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An NPN transistor is fabricated such that the collector has a uniform doping of 5 x 10¹⁵

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Step 2

Given:

An NPN Transistor is fabricated such that the collector has a uniform doping of 5 x 10¹⁵cm^-3

The emitter and base doping profiles are given by

$N_{d E} = 10^{20} e^{- x / 0.106} c m^{- 3}$

$N_{a B} = 4 \times 10^{18} e^{- x / 0.19} c m^{- 3}$

Step 3

A)
Find where $N_{d E} (x) = N_{a B} (x) = 5 \times 10^{15} c m^{- 3}$ in order to obtain the first junction $N_{d E} = 10^{20} e^{- x / 0.106} c m^{- 3}$

We know, $5 \times 10^{15} c m^{- 3} = 10^{20} e^{- x / 0.106} c m^{- 3}$

Solving above equation, we get,

$x_{1} = 0.77 μ m$ .

Similarly, to obtain the second junction equate N_aB(x) to the background concentration $N_{a B} = 4 \times 10^{18} e^{- x / 0.19} c m^{- 3}$

Again, we know, $5 \times 10^{15} c m^{- 3} = 4 \times 10^{18} e^{- x / 0.19} c m^{- 3}$

Solving above equation, we get,

$x_{2} = 1.27 μ m$ .

Therefore the base width is $Δ x = x_{2} - x_{1} = 0.5 μ m$ .

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An NPN transistor is fabricated such that the collector has a uniform doping of 5 x 1015