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# How to design a LDO regulator with specific requirements ??

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#### kvinod423

##### Newbie level 3
Design a LDO regulator with the following parameters
Input voltage 8-10 V, Output voltage 4.5 V, Output current 300 mA

There are literally hundreds of LDOs available by at least a dozen IC manufacturers.

so before we can provide suggestions, we require at least a pair of additional pieces of information:

What IC vendors are available in your geographic location? Many times I've suggested a device, and the poster comes back with something like: "Fairchild is not availble where I live".
Can you solder SMT components? Again, same scenario. No use suggesting a device that you can't solder. By the way, MOST LDOs nowadays come in SMT packages.

With your minimum input of 8V and your output of 4.5V then you do not need a Low Dropout Regulator (LDO) since an ordinary regulator is fine with an in-out of 3.5V.
When the input is 10V and the output is 4.5V/300mA then the regulator heats with (10V - 4.5V) x 300mA= 1.65W. Then you need a regulator with a metal tab.

Why do you want to design a regulator? Schoolwork?

As per your specifications, you are going to waste so much of power as heat distribution. So dont use LDO for this purpose. Use adjustable voltage reagulator along with heat sink.

Designing an LDO assumes you need low dropout voltage not low power loss.

When the dropout is not low, design of thermal power loss is key.
Pin = 10V*300mA = 3W
Pout = 4.5* 300mA = 1.35W
Ploss in LDO = Pin-Pout= 1.75W
Efficiency= Pout/Pin=45%

No heatsink on a TO-220 is...
Rja = 60 deg C/W
Rjc = 5 deg C/W

With no heatsink Rja*W= 60*1.75 is 105'C above ambient (25C) is too hot for a TO-220 case

Rca( heatsink) + Rjc(5) = Rja since the resistance is in series from ambient to case to junction.

So a fairly large mass heatsink is needed even for a small power of 1.75W

Any regulator such as LM317 adjustable can do the job, but this is considered inefficient.

- - - Updated - - -

An efficient regulator using TI.com's design site would look like this.

BOM Cost \$1.04
94% efficient
86mm² area
416kHz 21% duty cycle

Output current - need to comprehend a minimum as well
as a maximum (min could be zero, but that makes the
stabilization much trickier the closed you get to a cut-off
pass device).

Load step response also is key to goal-setting and the
optimization of error amp * final stage compensation

Any concern with process compatibility / capability,
does this need to play with other functions or is is a
standalone part (or simply an exercise)?

As Dick indicated any regulator design ought to have a spec for Step Load response ( both dynamic and steady state) and No load stability as well as full load dissipation.

It should also include permitted ripple, thermal drift or wide temp range thermal error and initial tolerance. Common specs are 1% initial and 1% temp tolerance. LDO's have low ripple, SMPS have 1% ripple more or less.

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