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if you have a 200mA LDO all that means is that this is what it is capable of outputing. So if your current demand is only 150mA, the LDO will internally adjust to deliver that. In reality there will be some additional current that is used to power up the LDO itself but this will just be a few mA.
Yes, no remainder current. LDO works like adjustable resistance.
Imagine 5v source and two resistors connected in series
from 5V to ground. Medium point is LDO output. First resistor
from 5V is LDO internal adjustable resistance, second resistor
is your load resistance.
For example input 5V, output 3.3V and your load current is 0.15A.
On first resistance voltage drop is 1.7V, on second(it is your load)
3.3V. Load resistance is 3.3/0.15 = 22ohm, well LDO adjusts
internal resistance to value 1.7V/0.15 = 11.3ohm
Now you have 3.3V and 0.15A load current and no additional currents.
Power dissipated LDO is 1.7V*0.15 = 0.255W on internal
LDO resistance. There is no reminder current with
LDO, because any additional current means
more power drop on LDO. LDO not likes dissipation
of power on it.
LDO must provide 3.3V to you, then 1.7V ALWAYS
will drop on LDO independent from load current.
When load is 10ma LDO losses will 0.01*1.7 = 0.017
when load is 100ma LDO losses will 0.1*1.7 = 0.17
LDO do not want dissipate any additional power and
make an effort to decrease this power to minimum
without external help. Be sure,there is now
useless currents when LDO works.