Field Effect Transistors Questions and Answers

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.

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.

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.

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.

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.

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.

Answer: A

An exponential characteristic is far steeper than a square-law one, so a bipolar device gives about forty millisiemens per milliamp while a field effect device gives a few. The absence of gate current is the reason for the high input impedance, not the low transconductance. Channel resistance and capacitance affect other things. One difference in the underlying law produces the family's central trade: enormous input impedance bought with much less gain per milliamp.

Answer: B

The maximum transconductance follows from the square law's slope:

\[ \begin{aligned} g_{m0} &= \frac{2 I_{DSS}}{|V_{P}|} \\ &= \frac{16}{4} \\ &= 4 \ \mathrm{mS} \end{aligned} \]

giving 4 millisiemens, against the forty a bipolar device would give at eight milliamps. The factor of two in the numerator comes from differentiating the square, and omitting it is the usual slip here.

Answer: C

Transconductance falls linearly towards pinch-off:

\[ \begin{aligned} g_{m} &= g_{m0}\left(1 - \tfrac{-2}{-4}\right) \\ &= 4 \times 0.5 \\ &= 2 \ \mathrm{mS} \end{aligned} \]

so 2 millisiemens. Halving is right here because the transconductance is the derivative of a square and so is linear in gate voltage, even though the current itself falls to a quarter. Knowing which quantity is linear and which is squared avoids applying the wrong factor.

Answer: D

Beyond pinch-off the current is set by the gate and barely moves with drain voltage, which is exactly what a current source requires - a single junction field effect transistor with its gate tied to its source gives a two-terminal source of the saturation current. The absence of gate current is convenient but separate, the transconductance varies, and the ohmic region is the wrong one. A flat characteristic is a high output impedance, and that is all a current source is.