Bipolar Junction Transistor Base-Emitter voltage

Question

I've seen several datasheets for Bipolar Junction Transistors providing the value of the Base-Emitter voltage at saturation with values way above 0.6~0.7V (Example). What exactly does this entail? I had the notion that the voltage drop from base to emitter was similar to a forward biased diode, ~0.7V, independently of the Base-Emitter voltage.

Answer 1

This is a guaranteed maximum number, with a relatively heavy base and collector current (although only at 25°C). There is no typical given.

Compare these typical curves for the common MMBT4401 transistor:

At the Ic=500mA, Ib=50mA level, the \$V_{BE}\$ is typically more than 1 volt.

The manufacturer chooses to guarantee it will be less than 2V (compare to the '4401 where the guarantee is 1.2V). That's about all you can say about it.. it's probably closer to 1V than 2V typically, and certainly much less than 2V (and actually well under 1V) at more reasonable currents such as 10mA base current/100mA collector current.

It's not unusual for manufacturers to have loose guarantees.. leakage currents are often guaranteed at 1uA levels when actual typical leakage may be more like 1nA or 100fA. That speeds up the testing. In the case of the Vbe, it may reduce false fallouts due to bad connections in their test jigs or whatever.

Answer 2

The notion that "a forward conduction SI diode drops 0.7V" is a very rough estimate, which works quite well for LOW currents (compared to the diode's maximum). Look at this graph from an 1N4148 datasheet (the archetypical small-signal diode). At higher currents the voltage gets close to 2V. Your BE junction is behaving as a good approximation of a diode, it is your assumption of a diode's behavior that needs a some adjustment.