i learn something formula of Bridge and full wave rectifier from my book

  1. DC voltage \$V_{DC}=V_m-\frac{V_{r_{(pp)}}}{2}\$,\$V_m\$ is the amplitude of sine wave voltage, \$V_{r_{(pp)}}\$ is the peak to peak voltage of ripple wave

  2. The RMS voltage value of ripple \$V_{r_{(rms)}}=\frac{V_{r_{(pp)}}}{2\sqrt{3}}\$

  3. If \$f=60Hz,V_{r_{(rms)}}=\frac{2.4V_{DC}}{R_L C}\$

  4. Ripple factor \$ =\frac{2.4}{R_L C} \times 100 \% \$

i have some questions about the formula above


In the 2. formula,how is the \$2\sqrt{3}\$ calculated?i mean,based on what reason,so we can say \$V_{r_{(rms)}}=\frac{V_{r_{(pp)}}}{2\sqrt{3}}\$,not \$V_{r_{(rms)}}=\frac{V_{r_{(pp)}}}{any \ number}\$?


In the 3. and 4. formula,how is the \$2.4\$ calculated? i mean,based on what reason,so we can say \$V_{r_{(rms)}}=\frac{2.4V_{DC}}{R_L C}, \$and Ripple factor \$=\frac{2.4}{R_L C} \times 100 \% \$ not \$V_{r_{(rms)}}=\frac{any \ number \times V_{DC}}{R_L C} \$ or Ripple factor \$=\frac{any \ number}{R_L C} \times 100 \% \$ ?


1 Answer 1

  1. The ripple can be approximated by a sawtooth. The ratio of peak to RMS of a sawtooth (proof involves a bit of calculus and is left as an exercise) of a sawtooth is Vr(pp)/(2\$\sqrt{3})\$. Keep in mind that the ripple goes plus Vr/2 and minus Vr/2 so you need only integrate one half of the triangle.

  2. If you assume a constant current from the load and the period of a half cycle, and the above formula you get a constant of 2.41 milliseconds. That should give you a hint of what units to use for C.


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