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}.
From Equation (12.63), the minimum open-emitter junction breakdown voltage must be
B V_{C E O}=\frac{B V_{C B O}}{\sqrt[n]{\beta}} (12.63)
B V_{C B O}=\sqrt[n]{\beta} B V_{C E O}
Assuming the empirical constant n is 3 , we find
B V_{C B O}=\sqrt[3]{100}(15)=69.6 \mathrm{~V}
From Figure 7.15, the maximum collector doping concentration should be approximately 7 \times 10^{15} \mathrm{~cm}^{-3} 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.
