Answer :
To obtain the closed-loop transfer function for each converter individually, we use the small-signal model with a voltage-controlled feedback loop.
The buck converters used in this instance are commonly stable in normal conditions. However, as shown in the Mini-project, constant power loads may destabilize the system. Even if the individual buck converters are stable, the resulting system can become unstable when not correctly regulated when two converters are cascaded.
Given the parameter values provided in Table 1, two cascaded buck converters are used in the following tasks: Vin = 48 V, Vout1 = 12 V, Vout2 = 5 V, L1 = 293 μH, L2 = 184 μH, and C = 47 µF.
Since the buck converters are essentially DC-DC converters, they are controlled by Pulse-Width Modulation (PWM). The PWM controller's duty cycle will change, resulting in the output voltage of the converter changing, depending on the input voltage and load characteristics. When calculating the transfer function, the small-signal model can be used, in which the system's nonlinear behavior is ignored and only its linear properties are taken into account. When calculating the closed-loop transfer function, the output voltage, Vout, is the feedback voltage (Vf).
The transfer function of the buck converter is given by the following expression: [tex]$$V_{out} =\frac{D}{1-D}\cdot V_{in}$$[/tex] where D is the duty cycle and it is given as: [tex]D = 1- Vout/Vin[/tex]
To derive the small-signal model of the Buck converter, the two-port network model is employed: [tex]$$\frac{V_o}{V_s} =\frac{-D}{1-D} \cdot \frac{1}{1+sL/R}$$[/tex]
This equation is obtained by substituting Vout= Vf and Vout is the output voltage of the buck converter and Vs is the input voltage, which is equal to Vin. L is the inductance of the buck converter and R is the equivalent resistance of the switch and inductor. In this instance, the switch is an ideal switch with zero resistance. Therefore, R can be represented by the on-state resistance of the power MOSFET, which is negligible compared to the inductor's resistance.
Since the buck converter's transfer function is a ratio of two polynomials, the closed-loop transfer function of the buck converter can be derived using the following equation:[tex]$$\frac{V_o}{V_s} = \frac{-D}{1-D}\cdot \frac{1}{1+sL/R}$$[/tex] where the transfer function can be expressed as:[tex]$$\frac{V_o}{V_s}=\frac{-D}{1-D}\cdot\frac{1}{1+sL/R}=\frac{-D}{1-D+\frac{sL}{R}(1-D)}$$[/tex]
Thus, the transfer function of the Buck converter can be expressed as: [tex]$$\frac{V_o}{V_s}=\frac{-D}{1-D+\frac{sL}{R}(1-D)}$$[/tex]
The transfer function of the second buck converter is represented by the following equation: [tex]$$\frac{V_{o2}}{V_{s2}}=\frac{-D_2}{1-D_2+\frac{sL_2}{R_2}(1-D_2)}$$[/tex] where [tex]$D_2 = 1 - V_{o1}/V_{in}$[/tex] is the duty cycle of the second buck converter.
The transfer function of the cascaded system of buck converters is given by: [tex]$$\frac{V_{o2}}{V_{s2}}=\frac{-D_2}{1-D_2+\frac{sL_2}{R_2}(1-D_2)}=\frac{-D_2}{1-D_2+\frac{sL_2}{R_2}(1-D_2)}\cdot\frac{V_{o1}}{V_{s1}}$$[/tex]
Substituting [tex]$D_2 = 1 - V_{o1}/V_{in}$[/tex] we get:[tex]$$\frac{V_{o2}}{V_{s2}}=\frac{-D_2}{1-D_2+\frac{sL_2}{R_2}(1-D_2)}\cdot\frac{V_{o1}}{V_{s1}}=\frac{V_{in}-V_{o1}}{V_{in}}\cdot\frac{-D_2}{1-D_2+\frac{sL_2}{R_2}(1-D_2)}$$[/tex]
Thus, the closed-loop transfer function of the cascaded system of Buck converters is given by:[tex]$$\frac{V_{o2}}{V_{s2}}=\frac{-D_2}{1-D_2+\frac{sL_2}{R_2}(1-D_2)}\cdot\frac{V_{o1}}{V_{s1}}=\frac{V_{in}-V_{o1}}{V_{in}}\cdot\frac{-D_2}{1-D_2+\frac{sL_2}{R_2}(1-D_2)}$$.[/tex]
This is the final result of the closed-loop transfer function for each converter individually, using the small-signal model with voltage controlled feedback loop.
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