Electronic Devices - BJT Devices - Discussion

Discussion Forum : BJT Devices - General Questions (Q.No. 1)
1.
How much is the base-to-emitter voltage of a transistor in the "on" state?
0 V
0.7 V
0.7 mV
Undefined
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
10 comments Page 1 of 1.

Pooja thakur said:   8 years ago
In active mode BE junction in forward bias to overcome this Barrier potential we need to apply voltage greater than 0.6.

Pavankalyan said:   9 years ago
The transistor is in on state above 0.7v need to overcome the potential barrier.

Madhu said:   10 years ago
Has silicon breakdown 0.7 in forward bias so based on this.

Deva said:   1 decade ago
Very simple. If we provide the voltage just exactly above the 0.7v (for p-n diode) then it will operate as it designed.

Ramesh R said:   1 decade ago
For a transistor to be in ON state it need to overcome its barrier potential which is 0.7V.

Saif said:   1 decade ago
Because always from base to emitter its working is like silicon diode so it hold 0.7v.

Neelaji said:   1 decade ago
In active mode base-emitter junction should be in forward biased so, to over come the barrier potential we have to provide voltage >0.6v

Subhajit ghosh said:   1 decade ago
To be in on state both BE junction and CE junction has to be forward biased.since cut in voltage of a diode is 0.7 v to be in forward biased, BE junction, which is also a pn junction, applied voltage should be greater than 0.7 v.

Dhandapani said:   1 decade ago
The answer is 0.7mv

Sri said:   1 decade ago
because in active mode base-emitter junction in forward biased so, to over come the barrier potential we have to provide voltage >0.6v

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