Buck Converter Output Power

Question

I'm having some doubts in the definition of the Buck Converter output power with a RL+EMF load.

Some books just consider the output power to be

$$P_o = E_o I_o$$

i.e, average voltage times average current. But, why? The power dissipated in the resistor is

$$P_R = E_RI_{orms}^2$$

isn't it? Why I can ignore the current harmonics in this calculus. I know that the average current is way bigger than the harmonics. But anyway, is this an approximation? Most books just don't bother explaining this.

Edit: Shoudn't the power in the load be

$$P_o = E_bI_o+E_RI_{orms}$$

Answer 1

If you multiplied a sinewave by a DC value you get another sinewave with a different amplitude. This is equivalent to multiplying a sinewave current by a DC voltage. The average power (V x I) is zero. That's what power is - it's an average of V x I.

What this tells you is that current taken from a battery (for example) that is sinusoidal (or has sinusoidal content), creates no average power i.e. the average of \$V \times I_{SINE} = \text{zero}\$.

Now it can be one sinewave or any number of sinewaves or any shape of signal you want because, the pure AC content of the current multiplied by the DC voltage is always having an average value of zero.

Answer 2

There is no measure of output power in your schematic. Edit: Your Query" Edit: Shouldn't the power in the load be \$P_o=E_bI_o+E_RI_{orms}\$"

No : Power is a function of initial conditions and time or frequency for reactive parts as each has an effect on voltage and current due to Z(f).

If Eb were a supercap with some initial voltage to model a rechargeable battery.

Then the instantaneous energy stored in the capacitor \$E_C=½ CV^2\$ could be added with any energy stored in the choke \$E_L=½ LI^2\$ and with losses \$P_D=\int I(t)^2R ~dt\$

The ratio of harmonics to signal is called THD in [dB] but in a DCDC regulator these harmonics could be reactive stored harmonics and not necessarily lost energy depending on the power factor or current phase shift.

In any case the Fourier response of the output power or power spectral density can also indicate how far down each harmonic is from the DC power.

Then Pout would be an RL added to the Vbat in parallel.