- #1
Razvan
- 53
- 0
I would like to know how carriers flow in a bipolar junction transistor when working in saturation mode.
Normally, electrons would flow from the emitter to the base. Then, from the base, some recombine with holes nad leave, forming the base current, and the others (the majority) diffuse in the collector and are swept because of Vcc (forming the collector current). But when saturated, the collector base junction is forward biased, so some electrons should be moving from the collector to the base. Is this right? Or there are so many electrons in the base and because the base is so thin they diffuse into the collector and are still "attracted" by the "positive" potential of Vcc (even though the normal electric field between the base and the collector would push electrons the opposite way if it were for a wire instead of a pn junction)?
What is the correct flow of charge carriers when the transistor is saturated?
Normally, electrons would flow from the emitter to the base. Then, from the base, some recombine with holes nad leave, forming the base current, and the others (the majority) diffuse in the collector and are swept because of Vcc (forming the collector current). But when saturated, the collector base junction is forward biased, so some electrons should be moving from the collector to the base. Is this right? Or there are so many electrons in the base and because the base is so thin they diffuse into the collector and are still "attracted" by the "positive" potential of Vcc (even though the normal electric field between the base and the collector would push electrons the opposite way if it were for a wire instead of a pn junction)?
What is the correct flow of charge carriers when the transistor is saturated?