12 VDC 500mA Wall Wart Transformer Turnout Control Capacity Question

Hardcoaler May 15, 2018

  1. Hardcoaler

    Hardcoaler TrainBoard Member

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    The N Scale railroad that I'm planning will have 24 Kato turnouts, each powered by a DPDT switch wired to a 1000 mfd capacitor and two 5mm LEDs, only one of which will be lit at any one time. 2.2K Ohm resistors are wired with the LEDs. I'm using a common circuit design.

    Will a simple 12 VDC 500mA "Wall Wart" Transformer be able to support this assignment? I'm thinking that it might be weak, but then again, it's output is 18 VDC unloaded.

    24 LEDs * 20 mA Ea = 480 Total
    24 1000 mfd Capacitors * ??? = ??? Total (Max draw at startup as the caps charge)

    Any advice is welcome. Thank you.
     
  2. SP_fan_1951

    SP_fan_1951 TrainBoard Member

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    First of all, you are over estimating the current through your LED's. Assuming red LED's, with a forward voltage drop of 1.8V, the current through each LED will be 12V-1.8V/2200 = 4.7 mA, instead of the 20mA. This gives you just about 100 mA total. The current required for charging the caps will be limited by the internal impedance of the wall wart and the equivalent series resistance (ESR) of the caps. What is your intended use of the DPDT switches? The Kato turnouts are intended for momentary pulsing. I suggest you look at the thread here:.
    http://www.trainboard.com/highball/index.php?threads/kato-turnout-throw-current.74277/
     
  3. Hardcoaler

    Hardcoaler TrainBoard Member

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    Thank you for the clarification on power consumption -- good to know this. The circuit uses the capacitor to provide a momentary supply of power, then shuts the power off. Various circuits by Rob Paisley and George Stillwell provide this and they work well. They're pretty neat. I've had fun building several variants and testing them with analog and digital VOMs to assure the power shuts off instantly and completely (using a DPDT non-momentary contact switch). One side of the switch controls the turnout power and the other side, the LEDs.
     
    Last edited: May 15, 2018
  4. sachsr1

    sachsr1 TrainBoard Member

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    I think it really would depend on how often you are throwing the turnouts. The capacitors should hold enough power to throw the turnout, but will draw power while recharging. If you have a route that requires rapidly throwing a couple turnouts it might not have enough current. It might be easier and cheaper to just buy or find another power supply with more current. I used a few wall warts around my layout, but now I use old computer power supplies. I salvaged them out of old desktop computers, and they have 12, -12, 5, and 3 volts outputs. You can get a 2 amp power supply on Amazon for under $8.
     
  5. Hardcoaler

    Hardcoaler TrainBoard Member

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    Thanks sachsr1. I won't be rapidly throwing a lot of turnouts, but when I plug in the layout after a long recess, all of those caps will be drawing current at once and I might have trouble.

    I have a surplus laptop power supply that produces 18VDC @ 1.1 Amps, but I'm afraid that the voltage might be too high for my circuit. I might buy a cheap "buck converter" to drop the output a bit, but per your suggestion, a new 12 VDC supply with higher amperage might be easily bought. I didn't know that power supplies were so cheap these days. I'll check around, and thank you again.

    I guess I ought to protect these supplies with fuses or are most self-protected would you guess?
     
  6. SP_fan_1951

    SP_fan_1951 TrainBoard Member

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    Its always a good idea to provide a fuse for safety when using an OEM supply. While they are generally current limited and short circuit protected, that does not provide protection against internal component failure. The old fashioned linear supplies were more robust than the new switching supplies in that regard. If the switching device fails shorted, the only limitation to the current draw is the breaker on the distribution panel and that can allow too much current to prevent a fire.
     
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  7. sachsr1

    sachsr1 TrainBoard Member

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    tjlaswell likes this.
  8. Hardcoaler

    Hardcoaler TrainBoard Member

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    Whoa -- that's pretty neat! Doesn't get much easier than that and free shipping too. Thanks.
     
  9. Hardcoaler

    Hardcoaler TrainBoard Member

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    Good point and at a minimum I want to protect the components in my circuits. I' cry if I had to rebuild 24 controllers, with caps, resistors and LEDs.
     
  10. Hardcoaler

    Hardcoaler TrainBoard Member

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    I've recently come across a 12VDC 3 AMP power supply "brick" that should provide more than ample power for my turnout circuits. Because I'm not sure if the power supply is fuse protected, I'd like to add protection against a short. If I were to add a simple 2-1/2 Amp fuse like this upload_2018-6-25_16-8-59.png , would it suffice? There's no way my circuits will draw even 2-1/2 Amps, so I figured to stay on the safe side to protect a 3 Amp supply.

    I've tried reading about fuses on the Internet and am more confused than ever. Thank you.
     
  11. sachsr1

    sachsr1 TrainBoard Member

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    That should work but I think there are fast burn and slow burn fuses. You probably want the one that pops fast.
     
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  12. Hardcoaler

    Hardcoaler TrainBoard Member

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    I bought an inexpensive 12VDC 3A "switching" power supply like those normally used to power modems and the sort. As you know, these are sealed in a plastic case and seem to produce very little heat when compared to a "wall wart" type simple transformer.

    Question: Does my switching power supply draw only enough amperage to satisfy the demand of its load or does it pull 3 Amps all the time?

    I hope to power about 50 Ea 20mA LEDs (with dropping resistors) with this power supply and am trying to figure what kind of amperage the combined load of the LEDs and dropping resistors will draw from this power supply. SP_fan_1951's reply indicates that the resistors actually reduce the amp draw? I'd have assumed that they consume energy and waste it in heat? I'm confused.

    Thanks y'all.
     
    Last edited: Aug 11, 2018
  13. wvgca

    wvgca TrainBoard Member

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    yes, the resistors consume energy, that's why you select a dropping resistor , among other things, on power draw ..
    but..... the TOTAL amp draw is LESS .. by the way, 20ma may be too bright, try running them at 10ma, or less
     
  14. Hardcoaler

    Hardcoaler TrainBoard Member

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    I recently noticed the brightness you wrote about and it's indeed too much. I'm playing with tinted plastic diffuser caps that fit over the LEDs combined with various higher value resistors to reduce and even out the level of light across the colors. Yes, I want them to softly glow, not blind the operator.

    So then the resistors have nothing to do with amp draw? If so, would 50 Ea 20 mA LEDs running with dropping resistors at 10mA draw 1/2 Amp?
     
  15. wvgca

    wvgca TrainBoard Member

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    yup...increase the dropping resistor to give 10ma power draw per LED
     
  16. RBrodzinsky

    RBrodzinsky Staff Member TrainBoard Supporter

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    A little bit more. Each resistor does draw current, which it then gives off as heat.
     
  17. Hardcoaler

    Hardcoaler TrainBoard Member

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  18. Hardcoaler

    Hardcoaler TrainBoard Member

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    That's what I figured too Rick, that the LED uses current to give off light and the resistor uses the excess current and gives it off as waste heat, albeit tiny amounts. The higher the resistor value, the higher the heat. I thought that by adding the amp draw of each together, I'd approximate total amp draw.

    This is an interesting discussion and thanks for your conclusions too wvcga. There's an answer somewhere in thus continuing dialog.
     
    Last edited: Aug 12, 2018
  19. SP_fan_1951

    SP_fan_1951 TrainBoard Member

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    A more exact description of the operation of LED's is as follows: LED's, or any diode for that matter, have a non-linear current vs voltage curve. Once the diode is biased greater than the forward band-gap voltage of the semiconductor (depends on the actual material) the only limit to current through the device is the very small intrinsic bulk resistance of the semiconductor. To prevent drawing so much current the diode will literally melt, a limiting resistor is placed in series. In a series circuit, the same current flows through each component. The turn-on voltage for LED's varies from about 1.8V for red, up to 3.3V for some blue or white LED's. Ohm's law states Voltage = Current x Resistance, or rearranged Resistance = Voltage divided by current. So if you take the supply voltage, and subtract the LED turn-on voltage, divide that by the current you desire you get the value for the current limiting resistor. The total power dissipation in the circuit is equal to the supply voltage divided by the current. The total current for multiple parallel circuits is simply the sum of their individual currents. If you want to use smaller resistors, put several LED's in series to increase the total turn-on voltage. Because the same current flows through several LED's, you get more light for the same power used.
     
  20. Hardcoaler

    Hardcoaler TrainBoard Member

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    I made up a little Excel sheet SP_fan, but I get lost in the final calculation. You wrote that "The total power dissipation in the circuit is equal to the supply voltage divided by the current". Is power dissipation the same as power consumption? Using your narrative, I calculate 0.53 mA dissipation and an on line calculator shows power consumption of 18.9mA. Am I messing up somewhere? Math isn't my strong suit. :oops: Thank you.

    Calculator:
    Supply Voltage: 12.3 Volts
    LED Turn On Voltage: 1.7 Volts
    Excess Voltage: 10.6
    Desired Current: 20 Milliamps
    Limiting Resistor Should Be: 530 Ohms
    Total Power Dissipation: 0.53
    Total Power Consumption: 18.9 mA per an on line calculator. How is this number calculated?
     

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