Determine the increase in pE0 in the emitter due to bandgap narrowing. Consider a silicon emitter at T = 300 K. Assume the emitter doping increases from 10^18 to 10^19 cm^-3. Determine the new value of pE0′ and determine the ratio pE0′ /pE0.

Question

Determine the increase in [latex]p_{ ext {E0 }}[/latex] in the emitter due to bandgap narrowing.

Consider a silicon emitter at [latex]T=300 \mathrm{~K}[/latex]. Assume the emitter doping increases from [latex]10^{18}[/latex] to [latex]10^{19} \mathrm{~cm}^{-3}[/latex]. Determine the new value of [latex]p_{\mathrm{E} 0}^{\prime}[/latex] and determine the ratio [latex]p_{\mathrm{E} 0}^{\prime} / p_{\mathrm{E} 0}[/latex].

Accepted Answer

For emitter doping concentrations of [latex]N_{E}=10^{18}[/latex] and [latex]10^{19} \mathrm{~cm}^{-3}[/latex], we have, neglecting bandgap narrowing,

[latex]p_{E 0}=\frac{n_{i}^{2}}{N_{E}}=\frac{\left(1.5 imes 10^{10} ight)^{2}}{10^{18}}=2.25 imes 10^{2} \mathrm{~cm}^{-3}[/latex]

and

[latex]p_{E O}=\frac{n_{i}^{2}}{N_{E}}=\frac{\left(1.5 imes 10^{10} ight)^{2}}{10^{19}}=2.25 imes 10^{1} \mathrm{~cm}^{-3}[/latex]

Taking into account the bandgap narrowing effect shown in Figure 12.26, we obtain, respectively, for [latex]N_{E}=10^{18}[/latex] and [latex]10^{19} \mathrm{~cm}^{-3}[/latex],

[latex]p_{\mathrm{EO}}^{\prime}=\frac{n_{i}^{2}}{N_{E}} \exp \left(\frac{\Delta E_{g}}{k T} ight)=\frac{\left(1.5 imes 10^{10} ight)^{2}}{10^{18}} \exp \left(\frac{0.020}{0.0259} ight)=4.87 imes 10^{2} \mathrm{~cm}^{-3}[/latex]

and

[latex]p_{E 0}^{\prime}=\frac{\left(1.5 imes 10^{10} ight)^{2}}{10^{19}} \exp \left(\frac{0.080}{0.0259} ight)=4.94 imes 10^{2} \mathrm{~cm}^{-3}[/latex]

Taking the ratio of [latex]p_{\mathrm{EO}}^{\prime} / p_{\mathrm{EO}}[/latex] for [latex]N_{E}=10^{18} \mathrm{~cm}^{-3}[/latex], we find

[latex]\frac{p_{E 0}^{\prime}}{p_{E 0}}=\exp \left(\frac{\Delta E_{g}}{k T} ight)=\exp \left(\frac{0.020}{0.0259} ight)=2.16[/latex]

and for [latex]N_{E}=10^{19} \mathrm{~cm}^{-3}[/latex], we find

[latex]\frac{p_{E 0}^{\prime}}{p_{E 0}}=\exp \left(\frac{0.080}{0.0259} ight)=21.95[/latex]

Comment

If the emitter doping concentration increases from [latex]10^{18}[/latex] to [latex]10^{19} \mathrm{~cm}^{-3}[/latex], the thermal-equilibrium minority carrier concentration actually increases rather than decreasing by a factor of 10 as would be expected.

Question 12.9: Determine the increase in pE0 in the emitter due to bandgap ...

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