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In steady state operation the average capacitor current in a buck converter is zero, which means that the rippling part of the inductor current charges and discharges the capacitor, while the "DC" part of the inductor current is the load current. I am not sure how this "DC" component is equal to average inductor current.

How do I find the mathematical relationship between load current of a buck converter and average inductor current? How do I relate load current (which in itself does not depend on inductor/inductor current) to the average value of the inductor current?

Edit:

I get that steady-state operation implies zero capacitor current (average). KCL at the output would yield Io = IL_AVG. But by evaluating the average value of the inductor current waveform, I couldn't find any way to equate the output current (Vo/Ro) to Average Inductor current (Ipeak/2). The output current does not depend on either the value of inductor or inductor current ripple. How then would you relate the average inductor current to load current?

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  • \$\begingroup\$ Because RL is in series with the inductor the average inductor current must be equal the the average load current (DC load current). So you are asking how to find the average value of a triangle waveform? \$\endgroup\$
    – G36
    Jul 16, 2021 at 18:52
  • \$\begingroup\$ I'm asking how to equate the average of the triangular waveform to output current. \$\endgroup\$
    – JGalt
    Jul 16, 2021 at 18:55
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    \$\begingroup\$ Is the KCL and L and R are connected in series this is not enough? \$I_L - I_C - I_{LOAD} = 0\$ \$\endgroup\$
    – G36
    Jul 16, 2021 at 19:01
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    \$\begingroup\$ And in steady-state, the capacitor average current is 0A (charging and discharging currents are equal ). \$\endgroup\$
    – G36
    Jul 16, 2021 at 19:03
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    \$\begingroup\$ First, we find the output voltage (Vo) using Volt·second balance "law" and assuming continuous current mode of operation. Next, we used KCL to find that \$I_L = I_{Load}\$ And this is the end. \$\endgroup\$
    – G36
    Jul 16, 2021 at 19:20

4 Answers 4

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I am not sure how this "DC" component is equal to average inductor current.

First, "DC component" is an exact synonym for "signal average". The latter term is the time-domain math explanation (you integrate the signal over time and divide by the interval). The former is the frequency-domain math explanation (you take the Fourier transform of all the components, then divide by the interval -- but the "DC component" is the zero-frequency component, which is the integral of the signal over time).

The only place that the charge from the inductor can go is into the output capacitor or the load. If the average current in the inductor doesn't match the average current into the load, then the excess goes into (or comes out of) the capacitor.

Capacitors integrate current into voltage -- so if the net current of the capacitor isn't zero then its voltage will change. That'll change the current from the inductor and change the current into the load.

Let the inductor current be \$i_L\$, and the output current be \$i_O\$. Then, by Kirkoff's current law, the capacitor current has to be \$i_C = i_L - i_O\$.

If the average capacitor current is anything but zero, then the capacitor voltage will continually climb or decrease. Since the capacitor current is the difference between the inductor current and the output current, if the average capacitor current is zero, then the average inductor current must equal the average output current.

So, just by the fact that the output voltage is steady and there's a cap there, the inductor average current has to match the load current.

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  • \$\begingroup\$ Thanks, I understand that. Since the converter is in steady state, the DC part of inductor must flow through load. But I am not convinced that this DC component equals the average of inductor current waveform. If I try to find a mathematical average for the waveform, I have no way of relating it to output current! \$\endgroup\$
    – JGalt
    Jul 15, 2021 at 23:55
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    \$\begingroup\$ "DC component" is an exact synonym for "average value". Perfectly exact. As in -- they're just two different words for the same thing. \$\endgroup\$
    – TimWescott
    Jul 16, 2021 at 0:20
  • \$\begingroup\$ Does it not refer to mathematical average of the triangular waveform in this context? \$\endgroup\$
    – JGalt
    Jul 16, 2021 at 0:35
  • \$\begingroup\$ You are confusing yourself somehow. What context? In all contexts both the "DC component" and the "average" of a signal (or current, or voltage, or whatever) is the mathematical average of that signal. If you mean "the mathematical average of the trianglular part of the inductor current waveform" -- no. It is the mathematical average of the entire inductor current waveform, whatever it is, whether triangular, piecewise exponential, piecewise sinusoidal, or weirder. \$\endgroup\$
    – TimWescott
    Jul 16, 2021 at 14:05
  • \$\begingroup\$ Sorry, my statement wasn't very clear in the comment. What I meant is that, there is some DC component to the inductor current (which, as you say, is the average of the waveform). But if you were to evaluate this average, how would you equate it to the output current? I get that this is implied in the steady-state operation, but I couldn't find any way to equate the output current (Vo/Ro) to Average inductor current (Del iL /2) \$\endgroup\$
    – JGalt
    Jul 16, 2021 at 18:25
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Simply put $$I_C = I_L - I_R$$

Depending on the impedance ratio of the load to reactance ratio for L(f),C(f) values at the switching rate (f) determines the attenuation of current ripple. This also determines the Q of the filter and attenuation from resonance to switching rate. Too big a ratio for attenuation of ripple disrupts closed-loop step-load stability from too much energy stored.

This means you must follow the app note for peak current to prevent saturation which also takes into account the Cap ripple current rating and the Inductor saturation margin and loop stability.

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Buck converter basic circuit: -

enter image description here

Current from the inductor flows into C1 and RL. The AC part of that current mainly flows into C1 (by design) but the average (DC) part of the inductor current cannot flow in C1 hence....

this "DC" component is equal to average inductor current

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  • \$\begingroup\$ I get that steady-state operation implies zero capacitor current (average), but by evaluating the average value of the inductor current waveform, I couldn't find any way to equate the output current (Vo/Ro) to Average Inductor current (Del IL/2). \$\endgroup\$
    – JGalt
    Jul 16, 2021 at 18:41
  • \$\begingroup\$ Well, I'm unable to guess where your calculation failed. \$\endgroup\$
    – Andy aka
    Jul 16, 2021 at 18:45
  • \$\begingroup\$ Andy, from what I understand, the load current does not depend on the inductor/inductor current. It is simply Vo/Ro. I am just stuck in relating these two parameters. Suppose I get an expression for average inductor current in terms of inductor ripple current, how would you relate it to output current mathematically? \$\endgroup\$
    – JGalt
    Jul 16, 2021 at 18:58
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    \$\begingroup\$ Listen, it's quite simple; the DC load current dictates what the average current through the inductor is. No magic; just simple cause and effect. \$\endgroup\$
    – Andy aka
    Jul 16, 2021 at 19:58
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All of the charge (current) flowing out of the inductor must eventually flow through the load, it's just that some of this charge (current) coming from the inductor flows to the load via the capacitor.

When the buck converter's switch closes, the inductor's current will start to rise. The inductor's current will rise to a level which will be greater than that being drawn by the load and at this point the capacitor will start to take charge (current) and it's voltage (and that of the load) will rise slightly. Next the buck converter's switch opens and the inductor's current will start to decrease. At some point the inductor's current will be less than the current required by the load and at this point the capacitor will start to contribute to the load current and the capacitor's voltage (and load voltage) will fall slightly as the capacitor discharges. So we can see how a small load ripple voltage is created. Note how the charge (current) which the capacitor supplies to the load during the switch-off period has previously been supplied to the capacitor by the inductor during the switch-on period

No charge can pass through the capacitor to ground apart from a tiny leakage current.

Therefore the average current in the inductor must equal the average current in the load when calculated over a time period which is large compared to the time period of the buck converter's switching frequency.

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