I’m just waiting for more info from Bruce at Litchfield Station, but thought somebody in the forum might have a explanation for a beginner to the circuit that is used to eliminate the glow when using LEDs in place of light bulbs. I am no wiring expert but what I have seen from Bruce’s diagram is a kind of “loop” of a capacitor and a resistor.
Now, I’m not sure if I am seeing this right…a “loop” or is it connected differently than I think in this diagarm…http://litchfieldstation.com/lobby/s_KATO_SD70MAC.htm
Anyone have some further help? Thanks.
The loop is just a way to draw the picture without having to draw non-connecting wires crossing each other. The idea of the circuit is very simple. A 1K resistor and a 0.1 micro-farad capacitor are placed in parallel with the LED. These suck up the power that would normally make the LED glow when it isn’t supposed to.
So one lead from the resistor (pink one), one lead from the capacitor(also pink), and one lead of the LED(orange) all get connected togther. Then the other leads all get connected together. Then these three components are put in series with the normal current limiting 1K resistor (white) for the LED.
Ok, I’m just being dense, but I think I need to see a diagram with true symbols for non-connecting wires. I still don’t get it. I can solder well enough and connect things, but I am not knowledgeable about parallel and series circuitry.
So one lead from the resistor (pink one), one lead from the capacitor(also pink), and one lead of the LED(orange) all get connected togther. (So far so good. ) Then the other leads all get connected together. (You mean just the two leads from the opposite ends of the resistor and capacitor?) Then these three components are put in series with the normal current limiting 1K resistor (white) for the LED. (ok now I’m at sea. What does this last line mean in terms of making connections?)
I do learn, however, and once I have this explained I will be able to do all the other conversions more easily. I know enough that the simple diagram did not look right.
The second resistor and capacitor are placed in parallel with the LED. This creates a time delay circuit that takes the brief pulses of power that for whatever reason the Soundtraxx decoder used in that case puts out when the LED is supposed to be off. This isn’t needed with other decoders, so unless you are using a Soundtraxx DSD, don’t worry about it. I use LEDs with many different decoders and they don’t have the ‘glow’ problem.
–Randy
Yes, and the other side of the LED too. So there are now three components but only two leads.
Connect the two leads of all three of these things connected together(as above), where the LED would have been connected if it was alone.
In other words one side goes to the other resistor (white), and the other back to the locomotives circuit board.
Basically, the capacitor is there to short out the LED. A capacitor will pass AC signals, not DC, so by putting it in parallel with the LED, there is an alternate path for the AC signals that cause the LED to glow.
When there are pulses on the line, the capacitor (which in this case is the LED’s junction capacitance plus the capacitor) is charged. The resistor is added to bleed off the charge when there is no pulse. The charge in the capacitor will flow through the resistor until the next pulse comes along.
Without it, the pulses would simply charge the junction in the LED until the charge is high enough to cause the diode to conduct. Since this is happening very rapidly, it appears to glow.
When the LED is switched on the normal way, it’s resistance is much less than the resistor in parallel with it, so it works as it should.