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I am developing a circuit for somebody else to control 3 analog DC model trains. I am unsure how I should make the short circuit part. There are more ways to achieve this.

My idea is as follows: enter image description here

1 single 12V DC power supply is used to supply power to 3 analog controllers. One such a controller exists out of TIP120 Darlington transistor, a potentiometer, a DPTS switch (for direction) and a couple of flyback diodes.

The Tip120 in TO220 casings will be connected to a large metal plate and the trains are N-gauge so the current will most likely not exceed ~300mA.

The short circuit part (left side of the schematic) exists out of 3 Bc547 tranistors and a N-channel MOSFET which acts as the primary power switch (can be replaced by a relay as well).

The idea here is that under normal circumstances transistor Q1 is in conductive state. This causes Q2 to be off and Q3 (N-channel MOSFET) should also be high as the gate is pulled up to ~12V via R12.

All current of the tracks flows to shunt resistor R3. If the current exceeds 1.4A the voltage over this resistor will be 0.7V which should cause transistor Q4 to conduct. This action should cause Q1 to go out and Q2 to go on. If Q2 is on, it should pull the gate of Q3 to ~0V via R13 (100R). And this action should cut all power to the tracks. That is atleast what I want to happen.

It may be (though it is unlikely) that the starting current of a train's motor may exceed this 1.4A. In an attempt to prevent a shut-off I added C1 and R13 which should keep Q3 in a conductive state for about 0.1s. So for 0.1s I want to allow a short circuit. I believe that this should prevent the Tip120s from frying. This is the part of which I am least certain it would actually work.

After the power is cut off, one can reset the power by pressing SW1a. This should disable Q2 which enables Q1 and Q3 again.

I copied the idea from enter image description here ElecCircuit.com The transistor Q4 was an idea of my own. I thought it would work better than the schottky diode in this example because it would also limit the maximum current.

  • I would like to use as few components as possible.
  • I would like not to use 'exotic' components which are hard to find.
  • I would like that the circuit does not die, not ever.
  • I would like to allow a short circuit for no longer than 100ms
  • All tip120 transistors as well the as the N MOSFET will be physically connected to a large sheet of metal for heat dissipation.

My questions:

  1. Would this circuit work?
  2. Is it a good idea to put the short circuit protection in front of the train controllers?
  3. Is it likely that this circuit won't fry and die?
bask185
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  • These designs are very "1980s style" that's OK but things have moved on. I would consider using switched mode regulator modules, these can be bought cheap on Ebay etc. You would save the money for a heatsink. Many modules have a trim-pot to regulate the output voltage, you can replace it with a standard potmeter. Modules with an additional maximum current setting also exist, that is even better than short circuit protection. Also these modules use switching ICs which have built in temperature protection. – Bimpelrekkie Sep 21 '20 at 11:51
  • I'm talking about something like this: https://www.ebay.nl/itm/LM2596-DC-DC-Step-down-Adjustable-CC-CV-Power-Supply-Module-LED-driver-CK/191936556464?hash=item2cb04f6db0:g:LA8AAOSwaB5XoDzX These have a minimum output voltage of 1.25 V, if that is an issue I would suggest to put two diodes in series with the output to drop roughly the same voltage. – Bimpelrekkie Sep 21 '20 at 11:54
  • The metal sheet is the control panel itself. So it is there anyway. It will be 40 x 30cm big – bask185 Sep 21 '20 at 11:54
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    Have you simulated it? – user253751 Sep 21 '20 at 11:56
  • Even without the heatsink these modules are cheaper than all those discrete components. Also, modules "just work". Also, in the circuits you propose, a button needs to be pressed to reset the protection. I would find it more convenient if the voltage is automatically restored as soon as the short circuit is removed. – Bimpelrekkie Sep 21 '20 at 11:56
  • I have not simulated it. I tried that once on circuitOnline but I did not understood it. Also that buck converter modules supplies atleast 1.25V. How do you solve that problem? – bask185 Sep 21 '20 at 11:58
  • Also I am still learning about Kicad, so when I can simulate I will – bask185 Sep 21 '20 at 11:59
  • *Also that buck converter modules supplies atleast 1.25V. How do you solve that problem?* Like I mentioned: 2 diodes in series with the output. That is the most simple solution. There are more elegant solutions but then you need a more complex module as usually a negative voltage will then be needed. For your application, I would use the diodes as that's cheap and good enough. – Bimpelrekkie Sep 21 '20 at 11:59
  • Simulate before building and welcome to the 21st century. By the way, your title suggests you are trying to protect the model trains from over-current but I don't think that is what your circuits do. – Andy aka Sep 21 '20 at 12:02
  • I altered the title. In al likeliness it is going to take a lot more time to learn how to get the simulator working in Kicad than to solder and test the circuit. I tried it in circuits online, but it appears it cannot continuously run a simulation. + you can only use that webpage for a limited amount of time – bask185 Sep 21 '20 at 13:29

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