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Can someone explain how they got equation 5.1. Why is the output voltage of their buck look like an equation for an non-inverting op-amp amplifying a Vref?


1 Answer 1


If you have a resistor potential divider with R1 and the top and R2 down to ground and, Vout was applied at the top, the divided voltage (call it Vref) can be found this way: -

$$V_{REF} = V_{OUT}\cdot\dfrac{R_2}{R_1 + R_2}$$

Then, if you rearranged it to find Vout you'd get this: -

$$V_{REF}\cdot\dfrac{R_1+R_2}{R_2} = V_{OUT}$$

And this clearly is: -

$$V_{REF}\cdot\left(1 +\dfrac{R_1}{R_2}\right)= V_{OUT}$$

In other words, it's the standard formula for pretty much any linear or switching regulator that can be adjusted via a resistor divider. And yes, it's the same basic equation as a non-inverting op-amp because the error amplifier inside the buck converter (or linear regulator) is an op-amp acting as a non-inverting gain amplifier with a bunch of other circuits on its output but, once the loop is closed, it behaves exactly like a non-inverting op-amp amplifier: -

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Highly modified version of image found here.


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