I am struggling with output voltage ripple calculation of the buck converter below. Firstly, I don't understand the following statement from the lecture.
Can anyone explain it?
If the capacitor voltage ripple is small, then essentially all of the ac component of inductor current flows through the capacitor.
The images are from the lecture here (pages 39-40).
Answer
If the capacitor voltage ripple is small, then essentially all of the ac component of inductor current flows through the capacitor.
Paint a scenario and examine the currents: -
- Ripple is 50 mV p-p, nominally triangular and 100 kHz
- Output load is 10 ohm
- Output capacitor is 100 uF
The AC ripple current through the load is simply 5 mA p-p - this is the baseline for comparison. The RMS is the peak value (2.5 mA) x 0.577 = 1.443 mA
For the capacitor, we have to calculate the slope of the voltage. It rises 50 mV in 5 us so that's a rate of 10 kV/s. Going back to basics, Q=CV and differentiating we get: -
\$\dfrac{dq}{dt} = C\dfrac{dv}{dt}\$ which of course equals current.
Therefore current is 10,000 x 100uF = +/-1 amp and square in shape. RMS is 1 A.
1 amp is a lot bigger than 1.443 mA and "C" is always chosen to minimize ripple so, as C gets bigger, then the ripple gets smaller (hence a lower AC current through the load resistor). Ultimately the AC ripple current through the resistor tends towards zero and the ripple current in the capacitor remains at a constant.
Capacitor ripple current remains constant because it is defined by the inductor and input voltage to the regulator and, to make this analysis clearer it makes sense to consider the input voltage to be constant.
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