Field Effect Transistors Questions and Answers

Q: Why does a field effect transistor's bias need to accommodate a wide parameter spread?

Both the saturation current and the pinch-off voltage vary two or three to one across a part number, so the operating point must be tolerant of a whole area of the transfer characteristic rather than a point. Thresholds are stable in service, transconductance does not follow the supply, and gate leakage is negligible either way. The spread is if anything worse than beta's, which is why graphical design against a maximum and minimum curve is conventional here. Working one numerical case through both cases side by side fixes the distinction better than any rule of thumb.

Q: How is that spread handled graphically?

Drawing the self-bias line across both extreme transfer curves shows the range of operating points the design will actually produce, and the source resistor is chosen so that range is acceptable. Averaging the curves hides exactly the information wanted. The output load line answers a different question. Designing against a range rather than a nominal value is what makes a circuit manufacturable, and here the graphical method makes the range visible. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Q: Which bias arrangement gives the tightest control of drain current?

A current source in the source lead fixes the drain current outright, leaving the gate voltage to settle wherever the device requires - which removes the parameter spread from the answer entirely. Self bias and drain feedback both reduce the spread without eliminating it, and a divider with no source resistor is the least stable of all. Fixing the quantity you care about directly, and letting the device choose the rest, is the most robust approach available.

Q: Why does a MOSFET make a good analogue switch?

The channel is a genuine resistance, conducting either way with no offset voltage, and the drain and source are interchangeable - which is exactly what passing an analogue signal requires, and precisely what a bipolar transistor with its junction drop cannot offer. The threshold does drift with temperature, and the gate draws no current at all. A switch that introduces no voltage of its own is the property that matters here. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

Q: What is the chief non-ideality of a MOSFET analogue switch?

The gate-source voltage, and hence the channel resistance, changes as the signal itself moves the source potential, so the on-resistance is signal-dependent and the switch distorts. There is no offset voltage, conduction is bidirectional, and the gate needs no current. The remedy is a complementary pair in parallel, whose resistances vary oppositely and largely cancel, which is why analogue switches are built that way. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Q: How does a complementary switch improve on a single device?

An n-channel and a p-channel device in parallel have on-resistances that rise and fall in opposite directions as the signal moves, so the parallel combination is far flatter. Halving the resistance is a side effect. Both devices conduct both polarities. Placing two components in parallel so that their opposite errors cancel is a general technique, and it is why a transmission gate is drawn with two devices rather than one.

Q: Why is a MOSFET preferred to a bipolar transistor in power switching?

A MOSFET is a majority-carrier device with no minority charge to sweep out, so it switches far faster and has no storage delay. Its conducting drop can exceed a bipolar device's saturation voltage at high current, since the channel is a resistance. It has no current gain in the bipolar sense, and its safe operating area is better in some respects rather than all. Speed rather than conduction loss is what won it the field.

Q: What does a MOSFET's on-resistance mean for its conduction loss?

A resistance dissipates current squared times resistance, so doubling the current quadruples the loss - unlike a bipolar device's roughly constant saturation voltage, whose loss is proportional to current. That difference decides which device wins at a given current: MOSFETs at moderate currents and low voltages, bipolar or insulated-gate devices where the current is very high. A loss law of a different order changes the answer as the operating point moves.

Q: What is gate charge a measure of?

Gate charge is the charge a driver must deliver to turn the device on, and it decides the drive current needed at a given switching frequency. It is not leakage, not channel charge, and not depletion charge. A device with no steady gate current still needs substantial pulses of it, which is why gate drivers exist and why switching loss rises with frequency even in a device that draws nothing when static. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Q: Why does a MOSFET's switching loss rise with frequency?

Each transition moves the gate charge, and that energy is dissipated in the driver and in the device, so the total is proportional to frequency. On-resistance and threshold do not depend on frequency, and the channel current is set by the load. That a static device with no steady input current still costs power to switch is the essential economics of every switching converter and every logic gate. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

41.

Why does a field effect transistor's bias need to accommodate a wide parameter spread?

A.Pinch-off voltage and saturation current both vary widely
B.Its threshold voltage drifts substantially in service
C.Its transconductance depends on the supply voltage used
D.Its gate leakage varies by orders of magnitude between devices

Answer: A

Both the saturation current and the pinch-off voltage vary two or three to one across a part number, so the operating point must be tolerant of a whole area of the transfer characteristic rather than a point. Thresholds are stable in service, transconductance does not follow the supply, and gate leakage is negligible either way. The spread is if anything worse than beta's, which is why graphical design against a maximum and minimum curve is conventional here. Working one numerical case through both cases side by side fixes the distinction better than any rule of thumb.

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42.

How is that spread handled graphically?

A.The transconductance is computed at the nominal operating point
B.A bias line is drawn across the maximum and minimum transfer curves
C.The average of the two extreme curves is used for the design
D.The load line is drawn on the output characteristics instead

Answer: B

Drawing the self-bias line across both extreme transfer curves shows the range of operating points the design will actually produce, and the source resistor is chosen so that range is acceptable. Averaging the curves hides exactly the information wanted. The output load line answers a different question. Designing against a range rather than a nominal value is what makes a circuit manufacturable, and here the graphical method makes the range visible. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Discuss in Forum

43.

Which bias arrangement gives the tightest control of drain current?

A.Drain feedback with a large gate resistor
B.A divider on the gate with no source resistor
C.A current source in the source lead
D.Self bias with a carefully chosen source resistor

Answer: C

A current source in the source lead fixes the drain current outright, leaving the gate voltage to settle wherever the device requires - which removes the parameter spread from the answer entirely. Self bias and drain feedback both reduce the spread without eliminating it, and a divider with no source resistor is the least stable of all. Fixing the quantity you care about directly, and letting the device choose the rest, is the most robust approach available.

Discuss in Forum

44.

Why does a MOSFET make a good analogue switch?

A.Its threshold voltage is independent of temperature
B.Its gate draws a small current that biases the channel
C.Its drain and source cannot be interchanged in the circuit
D.Its channel resistance is symmetrical and low when on

Answer: D

The channel is a genuine resistance, conducting either way with no offset voltage, and the drain and source are interchangeable - which is exactly what passing an analogue signal requires, and precisely what a bipolar transistor with its junction drop cannot offer. The threshold does drift with temperature, and the gate draws no current at all. A switch that introduces no voltage of its own is the property that matters here. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

Discuss in Forum

45.

What is the chief non-ideality of a MOSFET analogue switch?

A.Its on-resistance varies with the signal voltage
B.It introduces a fixed offset voltage in series with the signal
C.It conducts in only one direction through the channel
D.It requires a continuous gate current to stay closed

Answer: A

The gate-source voltage, and hence the channel resistance, changes as the signal itself moves the source potential, so the on-resistance is signal-dependent and the switch distorts. There is no offset voltage, conduction is bidirectional, and the gate needs no current. The remedy is a complementary pair in parallel, whose resistances vary oppositely and largely cancel, which is why analogue switches are built that way. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Discuss in Forum

46.

How does a complementary switch improve on a single device?

A.The pair needs only half the gate drive voltage of one device
B.The two on-resistances vary oppositely and partly cancel
C.The two devices share the current and halve the resistance
D.One device conducts positive signals and the other negative

Answer: B

An n-channel and a p-channel device in parallel have on-resistances that rise and fall in opposite directions as the signal moves, so the parallel combination is far flatter. Halving the resistance is a side effect. Both devices conduct both polarities. Placing two components in parallel so that their opposite errors cancel is a general technique, and it is why a transmission gate is drawn with two devices rather than one.

Discuss in Forum

47.

Why is a MOSFET preferred to a bipolar transistor in power switching?

A.A higher current gain at the switching frequencies used
B.A wider safe operating area at every voltage and current
C.No stored charge to remove at turn-off
D.A lower voltage drop when fully conducting at any current

Answer: C

A MOSFET is a majority-carrier device with no minority charge to sweep out, so it switches far faster and has no storage delay. Its conducting drop can exceed a bipolar device's saturation voltage at high current, since the channel is a resistance. It has no current gain in the bipolar sense, and its safe operating area is better in some respects rather than all. Speed rather than conduction loss is what won it the field.

Discuss in Forum

48.

What does a MOSFET's on-resistance mean for its conduction loss?

A.Loss rises in direct proportion to the current
B.Loss is independent of the current carried
C.Loss falls as the current rises towards the rating
D.Loss rises with the square of the current

Answer: D

A resistance dissipates current squared times resistance, so doubling the current quadruples the loss - unlike a bipolar device's roughly constant saturation voltage, whose loss is proportional to current. That difference decides which device wins at a given current: MOSFETs at moderate currents and low voltages, bipolar or insulated-gate devices where the current is very high. A loss law of a different order changes the answer as the operating point moves.

Discuss in Forum

49.

What is gate charge a measure of?

A.The charge needed to switch the device
B.The leakage charge lost through the oxide per second
C.The channel charge available to carry current
D.The charge stored in the drain depletion region

Answer: A

Gate charge is the charge a driver must deliver to turn the device on, and it decides the drive current needed at a given switching frequency. It is not leakage, not channel charge, and not depletion charge. A device with no steady gate current still needs substantial pulses of it, which is why gate drivers exist and why switching loss rises with frequency even in a device that draws nothing when static. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Discuss in Forum

50.

Why does a MOSFET's switching loss rise with frequency?

A.Its channel carries more current at higher frequencies
B.The gate must be charged and discharged each cycle
C.Its on-resistance rises as the frequency increases
D.Its threshold voltage falls as the frequency rises

Answer: B

Each transition moves the gate charge, and that energy is dissipated in the driver and in the device, so the total is proportional to frequency. On-resistance and threshold do not depend on frequency, and the channel current is set by the load. That a static device with no steady input current still costs power to switch is the essential economics of every switching converter and every logic gate. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Discuss in Forum

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