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In a reverse biased diode, when the applied voltage changes by `1V`, the current is found to change by `0.5muA`. The reversebiase resistance of the diode isA. `2xx10^5 Omega`B. `2xx10^6 Omega`C. `200 Omega`D. `2 Omega` |
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Answer» Correct Answer - B `R=(DeltaV)/(DeltaI)=1/(0.5xx1^(-6)) = 10^6/(1//2)=2xx10^6 Omega` |
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