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From: Tilmann Reh
Subject: Re: What is LDO ( regulator ) ?
Date: Sat, 11 Jan 2003 11:14:41 +0100
Organization: Autometer GmbH
NNTP-Posting-Date: 11 Jan 2003 10:13:56 GMT
X-Mailer: Mozilla 4.78 [de] (Windows NT 5.0; U)
> Sorry for a basic question.
> I guess it means = minimum voltage different between
> input and output at full load ?
> At high load current these IC require only low head room voltage, right ?
> 7809 require head room 2V
> LM317/350 require head room 5V
> LM1084 require head room 1.2V
> Am I correct ?
Basically yes. LDO means "Low Drop Out", with drop-out being
the minimum voltage between input and output for that regulation
can be maintained (regardless of the load).
Minimum drop out for the LM317 is significantly lower than 5V, BTW.
78XX needs about 2V, modern LDOs about 0.2V.
There is one important difference between "standard" regulators
and LDOs. Standard regulators have NPN output transistors, so the
base current of the power stage also flows into the load. This
leads to a rather constant supply current of the regulator itself.
LDOs normally have PNP output stages to achieve the low drop-out.
So the base current of the output stage flows into groud, which
leads to significantly increasing quescient current for higher
loads. There are only a few LDOs with P-MOS output stages where
supply current stays small even at high loads.
However, it all depends on the application. If you have a low-power
circuit running at 5V (say, 3 mA supply current) and look for a
good regulator, you'll find that the 7805 tykes about 4-5 mA for
its own supply, more than doubling total power consumption. Using
an LP2950 instead (which is an LDO with PNP output) reduces regulator
current to about 200 uA (since even the LDO needs very littly current
at this small load). For load currents of 100 mA, things however look
Dipl.-Ing. Tilmann Reh
Autometer GmbH Siegen - Elektronik nach Maß.
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