Re: RGB LED
- From: Rich Grise <richgrise@xxxxxxxxxxx>
- Date: Tue, 13 Jun 2006 16:05:25 GMT
On Mon, 12 Jun 2006 18:02:19 -0700, beananimal wrote:
Rich Grise wrote:
....The reason for the diode in the emitter lead is to bias the emitter up
by .7V so that when the switch is closed, the base would be reverse
biased. Without that, the transistor could still leak if it had only
the diode from emitter to base, i.e., there'd be one diode drop, and
the e-b junction would be biased right at the knee.
I also assume that your omission of dropping resistors in your
schematic means that you've either already accounted for them or the
LED(s) has(have) their own current-limiting built in.
I think I understand what is going in here. I can see that the
"steering" diodes are driving the base high when the switch is in an ON
state and thus preventing the transistor from conducting. I also see
that they prevent the current from traveling back to the opposite LED on
the switch. I also understand why the RED led lights when the switch
is in the off position becasue the transistors base is pulled low
relative to the Emitter via the 47K pulldown. Even after your
edxplanation the emitter diode puzzles me somewhat (excuse my ignorance
but I am 100% self taught when it comes to electronics).
After reading what your wrote several times, I am starting become
confused :)
I had assumed that providing 12V to all of the switches and then using aWell, you seem to have de-confuserized yourself, because circuit3 is
single diode to feed all of the emitters would provide the same thing.
perfect, except for one minor detail - it's upside down. By convention,
we put the + supply at the top, and ground at the bottom. This is from
TOOB days, but people have stuck with it.
And, this really should have been at .basics, but it's probably a little
late for that...
I have drawn (I think) what you have proposed in circuit3
http://i67.photobucket.com/albums/h304/beananimal/circuit3.jpg
And what I had thought would work in circuit4
http://i67.photobucket.com/albums/h304/beananimal/circuit4.jpg
I'm too lazy to trace all of the current paths, but that shouldn't be too
hard for you to do - conventional current flows in the direction of the
arrows, and remember Ohm's law and Kirkhoff's law, and that transistor
E-B junctions, as well as diodes, drop about .6-.7V when they're
conducting. And, to really know if this would work "as advertised", you
should prototype it. :-)
I am learning... but it is taking the patience of people like yourself.
Thank you for all of the help. Please feel free to give up any time,
as I fear that my lack of knowledge may be making this a frustrating use
of your time.
Well, you're asking intelligent qustion, and doing your own homework. In
general, we like questions of the type: "Here's what I've done on my own,
but here's what I'm stuck on...".
Hang in there!
Rich
.
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