How to connect a 11.1V to 5V vcc chip without LDO

Discussion in 'The Projects Forum' started by rakeshm55, Aug 16, 2011.

  1. rakeshm55

    Thread Starter Member

    Oct 19, 2010
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  2. praondevou

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    Jul 9, 2011
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    I don't see a need to limit the voltage at the REGIN input to 5V. I do see a need to limit the voltage at BAT voltage sense pin, it's maximum rating is 6V.

    Since it obviously is used to measure the battery voltage you could just use a resistive voltage divider. I didn't read through the whole datasheet, so I don't know how it's going to behave.

    This IC is actually made for a single cell.
     
  3. rakeshm55

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    Oct 19, 2010
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    Actually in the first prototype I applied 11.1V directly to REGIN pin. The LDO voltage o/p (Vcc,V2_5) for this level of input voltage was 3V. I am facing the same problem with three of my boards. One board i have preserved to which I applied only recommended condition. It is working fine. i am out of samples so as yet i have not concluded that applying 11.1V would harm the Device .

    Please refer to the attached application note . Ti suggests to use an LDO for multicell application (Page 2).

    Now how to surmount the usage of LDO. It naturally draws contiuous quiescent current . and when it is from 11.1V the harm is multiplied three times. I was just trying a way out.
     
  4. SgtWookie

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    Jul 17, 2007
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    Why don't you power it from only the cell that is the most negative of the three? You will have about 3.7v available there without using a regulator.

    No, it won't. The supply to the IC will not be regulated.
     
    Last edited: Aug 16, 2011
  5. rakeshm55

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    Oct 19, 2010
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    I am using a battery pack. So no No access to individual cell voltages.


    Even if the voltage is not regulated this unregulated i/p will go to an Integrated voltage regulator.So in this case wont the scheme work.

    I am applying a stable voltage to the BAT pin anyways. SO there is no variation at the battery sense pin.

    The peak current drawn by the system is not more than 1mA (dynamic). Average maximum current is 0.1mA. Drop across 33Ohms will be at max 30mV. Internal LDO inherently will have some PSSR to suppress this though i couldnt decipher from the datasheet.
     
  6. praondevou

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    Ok . I thought this pin was for measuring purposes. Forget also my first post, as I've seen now the BAT-pin needs to be connected to a low impedance source to measure correctly, not a resistive voltage divider...

    1.Where should be the problem in applying a voltage that is HALF of the maximum rating?

    2. what's the 33R resistor for? If you just put the zener in series you drop the voltage to what you want. BUT, zener diodes have a minimum current for which the stabilization is effective. According to your datasheet the current can go down to 4uA in hibernation and 18uA in sleep mode. I don't know if a zener works still at this low current.

    3. let's say you drop 6.8V inside the LDO at maximum 1mA, gives you 8.6mW peak. Is that too much?

    4. If you want to make sure you have the voltage you want at the LDO input use a voltage regulator or a classical resistor/zener combination (which would be lossy)
     
  7. rakeshm55

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    Oct 19, 2010
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    Got your point my scheme wont work at all :(.

    Wat else I can do other that an LDO?? Will there be a low queiscent current LDOs at 11.1V i/p voltage levels. The issue with me was I will be continuously draining my battery by connecting the LDO that can never be turned off. Even If my system detects battery drain.
    I was of the notion that my "flawed scheme" will not drain more current

    Any ways thanks for spoting my mistake
     
  8. praondevou

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    Jul 9, 2011
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