# AC Power loss of inductor in a buck converter

I have a custom piece of equipment that gives a value for AC power loss (Pac) for an inductor plugged into an internal buck converter circuit. The custom equipment has an oscilloscope which is monitoring the voltage across the inductor and the current through it. I am trying to figure out how this value is calculated using this data. Since the circuit is a buck converter, the voltage across the inductor is a PWM shape, with the voltage around Vin-Vout while the PMOS is on, and about -Vout when the PMOS is off. The current is a ramp wave with a DC bias since the buck converter is running in CCM. I am also making sure to run the circuit so that the inductor is far from saturating.

I have tried the following to no avail:

Pac = (DT(Vin-Vout)*Iripple)-((1-D)TVout*Iripple)

where D is the duty cycle between 0 and 1, T is the inverse of the switching frequency.

How I came up with this is that for a perfect inductor, all AC power that is stored during the charging period would then be discharged during the discharge period. Since the inductor has some AC losses from eddy current and hysteresis, I took the power during the charging period and subtracted the power during the discharge period and what would be left is the loss. This is obviously not correct, as I am not receiving the same values as the equipment.

Any ideas?

• Perhaps you need to integrate the ramp waveforms or calculate the RMS value of the all the products before simple subtraction will work. – KalleMP Apr 9 '15 at 21:06
• @Brett, are you interested just in the part of losses caused by eddy currents and hysteresis, i.e. without losses in series resistance of L? I doubt any "custom piece of equipment" can find it out separately, in one step. I think that total losses can be found out the way I described in my answer below, then losses in the series resistance can be calculated from known (measured) resistance and current (Rser*Irms) and subtracted from the total. – Eric Best Apr 11 '15 at 21:29

## 2 Answers

Just at first sight, your formula gives an energy (Joule), not power (Watt)...

If the "...custom equipment has an oscilloscope which is monitoring the voltage across the inductor and the current through it...", then the inductor losses can be calculated right out of the measured values (i.e. right from definition of average power) as:

$P_{losses} = \frac{1}{T} \int_0^Tv(t)i(t)dt$, (average value of instant power during period)

where v(t) is the waveform of voltage across the inductor, i(t) is the waveform of current through it and T is period of these waveforms. Provided that the oscilloscope is a digitizing equipment, then, in principle, the corresponding voltage and current samples from within one period have to be multiplied, summed, multiplied by the sample interval and divided by the period (T) length.

For instance the trapezoidal integration method can be used:

If there are n equidistant samples (of $v_i, i_i$, i = 1 to n) covering one period T, then the losses can be calculated as:

$P_{losses}= \frac{1}{(n-1)} \cdot (\frac{{v_1} \cdot {i_1} + {v_n} \cdot {i_n}}{2}+ \Sigma_{i=2}^{n-1} v_i \cdot i_i)$

2015-04-12, $\textbf 1^{st}$ appendix

As I already stated in the very beginning, your formula is not okay. At first, the T in it is superfluous (it is already incorporated in the duty cycle, D). Let's have a look at it a bit more closely. It can be rewritten (omitting the T, of course) as:

$P_{AC} = [D \cdot (V_{IN}-V_{OUT})-(1-D) \cdot V_{OUT}] \cdot I_{RIPPLE} = (D \cdot V_{IN}-V_{OUT}) \cdot I_{RIPPLE}$,

but is it already okay?

You wrote "…Since the inductor has some AC losses from eddy current and hysteresis, I took the power during the charging period and subtracted the power during the discharge period and what would be left is the loss…".

In principle, this idea is right in my opinion, but:

1. The voltage across L during $t_{ON}$ (term with D multiplier) is: $V_{L\_on} = V_{IN}-V_{PMOS\_SWITCH\_ON}-V_{OUT}$,
not just $V_{IN}-V_{OUT}$ (the PMOS switch contribution isn't negligible).
2. The voltage across L during $t_{OFF}$ (term with (1-D) multiplier) is: $V_{L\_off} = -(V_{OUT}+V_{DIODE\_SWITCH\_ON})$,
not just $-V_{OUT}$ (neither the diode switch contribution is negligible).
3. If we presume both the above voltages as constants during their time intervals and the ripple current being "pure" sawtooth waveform, then value that must be used in the calculation on the place of current is $I_{RIPPLE}/2$ (i.e. its average value – it follows from the very first formula, because if v(t) = const., then it can be factored out the integral and the rest is the ripple current average value).

The resulting formula will be then:

$P_{AC} = [D \cdot V_{L\_on}+(1-D) \cdot V_{L\_off}] \cdot \frac{ I_{RIPPLE}}{2}$

( $V_{L\_off}$ is negative in relation to $V_{L\_on}$, we have to measure both the voltages the same way, that's why the "+" operator is used in the formula)

It is questionable, however, whether the speculated presumptions (3) are "sufficiently" valid/met and how much they affect accuracy of the result.

• The only thing I believe needs to be changed is VL_off should be equal to Vdiode_switch_on-Vout, and that got me within 5% of all their measurements. Thank you! – Brett Prudhom Apr 14 '15 at 21:30

The custom equipment probably simply multiplies instant voltage and instant current in the inductor to obtain instant power. When the PMOS is on, the inductor is receiving power, and when it is off, the inductor is giving out power. The custom equipment probably simply adds these values (assuming "received power" is positive and "given power" is negative), obtaining the total energy that stayed in the inductor - that is, energy that was lost.

This is an direct, experimental way of determining losses in any component, so you shouldn't doubt it. Now, if you could properly model the system, you should be able to obtain a formula for the losses. Your model, however, is inappropriate because you just can't say that the current when PMOS is Iripple. This only works for the perfect sawtooth wave case, and the losses will appear precisely by making the sawtooth wave seem more like this:

Sawtooth-y exponential decay wave http://img.deusm.com/planetanalog/2014/08/563378/Image-2.jpg

In this wave, the average values of the two parts of the cycle are different: one is higher, and the other is lower than the "middle value".