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Direct-Drive 445 Host






laser_freak said:
Or you could use a couple resistors is parallel to get a lower resistance value. The equation for the overall resistance is Rtotal=1/(R1 + R2 + R3 + etc...)
You can also use an online resistor calculator instead for the calculation such as this one.
Click to expand...

Correction, for multiple resistors the equivalent resistance is:

1/Req = 1/R1 + 1/R2 + ...

Thus, Req = 1/(1/R1 + 1/R2 + ...)

Or for two resistors, Req = R1*R2 / (R1 + R2)

Personally, I use Excel for my formulas, and I've got a "calculator" for a DDL driver (and others), to find both the ideal resistance for a given current, and for the exact current from a provided resistance, or several resistors in parallel. Saves a load of time, I can figure out the right resistance based on parallel resistors I have on hand in a matter of minutes.

I've attached the sheet if anyone wants it.
 

Attachments

rev0 said:
Correction, for multiple resistors the equivalent resistance is:

1/Req = 1/R1 + 1/R2 + ...

Thus, Req = 1/(1/R1 + 1/R2 + ...)

Or for two resistors, Req = R1*R2 / (R1 + R2)

Personally, I use Excel for my formulas, and I've got a "calculator" for a DDL driver (and others), to find both the ideal resistance for a given current, and for the exact current from a provided resistance, or several resistors in parallel. Saves a load of time, I can figure out the right resistance based on parallel resistors I have on hand in a matter of minutes.

I've attached the sheet if anyone wants it.
Click to expand...

Oops. Yes yours is the correct equation. I updated my original reply to be correct.
 
Wallydraigle said:
Can you expound on the direct drive thing, or link to somewhere that explains it? You're just connecting the laser diode directly to the battery and that's that?
Click to expand...

With direct drive, it would be very simple to measure current to the diode by just measuring the battery current draw...

And of course, the corresponding power reading and what lens used would be helpful information...

But obviously, the diode output will drop significantly as the battery drains. Unlike a FlexDrive or Micro BoostDrive build...

Edit: I saw that the current and power output is mentioned in the first post... So at only 600mA's, that means that the voltage of the battery must be instantly dropping as the load is put on it. And that is why it is only giving 600mA's.

And obviously, as the battery drains, the power will continue to drop...
 
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jayrob said:
But obviously, the diode output will drop significantly as the battery drains. Unlike a FlexDrive or Micro BoostDrive build...

I saw that the current and power output is mentioned in the first post... So at only 600mA's, that means that the voltage of the battery must be instantly dropping as the load is put on it. And that is why it is only giving 600mA's.

And obviously, as the battery drains, the power will continue to drop...
Click to expand...

The direct drive is possible because the voltage required by the diode matches up with the voltage of the battery (and the host has a good heatsink).

It draws 600ma because the voltage that the charged battery supplies equals the voltage required by the diode for that 600ma.

I checked the voltage drop over 10 minutes of continuous use.

It was only ~5% with the high energy 18650 batteries I am using.

That power difference is fine for my pointer usage.

Dr. Lava's Micro-Boost (with the new jumpers) is fine also.

LarryDFW
 
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What about 2 10440 lithiums. Would that work direct drive or will it be too much voltage?
 
yobresal said:
What about 2 10440 lithiums. Would that work direct drive or will it be too much voltage?
Click to expand...

You can use them directly if they're both in parallel. If they're in series, the maximum voltage they'll put out should be 8.4V, so use the diode's forward voltage of about 4.2, and your resistor needs to drop 4.2V in that circuit. If you're driving at 500mA for example, your resistor should be: V = I * R => R = 4.2 / 0.500 = 8.4 Ohms. This is a pretty inefficient setup though, because the power the resistor dissipates is I^2 * R = 0.25 * 8.4 = 1.05W, or 4.2W if you are driving at 1A.
 
rev0;

A buck driver might be a better solution,
since you are trying to drop a lot of voltage (8.4 to 4.5VDC)

LarryDFW
 
It's a shame the host is so expensive, its got a great heat sink I cant seem to find anywhere else, people like small slim builds to much.
 
More Power;

I have bought a large quantity of these hosts.

Send me a PM if you need one ($45 with 18650 cell ).

The finned heat sink is required for high power and duty cycles.

LarryDFW
 
$45 for the host and heatsink plus an
18650 hicap battery?

question 2, i can connect an 18650 battery directly to a heatsinked a130 diode and safely power it up?
 
nkgamer1990 said:
$45 for the host and heatsink plus an
18650 hicap battery?

question 2, i can connect an 18650 battery directly to a heatsinked a130 diode and safely power it up?
Click to expand...

1. Yes for host and 18650 battery (without machined heatsink).

2. Yes, with diode in heatsink.

LarryDFW
 
Just received some new 2800mah lithium ion cells.

They are a safer technology design.

They are also higher voltage designs, with low internal resistance.

My direct drive 445nm is now delivering ~675mw.

LarryDFW
 
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Just received some new 3000mah Samsung lithium ion cells.

They are a safer technology design.

They are also higher voltage designs (4.35VDC), with low internal resistance.

My direct drive 445nm is now delivering ~700mw.

LarryDFW
 
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Nice info on direct driving - if, like me, you don't really demand that the intensity stays at maximum until the batteries finally give up the ghost, it's apparently fine, especially if the Vb very nearly equals the Vf of the diode.

Thanks for the info!

Dave
 
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