The junction diode in the following circuit requires a minimum current of `1 mA` to be above the knee point `(0.7 V)` of its `I-V` characterstic curve. The voltage across the diode is independent of current above the knee point. If `V_(B)=5 V`, then the maximum value of `R` so that the voltage is above the knee point, will be

Explanation

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 Here the knee point voltage has been replaced by a cell connected in reverse bias.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`