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I was looking through previous posts and couldn't find anything that I was looking for on this topic. The waveform below is the input current to my system, which is also below. The question is, how do I go about getting the efficiency? Should I be taking the average or RMS value of this current waveform to get there?

Iin

Schem

Edit for question clarity: I was actually wondering if I should be using the RMS current for the input since this is not a pure sinusoidal waveform (the second half of the original post).

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    \$\begingroup\$ Do you know what efficiency means, how is it defined? Realize that in a toaster which runs on mains AC, the average current is zero. Does that mean the toaster does not dissipate power? \$\endgroup\$ Apr 21, 2021 at 14:43
  • \$\begingroup\$ typically power out divided by power in... which means having current and voltage waveforms at both input and output of smps... \$\endgroup\$
    – vicatcu
    Apr 21, 2021 at 14:45
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    \$\begingroup\$ @vicatcu I was asking OP, not you. Thanks for supplying a direct answer instead of making OP think about his question. \$\endgroup\$ Apr 21, 2021 at 14:47
  • \$\begingroup\$ @Bimpelrekkie right, should have phrased the question a bit differently. Eff = Pout/Pin. Pout in this case is just straight up dc values so Pout = Vdc*Idc, but Pin is what i am confused about. Should I be using RMS current to calculate the average input power to use in my efficiency calculations? \$\endgroup\$
    – justaguy
    Apr 21, 2021 at 15:17
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    \$\begingroup\$ See these answers. They apply to your case just as well. \$\endgroup\$ Apr 21, 2021 at 16:43

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To determine the efficiency of your power supply you have to deal with average power levels. The output power can be determined in various ways but if you are not sure of the ripple values, then the best is to perform a multiplication of instantaneous currents and voltages to obtain an instantaneous power \$p(t)\$ (note the lower case) that you will now average over a switching period to obtain the average power \$P\$.

The below graph shows the output voltage and current of a typical flyback converter. You can see some ripple in these waveforms. If you want the exact power, just multiply the current and the voltage signals to form the instantaneous power:

enter image description here

If you place the cursors to isolate a switching cycle in \$p_{out}(t)\$, then the average power is 22.55 W. You could also take the rms value of \$V_{out}\$ and square it to obtain \$P_{out}=\frac{V_{out, rms}^2}{R_L}=\frac{15^2}{10}=22.5\;W\$ over the 10-\$\Omega\$ load.

For the input power, if you have a stable dc source without ripple, then you can simply multiply the dc value of the source with the average input current and you'll have \$P_{in}\$. Below is the typical input current signature of the converter operated in DCM:

enter image description here

The average input current is measured along a switching cycle and you have 121.1 mA which multiplied by the 290-V dc source gives an input power of 35.12 W. The efficiency is thus \$\eta=\frac{P_{out}}{P_{in}}=\frac{22.5}{35.12}=64\$%. Oui, a poor performance but this is a simulation example :-)

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I don't know if LTspice has functions to calculate averages in the plot box, but that's not necessary.

Create a function current source that has the expression V(Vin)*I(Vin). So it outputs a current that is proportional to input power. Then you put a capacitor in parallel, and the voltage on this cap is equal to input energy.

You can do the same at the output, with V(Rload)*I(Rload) so you get the power in your load resistor, integrate it in a cap...

And then you plot the ratio of output energy to input energy.

You can also do a moving average over one switching period, which should be more interesting... simply by using this very clever trick, using a transmission line to delay the signal by one period, and substracting the current value from the delayed value.

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Power supply efficiency is calculated by the ratio of how much power is transferred to its output compared to the total energy consumed by the power supply. So the formula is

schematic

simulate this circuit – Schematic created using CircuitLab This is typically measured with an 80% of the total load . For Example: if I had a power supply that outputs 150W and it consumes 275W my efficiency is 150/275=0.54545 or 55%

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