Power Backup System with Linear Lithium-Ion Battery Charger

Discussion in 'General Electronics Chat' started by Fabián Romo, Mar 18, 2015.

  1. Fabián Romo

    Thread Starter New Member

    Mar 18, 2015
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    Hi.

    I want to build a backup power system with a lithium-ion battery that includes a charger.
    My system consumes 1 A during operation. The system works about a few seconds and then enter a low power mode and thus save energy.

    I made a conceptual image of the idea I have in mind of that system, please see the image at the following link:

    [​IMG]

    The system consists of a Voltage Boost Converter circuit and Lithium-Ion Battery Charger circuit.

    The Voltage Boost Converter circuit could work with input voltages from 2 volts to 10 volts and its output is 5 V.

    When an external voltage of 5V is present, the battery is charged by the Lithium-Ion Battery Charger circuit.

    When the external source of 5 V is removed from the system, the diode D (Schottky) allows the discharge of the battery and the system continues to provide 5V thanks to Voltage Boost Converter circuit.

    Do you think the concept is correct?

    Could there be a problem with that design?

    Any ideas or suggestions are welcome.

    Thanks
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,148
    1,791
    I'm not sure about the concept. What I see is that without the external supply, both the battery and the charger are supplying current at different voltages through the Schottky diode, to the input of the boost converter. This seems like a really bad idea.

    What exactly is a "3-A High Voltage Boost converter" that has a 5V output?
    The topology will have to be Boost-Buck if you wish to accommodate an input range of 3-10 Volts.

    Whatever happens, this system cannot create more power than is already there. So 5 Volts at 3 Amperes is 15 Watts. If the input is at 3.3 Volts and the conversion is 80% efficient then the input power will be 18.75 Watts which at 3.3 Volts requires approximately 5.7 Amperes. That is a pretty heavy discharge rate for a battery.
     
  3. Fabián Romo

    Thread Starter New Member

    Mar 18, 2015
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    0
    Hi.

    Thanks for aswering.

    In the "3-A High Voltage Boost converter" can work with 3 to 10V in the input, in my case when the external source is, there is 5V input and when the external source is removed, there are approximately 3.3V at the input. The output is always 5V.

    3 A is the maximum capacity that could supply the converter. In my case I only need 1 A.

    If the battery is supplying power, the approximate power is 5W (5V x 1A)

    Then the current battery supplies mores would be approximately 1.5 A (5W / 3.3V).

    All this is ideally and I know the efficiency of the converter is approximately 80% to 90%

    So I could assume that the current in the battery should be approximately 2 A.

    Why do you say 15 W?

    If my idea is wrong, please, can you provide me some proper design.

    Thanks
     
  4. Papabravo

    Expert

    Feb 24, 2006
    10,148
    1,791
    The words "High Voltage" suggests something on the order of 100 Volts or so, and using those words is misleading because there is no high voltage involved. Using the words "Wide input range" DC-DC Converter might be better.

    I thought 3A was the output of your converter, of which 1A was used externally and the rest supplied the charger. The problem is that you have power going around in a loop and you can't really do that. That pathway must be blocked when the external supply is on, and battery charging must be inhibited when the battery is supplying the converter. That's the way it needs to be done.
     
  5. Fabián Romo

    Thread Starter New Member

    Mar 18, 2015
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    'High Voltage' is the name that the manufacturer gives to the converter:

    http://www.ti.com/product/tps61175


    I am aware that when the external source is removed, the same battery can not be recharged itself. (It is obvious that the laws of physics do not allow).

    What's certain is that when the battery is powering the charger circuit energy is blocked in some way.
     
  6. Papabravo

    Expert

    Feb 24, 2006
    10,148
    1,791
    So the "High Voltage" refers to the maximum Vout of 38V which was not evident from your original post. I guess 38V is "High Voltage with respect to the minimum input voltage, but that is really stretching it IMHO.
     
  7. Roderick Young

    Member

    Feb 22, 2015
    408
    168
    This is the arrangement that would make sense to me. If you need 5 volts, no point in trying to put it through a boost converter that will basically not be boosting.

    charger.gif
    I suppose the Schottky diode at the top would be optional.
     
    Fabián Romo likes this.
  8. Fabián Romo

    Thread Starter New Member

    Mar 18, 2015
    4
    0
    Ok, thanks, think that's the best solution.
     
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