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Assume a simple circuit with a 12 V (RMS) AC supply and a 100 ohms load resistor. The power dissipated is P = V^2/R = 1.44 W.

Now assume that a full bridge diode rectifier (without filtering) is placed between the source and the load. Ignoring power losses in the rectifier, the power dissipated in the load resistor is the same as before. Side question: The output waveform now looks like a sine wave with its negative halves flipped over the x-axis. How would I calculate the power when the waveform looks like this? Is the RMS value still 12 V, so that P = V^2/R = 1.44 W is still correct?

Now assume that a smoothing capacitor is added to the output of the rectifier and that the resulting ripple voltage is small. Now the voltage across the capacitor and the load resistor takes the value of the peak voltage 12*sqrt(2) = 16.97 V. Then the power dissipated in the load is P = 16.97^2/100 = 2.87 W. Obviously the output power of the rectifier can't be higher than the input power, so what am I missing here?

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The RMS power from the 12V source will be the output power plus the losses in the diodes (and capacitor).

Instead of the power being a sort-of sinusoidal function of time, with a capacitor it will be very spikey, concentrated near the peaks of the AC waveform.

Here is the situation without a diode. I've increased the voltage and resistor value so the diode losses are less important, but the situation yields about the same resistor power ideally:

enter image description here

The resistor dissipation averages 1.42W.

Now let's add the 1000uF capacitor:

The resistor power is now average 2.85W (notice it's almost constant, the variation is due to the ripple voltage on the capacitor)

enter image description here

Now the power from the power supply with the capacitor fitted:

enter image description here

As you can see it averages about 2.9W, which includes the losses in the diodes.

Here's a closer look at the power from the supply during the spikes:

enter image description here

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Is the RMS value still 12 V, so that P = V^2/R = 1.44 W is still correct?

That is correct. The "S" in RMS automatically ensures that all polarities become positive due to the action of "S"quaring.

Obviously the output power of the rectifier can't be higher than the input power, so what am I missing here?

The only thing that you appear to be missing is that the input power has also risen and, the input current waveform into the ideal bridge rectifier has dramatically altered: -

enter image description here

Image from MathWorks Full-Wave Bridge Rectifier.

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  • \$\begingroup\$ In my case, the power source is a small wind turbine, whose input power is determined by the wind speed (as well as turbine parameters). So for a constant wind speed, the input power is constant. I just don't seee where this "extra power" comes from when the capacitor is added. \$\endgroup\$ Commented May 12, 2024 at 11:59
  • \$\begingroup\$ @leheim if the wind turbine cannot supply enough power, its output voltage will be lower and the resulting rectified and smoothed voltage across the load will also be lower. Either way, with an ideal bridge rectifier, average input power MUST equal average output power. There is no getting away from that fact. \$\endgroup\$ Commented May 12, 2024 at 12:13
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I think what you are missing is that the input power will also increase if the output power increases, so the input power and output power are equal.

The input power does not simply power the resistive load, but the capacitor too, by charging it at every peak.

The capacitor just powers the load while the supply does not, and the input power is zero.

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