capacitorcircuit analysiscircuit-designelectrolytic-capacitor
How do I calculate the ripple current a capacitor will experience for a given circuit?
For example, let's say I have a smoothing capacitor on the output of a full-wave bridge rectifier (120VAC, 60Hz) which leads into the primary winding of a flyback transformer (peak primary current is 0.775A).
How do I calculate what this capacitor's ripple current rating should be?
Assume the system is already precharged and operating in a steady state. The bridge has two discrete states: either the capacitor is charging (a diode pair is forward biased), or the capacitor is discharging. Call the period P, the charge time DP, and the discharge time (1-D)P.
During the charge cycle, we can approximate the current entering the capacitor as a triangle, starting at 0, and rising to a peak. $$ 1: I_{charge}(t) = \frac{t I_{peak}}{DP}\\ $$ Assume that the output capacitance is large enough that its voltage ripple is small, meaning the current out of the cap during the discharge time is fixed. $$ 2: I_{discharge}(t) = I_{load}\\ $$
Computing the RMS: $$ 3: I_{RMS}=\sqrt{\frac{\int_0^{DP}I_{charge}^2(t) dt + \int_{DP}^{P}I_{discharge}^2(t) dt}{P}} $$
Evaluating the integral: $$ 4: I_{RMS}=\sqrt{\frac{I_{peak}^2D}{3} + I_{load}^2(1-D)} $$
Since we're in a steady state, the total charge into the capacitor during the charge cycle must be equal to the total charge leaving the capacitor during its discharge time: $$ 5: Q_{charge}=Q_{discharge} $$
The total charge entering the capacitor is the area of the current triangle: $$ 6: Q_{charge}=\frac{I_{peak}DP}{2}. $$
The charge leaving the capacitor during the discharge cycle is the product of the fixed current and time: $$ 7: Q_{discharge} = I_{load}(1-D)P. $$
Which gives us: $$ 8: \frac{I_{peak}DP}{2} = I_{load}(1-D)P $$
Solve for peak current: $$ 9: I_{peak}=\frac{2I_{load}(1-D)}{D} $$
Substitute into equation 4: $$ 10: I_{RMS}=I_{load}\frac{\sqrt{D^3-5D^2+4D}}{D\sqrt{3}} $$
From this we see that the ripple current seen by the output capacitor is a function of the load current and the fraction of the AC period spent charging the capacitor. As D approaches 0, the ripple current approaches infinity. As D approaches 1, the ripple current approaches 0. Longer charge times reduce the ripple.
Consider the choke currents and capacitor voltages during a charge cycle: $$ 11: V_{choke} = L\frac{di}{dt}\\ 12: I_{cap} = C\frac{dv}{dt} $$
During the charge cycle, we have approximated the current through the choke into the capacitor as a triangle with a height of I_peak. The average current into the capacitor during the charge cycle is half this peak. The length of the charge cycle is DP. The voltage across the choke starts at 0, rises to a peak approximately equal to the ripple voltage dv, then falls back to zero. We can approximate the average voltage across the choke as half the ripple voltage. $$ di = I_{peak}\\ dt = DP\\ I_{cap} = \frac{I_{peak}}{2}\\ V_{choke} = \frac{dv}{2} $$ Substituting into 11 and 12: $$ 13: \frac{dv}{2} = L\frac{I_{peak}}{DP}\\ 14: \frac{I_{peak}}{2} = C\frac{dv}{DP} $$ Solve both equations for dv, then solve for D: $$ 15: \frac{2LI_{peak}}{DP} = \frac{DPI_{peak}}{2C}\\ 16: D = \frac{2\sqrt{CL}}{P} $$ Substitute into equation 10 to find the RMS current seen by the capacitor.
So the length of the charge cycle is twice the time constant of the LC resonant circuit. Increasing the size of the choke spreads the charge cycle over a longer time, reducing the RMS current (and improving line harmonics). Increasing the size of the capacitor lengthens the time the choke is forward-biased. And increasing the frequency (decreasing the period) means each charge pulse can be smaller and deliver the same current. Thus, three-phase rectifiers have lower ripple current on their output capacitors than single-phase. This math indicates that for a fixed capacitor ripple current, a three-phase rectifier run with a single-phase input can only run ~30% of the three-phase load current.
How do I calculate minimum capacity of the electrolytic capacitor based on current and voltage to have the smallest ripple?
Smallest ripple is when capacitance is theoretically infinite. You have to accept that there will be ripple and you have to decide how much this ripple can be: -
Taken from here
How do I know the wattage on AC line? I know that the DC wattage is around 25W (313V*0.08A).
The wattage taken from the AC power line is probably a few percent more than what is consumed in the DC circuits you have attached. Power in = power out + losses in rectifier and smoothing capacitor.
Best Answer
The determination of the bulk capacitor value in a full-wave rectification circuit depends on several parameters:
Determining the capacitance value requires a single equation which is given below. It an excerpt from a formula derived in my book on switching converters but you can also find it in a white paper I uploaded on my webpage in 2009.
Once you have your capacitance value, I recommend running a quick simulation using either a load resistance or a constant-power source if your downstream load is a dc-dc converter. You can then check your calculation and then update the capacitor value to the normalized value you will finally adopt.
As a final word, stay away from ready-made recipes like "3 µF per watt". In my opinion, it is much better to work the maths behind selecting a component as it teaches where the potential issues could be when writing equations and later when operating the component. I remember that I was taught the O.H. Schade's curves long time ago in university but I never used them and always derived the value I needed. You can also have a look at this article I published some years ago on the subject.