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Unformatted text preview: V OC A-B = V OC B-C = 10.5 V RMS = 14.8 V peak V OC A-C = 20.9 V RMS = 28.8 V peak In order for R test to dissipate 1 W, R tes t = V L 2 / P, R test = 390 V L RMS = 20.5 V To find R W : V A-C RMS V L RMS = 0.4 V drop across R TH , or 2*R W I = 20.5 / R test = 52.5 mA 2*R W = 0.4V / I R W = 3.8 Rectifier: Since we needed a V C of +24 V, we needed to use the full-wave bridge rectifier, as the center tapped can only produce a max of 14.8 V at V C . As a result, we are only using pins A and C of the transformer to generate the + 24V necessary at Vc. PIV = 28.8 V + 24 V = 52.4V V M = 28 V, V L = 23.8 V, R L = 390 , V r = 4.2 V, frequency = 120 Hz Therefore, we will de-rate to the 1N4935 diode, which has a PIV of 200 V, average current = 1 A and peak current = 30 A. C = 36 F, so we will de-rate to the 100 F capacitor....
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- Spring '07