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# Why reactive power is not considered for calculation of efficiency of SMPS?

#### Embedded_Geek

##### Full Member level 6
Hi Members,

Why the reactive power is not considered during calculation of efficiency of SMPS?
As per my understanding reactive power is generally not desired and isn't the reactive power due to the SMPS itself? Then why aren't we considering it in the calculation of efficiency by using apparent power like below:

Efficiency = Output power / Input apparent power

Thank you!

Efficiency is always related to AC real power respectively DC power. Power factor is a metric for ratio of real to apparent power.

### Embedded_Geek

Points: 2
Hi,

reactive power is often caused by the input capacitor.

But reactive power is nothing you can use to generate output power (real power).

Klaus

### Embedded_Geek

Points: 2
The whole idea of the DC-DC (SMPS) is to make clean
DC with hardly any AC. So, no reactive power except
what's thrashing in the middle, and "wallplug" eff% is
the name of the game. The input and output filters just
scrub all that back into the rails and it's accounted for

### Embedded_Geek

Points: 2
iIt’s a good question. Although reactive energy is just stored energy even though it took real energy to store it. But efficiency is averaged over time that neglects start surges and energy stored is discounted over time in both primary and secondary.

### Embedded_Geek

Points: 2
Hi,

reactive energy is NOT just stored energy.
reactive energy is CONTINOUSLY pushed forward and back energy .. in a way that the averaged power is zero.

One could start a own discussion about reactive energy.
Especially: usually the ENERGY is the integral of POWER over TIME. The result is REAL ENERGY.
But how to mathematically get reactive energy ... especially when involving overtones. Indeed I don´t know the answer.

Klaus

The reactive current does cause extra conduction losses in the mains distribution system......so in a way, it could be made into an efficiency type measurement.......but try drawing out the schematic for that...right back to the power station!.......then work out all the wire gauges and lengths and wire resistances........big calculation......

You don't pay for conduction losses between power station and your meter, unless you are a commercial customer.

Power quality regulations however limit the amount of reactive and harmonic current drawn from the grid.

The reactive current does cause extra conduction losses in the mains distribution system
True, but this loss has nothing to do with ans SMPS loss ... it happens outside of the SMPS.

Klaus

The whole idea of the DC-DC (SMPS) is to make clean
DC with hardly any AC. So, no reactive power except
what's thrashing in the middle, and "wallplug" eff% is
the name of the game. The input and output filters just
scrub all that back into the rails and it's accounted for
I am bit confused regarding the putting the reactive power onto the power rail back. How is this possible when there are diodes (rectifier)? The diodes will be reversed biased so bit confused here how energy flows back to the grid/rail.

Thank you!
--- Updated ---

Efficiency is always related to AC real power respectively DC power. Power factor is a metric for ratio of real to apparent power.
This means the efficiency given for a SMPS by the supplier is based on calculation of real power and output power. Then I am wondering why many suppliers/vendors do not specify the power factor. I mean we can say a SMPS is efficient enough (by not mentioning the power factor at all, but isn't it also a factor to consider when talking about efficiency in principle?)

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It is not considered significant for low PF < 100 W but is now mandatory for PF correction for >=100W which are active types with pf=1.

Its the fact that the waveform has current harmonics....and so the v x i product is sometimes pos , and sometimes neg.....neg means power flowing back to source, in spite of the diodes.
whenever i is not sine and in phase with v, then you get power flow back to source aswell....reactive power

Hi,

* a schematic
* input V, I, P diagrams

Klaus

Power Factor (PF) is the average of any integer number of cycles for the ratio of real to apparent. Apparent is the square root sum of real and reactive squared, also known as the orthogonal vector sum.

But efficiency is only about the ratio of useful real power to that which is supplied. Changing the reactance internal to a circuit can create effects to change efficiency. But to include reactive power in efficiency would lead to false estimates of power lost in heat = power dissipated. This is important when considering the cooling design.

Low PF is considered poor because the currents that just cycle in stored reactance power cause losses in the distribution to the load and transformer saturation margins or available real power in large transformers will be reduced measured as a utilization factor. Low pF is also contributed by non-linear parts (diodes) that interrupt the linear current. It may also occur from complex loads such as;
- high-frequency (HF) devices,
- over-excited synchronous motors, (OESM)
- resonant circuits,
- intermittent loads like welders, which are also HF, even if controlled by some current limit

Since low PF can be corrected by C banks, OESM, and harmonic filters in distribution, the maximum power margin loss is never included in efficiency calculations. Residential meters are designed to measure real power by the motor torque effect of eddy currents in a rotating aluminum disk on a magnetic "air-bearing" or by the real V(t)*Re{I(t)} = Pd calculations.

Thus, reactive power is not considered during the calculation of the efficiency of an SMPS.

VA Reactive or VAR power in large industrial sites can be measured independently and billed when appropriate.

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