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

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(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
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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
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Voltage Regulators
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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
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Additional Problems
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1. Find the value of the load resistance R L for the zener current of 19mA R L = 10K 2. A full wave rectifier is to be designed to produce a peak output voltage of 12 V and delivers 120mA to the load and produce an output with a ripple of not more than 5%. An input line voltage is 120 (Vrms), 60Hz is available. Find the load resistance, turns ratio, capacitance value and PIV. 𝑅 𝐿 = 12 120? = 100Ω 𝑃?𝑎? ?? 𝑠𝑢???𝑦 𝑉𝑃𝑠?𝑐 = 𝑉 0 (p) + 1.4 = 12 + 1.4 = 13.4V Vs(rms) = 13.4 2 = 9.48 𝑉 𝑉 𝑟(𝑝𝑝) = 0.05 ∗ 12 = 0.60𝑉 𝐶 = 12 2∗60∗100∗0.60 = 1667μF 𝑁 2 𝑁 1 = 𝑉 2 𝑉 1 = 9.48 120
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I D V S 3. Find i D analytically, given V S = 1.1 + 0.1 sin1000tV , nV T = 40mV and V D = 0.7V I D = 4 + 0.91sin1000t mA
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