The first diagram below illustrates the way that I currently set up signals and control panel LEDs with resistors attached.
But after reading a few recent threads, I wonder if there is a better way to wire this setup and either avoid or at least reduce the use of resistors.
Do some of the Tortoise connectors put out less power, thereby eliminating the need for resistors?
Someone recently suggested the wiring set up in the second diagram, but I haven't tried it.
I would be interested in a resistor-free wiring setup if that is possible.
I look forward to your advice and comment.
Rich
Alton Junction
Cut the yellow wire going fromt he DPDT tot he Tortoise. Insert the 2-leg bicolor LED there, without a resistor.
Kirchoff's Laws - the same current will flow through all devices in series. 15ma is about what flows through a stalled Tortoise motor, therefore 15ma will flow through the LED. No resistor needed.
For red/green signals, just use a red LED and a green LED wired back to back (cathode to anode, anode to cathode) - you are simply making your own bi-color LED out of two discrete LEDs. Wire the same way.
If your power supply is 12V, then 2 sets of LEDs will still leave enough voltage for the Tortoise. If you want more than 2 sets of LEDs in the chain, you may need to increase the voltage. Each colored LED will drop about 2.1 volts - so 2 sets of LEDs is about 4.2 volts, leaving about 7.8 volts for the Tortoise. If you up it to a 14V power supply, the Tortoise would get about 9.8 volts - still less than the 12V maximum.
The downside of series LEDs is that they will change indication as soon as you flip the toggle switch, and they will be dim while the Tortoise is moving, and brighten when it completes the movement.
This series wiring is what most of us have been talking about with respect to indications and Tortoises. The Tortoise is the perfect current limiter for LEDs. Also why the Cobalt is not as good, the Cobalt is around 30ma which is too much for most LEDs.
--Randy
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
Ah Kirchoff, been a while since io thought of that guy.
Just one addition to Randy's excellent answer: typically you would need 2 signals minimum per switch (diverging routes), plus 1 you want Green/Yellow on the facing point side. Plus 1 more for panel. That's a whopping 2.1x4=8.4v in series with the tortoise. So driving from a 12v supply is not an option. I would go at that point with a 16v supply.
The way I solved this is that I did not use an indicator for the facing point or the panel. To me, the position of the switch itself was enough for he panel.
Second point: I used 2x12v wall warts to make a +/-12v supply. That was $10 more but allowed me to use SPDT switches which saves so much wiring on your panel that it's worth it totally. If you have not already built the panel, consider this. What you have to do is co connect the + of one supply with - of the other. This becomes your 0V. You will be left with two more terminals from the supplies: a -12v one and a +12v one. Wire the 0v to pin 8 of all tortoises. Wire the + and - one to the two sides (throws) of the SPDT. The pole (center) of the SPDT goes to Pin 1 of tortoise.
NP.
Here is a visual representation of what Randy is talking about. you can put two LED's (the two LED's are in parallel but back to back) or a bipolar LED in series with the tortoise wires. This means you do not need the resistor as the tortoise motor performs that function when it is stalled. It limits the current to the LED's.
Colorado Front Range Railroad: http://www.coloradofrontrangerr.com/
I appreciate the replies so far, guys.
This is raising all sorts of questions in my mind.
First, and I should have mentioned this at the outset, I use MRC Railpower 1370 power packs for my tortoises, signals, LEDs, and DPDTs. The MRC Railpower 1370 is an 18 volt DC power pack. Is this higher voltage going to cause problems with these solutions?
Second, in a different thread, someone suggested that you must connect the LED to the #1 terminal on the Tortoise. In my illustration, the yellow wire that Randy suggests cutting is connected to the #8 terminal on the Tortoise. Is this a problem? The Tortoise instruction sheet does not list the power off of each terminal.
Third, if you connect one leg of the LED into the yellow wire without a resistor, where does the other leg of the LED connect?
Fourth, if I power my signals off of the #5 and #6 terminals of the Tortoise, do the LED wires for the green and red still need resistors.
Ugh, I have always been electronically illiterate.
HELP
Renegade's post shows what I am talking about. You cut the yellow wire but insert the bi-color LED to complete the circuit.
If you're using a measured 18V DC to drive the Tortoises per your original diagram technically you are driving the Tortoises with too much voltage. They are rated for 12V. If you insert 2 sets of LEDs in series with the Tortoise, it will still be getting over 13V even with the LED drops, so no, changing the circuit won;t do anything except reduce the voltage to the Tortoise to closer ot the rating. ANd it will run quieter. The sweet spot I've found seems to be about 9V, enough to mve them at a reasonable speed and hold, but MUCH quieter than the full 12V. Coincidently just about what you'd get using a common 13.8V (common because this si the real voltage of a "12V" car battery - so for devices that are meant for mobile use, a 13.8V wall wart is used to run them on house current - also it's good when runnign a 12V voltage regulator), and 2 sets of LEDs. 13.8 - 4.2, 9.6V.
Total current load per unit (A tortoise, plus 2 sets of bicolor LEDs) is still only 15ma, because the LEDs are in series. They way you have it set up now with parallel LEDs, yoy actually have a total current draw of 15ma for the Tortoise + whatever current is in each LED based on the resistor value you are using. So overall power requirements for your layout will actually drop using the series LEDs.
rrinker Renegade's post shows what I am talking about. You cut the yellow wire but insert the bi-color LED to complete the circuit.
The good news is that it works, cutting the yellow wire and inserting a bi-color led to complete the circuit.
The bad news is that I don't get it. If somewhere between 12 and 18 volts are coming up to the DPDT from the DC power supply, what keeps an unresistored LED from immediately burning out?
Our club uses a single, 20ma, 3mm green/red bipolar, two lead LED in series with one lead to the Tortoise. We use to use 12 vdc but lowered the voltage to 9 vdc for slower point operation.
We used a LM317 voltage regulator, two resistors and two caps to have regulated 9 vdc for the turnout motors.
Current is less than 20 ma when operating.
No resistor needed.
If you ever fall over in public, pick yourself up and say “sorry it’s been a while since I inhabited a body.” And just walk away.
richg1998 Our club uses a single, 20ma, 3mm green/red bipolar, two lead LED in series with one lead to the Tortoise. We use to use 12 vdc but lowered the voltage to 9 vdc for slower point operation. We used a LM317 voltage regulator, two resistors and two caps to have regulated 9 vdc for the turnout motors. Current is less than 20 ma when operating. No resistor needed. Rich
Rich, that is what I don't get. Since you r voltage is 9vdc, what keeps the unresistored LED from immediately burning out?
richhotrain rrinker Renegade's post shows what I am talking about. You cut the yellow wire but insert the bi-color LED to complete the circuit. The good news is that it works, cutting the yellow wire and inserting a bi-color led to complete the circuit. The bad news is that I don't get it. If somewhere between 12 and 18 volts are coming up to the DPDT from the DC power supply, what keeps an unresistored LED from immediately burning out? Rich
It doesn't matter if the LED's are on pin 1 or pin 8 of the tortoise. one leg of the LED will connect to the tortoise (pin 1 or pin 8) and the other will connect to your DPDT switch. Another wire will run in from the opposing pin on the tortoise to the other side of the DPDT switch. See my diagram above.
The tortoise swtich machine acts as your resistor. The switch machine is the current limiter that keeps your LED's from burning out. Remember LED'S are CURRENT driven devices. As Stated you potentially damaging your tortoises by running them at 18 volts since they are only rated at 12 volts.
Ohm's law will help you determine the right size resistor (which you don't really need if you are putting inline with tortoises). the formula is V= I*R where V=Voltage, R= Resistance and I= Current. Generally the max current you want to put thru a LED is around 15 milliamps (0.015 Amps). So if we rearrange the equation R = V/I. If you have an 18 volt power supply it would be R= (18 volts) / (0.015 Amps) which gives you 1200 Ohms for a resistor. This is the minimum size resistor you would want to use for an 18 volt power supply.
You can have higher voltages but you need the correct size resistor.
richhotrain richg1998 Our club uses a single, 20ma, 3mm green/red bipolar, two lead LED in series with one lead to the Tortoise. We use to use 12 vdc but lowered the voltage to 9 vdc for slower point operation. We used a LM317 voltage regulator, two resistors and two caps to have regulated 9 vdc for the turnout motors. Current is less than 20 ma when operating. No resistor needed. Rich Rich, that is what I don't get. Since you r voltage is 9vdc, what keeps the unresistored LED from immediately burning out? Rich
Rich:
The tortoise limits the current. Think of it as a big, green resistor if that helps.
Joe
JoeinPA The tortoise limits the current. Think of it as a big, green resistor if that helps.
OK, that makes sense.
Any way to eliminate the resistors on the signals in my earlier diagram?
Go back to my first post. It's one of Kirchoff's laws, all elements of a circuit ion series will have the exact same current flowing through them. A Tortoise is 15ma, ergo anythign in series with the Tortoise will also be 15ma.
To do the same witht he signal, just stick it in series witht he other LED and Tortoise motor wire. Or the other Tortoise motor wire, it really doesn't matter. For simplicity's sake - one wire fromt he DPDT to the Tortoise pin 1. The other wire from the DPDT to one side of the bicolr panel LED. The other side of the bicolor panel LED to one side of the signal LED pair. The other side of the signal LED pair to the Tortoise pin 8.
That's assuming the signal LED is actual a red and green LED hooked up back to back - like I said, making the same thing that a 2 lead bicolor LED is, just using individual LEDs. That's all those bicolor LEDs are, two LEDs in one case.
rrinker If you're using a measured 18V DC to drive the Tortoises per your original diagram technically you are driving the Tortoises with too much voltage. They are rated for 12V. The sweet spot I've found seems to be about 9V, enough to mve them at a reasonable speed and hold, but MUCH quieter than the full 12V.
If you're using a measured 18V DC to drive the Tortoises per your original diagram technically you are driving the Tortoises with too much voltage. They are rated for 12V.
The sweet spot I've found seems to be about 9V, enough to mve them at a reasonable speed and hold, but MUCH quieter than the full 12V.
I mentioned earlier in this thread that I was using an MRC Railpower 1370 to power my signals, Tortoises, DPDTs and control panel LEDs. I indicated that the power pack was 18 volts.
I was concerned after reading the comments about the recommended voltage to the Tortoise, so I used a volt meter to test the output of the MRC Railpower 1370. I got a reading of 13.25 volts. Good news for the Tortoise, but I wondered what happened to the alleged 18 volts which are the stated voltage on the MRC web site.
Turns out, in reading the operator manual, that the the voltage is less. Here are the stated voltages.
Output - 15V DC, 19V AC
Total Output - 18 VA
So, the DC voltage is actually 15, not 18.
When MRC says that Total Output is 18 VA, what are they referring to?
Why use the word "Total"?
And what does the A stand for in 18 VA ?
And, finally, why do I get a reading of 13.25 volts if the power pack says 15 volts?
18 VA is 18 Volt-Amps. It is similar to watts, though not exactly the same(volt-amps uses RMS values). For example, if the power pack can provide 18 VA at 12 volts, that would be approximately 1.5 amps(12 volts x 1.5 amps = 18 volt-amps).
If there was any load at all on the power pack when you measured the voltage, that would explain the difference in the rated voltage and what you measured. Since the power pack is not regulated, any load on it will cause a drop in the output voltage.
You cannot wire an LED without a resistor, but the Tortoise can be your resistor.
(Other wiring methods are known, but those are not what the OP asked about)
ROAR
The Route of the Broadway Lion The Largest Subway Layout in North Dakota.
Here there be cats. LIONS with CAMERAS
Well, so it's 15V not 18V, you'll still be fine with 2 sets of LEDs in series with the Tortoise, that gives you about 10.8 volts to the Tortoise, which is plenty.
Thanks CSX Robert for that explanation.
rrinker Well, so it's 15V not 18V, you'll still be fine with 2 sets of LEDs in series with the Tortoise, that gives you about 10.8 volts to the Tortoise, which is plenty. --Randy
Thanks, Randy. Sorry for unintentionally misleading you on the voltage.
BroadwayLion You cannot wire an LED without a resistor, but the Tortoise can be your resistor.
Is that absolutely true? Couldn't some LEDs operate without a resistor if the voltage is low enough?
Nope, LEDs operate on current, not voltage. They will quickly avalanche and run away of the current is not restricted. There is such a thing as a constant current source, using something like that set to a value below the LED's limit would allow the LED to be connected without a resistor.
So-called "12V LEDs" are just ordinary LEDs with a resistor already inside.
Essentially what happens is that once the LED conducts, it allows more current through. This allows it to conduct more..so more current goes through. Until it gets beyond the limit for the particular design and chemistry, and the junction is fried. Resistors in the circuit work because of, once again, Kirchoff's Laws. Two devices in series combine voltage, and two objects in sereis share current. LED (and regualr diodes too) have a set voltage drop, at all but the extremes of their operating range, this voltage drop is always the same (in reality, there is a very sharp curve once the limits are exceed, for the practical purposes that we are using here, we can assume it's a constant). This is part of the LED spec. Common for a T1 or T1 3/4 size red, green, or yellow LED is about 2.1 volts, but always check the specs on the ones you buy. Typcal white LEDs are more like 3.5 volts. By applying Kirchoff's Laws to our circuit of a power supply, LED, and resistor, we find that the sum of the voltage through the resistor and the voltage throught he LED must equal the power supply voltage. So if there is a 12V power supply, and the LED drops 2.1 volts, 9.9 volts must drop across the resistor. We also know that if we want 15ma in the LED, then the resistor must also have 15ma of current flowing through it. We now know the voltage across the resistor, and the current through it. Ohm's law provides us with the missing piece - how many ohms must the resistor be. Ohm's Law says E(voltage) = I (current) x R (resistence). Dust off that algebra, and solve for R, and you get R=E/I. We know E, 9.9 volts. We know I, 15ma, or .015 amps. (The other piece of the puzzle is that the units must match - to get Ohms you need Volts and Amps, not millivolts, or milliamps). Plugging in, we have R=9.9/.015, or 660 ohms.
Now you know more than you ever wanted to know about electronics. These are the basics that apply to any circuit, not just LEDs. You just have to know what is fixed and what is variable - with an LED or diode, the volts is fixed and the current can vary. With an incandescent bulb, the current is fixed and the voltage can vary, so you need to fix the voltage to something that won;t blow the bulb.
rrinker Now you know more than you ever wanted to know about electronics.
Now you know more than you ever wanted to know about electronics.
Randy, quite the contrary. I really appreciate your taking the time to explain this in such detail and so clearly. I find it all very fascinating. More importantly, though, it is important to know and understand at least the basics. When I first got into the hobby ten years ago, if I didn't understand how to wire something, someone told me how to do it and I just did it. But the more time passes, the more I feel a need to understand this stuff for my own good.