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> > > Prob. 3.21(b) that someone asked about today's class can be solved as following: > > - Case 1: D1 ON and D2 ON > > V_out = V_D,on for Iin>0 (Notice there is no current through R1) > > - Case 2: D1 OFF and D2 ON > > V_out = D_D,on for Iin>0 > > (This case is identical to Case 1) > > - Case 3: D1 ON, D2 OFF > > Vout = Iin x R + V_D,on (a) > > For D2 off, Vout < V_D,on (b) > > From (a) and (b), Iin < 0 but this cannot be true since D1 is On > > ==> This case can not be true > > - Case 4, D1 OFF, D2 OFF > > Iin cannot flow since it has no current path. Or Vout cannot be defined. > > Considering above four cases, only Case 1 is possible. > > Iin<0 cannot be possible since there is no current path. > > In short, this is a bad problem. > > A better problem would be adding a resistor (R2) in parallel with the current source. Then, > > Vout = V_D,on when Iin > VD,on / R2 or > > Vout = Iin x R2 when Iin < VD,on /R2 >
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