data given N=TURNS RATIO =4.73I out full load = 10 ATblocking= 6.3 meli secondv

Question

data given N=TURNS RATIO =4.73I out full load = 10 ATblocking= 6.3 meli secondv peak-peak ripple across cf =3.6vpp max alowedbeta B1=B2=25 B3= 80CF TOLRANCE = -10% +50%

Explanation / Answer

a) Nominal value of filter capacitor designed = ? Cf =? We know Vripple = Iload /f* C so C= Iload /f * Vripple Vripple = 3.6 f= 120 Hz narmally Iload = Iout = 10A so Cf = 0.023 farad b) The o/p rang will be max when there is max load therefore Vmax = Iload * Req Req = 0.3K+5% + 1.4K + 2% + 1k + 2% = 0.32K + 1.41K + 1.08 k = 2.81K theredore => Vmax = 10 * 2.81K = 28100 = 28.1 kV c) worst case Vout (range) = 0(in case of o/p s.circuit) to 28.1kV d)?1+?2+?3 = 200+125 = 325 Degrees e) Yes it is..They are mounted on the same heat sink needs a loarge block of hreat sonk that means the surface area exposed to ait or natural as well as forced cooling is available that will kepp these transistors tep in nominal range

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