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FrozenGate by Avery

Best Driver to Push 500mw Mitsubishi off single 3.7v Li-Ion Cell.






I'd like to know if Mohgasm round driver 1.25A if ok for this kind of diode.
Is it good to put this diode with 2 cells (2 x 16340)
 
Just4FunReally said:
Subject change, but does this confirm for me that the C6 host using an anodized heat sink as well is electrically isolated from the batteries? I've been meaning to make a C6 Mits laser but I feared that since the case pin is (-), if it wasn't isolated that it would short and poof. :beer:
Click to expand...

Just want to correct this (I know it's an old post). A Mits laser, presuming it's one of the 300mW or 500mW multi-mode 638nm diodes that we're all familiar with, IS NOT case pin (-). It is case pin (neutral).

djLscatt said:
I'd like to know if Mohgasm round driver 1.25A if ok for this kind of diode.
Is it good to put this diode with 2 cells (2 x 16340)
Click to expand...

Not really. It will work, if you use two cells, but it will be so inefficient that the heat it generates will be a problem, and much more quickly than it would be if using it for a 445.

Subtract the Vf of your diode (probably around 2.75V) from the Vin of the driver (as much as 8.4V) and the resulting voltage is what your driver will be dropping. You'll be generating (8.4 - 2.75) * 1.25 = ~7W of waste heat on your driver.

In comparison, using a Mohgasm to driver a 445 at 1.8A is going to generate (8.4V - 5V) * 1.8A = ~6.1W of waste heat.

So you're actually going to be creating more driver heat running your Mits diode at 1.25A than you would driving a 445 diode at 1.8.

My perspective (and others may disagree), is that there are essentially two wise options for driving the Mits 300 and 500 diodes:

1) A buck driver that pulls down the voltage of 1x or 2x lithium ion cells.
2) A linear driver with a low enough dropout that it can run the diode on 1x cell.

That's about it. Boosts don't really have a prominent place in the Mits driving field. In practice, the linears that fit into option #2 are going to be such low dropout, that there's probably not much practical difference in efficiency between them and a buck driver in option #1.
 
thank you for everything RHD.
about what happend with your 950mW 405nm 12x , I have the same problem with a moh driver 478 MA , I took a tailcap measurement, there was only 360MA of current with 2 cells and 478 MA with 3 cells . :thinking:
 
djLscatt said:
thank you for everything RHD.
about what happend with your 950mW 405nm 12x , I have the same problem with a moh driver 478 MA , I took a tailcap measurement, there was only 360MA of current with 2 cells and 478 MA with 3 cells . :thinking:
Click to expand...

That means the driver wasn't regulating with two cells. The voltage of the two cells wasn't high enough to overcome all the voltage drop in the circuit.
With three cells it was enough to regulate the 478mA.
The 405nm diode has a relative high Vf.
 
djLscatt said:
thank you for everything RHD.
about what happend with your 950mW 405nm 12x , I have the same problem with a moh driver 478 MA , I took a tailcap measurement, there was only 360MA of current with 2 cells and 478 MA with 3 cells . :thinking:
Click to expand...

Yep - 1085 based drivers are really only ideal for 445s. But even in that scenario, they're the steam engines to a buck driver's bullet train. It's pretty hard to derail a steam engine. They're big, beefy, and they can take a lot of abuse and keep on trucking. But steam engines require a lot more input to do the same amount of work, and they produce a lot more heat in the process.

I really wouldn't consider using a 1085 for anything other than a 445 at this stage. I like them for the 445s because they're probably the most well explored driver, we know they're safe, we know they are rock solid, and if you heatsink them, they'll run forever. But for 635s, you've got to go buck or super LDO linear, for 405s you need to boost or buck, and even for 445s, I'd probably trust the X-Drive as a buck, on par with the trust I'd place in a rock solid linear.
 
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Blord said:
Read this, http://laserpointerforums.com/f67/how-laser-diode-drivers-work-explanatory-thread-71513.html
It is a good start.
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Yes, but ignore the "reliability" figures, they're pretty absurd and arbitrary.

IE, suggesting that a buck-boost is more reliable than a buck driver, that's ridiculous. If you want to compare individual instances of particular types of drivers, then sure, you can evaluate reliability. But it's nonsense to say that moving from a boost-only driver topology, to one that also adds in buck functionality, would normally improve reliability. I think the rating was based on one particular (in fact the only) boost-buck driver that is frequently used around here, the Flexdrive. The fact is, that just happens to be a driver that (at least according to some people) is pretty good, and fairly reliable. However, the reliability in that instance is the result of good design, not any inherent advantage that a buck-boost has over a boost-only driver.

Those reliability ratings are kind of like saying "Blue cars go faster than green cars, because my blue Porsche is faster than my green Prius". In other words, they mistake characteristics observed of a few particular drivers, to be characteristics inherent in their respective driver topologies.
 
I am still in the testing mode and have some blitz bucks on the way again.. I'll agree with your math RHD I want to pull apart my mits500 atm because I cant get much more than a 2 min run on it. That tells me I am making heat.. I dont think the linear will do this diode any favors.. I want to use that linear for the 9mm diodes...

do the math at 5.6V

2.3A * (8v-5.2v-1.25) = 3.6W Heat in the regulator IC (case for 9mm at 2.3A)
1.8A * (8v-4.2v-1.25) = 4.6W Heat in the regulator IC (case for m140 at 1.8A)

Sorry found that edit button or is this Hogwash??
 
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Personally I use LT3080, I is adjustable up to 1.1A output. And under 1.1A out it only has 300mV drop out, drop out is even lower under lower out put. It can be as low as 80mV under 200mA out. Additionally it can be flowting in the device. It doesn't need a ground. And it's a positive regulator.
 
wee40811,

I was trying to understand this earlier but failed, how do you adjust current output on lt3080. I understand how one sets the voltage on out via resistor on set, but how is the current set? Also by changing the resistor but on Out pin or is it a constant 1.1A out as long as the power-source supports the current draw? In addition if the input voltage is greater than set output does the regulator just turn the rest into heat? I apologize if this question is silly and for being off-topic.

Thank you!
 
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It has a pot.. or potentometer or variable resistor whatever you want to call it... so when you turn the dial you chage the resistance thus changing the output...

Gosh I hope I didnt offend any english teachers out there with that.. I do try.. but I am a C- at best in english..

Also you can break some of these easily so dont turn after it stops you.. use a small scre driver and a light touch...
 
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Kizdawg said:
It has a pot.. or potentometer or variable resistor whatever you want to call it... so when you turn the dial you chage the resistance thus changing the output...

Gosh I hope I didnt offend any english teachers out there with that.. I do try.. but I am a C- at best in english..

Also you can break some of these easily so dont turn after it stops you.. use a small scre driver and a light touch...
Click to expand...

Thanks for the reply. But my question was not about how the current is adjusted on the driver withthis paricular IC. Instead how does IC itself adjust the current. Perhaps my question was rather unclear. my apologies for that. I had the working of op amp explained to m by a friend. So the question is resolved. Thank you.
 
gillza said:
Thanks for the reply. But my question was not about how the current is adjusted on the driver withthis paricular IC. Instead how does IC itself adjust the current. Perhaps my question was rather unclear. my apologies for that. I had the working of op amp explained to m by a friend. So the question is resolved. Thank you.
Click to expand...

The potentiometer is used to adjust the gain of the current sensing op amp.

For example, let's say that the op amp has a gain of... 2.

The resistance(s) in this example could be..say, 10k. This resistive divider creates a 2x gain in the output of the op amp.

Let's say the voltage reference of the actual main IC is 1.25 volts. We can figure out what the sensing voltage is by dividing 1.25 by 2, which is around 0.625 volts.

Now how does this relate back to the potentiometer?

By adjusting the potentiometer to a lower value, say, 5k, the gain is now 3. Dividing 1.25 volts by 3 now gives a voltage sense of 416 mV.

Does that make sense? o:
 
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