
The reverse saturation current of a silicon diode
A. Doubles for every \[\text{10 }\!\!{}^\circ\!\!\text{ C}\] increase in temperature.
B. Does not change with temperature.
C. Halves for every \[\text{1 }\!\!{}^\circ\!\!\text{ C}\] decrease in temperature.
D. Increases by 1.5 times for every \[\text{2 }\!\!{}^\circ\!\!\text{ C}\] increment in temperature.
Answer
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Hint: In a silicon diode, the current is produced due to the pair formation of charge carriers such as electrons and holes. A diode helps in the rectification of the current, that is, to direct the current to one side. Reverse saturation current of a diode is caused by the minority charge carriers of each region of the diode.
Complete step-by-step answer:
For creating a diode, two small crystals of a semiconductor are taken, e.g. Silicon. One crystal is doped with positive charge carriers (holes) and is termed the p-region and the other crystal is doped with negative charge carriers (electrons) and is termed the n-region. When these two crystals are brought together, some carriers combine and form a third region known as the depletion region. There are no carriers in the depletion region therefore it acts as a bridge between the two regions. Only when a higher electrical potential is applied on one side, the majority carriers of the side breaks the barrier and starts conducting.
The reverse saturation current in the diode is caused by the minority charge carriers of each region. Once the depletion region is reduced, a few electrons in the p-region flows to the n-region and few holes will move from n-region to p-region. Both movements will cause reverse saturation current. As temperature increases, the minority carriers in each region break out of the crystals and increase the reverse saturation current. The increase is estimated to be doubling the reverse current for a \[\text{10 }\!\!{}^\circ\!\!\text{ C}\] increase in temperature. Therefore, the correct answer is A.
Additional information:
A silicon crystal will have intrinsic charge carriers in itself. When we have impurities which are a trivalent or pentavalent compound to these silicon crystals, the process is called doping. Trivalent impurities will produce p-type semiconductor and n-type semiconductor is produced by pentavalent impurity. Even if we dope to form a p-type region, there will be few electrons in that region which forms the minority carriers of that region.
Ideal diode current equation can be written as,
\[I={{I}_{s}}\left[ {{e}^{\dfrac{qV}{\eta {{k}_{B}}T}}}-1 \right]\], where \[{{I}_{s}}\] is the reverse saturation current, \[q\] is the charge carrier, \[{{k}_{B}}\] is the Boltzmann constant, T is the temperature in Kelvin, V is the voltage across the diode and \[\eta \] is the ideality factor.
Note: Increase in electrical potential does not affect the reverse saturation voltage. The carrier lifetime \[(\tau )\] is reduced as temperature increases and this, in turn, increases the reverse saturation current. The same is true for germanium diodes, a 10°C increase in temperature doubles the reverse current. Doping produces extrinsic charge carriers.
Complete step-by-step answer:
For creating a diode, two small crystals of a semiconductor are taken, e.g. Silicon. One crystal is doped with positive charge carriers (holes) and is termed the p-region and the other crystal is doped with negative charge carriers (electrons) and is termed the n-region. When these two crystals are brought together, some carriers combine and form a third region known as the depletion region. There are no carriers in the depletion region therefore it acts as a bridge between the two regions. Only when a higher electrical potential is applied on one side, the majority carriers of the side breaks the barrier and starts conducting.
The reverse saturation current in the diode is caused by the minority charge carriers of each region. Once the depletion region is reduced, a few electrons in the p-region flows to the n-region and few holes will move from n-region to p-region. Both movements will cause reverse saturation current. As temperature increases, the minority carriers in each region break out of the crystals and increase the reverse saturation current. The increase is estimated to be doubling the reverse current for a \[\text{10 }\!\!{}^\circ\!\!\text{ C}\] increase in temperature. Therefore, the correct answer is A.
Additional information:
A silicon crystal will have intrinsic charge carriers in itself. When we have impurities which are a trivalent or pentavalent compound to these silicon crystals, the process is called doping. Trivalent impurities will produce p-type semiconductor and n-type semiconductor is produced by pentavalent impurity. Even if we dope to form a p-type region, there will be few electrons in that region which forms the minority carriers of that region.
Ideal diode current equation can be written as,
\[I={{I}_{s}}\left[ {{e}^{\dfrac{qV}{\eta {{k}_{B}}T}}}-1 \right]\], where \[{{I}_{s}}\] is the reverse saturation current, \[q\] is the charge carrier, \[{{k}_{B}}\] is the Boltzmann constant, T is the temperature in Kelvin, V is the voltage across the diode and \[\eta \] is the ideality factor.
Note: Increase in electrical potential does not affect the reverse saturation voltage. The carrier lifetime \[(\tau )\] is reduced as temperature increases and this, in turn, increases the reverse saturation current. The same is true for germanium diodes, a 10°C increase in temperature doubles the reverse current. Doping produces extrinsic charge carriers.
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