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Question about Iout max of LDO?

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letan

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How to calculate Iout,max of LDO?
And I have a question:
I have design a LDO with Vout=1.8v, Iomax=50mA. When I use this LDO with Rload=36ohm => Iload=50mA=Iomax, but if Rload < 36ohm => Iload > Iomax. Why? I'm wrong?
Can you please explain clearly for me?
Thanks.
 

u have Vout=1.8V and Iload=Vout/Rload=1.8/Rload so if Rload is less than 36 then Iload>50mA

Added after 1 minutes:

u have Vout=1.8V and Iload=Vout/Rload=1.8/Rload so if Rload is less than 36 then Iload>50mA
note here i assemed that Vout=1.8 which is fair assumption if the LDO loop is working (-ve feedback loop) which makes the range of Vout error =1/Loopgain which should be small in case of large loop gain
 

The summary is ... assuming that the loop is regulating ....(so taht output remains constant) current is inversly proportional to resistance!!( V=IxR)

now calculate...
 

yes it's right, necessarily if rload decrease then Iload increase.
you can calculate Iload from the expression I=V/R and if you use a transistor element to load your circuit then calculate Iload from the I transistor equation.
 

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