Using a reservoir capacitor as a temporary supply

Discussion in 'The Projects Forum' started by CircuitZord, Mar 29, 2014.

  1. CircuitZord

    Thread Starter Member

    Oct 8, 2012
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    I am using a Beaglebone Black development board (BBB) by Texas Instruments as part of a control unit for a race car at university.

    My task is to develop a circuit to activate the reset function on the dev board when the power is cut suddenly, because if it's not properly powered down (OFF button), the system gets stuck or hangs or something.

    This is essentially the schematic I've come up with:

    [​IMG]

    I'm thinking of using a NOR gate, with one input tied to Vcc (5V), and the other to ground. So basically when Vcc goes LOW, the output of the logic goes high and triggers RESET.

    The Beaglbone has the ability to charge an external battery to 3.7V, so this pre-charged battery will keep the logic gate powered.

    One thing I haven't indicated in the drawing is that I will also use a reservoir capacitor as a buffer between Input A and Vcc.

    I just want to check that my thought process here is correct. BBB uses roughly 60mA at 5V during the reset process. That's roughly 0.3W or 0.3J/s. I was informed that the reset operation is instantaneous, but as a worst case scenario I just assumed it would take 5 seconds for instance. If that's the case, then it would require 1.5J from a reservoir capacitor.

    Using E = 0.5 * C * V^2, I calculated C= 120mF

    To get a resistance model of BBB to calculate the RC time constant:

    R = 5V/60mA = 83 Ω

    Therefore τ = 83 x 120mF = 9.96 seconds

    Using the capacitor discharge equation, V = Vo.e^(-t/RC) and the fact that BBB will operate with at least 4.8V supply

    I calculated t = 0.407 seconds for the voltage across the capacitor to drop to 4.8V which I figure should be enough to allow the reset operation to complete, if not I can just bump up the capacitance rating.

    Thoughts?
     
  2. #12

    Expert

    Nov 30, 2010
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    What kind of race car?
     
  3. CircuitZord

    Thread Starter Member

    Oct 8, 2012
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  4. snav

    Active Member

    Aug 1, 2011
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    I have some NEC supercaps and they're rated for .047F@5.5v which let me take up to 60s to change memory backup battery in a transmitter we use. It's very small d=12mm h=8mm with leads 5mm c=c. I assume the battery holds up the BBB reset logic.

    Use a diode to isolate the Buffer cap to just the VCC of your failure detection logic regardless of size you choose.
     
    Last edited: Mar 29, 2014
  5. ronv

    AAC Fanatic!

    Nov 12, 2008
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    Here's my 2 cents worth.
    While the reset is not instantaneous it will be way faster than what you are worried about. Since the circuits in the BBB run on 3.3 volts plus the regulator drop (.2 v) I think I would just use a comparator (LM339) with a reference of say 4.5 volts (TL494) to form a power safe circuit.
     
  6. ronv

    AAC Fanatic!

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  7. CircuitZord

    Thread Starter Member

    Oct 8, 2012
    59
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    Thanks for that ronv! Had no idea such an IC existed, will definitely be cleaner than implementing something separately with logic gates, and probably quicker too.

    Any idea though of whether my idea would have worked in practice?
     
  8. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    I don't agree with your capacitance calculation. You can't use ALL the energy in the cap least you discharge it completely to zero volts.

    As always, I = C dV/dT Assume your BBB draws a constant current so we can assume this is linear. I'm going to assume you can tolerate a 1V drop during this reset. Solving for C:

    C = I * dT/dV = 60 mA * 5 seconds / 1 V

    C = 0.3 F

    That's farads. Not impossible these days but needs some special handling so it doesn't appear as a major short to the power supply.

    There's many makers of supervisory chips. Being a PIC used I know and like the ones from Microchip: they just sit on the power rails and provide an output so they fit in a SOT-23.
     
  9. CircuitZord

    Thread Starter Member

    Oct 8, 2012
    59
    2
    Sorry, I should have been a bit more clear. Yes, I do know that it won't use all that energy (1.5J) until the capacitor is fully discharged, but I took 5 seconds as a worst of the worst case scenario. I was just informed that the reset operation was instantaneous.

    So based on those 5 seconds which was pretty arbitrary I continued, and found out that the voltage in the capacitor will drop to roughly 4.8V in about 0.407 seconds, which I assume is enough for it to do the reset operation. If not, I can simply bump up the capacitance rating.

    Although, I can't tolerate a 1V drop, at most 0.2V drop, which means it would require a 1.5F capacitor.

    What do you mean by a major short to the supply?
     
  10. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    Put a discharged 1.5F cap across a 5V supply and see how much current it draws.

    Hint: if there is no current limiting in the supply expect to see the magic smoke.

    And I don't see where the 1.5F comes from. (I don't think your energy calculations are correct.) However, your numbers seem on the high side so it should work anyway.
     
  11. ronv

    AAC Fanatic!

    Nov 12, 2008
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    In principal it would probably work, but there isn't much room between the 5 volts falling and where the micro stops working. The switching level for the logic probably would not match that voltage.
     
  12. mattepps717

    Member

    Jan 20, 2014
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    CircuitZord, how much time are you needing for the discharge, so you get properly shutdown? Or rather, what would be your target time required for cap discharge?
     
  13. Papabravo

    Expert

    Feb 24, 2006
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    The problem with using a capacitor for energy storage is the exponential decay of voltage with time. Even with relatively large capacitors the needs to be limited to small currents in order to keep the circuits above their minimum VCC.

    If I were trying to do this, I'd be looking for a processor that could run at 1.2 volts and tolerate a Vcc of +5V. In other words I would maximize the voltage range over which the part would function. After dropping to 37% of the original VCC (1.85 volts), the discharge rate from the capacitor will slow down dramatically. Its like the Demon Drop; the initial acceleration will knock your socks off, but you will survive the deceleration.
     
    Last edited: Mar 30, 2014
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