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Feb 11, 2021 at 10:22 answer added Sil timeline score: 0
Aug 31, 2011 at 22:14 vote accept Saad
Aug 31, 2011 at 18:37 history tweeted twitter.com/#!/StackElectronix/status/108971727521976320
Aug 31, 2011 at 17:48 comment added Mike DeSimone If your test load is a 1k resistor, then LED2 + R4 is a more significant load: there will be a drop across LED2 (just below 2 V for red, just above for yellow-green, and around 3.5 V for true green, blue, or white) and the rest of the voltage is across R4. Assuming a red LED of 2.0 V drop, current will be 25 mA, which is really close to the limit for standard LEDs (2.0 mA is much more sane for a simple indicator). Compare this with 12 mA in the 1k resistor. This means 1/4 W will be dissipated in R4 and 50 mW in LED2; check the datasheets to see if this is okay. Same for LED1 and R3.
Aug 31, 2011 at 16:13 comment added Saad Kevin, I built the PCB based on the previous schematic. It seems that I should just trash the PCB and etch another one based on this, revised, schematic.
Aug 31, 2011 at 16:10 comment added Kevin Vermeer @saad - Thanks for cleaning up the schematic, it looks much better now. Did you build your PCB based on the old schematic (which had both material and stylistic errors), or build it on protoboard based on your intent for the schematic (as reflected in the current revision)?
Aug 31, 2011 at 15:44 history edited Saad CC BY-SA 3.0
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Aug 31, 2011 at 15:33 answer added Russell McMahon timeline score: 2
Aug 31, 2011 at 15:23 answer added Mike DeSimone timeline score: 2
Aug 31, 2011 at 14:52 comment added Saad Its IR2113. I'm using the footprint of a 14pin DIP, hence the label on the schematic.
Aug 31, 2011 at 14:51 comment added Mike DeSimone Schematic shows IC1 as 4561N but text says IR2113; which is it?
Aug 31, 2011 at 14:42 answer added Oli Glaser timeline score: 2
Aug 31, 2011 at 14:22 answer added Olin Lathrop timeline score: 2
Aug 31, 2011 at 14:02 history asked Saad CC BY-SA 3.0