A p-n jucntion when forward biased has a drop of 0.7 V which is assumed to be independent of current. The current in excess of 10 Ma through the diode produces a large joule heating effect which burns the diode. If we want to use 1.6 V battery to forward ibas the diode, the value of resistor used in series with the diode so that the maximum current does not exceed 6 mA should be

Options

A. `50Omega`
B. `200Omega`
C. `150Omega`
D. `250Omega`

Answer

C

Explanation

Here, voltage drop across p-n junction=0.7V, E.M.F. of battery used=1.6V,
Let R be the maximum value of resistance R used in series with the p-n junction so that the maximum current is 6 mA.
Using Ohm's law, we get
`lR+0.7=1.6`
or lR `1.6-0.7=0.9`
o `R=(0.9)/(l)=(0.9)/(6xx10^(-3))=150Omega`