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in this model can anyone explain me how is it going to work in CV mode and where is the potential divider and all, like where is the Vfb

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  • \$\begingroup\$ No voltage divider. The Zener diode and RFB resistor form a CV mode feedback. \$\endgroup\$ – G36 Sep 13 at 13:39
  • \$\begingroup\$ could you please explain me in detail about the cc and cv operation \$\endgroup\$ – Nitish Patel Sep 13 at 15:08
  • \$\begingroup\$ It's a two-loop system: when one loop operates, the other one is silent. The voltage loop uses the Zener diode and drives the FB path when active. When \$V_{out}\$ increases, it biases the Zener and the voltage across the FB resistor increases until the loop stabilizes to \$V_Z+V_{FB}\$ (a simple proportional system). When the current increases, the drop across \$R_{sense}\$ goes up and the op-amp biases the PNP which now biases the FB resistor and takes the lead to regulate the current. As \$V_{out}\$ goes down, the Zener stops conducting. \$\endgroup\$ – Verbal Kint Sep 14 at 7:28
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Here you have a link to the original Texas Instrument document where this drawing is from. There is also explained the principle of operation of this system.

http://www.ti.com/lit/an/snva829/snva829.pdf

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Well in VC mode we have this situation:

schematic

simulate this circuit – 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} > 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:

enter image description here

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 & op amp voltage regulator

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