Design a bipolar transistor to meet a breakdown voltage specification.

Consider a silicon bipolar transistor with a common-emitter current gain of $\beta=100$ and a base doping concentration of $N_{B}=10^{17} \mathrm{~cm}^{-3}$ . The minimum open-base breakdown voltage is to be $15 \mathrm{~V}$ .

Question

Design a bipolar transistor to meet a breakdown voltage specification.

Consider a silicon bipolar transistor with a common-emitter current gain of [latex]\beta=100[/latex] and a base doping concentration of [latex]N_{B}=10^{17} \mathrm{~cm}^{-3}[/latex]. The minimum open-base breakdown voltage is to be [latex]15 \mathrm{~V}[/latex].

Accepted Answer

From Equation (12.63), the minimum open-emitter junction breakdown voltage must be

[latex]B V_{C E O}=\frac{B V_{C B O}}{\sqrt[n]{\beta}}[/latex] (12.63)

[latex]B V_{C B O}=\sqrt[n]{\beta} B V_{C E O}[/latex]

Assuming the empirical constant [latex]n[/latex] is 3 , we find

[latex]B V_{C B O}=\sqrt[3]{100}(15)=69.6 \mathrm{~V}[/latex]

From Figure [latex]7.15[/latex], the maximum collector doping concentration should be approximately [latex]7 imes 10^{15} \mathrm{~cm}^{-3}[/latex] to achieve this breakdown voltage.

Comment

In a transistor circuit, the transistor must be designed to operate under a worst-case situation. In this example, the transistor must be able to operate in an open-base configuration without going into breakdown. As we have determined previously, an increase in breakdown voltage can be achieved by decreasing the collector doping concentration.

Question 12.11: Design a bipolar transistor to meet a breakdown voltage spec...

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