Building Tiny DDL Driver

[Wolfman29]

I am using a red diode anyway, so I figure I can probably run the laser on an 18650 and an LM1117... 1.2V drop at max, yeah, and about 2.2Vf for a red diode, right? That leaves me with .3V of run time, and it should be a nice, compact build!

 

[Leodahsan]

Maybe.

 

[rhd]

Wolfman - I don't think anyone was concerned about whether the driver could dissipate the difference in the voltage drop ?!?

I'm pretty sure the reason Leo pointed out the ~1.5 V drop of the chip, was a concern for whether you'd have enough left over for your diode.

If you used a chip with a 1.5V voltage drop, and cells that had 3.7V of charge, you'd only have 2.2V for your diode - which wouldn't be enough for a red.

That's all. I don't think anyone was worried about whether an LM317 / LM1117 could dissipate an extra quarter-volt here or there

 

[Wolfman29]

Anyway, with such a small excess voltage (.3V at full charge), I only really need to dissipate 3.7V*320mA = 1.184W, right? That's hardly anything, no?

^ OH... okay. I am sick, not thinking straight today.

Anyway, isn't the LM1117's max voltage drop around 1.25V? The datasheet tells me that that *is* the max for 500mA, which I will be running under.

 

[Leodahsan]

Dude.. Just try it. If the diode goes weak just put another battery on it.

 

[rhd]

No, you're not DISSIPATING 3.7 * 320mA. If you were dissipating that, you'd have ZERO power for your diode.

You're DISSIPATING the EXCESS voltage... which in your case (single cell) you're not going to have ANY of anyway.

That's my point. If you're planning to run the red at anything above 200mW, you'll need 3.0V or more for the diode.

So:
3.0V (Vf diode) + at least 1.2V (for the LM1117) = 4.2V

In other words, when the li-ion cell is absolutely fully charged, and giving you almost 4.2V, you'll still have just barely enough Voltage to run the diode at 200mW or under...

And that's a MAYBE:
- Some people have suggested that the current setting resistor adds some additional voltage drop
- Some li-ions might not fully charge to 4.2, but maybe just 4.1 under a load, or even 4V

The problem you'll face:
- Suppose your battery cell can only hit 4.1V fully charged, and suppose the LM1117 with current setting resistor factored in eats up 1.5V. That leaves 2.6V for your diode. If that's the case, your Voltage becomes the limiting factor, not your current. You may only be able to get 120mW or so out of that red diode.
http://laserpointerforums.com/f50/question-about-lpc-815-loc-forward-voltage-46572.html#post631700

There is a solution. Get a boost driver. They're not expensive if you avoid the Flexdrive / Microboost route (totally unnecessary here)
Compact Variable Current Boost Laser Diode Driver Board for Hand-held Laser 0-300mA+ - Detailed item view - OdicForce Lasers Online Shop
Variable Current Step-up (Boost) 405nm Laser Driver Board - Detailed item view - OdicForce Lasers Online Shop

 

[Wolfman29]

Well, I can't really do that yet, because I don't have any of the parts. Haven't even started building the driver, and I haven't even ordered the diode yet. =o

 

[Wolfman29]

Oh, yeah, that's completely true.

Hmm... I don't *really* want to use a boost driver unless I can build it myself, but I can't, so I will probably have to use a higher voltage battery, or just make it a bigger handheld... maybe 2 CR123As? Or... a 1/2AA? Saw it on Wiki, may not be that common, but it's the same diameter of a AA, but it is half the length and provides anywhere from 3-3.6V, and has anywhere from 850-1200mAh.

So I have another question, which I may as well put here: for a normal red diode, I know the pin out is like this:

...........Negative
Positive.............Not used

correct? And the negative is equivalent to the case pin, as well? In that case, for my design, is everything good? Just hook up negative/case to the driver as I usually would, but instead of hooking the input wire (to the driver) to a battery, just solder it to the metal host, and have the positive lead input to the driver be just normally hooking up to the battery plate positive end?

EDIT: I think I am going to do two RCR123As of the 3V 900mAh variety, that way I am not dissipating a huge amount of heat (only 1.8V*320mA = .576W, which is nothing and is easily handleable by a small heatsink, yes?

 

[jib77]

The SMD 1117 I use in my drivers, good up to 1.6A, 0.9V dropout:

NCP1117DTARKG ON Semiconductor Low Dropout (LDO) Regulators

Low ESR input and output caps are a must on LDOs like the LM1117. Follow the datasheet recommendations ... especially when it comes to the copper area needed for the thermal pad on SMD regulators.

Follow link in my sig to check out my driver.

 

[Wolfman29]

Pardon the stupid question, but why do we need caps at all for the LM1117? And do we need caps for the LM317?

 

[rhd]

Jib: Can you clarify something that has been touched upon in other threads?

The 0.9V dropout - do we have to also factor in a dropout associated with the current setting resistors? My understanding is not. So, for your SMD 1117:

Battery Voltage must at least = 0.9V (1117) + vF

But someone suggested that the equation is:

Battery Voltage must at least = 0.9V (1117) + 1.25V (resistor) + vF

Whats the reality here? Anecdotally, I think my understanding is the correct one - it seems to have held true throughout testing. But I don't really know.

 

[rhd]

Wolfman: To absorb voltage spikes and surges in the power. For battery supplies, I think this is mostly a concern at the very beginning (power up).

 

[Wolfman29]

Alright. So wouldn't we only need one cap? Or is the second cap at input to prevent the driver from blowing? Because I would think that, due to the nature of the LM1117/LM317, it could take the excess voltage, no?

 

[Cyparagon]

Your diode is backwards :undecided:

 

[Wolfman29]

Haha, yeah, I suppose it is. I just threw that together quickly =p

 

[jib77]

Jib: Can you clarify something that has been touched upon in other threads?

The 0.9V dropout - do we have to also factor in a dropout associated with the current setting resistors? My understanding is not. So, for your SMD 1117:

Battery Voltage must at least = 0.9V (1117) + vF

But someone suggested that the equation is:

Battery Voltage must at least = 0.9V (1117) + 1.25V (resistor) + vF

Whats the reality here? Anecdotally, I think my understanding is the correct one - it seems to have held true throughout testing. But I don't really know.

I'm more inclined to say its Vf of resistor (1.25V) + Vf of diode.

Pardon the stupid question, but why do we need caps at all for the LM1117? And do we need caps for the LM317?

A Input and Output cap is needed because the LM1117 is something called a "Low Drop Out" regulator aka LDO. Input caps are required only if the regulator is an LDO to keep it stable due to it's simplified internal circuit. Thus the LM317 does *not* need them because it is not a LDO, the ouput cap in DDL's circuit is only there as extra insurance for the Laser Diode not the driver.