This is a circuit for a signal system for a large scale outdoor model railroad that was designed by someone else.

I want to upgrade the system using 10mm LEDs and they have a different Forward Voltage than the 5mm as designed.

I do not understand how to calculate needed resistor values since the circuit has LEDs in parallel with resistors.

Please help me understand the circuit and how to calculate resistor value for new LEDs.

Power supply is 12V battery with solar charger to keep it up.

*** Original LEDs
Forward Voltage = 1.9-2.02V
Forward current = 20mA

*** New LEDs
Forward Voltage = 2.4V
Forward Current = 20mA

Thanks in advance - http://www.csppry.org

 

I'm not wild about this circuit because it overdrives the relay coils. They're rated for 5V, and what they'll get is 12 minus the voltage drops in the LEDs, so about 8V. Changing to LEDs with a higher voltage drop would actually improve this aspect of the circuit.

Now I've done a calculation which says that with the 2V LEDs, each LED carried 35mA and the resistor carried 13mA, so the LEDs are overloaded too. With 2.4V LEDs, it becomes 27mA and 16mA, which is a little better, but in both cases the LEDs, as well as the relays, have too much current by quite a large margin. Apparently that isn't enough to kill them, and the new parts should make things better rather than worse. The right thing to do may be just to replace the LEDs and do nothing else.

By my calculations, if you want to drive the components according to specifications, you should add a 68 ohm resistor in series with the relay, and use 270 ohm in parallel with the LEDs.

 

First question I have is why are the resistors in parallel with the leds, instead of series?
On the second drawing , green leds the resistors are in the correct place, but the others do not have any current limiting, other than the relay coils. Not sure how effective that is.

 

I'm not wild about this circuit because it overdrives the relay coils. They're rated for 5V, and what they'll get is 12 minus the voltage drops in the LEDs, so about 8V. Changing to LEDs with a higher voltage drop would actually improve this aspect of the circuit.

Now I've done a calculation which says that with the 2V LEDs, each LED carried 35mA and the resistor carried 13mA, so the LEDs are overloaded too. With 2.4V LEDs, it becomes 27mA and 16mA, which is a little better, but in both cases the LEDs, as well as the relays, have too much current by quite a large margin. Apparently that isn't enough to kill them, and the new parts should make things better rather than worse. The right thing to do may be just to replace the LEDs and do nothing else.

By my calculations, if you want to drive the components according to specifications, you should add a 68 ohm resistor in series with the relay, and use 270 ohm in parallel with the LEDs.

There are 2 relay coils in series, so the 12v is dropped across a 10 volt load.
I'm not sure I understand why its wired the way it is, but it has some quirks.

 

I don't think that's true. Yes, there are relays in the Master and Slave sections, but they don't seem to run in series. When nothing is operated, someone presses the left-hand "capture" button, and then current flows through the left-hand relay coil, the left yellow light, the right red light, the right normally-closed contacts 11-13 and then to Gnd. The left "hold" contacts 13-9 then latch the relay, and that's the stable situation until a "release" button is pressed. Or have I missed something?

This circuit could be used to protect a section of single track that can be entered from either end. When it's idle both sides see green. If "capture" is hit at either end, that side sees yellow (proceed with caution) and the other gets red, until the train is through the section and "release" is hit. Then it goes back to idle.

 

I don't think that's true. Yes, there are relays in the Master and Slave sections, but they don't seem to run in series. When nothing is operated, someone presses the left-hand "capture" button, and then current flows through the left-hand relay coil, the left yellow light, the right red light, the right normally-closed contacts 11-13 and then to Gnd. The left "hold" contacts 13-9 then latch the relay, and that's the stable situation until a "release" button is pressed. Or have I missed something?

This circuit could be used to protect a section of single track that can be entered from either end. When it's idle both sides see green. If "capture" is hit at either end, that side sees yellow (proceed with caution) and the other gets red, until the train is through the section and "release" is hit. Then it goes back to idle.

You're correct, I didn't look as close as I thought I did.

 

Thanks for the input guys.

I have tried the new LEDs in the circuit and they seem to work ok.

We do lose some LEDs from time to time but have attributed it to lightning, do you think it could be because we do not have enough resistance in the circuit?

I am a newbe to electronics so I do not understand the placement of the resistors in parallel with the LEDs - I thought you might add the value of the relay coil and the resistor in parallel with the yellow LED and that would equal the value needed for the red LED and vice versa - as I said I do not understand how to count the value of the resistors here???

 

Lightning might happen, but the LEDs are being overloaded.

This is all Ohm's law. You have to look at the resistance in the circuit and the voltage. The tricky but actually very easy part is that the "voltage" isn't 12, but 12 minus twice your LED voltage. For the original 2V LEDs, that comes out to 8, and for the newer ones it's 7.2. This voltage combined with the resistance of the relay coil determines the current, and that should be 30mA to keep the relay happy, but in fact it's more. That's why I say you should add an additional resistor to limit the current.

The reason for the parallel resistors adjacent to the LEDs is that the relay should be getting 30mA, but the LEDs only want 20mA, so the resistors should be diverting some fraction of the current around the LEDs. The way to calculate them is to take the total current and subtract 20mA, so what's left is the amount that the resistor should carry. Again you apply Ohm's law: the voltage (2V or 2.4V) divided by the current (which you just calculated) gives you the resistance.

 

Thanks John P I appreciate your explanation and I will try adding more resistor to the circuit.