switching but it's not really that high; just above the reference voltage yet still pulling the LED/Laser down from its positive rail.


I didn't see any connection out from the sense resistor.


Oct 26, 2015, 07:47 pm Last Edit: Oct 26, 2015, 07:53 pm by Shanjaq
Is the sim link loading an incomplete schematic?  Falstad has been known to do that on occasion..  If so, the image attachment should suffice, it's the 5-ohm resistor under the Power transistor to the far right (selected to pull up to the reference voltage ~114mA,) which is connected to the Base of the Control transistor that pulls the non-inverting input of the op-amp down when actively regulating.


Yes, I SEE the sense resistor. I'm saying that if were a real circuit , more than likely you would have a wire coming off that going to the analog input pin to measure the current.  I can see the
voltage drop across the sense resistor is used for the drive circuit.


Oct 26, 2015, 08:13 pm Last Edit: Oct 26, 2015, 08:19 pm by Shanjaq
Oh!  I see what you're getting at.  The circuit posted earlier has a static current setting based on the value of the sense resistor and the Vbe threshold of a small-signal transistor.  The op-amp is what uses the current-sense input internal to the circuit, not an external microcontroller.  (I would not trust the resolution of a uC's ADC or latency/stability of its software to servo the current fast enough and with enough precision to survive any unforeseen changes in supply voltage or thermal resistive changes in the diode, but that's a personal preference since I only have the older Arduinos & some super-slow 8-bit Atmel's..)


Vo = 1 if V+ > V -
Vo = 0 if V + < V-

For H = 5V,
V+ = 3.4V (due to voltage divider)
If current through R = 0.140 A, Vbe = 0.7V, and V+ = 0
thus the comparator output is driving the led driver transistor between 0.5  and 0.7 V.
If the the feedback transistor turns on completely, V+ will be pull below V- , shutting down
down the output of the comparator.

At least that's how I see it. If I am wrong, please correct me.

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