# B what minimum piv rating must the diodes have n 05 v

This preview shows page 33 - 40 out of 52 pages.

(b) What minimum PIV rating must the diodes have? n = 0.5 V p ( sec ) = nV p ( pri ) = 0.5*(100 V) = 50 V PIV = 2 V p ( out ) + 0.7 V = 2*(24.3 V) + 0.7 V = 49.3 V
Bridge Full-Wave Rectifier Operation V p ( out ) = V p ( sec ) = 1.4 V Determine the peak output voltage for the bridge rectifier in figure. Assuming the practical model, what PIV rating is required for the diodes? The transformer is specified to have a 12 V rms secondary voltage for the standard 120 V across the primary. V p ( sec ) = 1.414 V rms = 1.414*(12 V) = 17 V V p ( out ) = V p ( sec ) - 1.4 V = 17 V - 1.4 V = 15.6 V PIV = V p ( out ) + 0.7 V = 15.6 V + 0.7 V = 16.3 V
Voltage Regulators
Determine the ripple factor for the filtered bridge rectifier with a load. n = 0.1 V p ( pri ) = 1.414 V rms = 1.414*(120 V) = 170 V V p ( sec ) = nV p ( pri ) = 0.1*(170 V) = 17.0 V V p ( rect ) = V p ( sec ) - 1.4 V = 17.0 V - 1.4 V = 15.6 V Vr ( pp ) ≈ ( 1/ fR L C)*V p ( rect ) = 1/((120 Hz)(220 A)(1000 mF) *15.6 V = 0.591 V V DC = (1 (1/2 fR L C))V p ( rect ) = 1 1/((240 Hz)(220 A)(1000 mF)) =15.6 V = 15.3 r = V r ( pp ) / V DC = 0.591 V /15.3 V = 0.039