Calculate the change in collector current with a change in neutral base width , and estimate the Early voltage. Consider a uniformly doped silicon npn bipolar transistor with the following parameters: NB = 5 × 10^16 cm^-3 , NC = 2 × 10^15 cm^-3, xB0 = 0.70 μm, and DB = 25 cm^2 /s. Assume that xB0

Question

Calculate the change in collector current with a change in neutral base width, and estimate the Early voltage.

Consider a uniformly doped silicon npn bipolar transistor with the following parameters: [latex]N_{B}=5 imes 10^{16} \mathrm{~cm}^{-3}, N_{C}=2 imes 10^{15} \mathrm{~cm}^{-3}, x_{B 0}=0.70 \mu \mathrm{m}[/latex], and [latex]D_{B}=25 \mathrm{~cm}^{2} / \mathrm{s}[/latex]. Assume that [latex]x_{B 0} \ll L_{B}[/latex] and that [latex]V_{B E}=0.60 \mathrm{~V}[/latex]. The collector-base voltage is in the range [latex]2 \leq V_{C B} \leq 10 \mathrm{~V}[/latex].

Accepted Answer

Assuming [latex]x_{B 0} \ll L_{B}[/latex], the excess minority carrier electron concentration in the base can be approximated by Equation (12.15b), which is

[latex]\begin{aligned}\delta n_{B}(x) & \approx \frac{n_{B 0}}{x_{B}}\left\{\left[\exp \left(\frac{e V_{B E}}{k T} ight)-1 ight]\left(x_{B}-x ight)-x ight\} \end{aligned}[/latex] (12.15b)

The collector current is

[latex]\left|J_{C} ight|=e D_{B} \frac{d\left[\delta n_{B}(x) ight]}{d x} \cong \frac{e D_{B} n_{B 0}}{x_{B}} \exp \left(\frac{V_{B E}}{V_{t}} ight)[/latex]

The value of [latex]n_{B 0}[/latex] is found as

[latex]n_{B 0}=\frac{n_{i}^{2}}{N_{B}}=\frac{\left(1.5 imes 10^{10} ight)^{2}}{5 imes 10^{16}}=4.5 imes 10^{3} \mathrm{~cm}^{-3}[/latex]

For [latex]V_{C B}=2 \mathrm{~V}[/latex], we find (see the following Exercise Problem Ex 12.8)

[latex]x_{B}=x_{B 0}-x_{d B}=0.70-0.0518=0.6482 \mu \mathrm{m}[/latex]

and

[latex]\left|J_{C} ight|=\frac{\left(1.6 imes 10^{-19} ight)(25)\left(4.5 imes 10^{3} ight)}{0.6482 imes 10^{-4}} \exp \left(\frac{0.60}{0.0259} ight)=3.195 \mathrm{~A} / \mathrm{cm}^{2}[/latex]

For [latex]V_{C B}=10 \mathrm{~V}[/latex], we find (see the following Exercise Problem Ex 12.8)

[latex]x_{B}=0.70-0.103=0.597 \mu \mathrm{m}[/latex]

and

[latex]\left|J_{C} ight|=\frac{\left(1.6 imes 10^{-19} ight)(25)\left(4.5 imes 10^{3} ight)}{0.597 imes 10^{-4}} \exp \left(\frac{0.60}{0.0259} ight)=3.469 \mathrm{~A} / \mathrm{cm}^{2}[/latex]

We now can find, from Equation (12.45a)

[latex]\begin{aligned}\frac{d I_{C}}{d V_{C E}} \equiv g_{o}=& \frac{I_{C}}{V_{C E}+V_{A}}=\frac{1}{r_{o}} \ \end{aligned}[/latex] (12.45a)

[latex]\begin{aligned} \frac{d J_{C}}{d V_{C E}}=\frac{\Delta J_{C}}{\Delta V_{C B}}=\frac{J_{C}}{V_{C E}+V_{A}}=\frac{J_{C}}{V_{B E}+V_{C B}+V_{A}}\end{aligned}[/latex]

or

[latex]\frac{3.469-3.195}{8}=\frac{3.195}{0.60+2+V_{A}}[/latex]

The Early voltage is then determined to be

[latex]V_{A}=90.7 \mathrm{~V}[/latex]

Comment

This example indicates how much the collector current can change as the neutral base width changes with a change in the B-C space charge width, and it also indicates the magnitude of the Early voltage.

Question 12.8: Calculate the change in collector current with a change in n...

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