The steady state current flowing through the inductor of a DC-DC buck-boost converter is given in the figure below. If the peak-to-peak ripple in the output voltage of the converter is 1 V, then the value of the output capacitor, in μF , is _____. (round off to nearest integer)

The correct answer is 168.

1. Analysis of the Inductor Current Waveform:
From the given inductor current waveform in the image below:

• The minimum inductor current is $I_{\text{L,min}}=12\text{A}$.
• The peak (maximum) inductor current is $I_{\text{L,max}}=16\text{A}$.
• The ON-time (rise time of the current) is $t_{\text{on}}=DT=20\text{μs}$.
• The OFF-time (fall time of the current) is $t_{\text{off}}=(1-D)T=30\text{μs}$.

2. Calculations for switching parameters:
The total switching period $T$ of the converter is:

$T=t_{\text{on}}+t_{\text{off}}=20\text{μs}+30\text{μs}=50\text{μs}$

The duty cycle $D$ is:

$D=\frac{t_{\text{on}}}{T}$

3. Calculating the Average Inductor and Output Currents:
The average inductor current $I_L$ is the mid-point of the linear ramp:

$I_L=\frac{I_{\text{L,min}}+I_{\text{L,max}}}{2}=\frac{12+16}{2}=14\text{A}$

In a buck-boost converter, the relationship between the average load current $I_o$ and average inductor current $I_L$ is given by:

$I_o=I_L\times (1-D)$

Substituting the calculated values:

$I_o=14\text{A}\times (1-0.4)=14\times 0.6=8.4\text{A}$

4. Output Capacitor Calculation:
During the ON period (when the switch is closed), the diode is reverse-biased, and the output capacitor solely supplies the constant load current $I_o$ to the load. The charge $\Delta Q$ lost by the capacitor during this interval is:

$\Delta Q=I_o\times t_{\text{on}}=I_o\times DT$

This charge loss results in a peak-to-peak output voltage ripple $\Delta V_o$:

$\Delta V_o=\frac{\Delta Q}{C}=\frac{I_o\times DT}{C}$

Rearranging the formula to find the capacitance $C$:

$C=\frac{I_o\times DT}{\Delta V_o}$

Given that the peak-to-peak ripple in the output voltage $\Delta V_o=1\text{V}$:

$C=\frac{8.4\text{A}\times 20\text{μs}}{1\text{V}}=168\text{μF}$