Do you think you could get answers quicker if you just pull out a breadboard, transistor and resistors?If the collector-base terminal of the bjt is reverse biased then how is the collector current positive? Also what is the voltage of Vbc at which the bjt becomes saturated?
Because a BJT transistor is not just two diodes glued together. This is a common misconception that people new to transistors have and it is a very reasonable misconception that is how the usual pictures make it look and, indeed, under specific conditions that is how it behaves. But when operating in the active mode, the base-emitter junction (which behaves very diode-like) allows charge carriers to enter the very thin base region. The presence of these charge carriers in the region effectively makes a channel that allows charge to flow from collector to emitter through the base region. That's a very simplistic description, but hopefully it gets the idea across.If the collector-base terminal of the bjt is reverse biased then how is the collector current positive? Also what is the voltage of Vbc at which the bjt becomes saturated?
If Vcesat = 0.2 V and Vbe = 0.7 V, then Vbc = 0.5 V (we are talking NPN silicon transistors here). Yes, this is "forward biased", but by so little that even if it were a diode the current would be very low. With a Vcesat of 0.2 V, the diode current in the base-collector junction would be more than three orders of magnitude less than the diode current in base-emitter junction. When you then add in the transistor current gain, even in saturation, any nominal base-collector current that is superimposed on the collector-emitter current is completely lost in the noise.But never enough to forward bias the B-C junction?
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