# CC/CV Buck controller 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

• No voltage divider. The Zener diode and RFB resistor form a CV mode feedback. – G36 Sep 13 '19 at 13:39
• could you please explain me in detail about the cc and cv operation – Nitish Patel Sep 13 '19 at 15:08
• 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. – Verbal Kint Sep 14 '19 at 7:28

Well in VC mode we have this situation: 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: 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

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