
When the temperature changes, the operating point is shifted to:
A) Change in ${I_{CBO}}$
B) Change in ${V_{CC}}$
C) Change in the value of circuit resistance
D) None of the above
Answer
233.1k+ views
Hint: Keep in mind, with the change in temperature resistivity of the material also changes. As the resistivity changes, the resistance also changes. Change in resistance leads to the change in current.
Complete step by step solution:
${I_ {CBO}} $ is abbreviated as reversed leaked Current between Collector and Base while the emitter is open.
Therefore, we can say that ${I_ {CBO}} $ is the current obtained from the collector when the junction is reverse biased and the circuit is open. In this, the current goes from collector to base.
${I_ {CBO}} $ Can be calculated as
${I_ {CBO}} = \dfrac{{{I_{CEO}}}}{{1 + \beta}}$
Here, ${I_ {CEO}} $ is the collector-emitter current and is abbreviated as Reverse leakage Current between Collector and Emitter while the base is Open.
Therefore, we can say that ${I_ {CEO}} $ is the current obtained from the collector when the junction is reverse biased and the circuit is open.
As we all know, in a transistor, ${I_ {CBO}}$ is the current which flows due to minority charge carriers. So, the increase in temperature leads to an increase in minority charge carriers.
Hence, from here, we can say that, with the change in temperature ${I_ {CBO}} $ changes.
Therefore, when the temperature changes, the operating point is shifted due to change in ${I_ {CBO}}$.
Therefore, option (A) is the correct option.
Addition Information:
In addition to the reverse current through the junction, there exists a leakage current that flows around the junction and across the surfaces. New carriers may be generated by collision in the collector junction transition region.
Note: Now you might get confused about ${I_ {CBO}} $ and ${I_ {CEO}} $, as both are quite similar. Just remember the parameter contains $CBO$ and $CEO$. Here $CBO$ is the collector- base current when the circuit is open and $CEO$ is the collector-emitter current when the circuit is open. So, both are different parameters.
Complete step by step solution:
${I_ {CBO}} $ is abbreviated as reversed leaked Current between Collector and Base while the emitter is open.
Therefore, we can say that ${I_ {CBO}} $ is the current obtained from the collector when the junction is reverse biased and the circuit is open. In this, the current goes from collector to base.
${I_ {CBO}} $ Can be calculated as
${I_ {CBO}} = \dfrac{{{I_{CEO}}}}{{1 + \beta}}$
Here, ${I_ {CEO}} $ is the collector-emitter current and is abbreviated as Reverse leakage Current between Collector and Emitter while the base is Open.
Therefore, we can say that ${I_ {CEO}} $ is the current obtained from the collector when the junction is reverse biased and the circuit is open.
As we all know, in a transistor, ${I_ {CBO}}$ is the current which flows due to minority charge carriers. So, the increase in temperature leads to an increase in minority charge carriers.
Hence, from here, we can say that, with the change in temperature ${I_ {CBO}} $ changes.
Therefore, when the temperature changes, the operating point is shifted due to change in ${I_ {CBO}}$.
Therefore, option (A) is the correct option.
Addition Information:
In addition to the reverse current through the junction, there exists a leakage current that flows around the junction and across the surfaces. New carriers may be generated by collision in the collector junction transition region.
Note: Now you might get confused about ${I_ {CBO}} $ and ${I_ {CEO}} $, as both are quite similar. Just remember the parameter contains $CBO$ and $CEO$. Here $CBO$ is the collector- base current when the circuit is open and $CEO$ is the collector-emitter current when the circuit is open. So, both are different parameters.
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