Electrical – Why does the base-emitter voltage in NPN transistor become constant

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schematic

The setup above shows the experiment that I have been working with. During the experiment, the current \$i_c\$ and the voltage \$V_{be}\$ were measured for increasing \$V_{in}\$. At some point, the current through the collector will be constant, due to the barrier in the forward-biased base-emitter region being lowered as much as possible by \$V_{in}\$, meaning that its maximum value follows Ohm’s law.

As we further increase \$V_{in}\$, we will find that \$V_{be}\$ will go towards a constant value. Why is that? Shouldn’t it increase too? I have great trouble understanding this last part.

I will appreciate any help!

EDIT: The polarity on the \$V_{in}\$ is wrong. My mistake.

Best Answer

The base-emitter voltage generally follows the Shockley diode equation with some series resistance. There is a small effect from the collector current.

The ideal diode portion of the voltage approximately increases proportional with the logarithm of the ratio of current to a constant(the “saturation current”). A 10:1 increase in current only results in about 60mV of increase, as shown in the linked Wiki page.