Question on LDO basic concept

Status
Not open for further replies.
M

mehboobali

Junior Member level 1
Joined
Aug 27, 2013
Messages
17
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Location
Bangalore
Visit site
Activity points
90
LDOs also find use in applications where the regulator must maintain regulation with small differences between the input supply voltage and output load voltage, such as battery-powered systems.
What is meant by "small differences between input supply voltage and output load voltage"..?? Why it is so..??
 

LDO (Low-Dropout Regulator) is one of the Linear regulator.
In any regulator Efficiency is the most important criterion. Higher the Efficiency better the regulator is.
if
Vin= input supply (VDDA), Iin= input Current = IQ + Load Current
Vo= output, Io= Load Current
then
Efficiency = (Vo x Io) /(Vin x Iin) = Vo/Vin (Considering the In=IQ + Io is close to Io as Io is in mA range while Iq is in uA Range)

So we can see that : if Vin-Vo = dropout, if its LESS or say "Low" (Low Dropout) the efficiency will be good. Higher dropout will decrease efficiency.
 

Additional to voltage difference, LDO regulators can have small quiescent current. Also LDOs are linear regulators, that means poor efficiency compared with switcher regulators.


Best regards,
Peter
 

LDO is meant by LOW DROP Regulators.
Regulators gives fixed output voltage when the input voltage is greater than the output voltage.

For example, lets take 78L05 Regulator which is quite popular. This regulator will give fixed output voltage with minimum input of 7V. If you want fixed +5V output voltage, your input voltage should be greater than +7V. In this case, Drop across regulator is +2V, which is quite high in some applications and this drop causes the regulator to heat.

In the case of LDO, take for example, LP2950 which is a LOW drop regulator. Drop across LP2950 is about +380mV. which means that the output voltage will be fixed +5V, when the input voltage is reduced upto 5.380V.

Is this clear to you?

Best wishes
 
Reactions: tpetar

    tpetar

    Points: 2
    Helpful Answer Positive Rating


Little correction: 78L05 7805 minimal voltage difference si 2,5V. If you want 5V on output you need at least 7,5V not 7V on input.


Best regards,
Peter
 

LDO--> Low Drop Out Regulators is comes under the linear regulators. Its have pass transistor elements, this should work in saturation region, then only it regulates. pass transistor enter from the linear region to saturation region, drop out voltage is utilized

My understanding is, To reduce the Power dissipation, now a days, drop out voltages are becoming lower ranges.

Regards,
Prasad
 



Todays to reduce power dissipation you should use switcher regulators not linear. But for sure linear regulator can do the job.

If you have higher input voltage then output you will have huge waste of power even with LDO, but switcher can give good solution for that situation.



Best regards,
Peter
 

The "LDO" nomenclature is all relative to the older,
Darlington NPN pass-device regulators. These took up
to 2V of headroom (@ low temp). Today you see more
MOSFET pass elements but still some folks calling
common-emitter bipolar output stages "low dropout".

Now if you don't -need- the output to be within a
couple hundred mV of the input rail minimum, at full
load, then you don't need a LDO necessarily.

LDOs can also be useful for "cleaning" switcher
outputs if they have a decent HF PSRR, and the
composite of switcher and LDO can be better on
efficiency terms than a LDO alone, and quieter
than a switcher, though more expensive (duh)
than either in terms of BOM cost.
 
Reactions: tpetar

    tpetar

    Points: 2
    Helpful Answer Positive Rating
Hi All,

Thank you for replying.

I have one more question regarding this:
I got the concept of the efficiency...higher dropout decreases the efficiency, dropout should be less. If there is higher dropout (voltage) then where it is utilized and get effected and make the less efficient.??
For e.g. 7805 regulator, we need 7.5V input to give regulated output of 5V. There is 2.5V dropout, where this 2.5V is utilized...??


Thanks,
Mehboob Ali
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…