A p-n junction when forward biased has a drop of `0.6 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 `1.5 V` battery to forward bias the diode, what should be the value of resistor used in series with the diode so that the maximum current does not exceed 6 mA?
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Here, voltage drop across p-n junction, `=0.6V`, E.M.F. of battery used `= 1.5V` let R be the maximum value of resistance R used in sereis with the p-n junction so that the maximum current is 6 mA. Using Ohm's law, we get, `IR+0.6=1.5` or `IR=1.5-0.6=0.9` or `R=0.9/I=0.9/(6xx10^(-3))=150Omega`
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