I'm working some equations for a boost converter, and running into a contradiction I don't understand. Assume a steady state, perfect efficiency, and continuous conduction.

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Total instantaneous energy stored in the choke is \$\frac{1}{2}LI^2\$

The energy stored in the choke changes as the choke current changes, rising during the on-time, and falling during the off-time. This change in energy is the energy transferred to the load each cycle. Total energy transferred to the load in a given switching cycle is \$\frac{1}{2}L(I_{peak}^2-I_{trough}^2)\$

By difference of squares \$I_{peak}^2-I_{trough}^2 = (I_{peak}-I_{trough})(I_{peak}+I_{trough})\$

The average input current of the converter is the average of the choke current peak and trough, \$I_{peak}+I_{trough} = 2\hat{I}_{in}\$

Define \$\Delta I=(I_{peak}-I_{trough})\$

Total energy transferred per switch is \$\Delta IL\hat{I}_{in}\$

Power transfer \$P=f\Delta IL\hat{I}_{in}\$

Basic choke equation \$V=L\frac{di}{dt}\$

During the "on" part of a switching cycle \$V_{in}=L\frac{\Delta I}{t_{on}}\$

Solve for \$\Delta I=V_{in}\frac{t_{on}}{L}\$

Substitute \$P=ft_{on}V_{in}\hat{I}_{in}\$

On-time times frequency is duty cycle, and input voltage times input current is throughput power \$P=PD\$


I've just proved that every boost converter on the planet must, inevitably, explode. I'm reasonably confident this does not occur. But I can't find the flaw in my assumptions or my algebra. Help me?

  • \$\begingroup\$ When you said total energy transferred per switch did you not forget the 0.5 in front of the equation. \$\endgroup\$
    – Andy aka
    Jun 3, 2014 at 21:59
  • \$\begingroup\$ That .5 is canceled by the 2 in (I_peak + I_trough = 2I_in). \$\endgroup\$ Jun 4, 2014 at 12:29

1 Answer 1


Everything seems fine, except right at the end. You are equating the average power delivered by the inductor L, with the average power delivered by the power supply (\$V_{in}\$). As you found out, this is only true if \$D=1\$ (duty cycle of 100%).

Simply put, during \$T_{off}\$ the power supply adds its voltage to the voltage that L develops, and share the same current, so the energy delivered to the load HAS to be more than what the inductor delivers (which can't be more than what it accumulated during Ton).

In the general case, during the on-time (\$T_{on}\$) the power supply delivers an energy of \$V_{in}I_{avg}T_{on}\$ exclusively to the inductor L. All of this energy has to be delivered to the load during the off-time (\$T_{off}\$), because this is supposed to be perfectly cyclical. But during \$T_{off}\$, \$V_{in}\$ is still delivering current, with the same average as in \$T_{on}\$ (we're calling it \$I_{avg}\$), and that energy \$V_{in}I_{avg}T_{off}\$ is going to the load as well.

So in the end the load gets, per cycle, an energy of \$V_{in}I_{avg}(T_{on}+T_{off})\$, which means the average power \$P = V_{in}I_{avg}\$. But this is not the same as the average power delivered by the inductor, which is only \$V_{in}I_{avg}(\frac{T_{on}}{T_{on}+T_{off}})\$.

A way to look at it is that, per cycle:

  • [Total Energy received by the load] = [Total Energy delivered by power supply] = [Energy delivered by the power supply during Ton] + [Energy delivered by the power supply during Toff].
  • [Energy delivered by the power supply during Ton] = [Energy delivered to the inductor during Ton] = [Energy delivered to the load by the inductor during Toff] < [Total energy delivered to the load].

So your last equation is correct: \$ P_L=ft_{on}V_{in}\hat{I}_{in} \$

But it just talks about the average power delivered by the inductor, and is not all the average power delivered by the power supply: \$ P_{PS} = V_{in}\hat{I}_{in} \$

Well, only if \$ ft_{on}=1\$ (100% duty cycle), or Iavg=0 (0% duty cycle). But we wouldn't even be in the correct operating mode in those cases.

  • \$\begingroup\$ I don't think I agree. All the power delivered by the source must go through the inductor, and the power going through the inductor must come from the source. The two numbers must be identical, just like input power and output power are identical. Otherwise there would have to be a third source or sink for power. \$\endgroup\$ Jun 4, 2014 at 12:26
  • \$\begingroup\$ @StephenCollings apalopohapa is correct. And your latest comment starts out correct. The issue I think you're hung up on is your beginning premise of $$P_L= \frac{1}{2}LI^2$$ which is the power stored by the inductor. This has nothing to do with power going through the inductor. In DC, an ideal inductor can have infinite power passing through it, without storing any of it. It acts like a 0 ohm resistor. That is why you have stored power + supply power when the switch is off getting transferred directly to the load. \$\endgroup\$
    – horta
    Jun 4, 2014 at 14:09
  • 1
    \$\begingroup\$ OH! OH! I get it! The energy stored and lost by the choke isn't the only energy going to the load! There's the DC component, like you're saying, which is always carrying energy to the load, even though there's no change in the energy in the choke. If I add that in, my equations all balance out. \$\endgroup\$ Jun 4, 2014 at 14:21
  • \$\begingroup\$ @StephenCollings Sorry, I meant to say energy. Everything should be energy in my comment. Yup, you got it now! \$\endgroup\$
    – horta
    Jun 4, 2014 at 14:22
  • 1
    \$\begingroup\$ So there are two components to the power being delivered through the choke to the load: DC and AC. The DC component is \$V_{in}I_{in}(1-D)\$, the fraction of the choke current that makes it to the load times the input voltage. The AC component is \$V_{in}I_{in}D\$ as per my equations in the question. Add those together and you get a total power delivery of \$V_{in}I_{in}\$, exactly as expected. \$\endgroup\$ Jun 4, 2014 at 14:26

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