Bridge rectifier, the resistance of each diode in it is (Ohms 1). If the voltage supplied to the rectifier Vrms = 240v, and the load resistance is RL = 480 Ohms, then the values of the three currents are equal to O (Im = 0.9 A, ldc = 0.5 A, Irms = 0.25 A) O (Im = = 0.8 A, Idc = 0.35A, Irms = 0.45 A) D (Im = 0.7 A, Idc = 0.45 A, Irms = 0.35 A)
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- A full-wave bridge rectifier is constructed using 4 germanium diodes, each with a forward voltage drop of 0.3 V. The rectified waveform is described by the function vout(0) = V, sin 0 - 2 Vp where 8 = sin1 (2Vp/V,). Use integration to determine the exact average value of Vout for V, = 1.5, 2, 2.5, 3, 3.5, 4, and 4.5 V (using Excel or Matlab will speed up this process considerably). Then use the estimation formula (0.636 V, - 2 Vp) to determine the average value for each value of V, above and find the percent difference between the exact and estimated values for each V, value. At what value of V, does the percent error become greater than or equal to 5%? OVs = 2 V OVs - 1.5 V OVs - 2.5 V OVs - 3VThe resistance (k2) of a 1 cm pure silicon crystal.[n,-1x10" cm, He=1250 cm² V-'s', -350 cm2 V-ls] 390 39 347 34.7Q': Assuming the ideal diode model (VOn=0 V), calculate VoUT for the circuits in Figures (b) and (c) for the following input voltages (your answers should be in a table similar to that given below) 4k VOUT,b νουτε v20 v20 4k (b) (c) V1 V2 5
- Determine the voltage across the diode in Figure below, assuming silicon diode IM+ 5 V Μ 10 Ho 8 V -31 Ο 7V O ον Ο 4v ΟIn the given figure, each diode has a forward bias resistance of 309 and infinite resistance in reverse bias. The current I₁ will be: -> K 200 V 130 Ω www 130 Ω www 130 Ω www 20 Ω wwwThe resistance ( k2) of a 1 cm' pure silicon crystal.[n-1x1011 cm3, le-1250 cm? V-ls, h-350 cm? V-ls1] 39 34.7 390 347
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