I have the following problem.

enter image description here


The switch-mode power supply looks like it's a boost converter. I found that the DC-transfer functions for the boost converter to be:

\$\frac{V_C}{V_{in}}=\frac{1}{1-d} \: \: \$ and \$\frac{I_L}{I_{out}}=\frac{1}{1-d} \: \: \$ where I presume that \$ d\$ is the duty cycle.

My idea was first to find \$d\$ with the first equation: \$d=-\frac{V_{in}}{V_C}+1=\frac{-2.7 \text{V}}{5 \text{V}}+1=0.46 \$

From here, we can plug \$ d\$ into the second equation and find \$I_L=\frac{I_{out}}{1-d}=\frac{2.3 \text{A}}{1-0.46}=4.256 \text{A}\$

And since the \$V_{in}\$ and the inductor is in series, we conclude that \$I_{in}=4.256 \text{A}\$.

But is what I have found the \$\underline{peak}\$ input current or something else? I am in doubt, because I don't use the information about the switching frequency at all.

I hope someone can help me with this.

  • 1
    \$\begingroup\$ You have to establish if the circuit operates in CCM because your formula assumes CCM and, I suspect that it operates in DCM (different formulas) \$\endgroup\$
    – Andy aka
    May 18, 2020 at 13:09
  • \$\begingroup\$ Also consider that the text reads "100% efficient" so this looks like components are considered ideal. At t=0, what is the impedance of the 100nF cap? \$\endgroup\$
    – rdtsc
    May 18, 2020 at 13:17
  • \$\begingroup\$ In CCM the peak inductor current is $$I_{Lpeak} = IL_{avg} + \frac{\Delta I_L}{2}$$ and the \$\Delta I_L = \frac{V_{IN}}{L} \times T_{ON} \$ \$\endgroup\$
    – G36
    May 18, 2020 at 14:54

1 Answer 1


First thing to notice is that you have a a very small output capacitor, which is used to supply the load during the on phase (when the inductor is being charged and there is no power transfer between the primary and secondary sides).

The required capacitor to hold the output voltage long enough until the next cycle can be calculated through the following:

$$C=\dfrac{P_o\cdot D \cdot 2}{f_S\cdot (V_o^2-V_{o,min}^2)}$$


\$P_o\$ is the output power (\$5V\cdot 2.3A=11.5W\$)

\$D\$ is the duty cycle \$1-\dfrac{V_{i}}{V_o}=0.46\$

\$f_S\$ is the switching frequency

\$V_o\$ is the nominal output voltage

\$V_{o,min}\$ is the minimum output voltage

Plugging in your values, you are gonna find out that with your current capacitor, the output voltage drops to zero at every cycle because of the high load. This also means that at every off cycle, the output capacitor will have to be recharged from zero again, as it was completely discharged during the on phase.

Anyway, disregarding that for the moment (assuming that indeed there is a stable output voltage \$V_o=5V\$ and output current \$I_o=2.3A\$) you can use this application note from Texas Instrument to calculate what you need. Here is a summary using your values:

The duty cycle is given by:

$$D=1-\dfrac{V_i}{V_o} = 0.46$$

The ripple current of the inductor can then be calculated:

$$\Delta I_L=\dfrac{V_i \cdot D}{f_S \cdot L}=936mA$$

The maximum peak current is then given by:

$$I_{SW,MAX}=\dfrac{\Delta I_L}{2} + \dfrac{I_o}{1-D}=4.727A$$


Here is a small simulation just to double check it. The output capacitor was increased in order to account for the aforementioned problem regarding the high load:


  • 3
    \$\begingroup\$ Good answer and also very valid point regarding the output capacitor \$\endgroup\$
    – Big6
    May 18, 2020 at 16:02

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