The voltage across the capacitor in this buck boost converter is V_c. The differential equation for this variable is: 

Explanation:

The voltage across the capacitor in a buck-boost converter, denoted as \( V_c \), is an important parameter in the analysis of the converter's behavior. Understanding the differential equation governing this voltage is crucial for the design and analysis of such converters.

Buck-Boost Converter Overview:

A buck-boost converter is a type of DC-DC converter that can step up (boost) or step down (buck) an input voltage. The basic components of a buck-boost converter include an inductor, a capacitor, a switch (typically a transistor), and a diode. The operation of the buck-boost converter can be divided into two modes: when the switch is on and when the switch is off. The inductor stores energy when the switch is on and releases it to the capacitor and the load when the switch is off.

Derivation of the Differential Equation:

To derive the differential equation for the voltage across the capacitor \( V_c \), we need to analyze the behavior of the circuit in both modes of operation. Let's consider the following assumptions for simplicity:

• The converter operates in continuous conduction mode (CCM), meaning the inductor current never falls to zero.
• The capacitor voltage \( V_c \) is relatively stable and changes slowly compared to the switching frequency.

When the switch is on, the inductor is connected to the input voltage \( V_s \), and the inductor current increases. The voltage across the inductor \( L \) is \( V_s \), and the current through the inductor \( I_L \) increases linearly. The capacitor \( C \) is disconnected from the input, and the load is supplied by the capacitor.

When the switch is off, the inductor is connected to the capacitor and the load. The inductor current decreases, and the energy stored in the inductor is transferred to the capacitor and the load. The voltage across the inductor is \( V_o - V_c \), where \( V_o \) is the output voltage.

The differential equation for the capacitor voltage \( V_c \) can be derived by applying Kirchhoff's current law (KCL) at the capacitor node. The current through the capacitor \( I_C \) is given by:

\[ I_C = C \frac{dV_c}{dt} \]

The current through the load \( I_L \) is given by:

\[ I_L = \frac{V_o}{R_L} \]

Applying KCL at the capacitor node, we get:

\[ I_C = I_L \]

Substituting the expressions for \( I_C \) and \( I_L \), we get:

\[ C \frac{dV_c}{dt} = \frac{V_o}{R_L} \]

Dividing both sides by \( C \), we get the differential equation for the capacitor voltage \( V_c \):

\[ \frac{dV_c}{dt} = \frac{V_o}{C R_L} \]

Correct Option Analysis:

The correct option is:

Option 1: \(\rm \frac{dV_c}{dt}=\frac{V_o}{CR_L}\)

This option correctly represents the differential equation for the voltage across the capacitor \( V_c \) in a buck-boost converter, derived from the principles explained above.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: \(\rm \frac{dV_c}{dt}=\frac{V_s}{CR_L}\)

This option incorrectly uses the input voltage \( V_s \) instead of the output voltage \( V_o \). The differential equation for the capacitor voltage depends on the output voltage \( V_o \), not the input voltage \( V_s \).

Option 3: \(\rm \frac{dV_c}{dt}=\frac{V_S}{C}\)

This option is incorrect because it omits the load resistance \( R_L \), which is a crucial factor in the differential equation. The load resistance \( R_L \) affects the current through the capacitor and, consequently, the rate of change of the capacitor voltage.

Option 4: \(\rm \frac{dV_c}{dt}=\frac{V_o}{R_L}\)

This option is incorrect because it does not consider the capacitance \( C \). The capacitance \( C \) is essential in determining the rate of change of the capacitor voltage. The correct differential equation must include \( C \) in the denominator.

Conclusion:

Understanding the derivation of the differential equation for the voltage across the capacitor in a buck-boost converter is crucial for the analysis and design of such converters. The correct differential equation is \(\rm \frac{dV_c}{dt}=\frac{V_o}{CR_L}\), which accounts for the output voltage \( V_o \), the capacitance \( C \), and the load resistance \( R_L \). This equation helps in predicting the behavior of the capacitor voltage and designing the converter for desired performance.