Question

Base thickness of modern bipolar transistors can be as small as
1. 1 μm
2. 0.1 μm
3. 10 nm
4. 10 A

Answer 1

Correct Answer - Option 3 : 10 nm

Explanation:

• Conductivity is increase with an increase in doping. So if we highly doped base region so more current passing to base terminal and output current ( collector current ) is reduced. Because of this reason base is lightly doped.
• Thus, more current passing to the collector side or an output side. Also, we know only 5 % of emitter current transfer to base and 95 % current passing to the collector.
• Base region doping of the transistor is one of the major factors in deciding the amplification. If the doping in the base region will be high than recombination rate also will be high in the base region, if recombination rate is high than base current will also be more, But for amplification, we need base current as small as possible. That’s why in all the methods we always reduce the base current like less doping of the base region, the high resistance value of base resister and high beta value.
• As per above, we saw base is lightly doped. The width of the region depends upon their doping. The base region is lightly doped so that width of the base is less and it is kept around 10 nm. Collector region of a transistor is highly doped so collector region has more width than other two.

BJT:

• BJT's are current-driven devices.
• The current through the two terminals is controlled by a current at the third terminal (base).
• It is a bipolar device (current conduction by both types of carriers, i.e. majority and minority electrons and holes)
• It has a low input impedance.