Electrical Engineering is a free introductory textbook to the basics of electrical engineering. See the editorial for more information.... |
![]() |
Home ![]() ![]() |
|||
![]() ![]() ![]() ![]() ![]() ![]() ![]() |
|||
Magnitude of Ripple in the Output Voltage of RC LoadAuthor: E.E. Kimberly The voltage variation across the terminals of an RC load on a rectifier depends on the loss of charge on the condenser between charging impulses, and so depends on the magnitude of current drain to the follower circuit and also on whether the rectifier is half-wave or full-wave. With either half-wave or full-wave rectifier, sufficient accuracy may be had by using the simplifying assumption that the ripple is saw-toothed instead of having sinusoidal and exponential components. This comparison is shown in Fig. 27-9 for a full-wave rectifier.
Example 27-1. - Suppose that a condenser in an RC load on a full-wave rectifier is required to deliver an average current of 100 milliamperes at 1000 volts (average) with a deviation therefrom of only 10 volts. How many microfarads of capacitance are required? The frequency is 60 cycles per second.
Solution. - Under the simplifying assumptions the condenser discharges from 1010 to 990 volts in one half-cycle or 1/120 sec. If the discharge continued at that rate, the time t required for the condenser to discharge would be The initial discharge current is The initial power of discharge is then
The condenser voltage and current decrease together and their product is inversely proportional to the square of the time of discharge. The average value of this square-law function is one-third of the initial discharge rate. Therefore,
If the initial charge is Q1 and the charge remaining 120 sec later is Q2, then
and In a condenser, Q = EC. Therefore,
By solution of either of these equations, C = 0.014115 farad or 14,115 mfd.
|
|||
Home ![]() ![]() |