How to choose a inductor for a buck regulator circuit?
Here is a quick and somewhat dirty way to calculate an inductor value for buck regulators operating in constant conduction mode (CCM).  It will result in an inductance that will be close to what you would get with a more exact calculation, and will not get you into trouble.

What you need to know to calculate inductance:


Output Voltage, \$V_o\$  
Output Current, \$I_o\$  
Switching Frequency, \$F_{\text{sw}}\$  
L = \$\frac{\text{$\Delta $t}V_o}{\text{$\Delta $I}}\$  


Make a couple of assumptions:


\$\text{$\Delta $I} = \frac{I_o}{10}\$
\$\text{$\Delta $t} = \frac{1}{F_{\text{sw}}}\$


so

L = \$\frac{10 V_o}{I_o F_{\text{sw}}}\$ 

for \$I_o\$ = 1A and \$F_{\text{sw}}\$ = 2.2 MHz  

L = 22.7 \$\text{$\mu $H}\$

When choosing the inductor:


Find one that is rated for 1.4 to 2 times the output current.  In this case 1.4A to 2A.  Most standard inductors are specified for 40C heat rise with rated current, which is kind of hot.  Conductive losses scale by the square of the current.  Using a current rating of 1.4 \$I_o\$ will reduce that heat rise by half, and a current rating of 2 \$I_o\$ will reduce heat rise to 1/4.  
Make sure the series resonant frequency (SRF) is at least a decade higher than the switching frequency.