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Issue #214 May 2008
Problem 3 —Assuming a constant-current load for simplicity, describe the voltage and current waveforms in the output capacitor C2 in qualitative terms. How do they change, if your have a constant-power load, such as a secondary switching regulator?
Answer 3 —The charging current is the rectified sine wave shape. The constant load current shifts this downward, and the voltage feedback to the switching controller causes this shift to be such that the net current through C2 over the long term is zero, which is the same as saying that the voltage is held constant. Another way of looking at it is that the net charge in and out—the areas enclosed by the current waveform above and below zero—are equal (see the upper trace in Figure 3).
The instantaneous capacitor voltage is the integral of this waveform, as shown in the second trace. The magnitude of this ripple is the peak-to-peak current value divided by the value of the capacitor and the ripple frequency. For example, if the output voltage is 400 V and the load is 40 W, the average current is 100 mA. The peak charging current is p/2 times the average value, or 157 mA. If C2 is 1,500 µF and the ripple frequency is 120 Hz, then the ripple voltage will be about 0.87 VP-P.
With constant-power load (e.g., switching buck regulator), the load current includes a ripple component that is the inverse of the voltage ripple. This distorts the net current waveform slightly, which in turn affects voltage waveform as well. Using the same example, the current ripple is only about 0.25% of the average value, so the effect is very tiny. The waveforms are still basically sine wave-like—no significant additional harmonics are created.
Bonus question —Note that with a 40-W load, the 120-VAC line is going to be supplying an RMS current of 300 mA, which has an average value of 270 mA. How do you account for the difference between this number and the 100- mA average current that the load sees?
Bonus answer: At first glance, it might seem that the average current should be the same in both cases, but you have to remember to account for the current through the switch during the time that the coil is being “charged.” This current does not flow on the output side of the PFC circuit.
This current does not, however, represent a loss of efficiency. The energy associated with this current is stored in the magnetic field of the coil, and when the switch opens, it is used to boost the voltage to the output level.
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