# In BJT Ie=Ib+Ic ; Ie-emitter current; Ib-base current; Ic-collector current

1. Apr 8, 2012

### kpraneethin00

I have three doubts in BJT:
1). in BJT Ie=Ib+Ic ; Ie-emitter current; Ib-base current; Ic-collector current
then i find in many circuits, collector current is given to load, if the emitter current is greater than the collector current, we can connect the emitter terminal to the load, to get more current...

2). why the collector region is large in BJT

3). why the collector region is moderately doped when compared to highly doped emitter

2. Apr 8, 2012

### Jony130

Sometimes we connect the load to emitter, Google emitter follower. But then we lose the voltage gain.
Also kept in mind that the emitter current is greater than the collector current just by a small amount of a base current. So for BJT that have a large current gain β the difference is not significant. Ic/Ie = β/(β+1) for β = 100 -->Ic/Ie = 100/101 = 0.990099009

3. Apr 8, 2012

### dlgoff

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/trans2.html

4. Apr 8, 2012

### yungman

You have to be careful in how the BJT works. Collector current IS CONTROLLED by the emitter current. You design the circuit to setup the emitter current to get the collector current. Collector has very high output impedance where emitter is very low output impedance. Collector is a controlled current source and emitter is a voltage source. Collector current is controlled by the Vbe drop. Yes, Ie=Ib+Ic, but

$$I_c=I_s e^{[\frac{V_{BE}}{V_T}]}$$

Emitter being a voltage source, the current is govern by the voltage drop across the resistor connects to the emitter. Collector current reflects the emitter current.

Last edited: Apr 8, 2012