Since R1 = 20 kΩ and R2 = 10 kΩ, Equation 2 can be simplified to:
VOut – Vr = 3(Vfb – Vr)(3)
or:
VOut = 3 Vfb – 2 Vr(4)
R3 and R4 have the same value, 10 kΩ, so amplifier U2’s output voltage is:
Vr = 2 VRef – VC(5)
where VRef is the reference voltage generated by U3 after resistor divider R7/R8.
Combining Equation 4 and Equation 5:
VOut = 3 Vfb – 4 VRef + 2 VC(6)
To simplify Equation 6, choose components that make:
3 Vfb = 4 VRef(7)
Then Equation 6 becomes:
VOut = 2 VC (8)
The internal voltage reference of U1 is 0.8 V. By choosing R7 = 10 kΩ and R8 = 3.16 kΩ, VRef = 0.6 V, satisfying Equation 7. Finally, C1 lowers U2’s output impedance at high frequencies, maintaining the stability of U1’s feedback loop.
The added circuitry allows users to control the buck converter’s output voltage, VOut, in the range of 0 to 5 V with a control voltage, VC, in the range of 0 to 2.5 V. Similar circuitry can be designed for use with a boost converter, or any other dc-dc converter, as long as its feedback voltage pin is accessible.