Field Effect Transistors Questions and Answers

Q: A power MOSFET has \( R_{DS(on)} = 20 \ \mathrm{m\Omega} \) and carries \( 10 \ \mathrm{A} \). Its conduction loss is

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.

Q: Doubling that current changes the conduction loss to

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.

Q: Why is an insulated-gate bipolar device used instead of a MOSFET at high voltage?

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.

131.

A power MOSFET has \( R_{DS(on)} = 20 \ \mathrm{m\Omega} \) and carries \( 10 \ \mathrm{A} \). Its conduction loss is

A.20 W
B.0.02 W
C.2 W
D.0.2 W

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.

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132.

Doubling that current changes the conduction loss to

A.4 W
B.2 W
C.16 W
D.8 W

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.

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133.

How does paralleling two such devices affect the total conduction loss?

A.It halves the loss for the same total current
B.It leaves the loss unchanged, since the current divides
C.It doubles the loss
D.It quarters the loss, by halving both current and resistance

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.

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134.

Why is an insulated-gate bipolar device used instead of a MOSFET at high voltage?

A.Its input impedance is higher than a MOSFET's gate offers
B.Its conducting drop grows more slowly with current
C.Its gate charge is smaller than a MOSFET's of the same rating
D.It switches faster than a MOSFET at every voltage rating

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.

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Field Effect Transistors Questions and Answers

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Field Effect Transistors Questions and Answers

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