Question for the electronic gurus here. I'm wanting to use a bi-polar (red/green) 3-lead LED to tell direction of track current. i.e., it's green one way, when you switch the direction of the loco it turns red. I thought I had this thing figured out, but it's got me stumped.
I can connect -lead to the center and touch the +lead to the left or right and they work fine (red/green respectively). But when I switch the direction on the throttle and try it again nothing works. I've tried the resistor on the center lead (-) and I've tried 2 resistors on the positive leads but nothing. Am I barking up the wrong stack?
Mike
Wrong type! You want one with 2 diodes hooked in parallel in opposite polarities, total of 2 leads.
You have ones with a common cathode (or anode).
You basically "cannot get there from here".
By the way, NEVER test LEDs without a dropping (current limiting) resistor.
Regards, Greg
QuoteWrong type! You want one with 2 diodes hooked in parallel in opposite polarities, total of 2 leads.
You have ones with a common cathode (or anode).
You basically "cannot get there from here".
Actually, Greg, there
may be a way for him to "get there from here" with a 3-lead LED.
Assuming that the common lead is
EITHER anode or cathode, if the OUTER 2 leads are tied together, that would effectively give him a "bipolar" LED (just in a slightly less convienient package!). ;)
You'll still need a dropping resistor, as Greg recommended. If you have a "spec sheet" on the LED showing it's maximum recommended current drain, & know what the maximum voltage output of your power pack is (a digital voltmeter - "DVM" - is the most accurate way to determine that), you can use Ohm's law to calculate the value of the dropping resistor. Subtract 1.2 volts (the typical voltage drop through an LED) from your maximum power supply voltage, then divide that voltage number by the LED's max current rating in Amps (it's probably rated in milliamps - "ma." so that would probably be in the order of a milliamp or so -".001 amp"). That will give you a "ballpark" value for your dropping resistor in Ohms. If the figure is over 1000 ohms, that would be "Kilo Ohms" (K). Using 12 VDC as an example there, subtract 1.2 v. = 10.8 V. Divide that by your LED's max current rating (let's use 1 ma - .001 amp - as an example). That gives us a resistor value of 10,800 ohms, or 10.8 K. Use that calculated value as a "starting point" for your dropping resistor, with the closest common value you could find (probably 10K). Go up or down slightly in value from there if you want a brighter or dimmer LED, just try to stay within the max current limits or you will have a VERY bright LED for a VERY short time! :o
;) Tom
DUH! I just had second thoughts on that! :o - Just realized that rather than OPPOSITELY polarized LED's, they would be polarized in the SAME direction! Greg's right - if it's that style, "you CANT get here from there"! (Must've been having a "senior" (tech) moment! ???). :D Tom
Thanks guys. It's as I feared. Back to Hobbyland for the 2-legged variety.
Thanks again,
Mike.
Mike, all is not lost if you have some diodes around. Just connect them up as shown below. Just about any kind of diode will do. Connect the Rail A and Rail B wires to your track.
(http://members.shaw.ca/sask.rail/Diodes.jpg)
If you do go back to the hobby shop, you might ask for bicolor LEDs. Bipolar LEDs are only good if they are in their manic phase. In their depressive phase, they become DEDs. As you may know, DEDs, or Dark Emitting Diodes, suck all the light out of a room when you turn them on. If you accidentally turn them on outside in bright sunshine, they suck in so much light that they blow up with a loud pop and a blue flame.
Seriously, three lead devices are usually called bicolor LEDs to differentiate them from tricolor LEDs which have two leads. Tricolor LEDs produce red on one polarity, green on the other, and yellow on ac. Now how come you believe this when you wouldn't believe me about the DEDs?
Dear Jim,
Very droll! I was wondering about those crazy lights, too!
Have been reading your posts with enthusiasm.
How are things in the frosty north country?
Best Wishes,
Jack
Thanks Jim, the diagram helps. And a bit of "light" humor Coitenly doesn't hoit' either! Nyuk! Nyuk!
Mike
Jack, the frozen north is slowly unfreezing. So far, about 50% of the outdoor railroad is visible again after going into hiding (under the snow) last October. I can hardly wait to start running outdoors again.
Mike, Very Punny. groan. But I am glad the diagram helps.
Jim
We the occupants of the province of Albert have decided that the warm air will be held in Alberta until at such time you pay the interprovinvial tax or we no longer have need of it. ( could be called ransom )
Don ;)
And here I was thinking all our hot air was in Ottawa.
Dear Jim,
Years ago in the PITT geology dept there was a hermit-like character who took forever to get a Ph.D. His name was Henry something, and he was from Ottowa, excpet he pronounced it Oh-Tau'-wah. :D
Best Wishes,
Jack
P.S. I probably shouldn't say this, but someone just e-mailed me a Monica Lewinsky campaign button that stated she was voting Republican because the Democrats left a bad taste in her mouth. Sorry, Mr. B.! ;D