# Transistor base-emitter voltage needed for a current to flow

what base-emitter voltage is needed for a current to flow through the trasnsistor?

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what base-emitter voltage is needed for a current to flow through the trasnsistor?
One forward p-n junction drop, about 0.65 volts, as the b-e junction is just a forward biased diode. This varies with temperature and current. At high temp and low current, the Vbe may fall to 0.50 V, and may exceed 0.80 V at low temp and high current. But, 0.65 V is a good first order approximation to use as a starting point. I hope this helps.

Claude

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0.65 V is a good first order approximation to use as a starting point.
This was discussed in stewarcs's reference (see Base-Emitter Junction Details)

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CEL
In reality, any forward voltage will make current flow from base to emitter, but low voltages will originate very small currents.

In reality, any forward voltage will make current flow from base to emitter, but low voltages will originate very small currents.
No matter what answer I give, someone can always "one up" me and add something I omitted. So allow me to return the favor. Even with a base-emitter forward voltage drop of *zero* volts (b-e short-circuited), some collector current will indeed flow. This is "Iceo", and if the base-emitter junction is open, the current is "Icbo". These are leakage currents from the reverse biased collector to base and emitter terminals.

I interpreted the OP's question from a practical utilitarian sense that one uses a transistor to obtain power gain greater than unity. If a low powered signal source such as a microcontroller, transducer, etc. is used to actuate a higher powered device such as a solenoid, lamp, motor, etc., operating a bjt with Ib = 250 nA, & Vbe = 0.40 V is of no practical use. It was understood by me that the OP intended for the bjt to do something useful. If the bjt application is that of linear amplification, the dc bias point, or "quiescent point", is generally in the neighborhood of 0.10 to 10 mA, and 0.60 to 0.70 V.

Unless one is using the bjt as the logging element in a logarithmic amp/converter, Vbe = 0.65 V give or take a tad, twice that for a Darlington, is where it operates. When using a bjt as a switch or an amplifier, the 0.65 V value for the Vbe forward drop is a *darn good* first order approximation. Peace and BR.

Claude

Last edited:
CEL
No matter what answer I give, someone can always "one up" me and add something I omitted. So allow me to return the favor. Even with a base-emitter forward voltage drop of *zero* volts (b-e short-circuited), some collector current will indeed flow. This is "Iceo", and if the base-emitter junction is open, the current is "Icbo". These are leakage currents from the reverse biased collector to base and emitter terminals.

I interpreted the OP's question from a practical utilitarian sense that one uses a transistor to obtain power gain greater than unity. If a low powered signal source such as a microcontroller, transducer, etc. is used to actuate a higher powered device such as a solenoid, lamp, motor, etc., operating a bjt with Ib = 250 nA, & Vbe = 0.40 V is of no practical use. It was understood by me that the OP intended for the bjt to do something useful. If the bjt application is that of linear amplification, the dc bias point, or "quiescent point", is generally in the neighborhood of 0.10 to 10 mA, and 0.60 to 0.70 V.

Unless one is using the bjt as the logging element in a logarithmic amp/converter, Vbe = 0.65 V give or take a tad, twice that for a Darlington, is where it operates. When using a bjt as a switch or an amplifier, the 0.65 V value for the Vbe forward drop is a *darn good* first order approximation. Peace and BR.

Claude
I was not nitpicking your post. The 0.6 to 0.7 V figure is in any elementary book about transistors, so I don't think anyone would ask such a question in this forum. I supposed that the OP was interested in knowing if a very small voltage would produce a base current independently of any practical application.