user14983: I want to build a simple voltage regulator  3.8-4.2V with atleast 2A Peak Current Capability for my GSM Shield (It doesn't have regulated power supply), I don't want variation or it may burn the board.
Using voltage divider won't help here as there can be variations.
Somewhere i read LDO are for for this task but i am able to get only 3.3v regulator with 800mA supply.
Can i use 3.3v to power the same ?

user3917: Consider using a buck converter for this.  Search ebay for "buck converter" and you will see items like "Buck Converter Step-Down Adjustable Converter Power Module Regulator LM2596.  That title indicates several things:  the converter is adjustable, it steps voltage down, it's a regulator, and it's based on TI's LM2596 chip.  It's rated at 2 A continuous, 3 A peak.
A buck converter has some advantages over a linear regulator:

Lower dropout (in general), ie, the input voltage has to be at least as much as the output voltage, possibly plus a small margin for one FET loss, but doesn't have to have the 1.1 to 1.3 volt margin typical of an LDO regulator.
Higher efficiency (in general).  For example, 95% is not uncommon.  (The ebay item mentioned above shows 92%.)   A linear regulator, whether LDO or not, turns the input-to-output voltage difference into heat.  A buck converter, on the other hand, uses an inductor and a capacitor to store energy; it refills the capacitor at a controlled rate via the inductor as often as necessary to deliver the required load.  The inductor's DC resistance wastes energy, but that resistance can be kept small.
Smaller parts due to high frequency switching operation.  As operating frequency is increased the inductor can be made smaller.  On the other hand, higher frequencies lead to higher switching losses, so there's a tradeoff.  Note, some SOT3-sized (~1.4 mm x 2.9 mm) transistors eg NTR4503 are rated over 2 A at 25°C and might be usable as an output FET for a buck converter chip, and inductors and capacitors also are available in sizes probably as small as you are looking for, but buying them and building your own converter is unlikely to be cost effective.


user612: 
I planned to use 3.7v LIPO battery

That's what your SIM800 module is intended to be directly powered by, without any intervening boost or buck converter.
Behind all the misdirection of a title that doesn't describe what you are actually trying to do, your question is an exercise in going around in a circle right back to where you started.  Don't try to convert energy from the lithium cell to some other voltage and then from there to the allowed input range of the SIM800 (which is very intentionally that of a single lithium cell), rather use the parts as they were intended to be used.

i guess it has current 800mA..

No, it has a capacity of 800 milliamp hours, not 800 milliamps.
A phone or RC model type lithium cell you would find today would have a "C" rating of 10 or better, but even if we assumed one of only 5, that would predict that your cell should be expected to be able to support pulse loads of 5 times the one-hour draw rate, which would be .8 * 5 = 4 amps.

14983: <p>I want to build a simple voltage regulator  3.8-4.2V with atleast 2A Peak Current Capability for my GSM Shield (It doesn't have regulated power supply), I don't want variation or it may burn the board.</p>

<p>Using voltage divider won't help here as there can be variations.
Somewhere i read LDO are for for this task but i am able to get only 3.3v regulator with 800mA supply.</p>

<p>Can i use 3.3v to power the same ?</p>
 3917: <p>Consider using a <a href="https://en.wikipedia.org/wiki/Buck_converter" rel="nofollow">buck converter</a> for this.  Search ebay for "buck converter" and you will see items like "<a href="http://www.ebay.com/itm/261956897110" rel="nofollow">Buck Converter Step-Down Adjustable Converter Power Module Regulator LM2596</a>.  That title indicates several things:  the converter is adjustable, it steps voltage down, it's a regulator, and it's based on TI's <a href="http://www.ti.com/product/lm2596" rel="nofollow">LM2596</a> chip.  It's rated at 2 A continuous, 3 A peak.</p>

<p>A buck converter has some advantages over a linear regulator:</p>

<ol>
<li><p>Lower dropout (in general), ie, the input voltage has to be at least as much as the output voltage, possibly plus a small margin for one FET loss, but doesn't have to have the 1.1 to 1.3 volt margin typical of an LDO regulator.</p></li>
<li><p>Higher efficiency (in general).  For example, 95% is not uncommon.  (The ebay item mentioned above shows 92%.)   A linear regulator, whether LDO or not, turns the input-to-output voltage difference into heat.  A buck converter, on the other hand, uses an inductor and a capacitor to store energy; it refills the capacitor at a controlled rate via the inductor as often as necessary to deliver the required load.  The inductor's DC resistance wastes energy, but that resistance can be kept small.</p></li>
<li><p>Smaller parts due to high frequency switching operation.  As operating frequency is increased the inductor can be made smaller.  On the other hand, higher frequencies lead to higher switching losses, so there's a tradeoff.  Note, some SOT3-sized (~1.4 mm x 2.9 mm) transistors eg NTR4503 are rated over 2 A at 25°C and might be usable as an output FET for a buck converter chip, and inductors and capacitors also are available in sizes probably as small as you are looking for, but buying them and building your own converter is unlikely to be cost effective.</p></li>
</ol>
 612: <blockquote>
  <p>I planned to use 3.7v LIPO battery</p>
</blockquote>

<p>That's what your SIM800 module is <em>intended</em> to be <em>directly</em> powered by, <em>without</em> any intervening boost or buck converter.</p>

<p>Behind all the misdirection of a title that doesn't describe what you are actually trying to do, your question is an exercise in going around in a circle right back to where you started.  Don't try to convert energy from the lithium cell to some other voltage and then from there to the allowed input range of the SIM800 (which is very intentionally that of a single lithium cell), rather use the parts as they were intended to be used.</p>

<blockquote>
  <p>i guess it has current 800mA..</p>
</blockquote>

<p>No, it has a capacity of 800 milliamp hours, not 800 milliamps.</p>

<p>A phone or RC model type lithium cell you would find today would have a "C" rating of 10 or better, but even if we assumed one of only 5, that would predict that your cell should be expected to be able to support pulse loads of 5 times the one-hour draw rate, which would be .8 * 5 = 4 amps.</p>
