Field Effect Transistors Questions and Answers

Answer: A

The first two are correct and the third is essentially false: the oxide is an insulator and the gate current is immeasurably small, which is both the device's strength and the reason static charge accumulates on it. A junction field effect transistor's gate does leak a little, since it is a reverse-biased junction, and confusing the two families on this point is the trap. A limit that survives making every external component ideal is an intrinsic one, and worth knowing for that reason.

Answer: B

The pairing is A-P, B-Q, C-R, exactly parallel to common emitter, common base and common collector in the bipolar family. Common source inverts and gives the gain, common gate offers a low input impedance and a wide bandwidth because no inverting gain bridges its terminals, and common drain buffers at unity gain. That the correspondence holds property for property means every intuition about impedance reflection and the Miller effect carries straight across, which halves the work of learning the second family.

Answer: C

The pairing is A-P, B-Q, C-R. Each parameter presupposes a structure: a channel that can be pinched off, an inversion layer that appears above a threshold, or a base current that is amplified. Recognising which structure a parameter assumes is a quick way to place an unfamiliar specification in the right family.

Answer: D

The first three are correctly paired and the fourth is not, so three match. Static damage is a field effect weakness rather than a strength, and follows directly from the insulated gate that gives the family its high input impedance. That the same structural feature produces the greatest advantage and the greatest fragility is worth holding together.

Answer: A

Reading the table, the field effect device loses on gain by twenty to one but wins on input impedance by two thousand, so it belongs wherever the source must not be loaded. The other three applications all favour the bipolar column. A comparison table becomes a selection guide once the application names which row is binding.

Answer: B

This is the pinch-off voltage. Threshold is the enhancement MOSFET's equivalent with the opposite sense, breakdown is a destructive limit, and saturation describes the region beyond pinch-off rather than a voltage. Four terms for turning points in a characteristic, and attaching each to its own device and mechanism is what keeps them from being interchanged.

Answer: C

Current gain has no meaning for a device whose input current is zero - there is nothing to take a ratio to. The other three are standard field effect specifications describing the transfer characteristic, its slope and the conducting channel. A ratio whose denominator is zero is a reliable sign that a parameter has been carried across from the wrong family.

Answer: D

Applying the enhancement-mode square law to the overdrive voltage:

\[ \begin{aligned} I_{D} &= k(V_{GS}-V_{T})^{2} \\ &= 0.5(3)^{2} = 4.5 \ \mathrm{mA} \end{aligned} \]

giving 4.5 mA. Using the full five volts rather than the three volts of overdrive gives 12.5 mA, which is the standard error. Subtracting the threshold before squaring is the one step that matters, and it is easy to overlook when the numbers are convenient.

Answer: A

Differentiating the square law at the operating point:

\[ g_{m} = 2k(V_{GS}-V_{T}) = 2(0.5)(3) = 3 \ \mathrm{mS} \]

giving 3 millisiemens. Because the transconductance is the derivative of a square, it is linear in the overdrive voltage while the current is quadratic in it - so doubling the overdrive doubles the transconductance but quadruples the current. Keeping track of which quantity is which power avoids applying the wrong factor.

Answer: B

Multiplying transconductance by load:

\[ A_{v} = -0.003 \times 3300 \approx -10 \]

giving about -10. A bipolar stage at the same 4.5 milliamps would give a transconductance of 180 millisiemens and a gain of nearly 600, which puts the family difference in perspective. Working the same load through both device types is the quickest way to feel how large that difference is.