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

[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
Please help me in this

 

[Jony130]

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...

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

 

[dlgoff]

2). why the collector region is large in BJT

3). why the collector region is moderately doped when compared to highly doped emitter
Please help me in this

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

 

[yungman]

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...

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.

 

[alphysicist]

I would like to address your doubts in BJT and provide some explanations.

1) In BJT, the equation Ie=Ib+Ic represents the principle of current conservation, where the total current entering the emitter (Ie) is divided into two parts - the base current (Ib) and the collector current (Ic). This is due to the fact that the base-emitter junction acts as a forward biased diode, allowing a small current (Ib) to flow from the base to the emitter. The majority of the current (Ic) flows from the collector to the emitter, making it the dominant current in most cases. Therefore, in circuits where the collector current is used to drive a load, it is often larger than the base current.

2) The size of the collector region in a BJT is typically larger than the base and emitter regions because it is responsible for collecting the majority of the current (Ic) flowing through the device. This region is designed to have a larger area to increase the chances of current carriers (electrons or holes) reaching the collector and reducing the chances of recombination. This is important for maintaining a high collector current and improving the overall performance of the BJT.

3) The collector region is moderately doped compared to the highly doped emitter region for several reasons. First, a moderate doping level allows for a good balance between conductivity and avoiding the formation of a reverse-biased p-n junction at the base-collector junction. This ensures that the collector region can efficiently collect the majority carriers without interfering with the operation of the device. Additionally, a moderate doping level also helps to reduce the chances of recombination in the collector region, which would decrease the overall current gain of the BJT.

I hope this helps to clarify your doubts and provides a better understanding of BJTs. Keep exploring and asking questions to deepen your understanding of this important electronic device.