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    Why when the load current of LDO regulator goes up the output voltage goes down ?

    When the load current of an LDO regulator goes up the iutput voltage goes down in the transient response...It is not very clear to me...It is ok that LDO is like a voltage source but I would like some better answer..Please help me
    thank you

    •   Alt3rd June 2008, 10:47

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    LDO behavior

    The transient response of a voltage regulator is slow because the speed of rthe error amplifier is slow. if its speed if faster then it would oscillate.

    When the load current suddenly increases then the output voltage drops for a moment until the error amplifier catches up.

    The opposite occurs when the load current is suddenly reduced. The output voltage rises for a moment until the error amplifier catches up.

    Datasheets for voltage regulators show graphs of the output transient and how an output capacitor to ground improves the transient response by smoothing the output voltage spikes.



    •   Alt4th June 2008, 02:15

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    Re: LDO behavior

    you must check from requird specs that the max spark amplitude is in your specs



    •   Alt4th June 2008, 08:16

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    Re: LDO behavior

    Yes but Why does the voltage drop?



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    Re: LDO behavior

    The voltage at the output of a voltage regulator drops permanently when the load current increases because the series pass transistor has a certain amount of resistance. The resistance causes the voltage at its output to drop when the current through it increases.
    When the error amplifier catches up and detects the voltage drop at the output then it turns on the series pass transistor harder to increase the output voltage to where it is supposed to be.

    The error amplifier is not perfect, it does not have infinite gain so the voltage regulator has a spec of a certain amount of voltage drop. A uA7805 voltage regulator has a voltage drop of typically 0.015V when its load current is increased from 5mA to 1.5A.



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    Re: LDO behavior

    *For transient : You can imagine the pass-element for very fast changes seen as a current source [Ref Dr Millikan paper for cap free ldo](as before EA catch-up the EA output voltage is constant ) & so varying the load can be seen by two points of views :
    1) Varied resistance :for ex. load resistance varied from high resistance to low resistance,& so the current source sees a low resistance while passing the same current & so the voltage drops down.And vice versa.
    2)Varied Current source:For ex. the driven current varied from light load(low current) to heavy load(high current) ,in this case you can get calculate the output voltage as the current passes through the voltage divider times its total series resistance so here as the pass-element current is constant the increase in load current accompanied by decrease in volt. divider current which means output voltage drop.And vice versa.
    [Out cap value smoothing this process by delivering some instantaneous charges that overcome fast voltage changes.]


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    •   Alt4th June 2008, 18:56

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    Re: LDO behavior

    Well to make it even simpler...
    You can think of an ldo as a resistive voltage divider... so what ever be the input voltage the output is the divided value of it.
    vin=I*(R1+R2) and vout= (R2*I)/(R1+R2). So every time you change 'I' the output goes up or down.

    Now to keep the voltage constant the only way is to vary the R1... such that the drop across it is same what ever be the value of 'I' .... and this task is done by your error amplifier...

    Note: for error amp to work first some error isto be generated...and that error is the droop or overshoot inthe output voltage... so u will always get a change in output voltage whenthe load current is changed ...and this change will be corrected by ur feedbak loop via error amplifier.

    hope I answered , the way u wanted...!!



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