Field Effect Transistors Questions and Answers

Q: What separates the gate from the channel in a MOSFET?

A MOSFET's gate sits on an insulating oxide, so no gate current flows at all, in either direction and at any voltage - unlike a junction field effect transistor, whose gate is a reverse-biased junction and does leak a little. That insulator is why the input resistance is measured in gigohms, and also why the device is destroyed by static discharge. One structural difference accounts for both the family's greatest strength and its greatest fragility.

Q: Why is a MOSFET vulnerable to static discharge?

The oxide is only tens of nanometres thick, so a few tens of volts across it punctures it irreversibly, and since the gate draws no current there is nothing to bleed the charge away either. The third option states a contributing fact rather than the mechanism, and the threshold and capacitance are beside the point. A structure that provides perfect isolation also provides no path for accumulated charge, which is the whole handling problem.

Q: What precaution protects a MOSFET input in service?

Integrated clamp diodes to the rails divert any excess charge before the oxide is stressed, and most MOSFETs are supplied with them built in. A series resistor limits the current after breakdown has begun rather than preventing it, and neither a capacitor nor a bead limits the voltage. Protecting against an overvoltage means providing a path at a safe voltage, which is what a clamp does and what the alternatives do not.

Q: What distinguishes an enhancement-mode MOSFET?

An enhancement device has no channel at zero gate voltage; the gate must exceed a threshold to induce one, which makes it a natural switch and the basis of all digital logic. A conducting channel at zero gate voltage describes a depletion device. Negative gate bias and falling current describe a p-channel or depletion device. Whether a device is normally on or normally off decides both its bias arrangement and the applications it suits.

Q: What does a field effect transistor's transfer characteristic plot?

The transfer characteristic shows drain current against gate voltage, and for a junction device it is the square-law curve from the pinch-off voltage up to the saturation current. Drain current against drain voltage is the output family. Gate current is essentially zero, so plotting it would be pointless. Two characteristics describing one device, one relating output to input and one relating output to output, is the same arrangement as for a bipolar transistor.

Q: What relation does a junction field effect transistor's transfer characteristic follow?

The relation is a square law: the current falls as the square of one minus the ratio of gate voltage to pinch-off voltage. The exponential law belongs to bipolar devices and is why their transconductance is so much larger. Neither linear nor logarithmic describes it. That one family is exponential and the other square-law accounts for nearly every performance difference between them.

Q: A device has \( I_{DSS} = 10 \ \mathrm{mA} \) and \( V_{P} = -4 \ \mathrm{V} \). At \( V_{GS} = -2 \ \mathrm{V} \) the drain current is

Applying the square law:\[ \begin{aligned} I_{D} &= 10 \left(1 - \tfrac{-2}{-4}\right)^{2} \\ &= 10 (0.5)^{2} = 2.5 \ \mathrm{mA} \end{aligned} \]giving 2.5 mA. Five milliamps would follow from a linear law, which is the commonest error: halfway to pinch-off gives a quarter of the current, not half. The squaring is what makes the relation between gate voltage and current so much gentler at the top of the range than at the bottom. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Q: For that device to carry \( 5 \ \mathrm{mA} \), the gate voltage must be about

Rearranging the square law for the gate voltage:\[ \begin{aligned} V_{GS} &= V_{P}\left(1 - \sqrt{\tfrac{5}{10}}\right) \\ &= -4(1-0.707) \approx -1.17 \ \mathrm{V} \end{aligned} \]so about -1.17 volts. Halving the current needs only about a quarter of the way to pinch-off, which is the square law working in reverse. Taking the square root rather than halving is the step this calculation turns on.

Q: What does a MOSFET's threshold voltage specify?

The threshold is the gate voltage at which an inversion layer appears and conduction begins, so it is the enhancement device's equivalent of the pinch-off voltage but with the opposite sense. Pinch-off describes where the channel closes at the drain end, saturation onset is a drain-voltage condition, and oxide breakdown is a destructive limit far higher up. Four voltages that all sound like turning points, and attaching each to its own mechanism is what keeps them apart.

Q: What does a field effect transistor's transconductance describe?

Transconductance is the slope of the transfer characteristic, in milliamps per volt or millisiemens, and it sets the voltage gain a given load resistor produces. Current per drain volt is the output conductance, gate charge is a switching parameter, and resistance per gate volt describes the ohmic region. Because it is a slope on the transfer curve, it depends on where on that curve the device is biased. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

11.

What separates the gate from the channel in a MOSFET?

A.A lightly doped drift region within the channel
B.A depletion region maintained by the drain voltage
C.A thin insulating oxide layer
D.A reverse-biased junction of the opposite type

Answer: C

A MOSFET's gate sits on an insulating oxide, so no gate current flows at all, in either direction and at any voltage - unlike a junction field effect transistor, whose gate is a reverse-biased junction and does leak a little. That insulator is why the input resistance is measured in gigohms, and also why the device is destroyed by static discharge. One structural difference accounts for both the family's greatest strength and its greatest fragility.

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12.

Why is a MOSFET vulnerable to static discharge?

A.Its channel carries no current until a threshold is reached
B.Its gate draws no current and so cannot dissipate charge
C.Its input capacitance is too small to absorb any energy
D.Its gate oxide is thin enough to be punctured

Answer: D

The oxide is only tens of nanometres thick, so a few tens of volts across it punctures it irreversibly, and since the gate draws no current there is nothing to bleed the charge away either. The third option states a contributing fact rather than the mechanism, and the threshold and capacitance are beside the point. A structure that provides perfect isolation also provides no path for accumulated charge, which is the whole handling problem.

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13.

What precaution protects a MOSFET input in service?

A.A pair of clamping diodes to the supply rails
B.A series resistor of a few hundred kilohms in the gate lead
C.A capacitor from gate to source to absorb transients
D.A ferrite bead in the gate lead to suppress high frequencies

Answer: A

Integrated clamp diodes to the rails divert any excess charge before the oxide is stressed, and most MOSFETs are supplied with them built in. A series resistor limits the current after breakdown has begun rather than preventing it, and neither a capacitor nor a bead limits the voltage. Protecting against an overvoltage means providing a path at a safe voltage, which is what a clamp does and what the alternatives do not.

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14.

What distinguishes an enhancement-mode MOSFET?

A.Its drain current falls as the gate voltage is increased
B.No channel exists until the gate exceeds a threshold
C.Its channel conducts fully with no gate voltage applied
D.Its gate must be biased negative with respect to the source

Answer: B

An enhancement device has no channel at zero gate voltage; the gate must exceed a threshold to induce one, which makes it a natural switch and the basis of all digital logic. A conducting channel at zero gate voltage describes a depletion device. Negative gate bias and falling current describe a p-channel or depletion device. Whether a device is normally on or normally off decides both its bias arrangement and the applications it suits.

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15.

What does a field effect transistor's transfer characteristic plot?

A.Gate current against gate-source voltage
B.Channel resistance against drain-source voltage
C.Drain current against gate-source voltage
D.Drain current against drain-source voltage

Answer: C

The transfer characteristic shows drain current against gate voltage, and for a junction device it is the square-law curve from the pinch-off voltage up to the saturation current. Drain current against drain voltage is the output family. Gate current is essentially zero, so plotting it would be pointless. Two characteristics describing one device, one relating output to input and one relating output to output, is the same arrangement as for a bipolar transistor.

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16.

What relation does a junction field effect transistor's transfer characteristic follow?

A.An exponential law like a bipolar transistor's
B.A linear law over the whole useful range
C.A logarithmic law above the threshold voltage
D.A square law between gate voltage and drain current

Answer: D

The relation is a square law: the current falls as the square of one minus the ratio of gate voltage to pinch-off voltage. The exponential law belongs to bipolar devices and is why their transconductance is so much larger. Neither linear nor logarithmic describes it. That one family is exponential and the other square-law accounts for nearly every performance difference between them.

Discuss in Forum

17.

A device has \( I_{DSS} = 10 \ \mathrm{mA} \) and \( V_{P} = -4 \ \mathrm{V} \). At \( V_{GS} = -2 \ \mathrm{V} \) the drain current is

A.2.5 mA
B.5 mA
C.10 mA
D.1.25 mA

Answer: A

Applying the square law:

\[ \begin{aligned} I_{D} &= 10 \left(1 - \tfrac{-2}{-4}\right)^{2} \\ &= 10 (0.5)^{2} = 2.5 \ \mathrm{mA} \end{aligned} \]

giving 2.5 mA. Five milliamps would follow from a linear law, which is the commonest error: halfway to pinch-off gives a quarter of the current, not half. The squaring is what makes the relation between gate voltage and current so much gentler at the top of the range than at the bottom. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

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18.

For that device to carry \( 5 \ \mathrm{mA} \), the gate voltage must be about

A.-0.5 V
B.-1.17 V
C.-2 V
D.-2.83 V

Answer: B

Rearranging the square law for the gate voltage:

\[ \begin{aligned} V_{GS} &= V_{P}\left(1 - \sqrt{\tfrac{5}{10}}\right) \\ &= -4(1-0.707) \approx -1.17 \ \mathrm{V} \end{aligned} \]

so about -1.17 volts. Halving the current needs only about a quarter of the way to pinch-off, which is the square law working in reverse. Taking the square root rather than halving is the step this calculation turns on.

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19.

What does a MOSFET's threshold voltage specify?

A.The drain voltage at which saturation begins
B.The gate voltage at which the oxide breaks down
C.The gate voltage at which a channel forms
D.The gate voltage at which the channel pinches off

Answer: C

The threshold is the gate voltage at which an inversion layer appears and conduction begins, so it is the enhancement device's equivalent of the pinch-off voltage but with the opposite sense. Pinch-off describes where the channel closes at the drain end, saturation onset is a drain-voltage condition, and oxide breakdown is a destructive limit far higher up. Four voltages that all sound like turning points, and attaching each to its own mechanism is what keeps them apart.

Discuss in Forum

20.

What does a field effect transistor's transconductance describe?

A.Drain current change per unit drain voltage change
B.Gate charge required per unit drain current delivered
C.Channel resistance change per unit gate voltage change
D.Drain current change per unit gate voltage change

Answer: D

Transconductance is the slope of the transfer characteristic, in milliamps per volt or millisiemens, and it sets the voltage gain a given load resistor produces. Current per drain volt is the output conductance, gate charge is a switching parameter, and resistance per gate volt describes the ohmic region. Because it is a slope on the transfer curve, it depends on where on that curve the device is biased. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

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