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In a bipolar transistor at room temperature, if the emitter current is doubled the voltage across its base emitter junction will
- a)Double
- b)Halves
- c)Increases by about 20 mV
- d)decreases by about 20 mV
Correct answer is option 'C'. Can you explain this answer?
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In a bipolar transistor at room temperature, if the emitter current is...
In BJT if emitter current is doubled VBE increases by about 20 mV.
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Bipolar transistors are three-layer devices composed of a P-type region sandwiched between two N-type regions or vice versa. The three layers are called the emitter, base, and collector. The behavior of a bipolar transistor is determined by the flow of charge carriers (electrons or holes) across the different regions.
When a current flows through a bipolar transistor, it is primarily determined by the emitter current. The emitter current is the current that flows into the emitter terminal and is primarily composed of electrons or holes injected from the emitter region.
The base-emitter junction is a crucial part of the bipolar transistor. It acts as a diode, allowing current to flow in one direction but blocking it in the reverse direction. The voltage across the base-emitter junction determines the behavior of the transistor.
When the emitter current is doubled, it means that twice as many charge carriers are being injected into the transistor. This increased injection of charge carriers leads to an increase in the current flowing through the base-emitter junction.
Now, let's consider the behavior of the base-emitter junction. The base-emitter junction has a forward voltage drop, typically around 0.6 to 0.7 volts for silicon transistors. This voltage drop is necessary to overcome the barrier potential and allow current to flow.
As the current through the base-emitter junction increases, the voltage drop across it also increases. This is due to the relationship between current and voltage in a diode-like junction. The voltage across the base-emitter junction increases by about 20 mV for every tenfold increase in current.
Therefore, when the emitter current is doubled, the voltage across the base-emitter junction also increases. The increase is approximately 20 mV. This is because the increased injection of charge carriers leads to a higher current flowing through the junction, resulting in a higher voltage drop.
In conclusion, when the emitter current of a bipolar transistor is doubled, the voltage across its base-emitter junction increases by about 20 mV. This increase in voltage is a result of the increased current flowing through the junction.
When a current flows through a bipolar transistor, it is primarily determined by the emitter current. The emitter current is the current that flows into the emitter terminal and is primarily composed of electrons or holes injected from the emitter region.
The base-emitter junction is a crucial part of the bipolar transistor. It acts as a diode, allowing current to flow in one direction but blocking it in the reverse direction. The voltage across the base-emitter junction determines the behavior of the transistor.
When the emitter current is doubled, it means that twice as many charge carriers are being injected into the transistor. This increased injection of charge carriers leads to an increase in the current flowing through the base-emitter junction.
Now, let's consider the behavior of the base-emitter junction. The base-emitter junction has a forward voltage drop, typically around 0.6 to 0.7 volts for silicon transistors. This voltage drop is necessary to overcome the barrier potential and allow current to flow.
As the current through the base-emitter junction increases, the voltage drop across it also increases. This is due to the relationship between current and voltage in a diode-like junction. The voltage across the base-emitter junction increases by about 20 mV for every tenfold increase in current.
Therefore, when the emitter current is doubled, the voltage across the base-emitter junction also increases. The increase is approximately 20 mV. This is because the increased injection of charge carriers leads to a higher current flowing through the junction, resulting in a higher voltage drop.
In conclusion, when the emitter current of a bipolar transistor is doubled, the voltage across its base-emitter junction increases by about 20 mV. This increase in voltage is a result of the increased current flowing through the junction.
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In a bipolar transistor at room temperature, if the emitter current is doubled the voltage across its base emitter junction willa)Doubleb)Halvesc)Increases by about 20 mVd)decreases by about 20 mVCorrect answer is option 'C'. Can you explain this answer?
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