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I am trying to understand inductor energy storing during ON time in buck converter.

Energy stored in inductor of buck converter is (Input Power -Output Power)*Ton

Lets say output power is 5 W and converter efficiency is 90% then input power is 5.55 W.

Then difference is 0.55 W and energy stored during on time 0.55W*3.3 us =1.8 uJ. (SW Frequency is 100 kHz.)

But if efficiency is 100% then inductor will not store any energy during on time. Is that right thinking?

I was referring APEC 2023 SMPS presentation by Robert White enter image description here

following is the excerpt from Sanjay Maniktala book Switching Power Supply A-z

enter image description here

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  • \$\begingroup\$ 0.55W*3.3 us where's that 3.3 us coming from? What's the operation? DCM or CCM? \$\endgroup\$ Apr 15 at 12:23
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    \$\begingroup\$ Efficiency is about the power that is not stored in the inductor, because it is consumed by other parts of the circuit or converted to heat by resistive losses in the components and wiring. \$\endgroup\$
    – Dave Tweed
    Apr 15 at 12:44

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But if efficiency is 100% then inductor will not store any energy during on time. Is that right thinking?

No. If the inductor does not store any energy then what will supply the load when the switch is off?


Energy stored in inductor of buck converter is (Input Power -Output Power)*Ton

No. The inductor stores energy during the switch-on-time, and releases (to supply the load) during the off-time. But you should work out the details using the energy integral:

$$ E = \int_T P\ dt $$

\$P\$ is the power, obviously. During the on-time the voltage across the inductor is positive, so the inductor "stores" energy. And during the off-time, the voltage across the inductor becomes negative so the inductor "releases" energy.

If we assume the input and output voltages are constant (i.e. not varying over time), then the voltage across the inductor will be constant during the on-time:

$$ E = (V_{in}-V_{out})\int_{T_{on}} i_L\ dt $$

where \$i_L\$ is the inductor current which is a ramp-up. If the operation is DCM the ramp starts from zero but if it's CCM then there's going to be a DC offset which will be added to the stored energy.

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At a base level, think of the inductor as a bucket for energy. You put the energy into the bucket by charging up the current, which is 1/2LI². It's stored here in the magnetic field of the inductor. The rate you charge up the current is V/L, which in the case of a buck is IN-OUT. So for the main switch ON time, you're filling up the bucket.

During your OFF time, you then pour the energy out into the output capacitor. The inductor current goes down, and the energy enters the capacitor as it leaves the inductor. This is the fundamental way bucks, boosts, etc. all work. In a perfect world, you could do this 100% efficiently, if it wasn't for things like core loss, resistance, and gate capacitance losses.

The thing to note here is that for an ideal system, the efficiency does not depend on how much you fill your bucket (inductor) on a cycle by cycle basis. You could fill it a little bit very frequently, or you could fill it a lot, much less frequently. In practice, there's a balance somewhere in the middle where your ripple, efficiency, and component sizes are optimal (though often at expense of each other).

The point Maniktala is making here is that for a buck, your ON time charges your output capacitor too as your inductor charges, rather than just during your OFF time like it does with a boost converter. This is a function of your L, C, load current, and importantly duty cycle, but NOT your efficiency.

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Energy stored in inductor of buck converter is (Input Power -Output Power)×Ton

and

But if efficiency is 100% then inductor will not store any energy during on time. Is that right thinking?

No, that is not right thinking. Energy stored in the inductor that is passed through to the output is that energy needed to maintain the correct output voltage under load.

That means either

  • All the energy stored in the inductor is passed to the output to replenish it (discontinuous current mode) or,
  • If operating in continuous current mode (CCM) then some of the energy in the inductor is used to replenish the output.

The energy stored in the inductor has little to do with power efficiency. The energy stored is that needed by the output to keep the voltage steady.

This means that the assertion made about input power, output power and Ton is wrong.

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Energy stored in inductor of buck converter is (Input Power -Output Power)*Ton

Yes, but your definitions of "input power" and "output power" are wrong. What changes the energy stored into the inductor is power fed or drawn from the inductor, not the whole converter, not power stored or drawn from caps, also losses, etc.

Suppose a buck converter with the following characteristics: \$V_{in}\$, \$V_{out}\$, \$I_L\$, \$T_{on}\$, \$T_{off}\$, \$T=T_{on}+T_{off}\$.

During \$T_{off}\$, no current is drawn from the input, so no power is drawn from the input. Instead, the inductor provides power \$I_L V_{out}\$ to the output and stored energy decreases by \$T_{off} I_L V_{out}\$.

During \$T_{on}\$, \$I_L\$ is drawn from the input, so the inductor receives power \$I_L (V_{in}-V_{out})\$ and stored energy increases by \$T_{on} I_L (V_{in}-V_{out})\$

On average, in "steady state" mode, energy stored in the inductor stays the same. It increases during Ton, decreases during Toff, ends up where it started, and goes through another cycle. Therefore the amount of energy stored during \$T_{on}\$ is the same as the amount of energy released during \$T_{off}\$

\$T_{off} I_L V_{out} = T_{on} I_L (V_{in}-V_{out})\$

\$T_{off} V_{out} = T_{on} V_{in} - T_{on} V_{out}\$

\$(T_{off}+T_{on}) V_{out} = T_{on} V_{in}\$

\$ V_{out} = \frac{T_{on}}{T_{off}+T_{on}} V_{in}\$

...and we get to the usual formula for the duty cycle in continuous conduction mode.

Note my calculation assumes \$I_L\$ is constant, this is purely because I am too lazy to write integrals so I wrote \$ Energy = V.I.T \$ instead of \$ Energy = \int{v.i.dt}\$. So feel free to add the integrals while reading. The calculation also assumes \$I_L\$ is variable, since the energy in the inductor changes. But hey it's engineering.

In your question you are confusing average power, and instantaneous power. Average power is indeed zero since the inductor stores and releases the same amount of energy on each cycle, so your conclusion is correct on average/steady state. But it is not correct when considering instantaneous power during \$ T_{on} \$ and \$ T_{off} \$.

You are also confusing losses (0.55W) with stored energy. Losses are not stored, they simply turn into heat.

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