Field Effect Transistors Questions and Answers

Answer: C

Applying the resistive loss law:

\[ P = I^{2}R = 100 \times 0.02 = 2 \ \mathrm{W} \]

giving 2 watts, which needs a heatsink in most enclosures. Note that the loss is quadratic in current: at twenty amps it would be eight watts rather than four. A bipolar device with a one-volt saturation drop would dissipate ten watts at ten amps but only twenty at twenty, so its loss grows more slowly - which is why the two families cross over as the current rises, and why the crossing point has to be computed rather than assumed.

Answer: D

The loss goes as the square of the current:

\[ P = 400 \times 0.02 = 8 \ \mathrm{W} \]

so 8 watts, four times the original rather than twice. Four watts would be the answer if the loss were proportional to current, which is the bipolar saturation-voltage case rather than this one. The order of a loss law decides how a design scales, and a quadratic law means headroom disappears far faster than intuition suggests as the current rating is approached.

Answer: A

Each device carries half the current through the same resistance, so each dissipates a quarter as much, and two of them together dissipate half the original figure. That the saving is a factor of two rather than four is the point worth being careful about: the quadratic law helps, but there are two devices sharing. Leaving the loss unchanged would be true of a constant-voltage-drop device instead, which is another way the two families differ. And self-correcting current sharing is what makes the arrangement safe to attempt in the first place.

Answer: B

A high-voltage MOSFET needs a thick drift region, and its on-resistance rises steeply with the voltage rating, so the quadratic loss law becomes punitive - whereas a bipolar-output device keeps a roughly constant drop of a couple of volts. Its gate charge is comparable or larger, it switches more slowly rather than faster, and its gate is a MOSFET gate so the input impedance is the same. Combining a MOSFET input with a bipolar output takes the voltage-driven gate and the constant conducting drop together, which is exactly the compromise high-voltage switching wants.