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Next: Inductance loading Up: Power supplies Previous: Capacitor and inductance loadings

Capacitive loading

It's now time to compromise : we will assume in the following section that we will be working with a capacitive loading schema in which capacitor has a sufficient value so that voltage at its terminals is nearly constant. If we assume too that there is no voltage drop across the rectifier (every situation is reducible to this one simply by adding voltage drop to output voltage), current will flow thru it only when input voltage will excedd tex2html_wrap_inline159. This occurs in the time intervall tex2html_wrap_inline167 as depicted in the figure below :

The intervall tex2html_wrap_inline167 can also be represented by an angle tex2html_wrap_inline171 :
displaymath173
tex2html_wrap_inline171 is called the opening angle of the rectifier. Instantaneous current in the load is then computed easely as :
displaymath177
which yields, by an elementay integration to the following formula for mean current :
displaymath179
(we recall that n is the number of phases). tex2html_wrap_inline183 is mean current flowing thru the load and is generally given in specifications of the power supply : it is then possible to find tex2html_wrap_inline171 knowing that current, and, since tex2html_wrap_inline159 is specified too, to compute unloaded voltage tex2html_wrap_inline157. By putting tex2html_wrap_inline191, one has the short circuit current :
displaymath193
yielding an equivalent expression for mean current :
displaymath195
Since Ohm's law gives tex2html_wrap_inline197, it comes :
displaymath199
Finally, using tex2html_wrap_inline171 definition, namely tex2html_wrap_inline203, one has the following formula :
displaymath205
This expression can be used to compute tex2html_wrap_inline171 knowing power supply and load internal resistances.

One needs now to know if filtering is efficient enough. For that, current's fundamental wave amplitude tex2html_wrap_inline209 will be enough : since a rectifier is almost never used without an auxilliary low-pass filter, one can admit that harmonics have a negligeable contribution to total hum. A fourier transform gives :
displaymath211
Which is for most usual values of n :
equation52

From the amplitude tex2html_wrap_inline209, residual waving at capacitor terminals is obtained by multiplying with its impedance :
displaymath217
where f is mains frequency. This formula can be rewritten in the case of a 50Hz mains (correct that if you have a 60Hz mains!) :
equation82
In all previous computations, we have assumed (without saying it) that capacitor impedance is low when compared to that of the load, making possible to admit that all hum current is flowing thru the capacitor. In practical applications, if a RC filtering is used (see next sections), this hypothesis can be false and computed hum can be slightly lower than observed one (several percents). It is almost never a problem, but for some critical applications accuracy is needed : on can then compute total load impedance Z knowing auxilliary filter impedance tex2html_wrap_inline223 by : tex2html_wrap_inline225 thus the hum voltage (tex2html_wrap_inline209 expression is always valid) : tex2html_wrap_inline229.

Main characteristics of the power supply are now knowned. Nevertheless, mains transformer must still be designed, and for that purpose, RMS intensity of secundary current is needed. this value is easely computed by integration of the instantaneaous current :
displaymath231
or alternatively :
displaymath233
thus, using tex2html_wrap_inline235 and tex2html_wrap_inline183 relationship :
displaymath239


next up previous
Next: Inductance loading Up: Power supplies Previous: Capacitor and inductance loadings

Stephane Puechmorel
Tue Jul 15 10:57:31 WET DST 1997