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Old 07-06-2010, 04:54 AM #49
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Default Re: questions about micro boost drive

Hi.

I agree...
High current and heat could cause the aberration of the readings except the output power measurement.
I think A140 can't be killed with microboost or v5 if there is some heatsink.

I'm building high current test load with 1N5401s and 3W 1ohm.
And I wish there will be "v6" too.....

SHIN


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Old 07-06-2010, 05:20 AM #50
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Default Re: questions about micro boost drive

Quote:
Originally Posted by jayrob View Post
It works just fine with Li-Ion's...

It even put out high current for a tiny 10280 in one of my tests...

Then only problem I had was using a 3 volt supply with the phaser sound board. It worked fine without the phaser sound board using a 3 volt supply...
Please don't get me wrong Jay. I'm no electronics expert and I'm not trying to diss any of Dr.Lava's work at all, (I'd be an idiot to knock obvious talent and service to the community), but there suddenly seems to be a confidence gap between the Flexdrive, (which personally I would prefer to use with these new diodes), and the MicroDrive which, as far as I know, was specifically produced for the high voltage/high current BRs. I take Dr.Lava's tuition to use just the, "Red", selection on a commonly available test load to set the current for my new 445nm LDs but without my own testing my question still remains... If it's ok for these diodes, then why can't I set my other high voltage BRs on the, "Red", test load setting?


Quote:
Originally Posted by SHIN View Post
Hi.

I agree...
High current and heat could cause the aberration of the readings except the output power measurement.
I think A140 can't be killed with microboost or v5 if there is some heatsink.

I'm building high current test load with 1N5401s and 3W 1ohm.
And I wish there will be "v6" too.....

SHIN
I think maybe it's time for a new test load that is designed for the rising currents we require. I am almost constantly worried and burned by my old RCKSTR Dummy Test Load.

I was confused by a post giving the formula for working out the maximum power dissipation of a 1 Ohm resistor the other day and would welcome an explanation of my smoking diodes... , and just what power they are trying to dissipate driving 1A @ 4.1V.

Formulae for resistor Power; individual diode dissipation and total circuit dissipation would be welcomed here please.

M
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Last edited by Morgan; 07-06-2010 at 05:25 AM.
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Old 07-06-2010, 05:31 AM #51
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Default Re: questions about micro boost drive

Quote:
Originally Posted by Morgan View Post
Please don't get me wrong Jay. I'm no electronics expert and I'm not trying to diss any of Dr.Lava's work at all, (I'd be an idiot to knock obvious talent and service to the community), but there suddenly seems to be a confidence gap between the Flexdrive, (which personally I would prefer to use with these new diodes), and the MicroDrive which, as far as I know, was specifically produced for the high voltage/high current BRs. I take Dr.Lava's tuition to use just the, "Red", selection on a commonly available test load to set the current for my new 445nm LDs but without my own testing my question still remains... If it's ok for these diodes, then why can't I set my other high voltage BRs on the, "Red", test load setting?

I'm sure it's fine to use a red test load at lower currents if you want...

It's going to be very close to the same setting with a blu-ray test load using a FlexDrive or Micro BoostDrive...

Maybe just a couple of mA's difference at low current if any. You can try a test comparison with two different test loads...

I noticed a difference of maybe 6 or 8mA's at high current comparing a red test load vs a blu-ray test load...
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Last edited by jayrob; 07-06-2010 at 05:34 AM.
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Old 07-06-2010, 05:46 AM #52
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Default Re: questions about micro boost drive

Hi. Morgan

My specialty is that away from electronics...
Is it OK that test load for 3A can be built after the rkcstr test load, replacing 1N4001 by 1N5401?

SHIN
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Old 07-06-2010, 05:49 AM #53
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Default Re: questions about micro boost drive

^Well that's interesting. As I say, I'm no expert so that sort of comparison is good.

Recently I swapped out a stock driver on a green module for a Flexdrive. The module ended up failing, (It failed rather than me killing it!), but for the next hook up of a Flexdive set by test load I used a 1 Ohm resistor in series with the LD, (helpfully supply by Flaminpyro), and read just 6mA difference between a, 'Red', setting and LCC diode. I'll do the same for my 445nm set Flexdrive and connect the same resistor in series with my next green module that's in the post.

I'd still like to know why the difference is there... Detailed discussion for another time maybe.

[EDIT: Sorry, SHIN, my speciality is far from this close tolerance electronics too I'm afraid. I await guidance myself...]

M
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Last edited by Morgan; 07-06-2010 at 05:53 AM.
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Old 07-06-2010, 06:00 AM #54
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Default Re: questions about micro boost drive

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Originally Posted by SHIN View Post
I'm building high current test load with 1N5401s and 3W 1ohm.SHIN
This has been on my "build list" for a while, just haven't had the chance to place the order with digikey. The "N" series rectifier diodes, whether they are the 1N4001s or the 1N5401s, are still 0.7V forward voltage drop.
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Old 07-06-2010, 06:12 AM #55
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Default Re: questions about micro boost drive

Thank you. Morgan.
And thank you. Kevlar.

I can't wait for "buyable" high current test load...because my another A140 is on its way...

SHIN
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Old 07-06-2010, 06:18 AM #56
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Default Re: questions about micro boost drive

Maybe thats something you could make and sell here!!! I'd buy one!!
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Old 07-06-2010, 08:23 PM #57
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Default Re: questions about micro boost drive

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Originally Posted by Morgan View Post
Formulae for resistor Power; individual diode dissipation and total circuit dissipation would be welcomed here please.

M
Remember Ohm's law, V=I*R
Also, P=I*V


-power(P, in Watts) dissipated by resistor(R, in Ohms):
P=I*V
P=(V/R)V or P=I(I*R)
therefore
P=(V^2)/R or P=R*I^2 is the equation for the power dissipated by the resistor, where V is the voltage across the resistor.

-Power dissipated by the LD:
P=I*V. You will probably know what current your driver is putting out, and you can measure the voltage (V) across the LD. This will be the total power dissipated, meaning optical power + heat power.

-Total Circuit power usage:
again, P=I*V, but measure V and I at the battery while the circuit is in operation.


You'll see that all I'm doing is substituting and rearranging the bits and pieces of Ohm's law and the power equation. These two equations are very powerful, and the only confusing part is knowing where to pull the measurements for the variables from.
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