Battery Backup Circuit Design Help

Discussion in 'Power Electronics' started by Alex Fagard, Nov 27, 2016.

  1. Alex Fagard

    Thread Starter New Member

    Nov 27, 2016
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    Hey there! It is my first time posting to the forum so please let me know if I'm doing something wrong [​IMG]

    So recently I've been interested in designing a circuit that switches from one battery pack to another using one digital pin of my arduino micro (yes, I understand that both power supplies can be added in series but its more of a proof of concept).

    Anyways the specs of my batteries are as follows:
    Normal battery: 6v @ 2800mah
    Backup: 7.2v @2800mah
    Peak current draw of the circuit to be powered: somewhere around 850ma to 1amp (w.i.p.)

    Voltage regulator: https://www.pololu.com/product/2111 (7amp version)

    I have voltage sense circuitry that I read with my microcontroller to tell me when the normal battery drops below 5.5v (my regulator requires atleast 4.5v to output 5v) at which point the normal battery will switch off and the backup will turn on. Obviously this has to be done rather quickly (<50ms?) otherwise we loose power to arduino (bad).

    Temporarily I've been using a 5vdc relay, but it requires like a 1000uf tank capacitor on the reg input otherwise the power to the arduino drops out. I suppose as the circuit gets more and more finished towards the full 850ma to 1amp peak current that I mentioned above this value will have to increase. This solution seems rather clunky.

    So: I was thinking of using a couple of P-channel FETS and a logic inverter. Such that, when the arduino digital output is LOW to the circuit the FET for the normal battery is open. However, when the output is HIGH the FET for the backup opens and concurrently the FET for the normal battery closes.

    I don't want to use diodes, or ideal diode packages such as the LTC4414 or 4415.

    I've attached a schematic with some possible part types and layout. I'm not sure if it will work, infact, all I am sure about is that the logic inverter will work.

    The schematic can be viewed/edited here (I believe): https://easyeda.com/editor#id=c424c536b26a46c08d017fe005df420f
     
  2. AlbertHall

    AAC Fanatic!

    Jun 4, 2014
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    You don't need R3 and R4, remove them.
    The MOSFETs are connected backwards - swap the drain and source.
    There is a possible problem if, at the switchover, both MOSFETs are on at the same time, even briefly, as they will link the two voltage sources and a large current could flow.
     
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  3. Alex Fagard

    Thread Starter New Member

    Nov 27, 2016
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    Ah nice catch on the MOSFETs being backwards. Can you think of any way to remediate the brief voltage 'addition'? I don't think it will matter that much as its providing power to a 5v regulator that can go up to 24 volts on VIN.
     
  4. AlbertHall

    AAC Fanatic!

    Jun 4, 2014
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    The regulator won't mind, but the battery, main power source, or MOSFETs might. I can't at the moment think of a simple way of fixing this.
    There is another potential problem depending on the supply voltages and the actual MOSFET threshold voltages. The gate voltage needs to go sufficiently positive to ensure the MOSFET switches off. For M1 this can be done by connecting R2 to source1 instead of +5V. The gate of M2 should go up to the same voltage as source2, but it is directly connected to the digital output so that's not going to happen so whether that would work or not depends on the particular MOSFET and the voltage of source2.

    And it won't work with the MOSFETs that way round either! The parallel diode in M2 will connect Vout back to the backup battery while it is running on main power.

    Thinking about this, I can see why they made the LTC4414.
     
  5. Alex Fagard

    Thread Starter New Member

    Nov 27, 2016
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    I recently stumbled upon this: http://www.vishay.com/docs/71110/71110.pdf

    I found the schematic on the first page identical to what I want to accomplish. But I think it has the same pitfall with the possible voltage 'addition' on switching?
     
  6. Alex Fagard

    Thread Starter New Member

    Nov 27, 2016
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    Scratch that, its a make before break switch so the output cap provides the power while the inputs switch
     
  7. AlbertHall

    AAC Fanatic!

    Jun 4, 2014
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    It is break before make but that is what you need to prevent the linking of the two supplies during the switch. The switching time is very small so the capacitor needed would be small.

    It has two problems for your application.
    1. It needs a +12V supply for the MOSFET gates which you don't have.
    2. It requires the two supply voltages to very close in voltage (<1V difference) to avoid the main supply back-feeding the backup supply. Your voltages are too different for that and so the main supply would feed via a diode back to the battery.
     
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