# Low drop out voltage regulator.

Discussion in 'General Electronics Discussion' started by beanman, Aug 17, 2012.

1. ### beanman

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0
May 23, 2012
I am currently in the R&D stages of a Altoids tin iPhone charger but the voltage regulator I am using is much to inefficient. If anyone could tip me off on a voltage regulator that takes 9v down to 5v and has a drop out voltage of .7v (but a lower dropout voltage would be better) it would be very much apricated.

-Ben

2. ### davennModerator

13,721
1,913
Sep 5, 2009
Hi Ben,

Have you googled low drop out regulator ?

there are many to choose from, then look through the data sheets and find one with the lowest dropout voltage

cheers
Dave

3. ### beanman

7
0
May 23, 2012
I have but I have been haveing troubble finding one. Maybie I was being to specific in my searches or something.

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,448
2,809
Jan 21, 2010
Try using digikey. Start here. That's 34,000 low drop out regulators to choose from.

5. ### Harald KappModeratorModerator

10,808
2,438
Nov 17, 2011
Where's the catch?
A low drop regulator is one that can operate at a very low drop out voltage, but the voltage that really drops is the difference between input voltage and output voltage.So in order to regulate 5 V from a 9 V battery the regulator needs to drop 4 V.
Why the need for 0.7V drop out voltage? I guess you want to quench the last drop of juice out of the battery. But a linear regulator (even low drop) is the wrong approach. Depending on the current drawn it will very effectively convert the excess voltage from the battery to heat. In that case a switch mode regulator would be much more efficient.

Last edited: Aug 17, 2012
6. ### zalmonox

2
0
Aug 19, 2012
Indeed tehre is something about the undestanding of the problem. A low drop-out regulator is a regulator that can potentially use an input voltage very close to the desired output voltage (so close as the low drop-out value).
Ex. if you want 5V on output and your DC-DC has a drop-out of 0.5 it basically means you SHALL power it with at least 5.5V to work.
As much for the model (to get 5 from 9) this is another story. Pay attention with the LDOs ... they will dissipate a lot.
In the case above and for a current of 100mA (not much!) they will dissipate:
(9-5)*1/10 = 0.4 W which is already a lot.
You might want to use a switching regulator in this case.
Look at this on www.Digikey.com
296-12290-2-ND
296-18037-2-ND

Good luck

Last edited by a moderator: Aug 19, 2012
7. ### gorgon

603
24
Jun 6, 2011
You don't say anything about your charging current or the max voltage input range.
The efficiency you have is also not listed.

I stipulated the following:
Vinmax = 12V
Vinmin = 3V
Vout =5V
Ioutmax = 1A

Just for fun I put this into webench designer from TI/NS and came out with the following SEPIC design:
The efficiency of the converter is 80-84%

TOK

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