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FrozenGate by Avery

Nugm02 circuit resistance

A

Abel

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Feb 16, 2020
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Hello everybody,
I have a nugm02 diode, and I want to make a circuit for it, but I don't really know how. if I use a 9v battery (or power supply), and I want to put 4,86v at 2A into the laser what kind of resistor do I need? Because I am very confused...

Thanks for helpingšŸ˜
 





You need to regulate the current going into the LD, this can be done by either wasting all the excess power of the battery through use of a resistor in parallel (BAD idea since heat), rapidly adding/taking energy out of a capacitor using an LM317 voltage regulator or the likes (better, but also highly inefficient), or using an inductor to store the excess energy (highly efficient, but complicated)

For your purposes you are best off buying a purpose-built current regulator circuit from ebay or something (they use the inductor approach afaik):
www.ebay.com

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gazer101 said:
You need to regulate the current going into the LD, this can be done by either wasting all the excess power of the battery through use of a resistor in parallel (BAD idea since heat), rapidly adding/taking energy out of a capacitor using an LM317 voltage regulator or the likes (better, but also highly inefficient), or using an inductor to store the excess energy (highly efficient, but complicated)

For your purposes you are best off buying a purpose-built current regulator circuit from ebay or something (they use the inductor approach afaik):
www.ebay.com

3.7V Current Driver Board for 445nm 447nm 450nm Blue 1w 1.4w 2W Laser Diode 9315509996213 | eBay

Find many great new & used options and get the best deals for 3.7V Current Driver Board for 445nm 447nm 450nm Blue 1w 1.4w 2W Laser Diode at the best online prices at eBay! Free shipping for many products!
www.ebay.com
Click to expand...
Sorry but "inductor storing excess energy" is just outright wrong.
 
I feel the need to add that a 9V battery is one of the worst possible choices for power. They simply can't provide the 2 amps you're after for more than a few seconds. You need Lithium ion, nickel MH or another rechargeable chemistry capable of high discharge current.
 
WizardG said:
I feel the need to add that a 9V battery is one of the worst possible choices for power. They simply can't provide the 2 amps you're after for more than a few seconds. You need Lithium ion, nickel MH or another rechargeable chemistry capable of high discharge current.
Click to expand...
Yea, from my experience, a 9v is capable of around 500mA of sustained output without too much drop.
 
Giannis_TDM said:
Sorry but "inductor storing excess energy" is just outright wrong.
Click to expand...
Please correct me! What purpose does the inductor actually serve?
Does it not store energy from the battery using rapidly switching mosfets which can regulate the amount of current put into it per unit time as per a pwm signal?

Is this video fake news?:
 
Last edited:
The inductor isn't 'storing excess energy'. It's acting more like a flywheel. The MOSFET initiates a current through the inductor and when the MOSFET switches off, the inductor keeps the current flowing. When the energy of magnetic field in the inductor has been spent, keeping the current going, the MOSFET switches back on. Lather, rinse, repeat....several 10s or 100s of thousands of times per second.
 
WizardG said:
The inductor isn't 'storing excess energy'. It's acting more like a flywheel. The MOSFET initiates a current through the inductor and when the MOSFET switches off, the inductor keeps the current flowing. When the energy of magnetic field in the inductor has been spent, keeping the current going, the MOSFET switches back on. Lather, rinse, repeat....several 10s or 100s of thousands of times per second.
Click to expand...
Yes that's what I initially meant by excess energy. Since the battery's current fully unleashed onto the LD it would fry it, the inductor stores the energy from the battery and releases it in controlled bursts which are smoothed out by a capacitor
 
Almost there. The inductor itself smooths out the current delivery. The capacitor at the end of the chain is generally of very small value and acts to clean up switching noise.

Think of it as a single cylinder engine. The inductor is the flywheel, the MOSFET is the piston, and the capacitor is the muffler. It's not a perfect analogy but I'm frikkin' tired right now

:sleep:
 
WizardG said:
Almost there. The inductor itself smooths out the current delivery. The capacitor at the end of the chain is generally of very small value and acts to clean up switching noise.

Think of it as a single cylinder engine. The inductor is the flywheel, the MOSFET is the piston, and the capacitor is the muffler. It's not a perfect analogy but I'm frikkin' tired right now

:sleep:
Click to expand...
Thanks! GN lol
 
Okay, if I use two 18650 battery's how much ohms of Resistance do I need?
 
Forget the resistor idea as a driver. You need a buck driver and two Li-ion batteries. That will insure that you regulate the current the diode needs and you won't have all that extra heat from a resistor. A buck driver is more efficient than a linear one would be too.

That video that gazer101 posted is only for a voltage regulator. The circuit is different for a current regulator.
 
paul1598419 said:
Forget the resistor idea as a driver. You need a buck driver and two Li-ion batteries. That will insure that you regulate the current the diode needs and you won't have all that extra heat from a resistor. A buck driver is more efficient than a linear one would be too.

That video that gazer101 posted is only for a voltage regulator. The circuit is different for a current regulator.
Click to expand...
The fundamental circuit is not any different, Only the feedback is, instead of an opamp comparing against a known voltage reference using an output voltage divider you are comparing said internal reference with the drop of a known value shunt that you have calculated will set the current at X. Only other difference between them is that the reference voltage for current regulators is typically much lower to minimize the required shunt size, in term minimizing the burden voltage and losses. A CV IC can be easily made CC with as few external components as a dual opamp, shunt, supporting resistors and a diode. This aforementioned using a CV ic and 'hacking' the feedback loop has been done on a multitude of drivers from X-wosse, laserrer and astralist. The only downside is that getting it to work correctly is very finicky from component selection to math so that's why I don't use such systems in mine, the less stuff that can go wrong, the better!
Also, I should mention that If you want decent current regulation characteristics with the aforementioned system you need to choose a very high slew rate opamp with a high bandwidth, discrete shunt amp IC is even better.
 
I never said the circuit was incredibly different. Only that they are not the same. And yes it is difficult to try to use a voltage regulator buck that you've tried to modify into a current regulator.
 
paul1598419 said:
I never said the circuit was incredibly different. Only that they are not the same. And yes it is difficult to try to use a voltage regulator buck that you've tried to modify into a current regulator.
Click to expand...
Who said it's difficult? It's extremely simple, getting it to perform well is the issue and it will almost never be as stable as native cc regulators, mainly due to them effectively thermally coupling their voltage reference to the rest of the die improving tempco.
 
There is no magic or "better than anyone else" type of laser driver. U gotta just understand basic science behind what you are looking for and find a driver that suits your needs

Yo on that note @Giannis_TDM, do you think I could use an ATX PSU's PCI-E output to drive a 12v laser array? Those things are cheap as hell and (if their advertising is to be believed) should have low ripple
 
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