Answer :
The DC transformer model of the buck-boost converter shows that the output voltage is related to the input voltage and duty cycle. The converter allows both voltage step-up and step-down functionality and is essential in various applications where the output voltage needs to be adjusted with high efficiency.
To derive the DC transformer model of the buck-boost converter, we'll consider the following assumptions:
1. Ideal switching elements (MOSFETs or diodes) with no losses.
2. Continuous conduction mode (CCM) operation.
The buck-boost converter consists of an inductor, a capacitor, and two switches (MOSFETs). During the ON state, the inductor stores energy, and during the OFF state, it releases energy to the load.
1. ON state (S1 closed, S2 open):
The input voltage (Vin) charges the inductor (L) and the output capacitor (C). The inductor current (IL) increases linearly with a slope of Vin/L.
2. OFF state (S1 open, S2 closed):
The inductor current (IL) discharges to the output capacitor (C) and the load. The inductor current decreases linearly with a slope of -Vout/L.
By applying the Kirchhoff's voltage law (KVL) for the inductor and capacitor, we can write the following equations:
For ON state:
Vin - Vout - L di/dt = 0
For OFF state:
-Vout - L di/dt = 0
Combining the equations and solving for the inductor current (IL) during one switching cycle, we get:
IL(t) = (Vin/L) * t, 0 ≤ t ≤ DT (ON state)
IL(t) = (Vout - Vin)/L * t + (Vin/L) * D*T, DT ≤ t ≤ T (OFF state)
Where D is the duty cycle (Ton / T), T is the switching period, and DT is the ON time.
Using the average inductor current (Iavg) over one switching cycle:
Iavg = (1/T) * ∫(IL(t) dt) = (Vin * D) / L
Finally, using the dc transformer model, we relate the input and output voltages:
Vout = Vin * (1 - D) / D
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