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2012-02-29
, 19:04
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Posts: 2,292 |
Thanked: 4,135 times |
Joined on Apr 2010
@ UK
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#12
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According to schematics, the current through the LEDs is controlled by the the 220R resistors. Replace them all with 110R resistors for maximum IR lols.
...Although, the fact they have used two transistors in parallel implies that the forward current is already quite high for them (the transistors). So, will doubling it cause any problems? If not, why use 2 transistors?? Can the LED even take any MOAR juice anyway?
Who will try to find out?
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2012-02-29
, 21:15
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Posts: 412 |
Thanked: 480 times |
Joined on Feb 2011
@ Bronx, NY
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#13
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Actually, make it a transmitter and a receiver, and I could turn the N900 into a "learning" remote.(And it might be useful for the N9 too, if it has USB host mode.)
I'm not a hardware guy, so I don't really do this sort of thing, but I think it'd be easy to rework the Pierogi back-end to tie it to other IR devices...
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2012-02-29
, 23:31
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Posts: 306 |
Thanked: 566 times |
Joined on Jan 2010
@ Romania
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#14
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AFAIK messing with transistors linked like this can cause problems.
Two transistors in parallel like that have to have the same resistance (with not much tolerance).
If there not one transistor will end up taking all of the load. This will destroy both transistors quickly.
Two transistors allow double the current of one.
Most 5mm 860nm infra-red LED's I have seen have a max forward voltage of 1.8V. 940nm LED's can have even lower voltage (1.2-1.6)
Increasing voltage above this will reduce light output due to increased temperature and eventually fry the LED.
I am certainly not going to try.
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2012-03-01
, 17:41
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Posts: 2,292 |
Thanked: 4,135 times |
Joined on Apr 2010
@ UK
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#15
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On to paralleling the transistors - yes, you can parallel them, but it isn't as simple as just attaching the leads together. Transistors experience a positive temperature coefficient in gain. So higher temperatures mean less voltage drop. With two in parallel, they won't be identical to start with so one will have a little less voltage drop, which means more current flows through it (current follows the path of least resistance). So more heat, which raises the temperature of that transistor, causing a greater imbalance, until that transistor is hogging enough current to overheat it and burn it up. Then the second transistor has all the load and also burns up.
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2012-03-01
, 20:37
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Posts: 306 |
Thanked: 566 times |
Joined on Jan 2010
@ Romania
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#16
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2012-03-01
, 21:19
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Posts: 2,292 |
Thanked: 4,135 times |
Joined on Apr 2010
@ UK
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#17
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2012-03-02
, 08:58
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Posts: 282 |
Thanked: 220 times |
Joined on Aug 2011
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#18
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2012-03-08
, 17:30
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Posts: 11 |
Thanked: 2 times |
Joined on Sep 2011
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#19
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2012-03-08
, 19:04
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Posts: 1,455 |
Thanked: 3,309 times |
Joined on Dec 2009
@ Rochester, NY
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#20
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The Following 2 Users Say Thank You to woody14619 For This Useful Post: | ||
According to schematics, the current through the LEDs is controlled by the the 220R resistors. Replace them all with 110R resistors for maximum IR lols.
...Although, the fact they have used two transistors in parallel implies that the forward current is already quite high for them (the transistors). So, will doubling it cause any problems? If not, why use 2 transistors?? Can the LED even take any MOAR juice anyway?
Who will try to find out?