How to wire 4-terminal LDO? (VDis)

Discussion in 'General Electronics Chat' started by TheLaw, Jul 14, 2011.

  1. TheLaw

    Thread Starter Member

    Sep 2, 2010
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    Hi,

    Finally got some of my biggest projects done, and I'm moving onto something new.

    I never took any EE classes, so this might be easy...but not for me.

    I would like to use this low drop out regulator: http://www.fairchildsemi.com/ds/KA/KA78R05C.pdf

    You will notice it is a 4 terminal device. I'm confused as to what exactly VDis is and what it does...

    Where does it get attached to in the power supply circuit?

    Thanks.
     
  2. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    It's to disable/ enable the output. Looks like if the voltage on this pin is higher than 2 V then output is active (datasheet under electrical characteristics)
     
  3. strantor

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    Oct 3, 2010
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    pin 4 is disable pin. you need 2-35V on pin 4 to get voltage out. if you leave it disconnected, you will never get anything out. If you don't need the disable function, I think you could just tie it to pin1. see figure 3 Vo Vs. Vdis
     
  4. strantor

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    stop doing that to me! lol :)
     
  5. Pencil

    Active Member

    Dec 8, 2009
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    Look at page 2 and 3 of datasheet you posted.

    Vdis absolute maximum input: 35V

    Disable voltage voltage high: minumum 2V=output active

    Disable voltage voltage low: maximum .8V=output disabled

    Summation: Disable pin fed between 2v-35v output active.
    Disable pin taken below .8v output disabled.

    Edit: Holy cow it's an answer storm. I'm too slow.
     
    Last edited: Jul 14, 2011
  6. TheLaw

    Thread Starter Member

    Sep 2, 2010
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    Okay thanks....but where exactly is it "measuring" the voltage?
     
  7. strantor

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    Oct 3, 2010
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    for pin 4? between pin 4 and GND (pin3)

    EDIT: don't be confused by the diagram where it shows a little battery @1.4V; that's just representative, not really inside the IC. It's just there to represent that you need >1.4V (specs say 2V) to switch the output on.

    :) had post the quick & dirty answer, then come back and expound with the EDIT so I wouldn't get beaten to the punch again.
     
  8. praondevou

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    ?? did I write something funny :confused:
     
  9. strantor

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    no, you remember back on this post where we were both typing the same answer at the same time and you beat me to the punch? I was just joking with you now about it happening again.
     
  10. praondevou

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    Jul 9, 2011
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    Actually, I look if you're online and only then I write something... :D Just kidding...
     
    strantor likes this.
  11. TheLaw

    Thread Starter Member

    Sep 2, 2010
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    Gahhh. I'm an idiot. :confused: Speaking of quick and dirty....exactly where does pin 4 get connected to?

    Thanks. Sorry if I seem ignorant or something.
     
  12. strantor

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    connected to pin 1
     
  13. TheLaw

    Thread Starter Member

    Sep 2, 2010
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    Right. Is there a reason why they aren't connected internally?
     
  14. praondevou

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    Jul 9, 2011
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    It's just an additional functionality of this particular IC. So you can disable the output. Could be quite useful I guess.
     
  15. strantor

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    pin 4 is what you would use to shut off the output from somewhere else in the circuit with a low voltage (or high voltage, up to 35V) signal. You would remove said signal to stop the output. I assume you don't need this feature, which is why I said connect it to pin1. Using this feature could save you a component by means of not having to have a component (transistor, relay, other) to disconnect your incoming voltage (pin1).

    for instance, if you used this with a microcontroller, the microcontroller could shut off power by dropping the signal on pin4 (draws 20μA). Otherwise you would need a seperate transistor.
     
  16. TheLaw

    Thread Starter Member

    Sep 2, 2010
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    Thanks. I suspected you would need a mcu that could use a serial signal or something.

    This regulator sort of reminds me of a transistor, imagine pin4 is the base.
     
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