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GATE ECE 1995 | Question-20

The break down voltage of a transistor with its base open is $\mathrm{BV}_{\mathrm{CEO}}$ and that with emitter open is $\mathrm{BV}_{\mathrm{CBO}^{‘}}$ then

  1. $\mathrm{BV}_{\mathrm{CEO}}=\mathrm{BV}_{\mathrm{CBO}}$
  2. $\mathrm{BV}_{\mathrm{CEO}}>\mathrm{BV}_{\mathrm{CBO}}$
  3. $\mathrm{BV}_{\text {CEO }}<\mathrm{BV}_{\mathrm{CBO}}$
  4. $\mathrm{BV}_{\mathrm{CEO}}$ is not related to $\mathrm{BV}_{\mathrm{CBO}}$

     

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