Step2. Ib is very small.

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01:21:48

Vre is (V-Vbe)/(Rb+Re) because Ib is very small. In this circuit, Vre becomes (5-0.7)/(1000+2000)=1.43mA. It seems that the transistor boost Ib to Ie as like as few hundreds times. And when the Re is very big, the Ve approaches the V.

V ... Supply Voltage for base of transister.
Vb ... The base voltage.
Ve ... The emitter voltage.
Vre ... The voltage drops at Re.
Vbe ... Voltage from Ve to Vb.
Rb ... The resister connected base.
Re ... The resister connected emitter.
Ib ... The current into base.
Ie ... The current from emitter.

published 6 years ago

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