DEAR ZORRO!
HOW CAN I FIND OUT THE TRANSFORMER SECONDARY VOLTAGE FROM THESE SHADE CURVES FOR THE ABOVE GIVEN DATA, CAN YOU PLEASE EXPLAIN....?
Hi, bila ljaved
Let me show how to use the curves in the case you presented.
Look at the figure on the second page (lebelled as Fig.14).
You must supply Idc=100mA at Edc=300V, so the load is Rload=Edc/Idc=3000 ohms.
Let's assume that the total series resistance (the equivalent resistance of the transformer plus the dynamic resistance of the two diodes in the case of the full-wave bridge) is Rs=30 ohms, so Rs/Rload=1%.
Now, if the frequency is 50 Hz and C=16 uF, ω*Rload*C=15.1 .
With this abscissa, for the curve corresponding to the found Rs/Rload=1% we get Edc/Et(max)=91% (ordinate at the left).
As Edc=300V, we obtain Et(max)=300V/0.91, but for silicon diodes we shuold add about 1.5V giving Et(max)=331V. This is the peak voltage, so Et(rms)=Et(max)/sqrt(2)=234V.
(Observe that if C is big enough we are in the right flat part of the curve ans Et would be independent of C. This happens because the ripple is much smaller than Edc.)
Now, with the figure at page 4 (labelled Fig.16) we obtain the ripple.
For the full-wave circuit and Rs/Rload=1% we must use the third curve counting from the bottom (see the table CURCUIT-PARAMETER).
For the abscissa ω*Rload*C=15.1, we read from the plot that the ripple is about 5.5%. So, the rms value of the ripple is 300V*0.055=16.5V .
I hope it is clear.
regards
Z