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100uF Capacitor

enter image description here I did an experiment on full-wave rectifier with capacitor, and trying to calculate the energy that the capacitor supplies to the output. The picture above is the graph that I got through an experiment (full-wave rectifier) using logger pro. The red line shows the normal output without capacitor, and the blue line shows the smoothed output with capacitor. The question I want to ask is that, is there a formula/equation for solving the energy supplied by the capacitor? I want to find the area bounded by 2 lines (Integral of blue line curve - red line curve).

The AC supply was 60hz and 6V. The capacitance for capacitor is 100uF

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  • \$\begingroup\$ Perhaps see the answers here electronics.stackexchange.com/questions/134249/… \$\endgroup\$ – Tyler Oct 24 '16 at 14:18
  • \$\begingroup\$ Are you trying to solve the integral and actually estimate the energy? Or are you looking for rules of thumb and design guidelines for computing the size of a capacitor used in such a design from estimated load, etc? \$\endgroup\$ – jonk Oct 24 '16 at 17:45
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I have also analyzed this in the past with Java App. You can edit any value

The relationships are non-linear decay and non-linear voltage drop on diodes with surge charge currents.

So it depends on what simple Rule of Thumb you wish to remember or transfer function you wish.

RC filter time constant to rectifier frequency time interval ratio is my favorite. But it always best to simulate it with no load , full load and worst case tolerances to check for Vdc swing, Ipk , cap ripple current RMS to rated limit etc. Transformer efficiency or source series impedance, diode ratings and large cap ESR often play a significant role in the result as well as frequency.

  • we know the full bridge is 2x input f.
  • we know peak DC voltage out depends on 2 diode drops from 0-Peak AC
  • we know old LDO's need a higher input Vdc_min is most important and not Vdc_avg

So my fave Rule of thumb for Load cap C and equivalent load R is to ensure the time constant RC is much greater than the pulse interval of 1/2f or 10ms for 50Hz.

  • For output Vdc with 10% Vac p-p ripple out use RC = 8T
    • for T = 1/2f or 10ms after 50Hz rectified. Thus RC= 80 ms
    • if you want 4% ripple (p-p) , you need to increase RC = 20T
  • then Vdc out average is ~85% of Vp AC input for the 20Vp ac range or 14.1Vrms(ac)

enter image description here

The X axis of the graph is in RC time constant values with peak charge current on the right Y-axis only. Note that the two curves for peak charge/discharge current ratio and % V ripple out are inversely related.

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