CC/CV Buck controller
Well in VC mode we have this situation:







simulate this circuit &ndash; Schematic created using CircuitLab

In constant voltage mode, the negative feedback loop provides a constant output voltage within a certain range of load currents.   

Due to negative feedback action, the voltage across \$R_{FB}\$ resistor is equal to \$V_{REF} = 2.5V\$ hence the output voltage is equal to: 

\$V_O = V_F + V_Z = 2.5V + 4.7V = 7.2V\$

If for example, the output voltage drops the \$V_F\$ voltage will also drop. 
For example, if the output voltage drops to \$7V\$ the \$V_F\$ will drop to \$2.3V\$.
And the voltage at the error amplifier output increase and the duty-cycle will also increase to bring back the output voltage to the \$7.2V\$ (\$V_F = V_{REF}\$).   

The CC mode takes control over the feedback loop when \$I_{LOAD} &gt; I_{SET}\$. 

This part of a circuit can be viewed as a current to voltage converter. 
When \$R_{sens}\$ is an output current to the voltage converter (Ohms law). 

See the example: 



Opamp wont to keep non-inverting input voltages equal to inverting input voltages.

Hence the voltage across \$R_A\$ resistor is equal to the voltage drop across the sens current resistor. 

And the voltage across \$R_B\$ is equal to:  

\$V_{R_B} = \frac{R_B}{R_A} \cdot I_L R_S\$ 

Because in your circuit due to overall feedback loop the \$V_{R_B}\$ voltage will be equal to \$V_{REF}\$ voltage. 

So the CC mode set current will be equal to: 

\$I_{SET} = \frac{V_{REF}R_{IN}}{R_{FB} R_S} \$   

But your CC circuit has a big flaw in this application. 
The Zener diode can also work as an ordinary diode. Therefore if we short the output to ground the diode will start to conduct a current and the load current will be larger than \$I_{SET}\$.  

Hence, you should look for a different design. 

Also if you are beginner try to read this 
Some questions about a series pass transistor &amp; op amp voltage regulator