Welcome to Laser Pointer Forums - discuss green laser pointers, blue laser pointers, and all types of lasers

LPF Donation via Stripe | LPF Donation - Other Methods

Links below open in new window

ArcticMyst Security by Avery

FREE DIY open source BOOST driver!!! Tested & working!!

foulmist

foulmist

0
Joined
Mar 29, 2011
Messages
1,056
Points
48
rhd said:
*Facepalm*

The spec is the same. The +125C spec is the same for all of the LM3410s we've been using. The only difference that "Q" gets you, is supposedly improved defect detection AT the spec (of +125C) at the factory.

Lol, this isn't some magical version of the IC that can handle more heat....
Click to expand...

Reliability qualification is compliant with the requirements and temperature grades defined in the AEC-Q100 standard?

what does that mean actually? somebody can explain it to me simplier please :D
 





tsteele93

tsteele93

0
Joined
Dec 27, 2011
Messages
2,062
Points
48
rhd said:
*Facepalm*

The spec is the same. The +125C spec is the same for all of the LM3410s we've been using. The only difference that "Q" gets you, is supposedly improved defect detection AT the spec (of +125C) at the factory.

Lol, this isn't some magical version of the IC that can handle more heat....
Click to expand...

I'm building maybe 10 of these. That's maybe $3. I can't believe they are the exact same IC just at a different price. I'm not thinking they are magic. I'm just hoping that I'm buying cheap insurance with a chip that is supposed to have a lower defect percentage and perhaps better heat tolerance.

Heck, I can get the Q's cheaper from Newark than the regular ones from mouser right now.
 
Last edited:
rhd

rhd

0
Joined
Dec 7, 2010
Messages
8,475
Points
0
tsteele93 said:
I'm building maybe 10 of these. That's maybe $3. I can't believe they are the exact same IC just at a different price. I'm not thinking they are magic. I'm just hoping that I'm buying cheap insurance with a chip that is supposed to have a lower defect percentage and perhaps better heat tolerance.

Heck, I can get the Q's cheaper from Newark than the regular ones from mouser right now.
Click to expand...

My understanding is that it's the same chip, but more rigorously tested at manufacture. So, instead of the final user having to test and discard 2 devices per XX,XXX that had a faulty LM3410s, they only have a failure rate of 1 per XX,XXX units. etc
 
rhd

rhd

0
Joined
Dec 7, 2010
Messages
8,475
Points
0
Lol :)

Current of a Ben Boost is given by 0.19 / Resistance.

Wattage needs to be at least 0.19 x Current (A)
 
tsteele93

tsteele93

0
Joined
Dec 27, 2011
Messages
2,062
Points
48
rhd said:
Lol :)

Current of a Ben Boost is given by 0.19 / Resistance.

Wattage needs to be at least 0.19 x Current (A)
Click to expand...

NEVER MIND, I see the choice for the 805 case.

I got the current, I just don't know which one of those resistors to order. I'm afraid that I'll get some giant thing that won't fit. In other words, I think that I'm ok on the theory, it is the practical part I'm not familiar with.

Or maybe I only have one choice is if I choose .19 x current?

I just want to get the right resistor package if that makes sense.
 
Last edited:
rhd

rhd

0
Joined
Dec 7, 2010
Messages
8,475
Points
0
tsteele93 said:
NEVER MIND, I see the choice for the 805 case.

I got the current, I just don't know which one of those resistors to order. I'm afraid that I'll get some giant thing that won't fit. In other words, I think that I'm ok on the theory, it is the practical part I'm not familiar with.

Or maybe I only have one choice is if I choose .19 x current?

I just want to get the right resistor package if that makes sense.
Click to expand...


:) chances are that you'll be able to get like a 1/2 W resister and that should be fine.

Also, it's not 0.19 x current = resistance
It's current = 0.19 / resistance such simplifies to:
0.19 / current = resistance

:)
 
tsteele93

tsteele93

0
Joined
Dec 27, 2011
Messages
2,062
Points
48
rhd said:
:) chances are that you'll be able to get like a 1/2 W resister and that should be fine.

Also, it's not 0.19 x current = resistance
It's current = 0.19 / resistance such simplifies to:
0.19 / current = resistance

:)
Click to expand...

Brain fart. Thanks, and I see the heavy lifting has already been done on the first post in this thread when it comes to the resistors.
 
Last edited:
M

Machinist Joe

0
Joined
Jan 1, 2012
Messages
462
Points
18
Hey guys, any suggestions on where to get started with learning electronics. I get the basic stuff. (I think)
 
lasersbee

lasersbee

0
Joined
Sep 20, 2008
Messages
17,622
Points
113
Machinist Joe said:
Hey guys, any suggestions on where to get started with learning electronics. I get the basic stuff. (I think)
Click to expand...

I use Google for anything I don't already know and then
practical experimenting to see if I did learn "whatever"
correctly...:beer:


Jerry

You can contact us at any time on our Website: J.BAUER Electronics
 
Last edited:
C

conundrum

0
Joined
May 29, 2010
Messages
88
Points
0
The general rule is 50% higher than the peak voltage.
This ensures long capacitor life, however depending on the application and presence/absence of clamping zeners you can get away with less.
A sensible margin in this case would be 20% > Vpeak

(who found this out the hard way after having 8 C*p*on capacitors blow their tops on his expensive 19" flat panel)
 
You must log in or register to reply here.




Top