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I am trying to study the significance of resister divider feedback circuit in buck regulator.

This is the TI document I am referring to.

As per TI doc, voltage at the feedback pin is as follows,

enter image description here

When I try to estimate the same feedback voltage, I feel the drop at the R1 resistor (value in the red box) should be omitted from the equation. What is the information I am missing here.

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The Thevenin equivalent resistance at the feedback node is Req = R1||R2 = R1*R2/(R1+R2), so the current into the FB node has an influence Ifb*Req.

So the equation is correct. For low values of R1 & R2 the effect of Ifb will be small, but it will be there.

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The situation looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

And to find \$V_{FB}\$ voltage we can use a superposition theorem.

First, we remove the current source (\$I_{FB}\$) thus we have:

schematic

simulate this circuit

$$V_1 = I*R_2$$ $$I = \frac{V_{DD}}{R_1 + R_2}$$

Therefore

$$V_1 =I*R_2 = \frac{V_{DD}}{R_1 + R_2} * R_2 = V_{DD}\frac{R_2}{R_1 + R_2}$$

Next, we turn off the voltage source and the circuit becomes:

schematic

simulate this circuit

And the voltage at \$V_{FB}\$ is now equal to:

$$V_2 = -R_1||R_2 \times I_{FB} = - \frac{R_1 \times R_2}{R_1 + R_2} I_{FB}$$

So the \$V_{FB}\$ is equal to \$V_{FB} = V_1 + V_2\$

$$V_{FB} = \frac{ V_{DD} R_2}{R_1 + R_2} - \frac{R_1 R_2 I_{FB}}{R_1 + R_2} = \frac{R_2(V_{DD} - I_{FB}R_1)}{R_1 + R_2}= R_2 \times \frac{V_{DD} - I_{FB}R_1}{R_1 + R_2}$$

So the equation gives by Ti is correct.

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