Connect with us

Reverse battery protection Diode

Discussion in 'Power Electronics' started by Akshatha Venkatesh, Nov 27, 2017.

Scroll to continue with content
  1. Akshatha Venkatesh

    Akshatha Venkatesh

    145
    0
    Jan 14, 2017
    Hi , I have an application circuit in which the Schottky diode is connected as below to a LDO and is used for reverse battery protection. How is the diode reverse biased when reverse battery is applied taking LDO into account , can anyone please explain .thank you.
     

    Attached Files:

  2. dorke

    dorke

    2,342
    665
    Jun 20, 2015
    An LDO is a 3 terminal creature,you left out the GND terminal,added in blue.
    Green -forward bias.
    Red -reverse bias.

    Kapish?

    IMG_20171127_160448812.jpg
     
  3. Akshatha Venkatesh

    Akshatha Venkatesh

    145
    0
    Jan 14, 2017
    What happens to the LDO when battery is connected in reverse polarity ?
     
  4. dorke

    dorke

    2,342
    665
    Jun 20, 2015
    What do you think happens?
    Try to answer yourself,with and without the diode.
     
    Akshatha Venkatesh likes this.
  5. Akshatha Venkatesh

    Akshatha Venkatesh

    145
    0
    Jan 14, 2017
    Actually what I don't understand is how is the diode reverse biased when the battery is reversed , when the LDO is present .
     
  6. dorke

    dorke

    2,342
    665
    Jun 20, 2015
    Try to replace the LDO (diode cathode to GND) with a resistor.
    Can you now understand ?
     
    Akshatha Venkatesh likes this.
  7. Akshatha Venkatesh

    Akshatha Venkatesh

    145
    0
    Jan 14, 2017
    No , when the battery is reversed in the circuit , the anode of the diode is connected to ground , but what happens to the cathode of the diode , the cathode of the diode is connected to the LDO.
     
  8. AnalogKid

    AnalogKid

    2,393
    665
    Jun 10, 2015
    No, it isn't.

    When the battery is reversed, the battery positive terminal is connected to the circuit ground, and the battery negative terminal is connected to the diode anode. That is what "reversed" means.

    The circuit ground is defined by the nature of the circuit elements, not by any particular battery terminal. In your circuit, the regulator GND pin defines the circuit reference potential.

    ak
     
    Akshatha Venkatesh likes this.
  9. Akshatha Venkatesh

    Akshatha Venkatesh

    145
    0
    Jan 14, 2017
    Okay, so when the battery is reversed the battery negative is connected to the diode anode , what is connected to the diode cathode ?
     
  10. AnalogKid

    AnalogKid

    2,393
    665
    Jun 10, 2015
    The diode cathode is connected to the LDO input, exactly as before. When you reverse the input voltage, only those two connections are changed. Everything else is unchanged.

    Also, there should be decoupling capacitors from the LDO input to its GND pin. The LDO datasheet will have recommendations.

    ak
     
    Akshatha Venkatesh likes this.
  11. dorke

    dorke

    2,342
    665
    Jun 20, 2015
    Let;s try it step by step.

    Can you explain,to yourself and to us,
    the forward/reverse of the diode in the circuit below?
    show us.

    @AnalogKid ,chewing the food for the OP doesn't really help him,it takes self work and understanding ;)

    d.JPG





    .
     
    Akshatha Venkatesh likes this.
  12. Akshatha Venkatesh

    Akshatha Venkatesh

    145
    0
    Jan 14, 2017
    So when the diode cathode is connected to the LDO input , how is it reverse biased ?
     
  13. Akshatha Venkatesh

    Akshatha Venkatesh

    145
    0
    Jan 14, 2017
    When the battery is reversed , the diode is reverse biased , but when the LDO comes into picture , how is the diode reverse biased when the battery is reversed ?
     

    Attached Files:

  14. dorke

    dorke

    2,342
    665
    Jun 20, 2015
    I shell try to explain in the simplest way possible(In steps):

    Forget about the LDO for now,lets look at the resistor case only.

    For the diode to be conducting(forward biassed or ON),
    lets assume a minimum voltage of Von(about 0.5V) is needed.

    Thus the voltage at the Anode should be greater than the voltage at the Cathode by Von to produce a forward ON current.
    .
    Va - Vk.>Von..

    Looking at the diagram I have put a constantly variable power supply instead of the battery.

    Can you answer the flowing questions:
    1. What is the diode's current at 10V power supply voltage ,is it ON?
    2. What is the diode's current at 1V power supply voltage ,is it ON?
    3. What is the minimum power supply voltage for the diode to be ON(forward biased) and what is it's current at that point?

    d.JPG
     
    Akshatha Venkatesh likes this.
  15. Akshatha Venkatesh

    Akshatha Venkatesh

    145
    0
    Jan 14, 2017
    1)At 10V, the diode will be on , the current will be 9.5mA.
    2) At 1V, the diode will be on , the current will be 0.5mA.
    3)The minimum voltage should be 0.6V to turn the diode on and the current will be 0.1mA.
     
  16. dorke

    dorke

    2,342
    665
    Jun 20, 2015
    Well,
    number 3 should be a bit above 0.5V,but basically it's o.k.

    Now,it is clear that if the power supply voltage is below(or equal) to 0.5 volts the diode is not conducting,
    it is Off and practically "reversed biased".
    Obviously if the power supply voltage is lowered further to 0V or below to negative values (same as reversing the battery polarity) the diode is OFF.

    Now replace the resistor with an LDO.
    It is basically the same thing only that the LDO is not a fixed value resistor but from the diodes point of view it is the same thing.
    Agree?

    d.JPG
     
    Akshatha Venkatesh likes this.
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-