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Power consumption of reverse-polarity protection diode

Discussion in 'General Electronics Discussion' started by nexaen, Sep 2, 2019.

  1. nexaen

    nexaen

    3
    0
    Sep 2, 2019
    Battery:
    I am using a battery pack for my project that has SB540, Schottky rectifier, connected across the battery cells' poles. This is to protect against reverse polarity connection, I assume.
    The battery pack consists of two primary cells, Tadiran SL-2780, connected in series (7.2V output). Each cell has its own diode. Therefore, two diodes exist in battery pack. Battery chemistry is lithium thionyl chloride inorganic electrolyte.
    [​IMG]

    What is the question?
    My project is running very low on power as mostly it is in deep sleep mode and consumes about 7uA.
    Q1. Considering this little consumption, and the chance of connecting the battery in wrong polarity is almost zero, is it a good idea to remove the diodes to prolong battery lifetime?
    Q2. Do diodes, in this configuration, use energy and dissipate as heat? (we know that ideal theories do not apply in reality and all diodes have leakage in reverse.) I would like to prevent any energy loss, even small ones in uA range.

    BTW, I have already seen this topic.
     
    Last edited: Sep 2, 2019
  2. Alec_t

    Alec_t

    2,691
    712
    Jul 7, 2015
    Welcome to EP!
    The diodes are there to prevent the cell with the lowest capacity getting charged 'backwards' in the event that the battery is drained excessively. Removing them would not be a good idea.
     
    nexaen and Cannonball like this.
  3. Ylli

    Ylli

    226
    46
    Jun 19, 2018
    You may really have an issue there. Attached excerpt from the data sheet. The SB540 is a 40 volt part, so you are operating it at about 9% of ratings. At 25°C, the leakage is around 6-7 uA. At elevated temperatures, it gets worse. If it is possible to change the diodes, you could probably find some with lower leakage. But as others have said, total removal of the diodes is not a good idea.
    Capture.JPG
     
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  4. nexaen

    nexaen

    3
    0
    Sep 2, 2019
    Thanks for all your answers.
    Alec_t, could you please elaborate on it? How is it going to prevent a cell from being discharged in backwards?
    My electronics basics are not that strong.
    The diode is just allowing charge to pass through one way, and not the other. Therefore, if the diode that is parallel to the cells gets removed, then there is no path there at all. I would be thankful if you explain it or give me some links to read.
     
  5. Alec_t

    Alec_t

    2,691
    712
    Jul 7, 2015
    Perhaps this simulation will explain. Two cells each initially at +3.7V are simulated by capacitors, with C1(C3) having greater capacitance than C2(C4), so C2(C4) represents the weaker cell. Without the diodes, C2 discharges from +3.7V down to a damaging reverse voltage of -1.2V. With the diodes the reverse voltage of C4 is limited (by the forward drop of a Schottky diode) to a non-damaging -0.3V.
    ReverseCharge.png
     
    Last edited: Sep 4, 2019
    nexaen likes this.
  6. nexaen

    nexaen

    3
    0
    Sep 2, 2019
    Thanks Alec for the figure, explanation and test you made.
    Regarding SB540, its datasheet shows around 250mV at 10uA sourcing. This forward drop does not seem that much to protect. Do you agree with this statement?
    Based on this article, it seems battery manufacturer could use either other types of diodes or Pmos FETs. Do you confirm?
     
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