How to prove (1) ID ? (2). Io (max) =cW² Up ot + 1/2 • I₂ (1+3* y) ? = 2/ ip Dic (a) www +1₁ Vo (4.31) iDay = 1₁ (1+√ √
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How to prove (1) ID ? (2). Io (max) =cW² Up ot + 1/2 • I₂ (1+3* y) ? = 2/ ip Dic (a) www +1₁ Vo (4.31) iDay = 1₁ (1+√ √
How to prove (1) ID ? (2). Io (max) =cW² Up ot + 1/2 • I₂ (1+3* y) ? = 2/ ip Dic (a) www +1₁ Vo
(4.31) iDay = 1₁ (1+√ √2V₁/V₂) To select an appropriate diode, we must find the peak value of the diode current. We do this by evaluating Eq. (4.23) at the onset of conduction (t=-At where t=0 is at the peak) and using Eq. (4.24): (4.32) iDmax=1₂ (1+27 √2V₂/V₁)