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Homework Help: How to switch power source to a backup power source in case of a failure?

  1. Mar 1, 2010 #1
    1. The problem statement, all variables and given/known data

    So this is for a fourth year engineering project, I'm on the power subsystem of a lunar penetrator. I'm not in electrical engineering, I'm in aero (mech), so forgive me for my lack of specialized knowledge. I want a system in place where if the main power source fails (RTG), the loads are automatically switched to a backup battery.

    From researching switches I have come to the (perhaps incorrect) conclusion that a relay switch would be too big/bulky for the penetrator (~7cm internal diameter "artillery shell"), so perhaps transistor switches are a better option. I have found that a BJT transistor can be used as a switch (programmed to "expect" a certain input voltage, and if it does not receive said voltage it can open/close a switch).

    So if these are used the transistor can be "programmed" to expect 4.5 V, and if the RTG fails, and the voltage drops to zero, it opens the switch. My question is, how do I get the switch on the backup battery to close? If I used the same theoretical transistor setup, there would be no change. 0 Volts, RTG fails, still 0 volts. So basically I don't know how to go about approaching this problem. It seems to me that I'm missing something very easy here, I mean it's pretty much the same idea as an alarm clock having a backup battery. I'm just having trouble finding info on this...

    edit: For instance, I find patents like this, but in the context of this project I think a more simple solution might be more practical considering it's a preliminary design. http://www.freepatentsonline.com/4528459.pdf
     
    Last edited: Mar 1, 2010
  2. jcsd
  3. Mar 1, 2010 #2

    berkeman

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    Staff: Mentor

    A simpler solution if it can be used, is to diod-OR the two power sources. So feed power from your power source to your load through a diode, and also feed power from your battery to the load through a diode. If you power supply normally operates with a little higher output voltage than the battery, then the battery diode is open (off) due to the reverse bias. If the power supply fails and starts to drop, the battery diode turns on and supplies the load. The power supply diode would then be reverse biased, so no current from the battery flows into the powered-off power supply.

    Choose power Schottky diodes for their lower forward voltage. Size them to handle the power (V forward diode drop * I current)
     
  4. Mar 4, 2010 #3
    Thanks for the tips! So from my understanding, I no longer need an actual switch with this setup, just the sized diodes? I uploaded a schematic diagram here:

    http://i828.photobucket.com/albums/zz204/acampbeb/diagram.png

    You might notice that the load setup probably doesn't make a lot of sense. That would be the next situation I'm having difficulty with. If you have multiple loads, and you only want certain ones on at certain times (want to be able to control when one load goes "on", and one load goes "off"), is this achieved with a simple sensor, capacitor, switch, and feedback loop? I still don't understand how the sensor would work if the switch is naturally open.
     
  5. Mar 4, 2010 #4

    berkeman

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    Staff: Mentor

    How do a 4.5V power supply and a 3.9V battery source the 12V bus?
     
  6. Mar 4, 2010 #5
    DC-DC converter. Did I use the wrong symbol for that?
     
  7. Sep 22, 2010 #6
    Thanks again, berkeman.
     
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