I know that, for a BJT, when VBE < VBE(ON) the transistor is in cut-off, and when VBE > VBE(ON) the transistor is either in linear of forward active depending on VCE, but what about when VBE = VBE(ON)?
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\$\begingroup\$ If VBE > VBE(ON), you have a dead transistor, whose base-emitter junction is open. Any VBE greater than about 1v (for silicon transistors) causes too much base current to flow, overheating it. \$\endgroup\$– glen_geekCommented Apr 14, 2017 at 7:16
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3 Answers
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For a BJT as you say if VBE < VBE(ON) the transistor is in cut-off, when you get to VBE = VBE(ON) the base and the emiter of the transistor behave like a diode, with the voltage drop of a diode (0.6-0.7V) it means you "can't" have more than VBE = VBE(ON), you only can have the voltage drop across the diode. If you try to increase the VBE you only increase the base current, and you can destroy the transistor.
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Well, the VBE(ON) you're talking about is specified in datasheets under specific conditions, you maybe are aware of. Cut-off does not mean that Ic=0A, it is typically in µA region or so.
Likewise, having VBE slightly lower/higher than VBE(ON) will only allow less/more current to flow through the collector.
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The voltage across base is a logarithmic function of the current. About 0.058 volts increase, per 10:1 current increase.
Expect 0.4v +-0.1v at 1uA.
Expect 0.058 * 3 (for 3 orders of magnitude) or nearly 0.6 volts at 1mA. Plus or minus.
Changes 2mv or 2.2 mv per degree Centigrade.
answered Apr 14, 2017 at 17:40