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Assume that silicon diode in the circuit Fig.requires minimum current of 1 mA to be above the knee point (0.7 V) of its I-V characteristics. Also assume that the voltage across the diode is independent of current above the knee point. If V_(B)=5V, what should be the minimum value of R so that the voltage is above the knee point? If V_(B)=5V, what should be the value of R to establish the current of 5mA in the circuit? What is the power dissipated in the resistance R and in the diode, when a current of 5mA flows in the circuit at V_(B)=6V. If R=1kOmega, what is the minimum voltage V_(B) required to keep the diode above the knee point? |
Answer» Solution :Here the DIODE is non-ideal diode. The equivalent circuit for it is shown in Fig. Here the knee point VOLTAGE has been replaced by a CELL connected in reverse bias.  Given, minimum current, `I_(min)=1mA=10^(-3)A`. If `R_(max)` is the maximum value of resistance used, then `R_(max)I_(min)=5V-0.7=4.3V` or `R_(max)=4.3/I_(min)=4.3/10^(-3)=4.3xx10^(3)OMEGA` Here, `V_(B)=5V`, knee voltage, `V_(k)=0.7V, I=5mA=5XX10^(-3)A`. From the equivalent circuit as given above `IR=V_(B)-0.7=5-0.7=4.3V` `R=4.3/I=4.3/(5xx10^(-3))=860Omega` Here, `V_=6V, I=5xx10^(-3)A` `R=(V_B-0.7)/I=(6-0.7)/(5xx10^(-3))=1060Omega` Power dissipated in resistance R is `=I^(2)R=(5xx10^(-3))^(2)xx1060` `=0.0265W=26.5mW` Power dissipated in diode is `=IV_(k)=(5xx10^(-3))xx0.7=3.5xx10^(-3)W` `=3.5mW` Here, `R=1kOmega=1xx10^(3)Omega`, `I_(min)=1mA=10^(-3)A` `V_(B)-0.7= RI_(min) or V_(B)=0.7+RI_(min)` `:. V_(B)=0.7+10^(3)xx10^(-3)=1.7V` |
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