Help please.
I have two reverse loops in the new layout I am building. each with a DPDT switch controlling the polarity. I have tried wiring-in a bi-coloured LED so that I have a visual indication when both the adjoining loops are the same polarity. Attached is my wiring diagram.
My problem is that the power to the track works but the LED
I would test your LED's if I were you.
I think you will find that with 12v on the track you will need a 1/2w 560 ohm resistor. 200 ohm (100 +100) may not have been enough and you may well have blown the leds!
But I am not an expert.
It is just a suggestion but you could just control the loop polarity via contacts on the loop point.
Brian Meldon
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Perhaps the method for wiring reverse loops as used by John Dimitrievich will be of interest.
https://www.youtube.com/watch?v=_Lbdod_dXJU
http://www.zelmeroz.com/album_model/members/klyzlr/DynamiteCanyon.pdf
CP
John Dimitrievich's system is quite clever but it is set up to work with dead frog 'slip' points.
I would strongly advise against using dead frog points in 009. But others may think differently.
As it is virtually impossible to convert an electrofrog point to a slip point you will need to switch the point anyway.
So you may as well use extra contacts on the point to switch the loop pollrity.
With a diode controlled dead sections at both ends of the loop before the point the loop will be bi-directional and it will not matter what way the point is set when approaching the loop.
You will however have to remember to reverse the controller before you switch the point, because if you do not the train will run back round the loop.
However to stop this happening you could have a second diode controlled dead section on the loop before the one at the exit.
Brian Meldon
You have blown the LEDs
Try using 1k Ohm (which will limit the current to around 10mA at 12V) and new LEDs
I wouldn't go much below 560 Ohms.
Remember that a LED is a current-controlled device - unlike a filament bulb, which is voltage-controlled.
So the brightness is determined by the current flowing through it. The data sheet for a given LED will tell you what the maximum current is that you should let through the device (typically 20mA for most indicator LEDs) I find 10mA is OK for most purposes.. The data sheet will also tell you what the forward voltage (Vf) is - typically 2 to 3 volts - it depends on the colour of the LED.
The simplest way to control the brightness of a LED is to put a resistor in series with it; the voltage difference between the supply voltage and the Vf of the LED is dropped across the resistor, and the current flowing through both resistor and LED is given (by Ohm's Law) as V(across resistor) / resistance.
With two LEDs back to back, you don't have to worry about exceeding the reverse voltage of either device (typically 5V) because one protects the other.
Hope this helps
Quote from: hughn on November 28, 2017, 12:08
You have blown the LEDs
Try using 1k Ohm (which will limit the current to around 10mA at 12V) and new LEDs
I wouldn't go much below 560 Ohms.
That is what I said, but I am glad you confirmed it Hugh as I am no expert!!
Quote from: B.Meldon on November 27, 2017, 15:36
I would test your LED's if I were you.
I think you will find that with 12v on the track you will need a 1/2w 560 ohm resistor. 200 ohm (100 +100) may not have been enough and you may well have blown the leds!
Brian Meldon
Tanks everyone for your help. I've odered the correct resistors! - Keith