Calculating the peak current of my boost converter
From Figure 4. Block Diagram in the datasheet, you can see pin 6 is compared with pin 7 of the IC.
Note that the + input of the comparator lies 200mV lower than Vcc.
When pin 7 becomes lower than (Vcc-200mV) the peak detection will be activated (and so, the Darlington transistor will be disabled).  

From your posted schematic, the (peak) current through the inductor causes a voltage drop across R1.
This voltage drop is divided by resistors R2 and R3.
So, the voltage on pin 7 will be:
\begin{equation}(V_{cc} - I * R_1 ) * R_3 / (R_2+R_3)\end{equation}

The peak detection will be activated when
\begin{equation}
V_{cc} * R_3/(R_2+R_3) - I * R_1 *R_3/(R_2+R_3) &lt; V_{cc}-200mV
\end{equation}
Note that this depends on the current through the inductor as well as the applied voltage Vcc.

R3 is unclear to me, but 0M18 = 180K makes sense, and pick Vcc = 15V
\begin{eqnarray*}
15V * 180/181 - I * 0.082E * 180/181  &amp;&lt;&amp; 14.8V\\
14.917V - I * 0.082E * 180/181  &amp;&lt;&amp; 14.8V\\
117mV &amp;&lt;&amp; I * 0.082E * 180/181\\
118mV &amp;&lt;&amp; I * 0.082E\\
118mV / 0.082E &amp;&lt;&amp; I\\
I &amp;&gt;&amp; 1.44A \\
\end{eqnarray*}
So, max peak cuurent through inductor is 1.44 A  

When e.g. Vcc = 10V, the same circuit limits the current to 1.78 A