Questions on the MOSFET as a switch

Question 1

Explain why the gate of a MOSFET draws no steady current, and state what a logic output must supply in order to switch one.

Accepted Answer

The gate sits on a thin insulating oxide layer, out of contact with the silicon beneath, so it behaves as a small capacitor rather than as a conducting terminal and no steady current can enter it (1). The driver supplies only the brief charging current of that gate capacitance when the state changes (1).

Question 2

State what happens to the drain current of a MOSFET below its threshold voltage and above it.

Accepted Answer

Below the threshold no proper channel forms, so the drain current is negligible at any drain voltage within the device's rating (1). Above it the channel opens and widens, and provided the drain voltage is large enough to keep the channel saturated, the current climbs approximately with the square of the excess gate voltage above the threshold (1). Saying the current 'rises steadily' misses both points: a complete answer names the channel, and the shape is a square law, not a straight line. Both halves are the idealised model: below threshold a small leakage still flows, and the square law is the saturated behaviour, not the behaviour at every drain voltage.

Question 3

A zener diode is used to hold a supply rail steady. State how the diode is connected and what the series resistor does.

Accepted Answer

The zener goes across the load, reverse biased, so that it sits in breakdown at its rated voltage (1). The series resistor sits between the raw supply and the diode and drops whatever the supply has left over, which is what sets the current and lets the output stay put when the supply wanders (1). Without the series resistor the supply is shorted through the diode.

Question 4

State how a photodiode is biased in a light meter, and state how its photocurrent depends on the illumination.

Accepted Answer

It is used reverse biased, so that almost nothing flows in the dark and every absorbed photon adds to the reverse current (1). The photocurrent is proportional to the illumination, so the response is linear and one constant converts current to illumination across the range, where a curve would be needed for a device that is not (1). That constant still comes from a calibration against a known source: linearity saves the shape of the graph, not the fixing of its scale. Its current does not depend on the applied pd: the characteristics are flat, so the illumination alone sets the current.

Question 5

State what a Hall probe measures, and explain why the probe is rotated before a reading is taken.

Accepted Answer

It measures magnetic flux density, the Hall voltage across the slab being proportional to B when the current through the slab is held constant (1). The Hall voltage is largest when the field is perpendicular to the face of the slab, so the probe is turned until the reading is a maximum and that value is taken (1).

Question 6

Give two reasons a photodiode is preferred to a light-dependent resistor at the receiving end of an optical fibre.

Accepted Answer

It responds in nanoseconds, while carriers in an LDR take milliseconds to build up and longer to disperse, so an LDR cannot follow anything flickering faster than a few hundred hertz (1). Its photocurrent is proportional to the illumination, whereas an LDR's resistance is not, so the photodiode reproduces the pulse shape faithfully (1).

Question 7

A MOSFET has a threshold voltage of 2.0 V and passes a drain current of 0.10 A when the gate-source voltage is 3.0 V. Its drain current follows I_D = k(V_GS − V_T)². Calculate k, and the drain current when the gate-source voltage is 4.5 V.

Accepted Answer

The excess above threshold is 3.0 − 2.0 = 1.0 V, so k = 0.10/1.0² = 0.10 A V⁻² (1). At 4.5 V the excess is 4.5 − 2.0 = 2.5 V (1), so I_D = 0.10 × 2.5² = 0.625 A (1). Square the excess above the threshold, not V_GS itself: squaring the gate voltage gives 2.025 A, three times too big.

Question 8

A MOSFET switches a 6.0 Ω heater across a 9.0 V supply. With the gate at 5 V the channel resistance is 0.12 Ω. Calculate the current, the pd across the heater and the power dissipated in the MOSFET.

Accepted Answer

Channel and heater are in series, so R = 6.0 + 0.12 = 6.12 Ω and I = 9.0/6.12 = 1.47 A (1). Heater pd = 1.47 × 6.0 = 8.82 V (1). MOSFET power = I²R = 1.47² × 0.12 = 0.26 W, against 13.0 W in the heater (1). The channel carries only 0.18 V, so the power in it is I²R and not the full 9.0 V times the current.

Question 9

A 5.1 V zener diode is fed through a 390 Ω series resistor from a supply of 15 V, and the load draws 12 mA. Calculate the current in the resistor, the current in the zener and the power dissipated in the zener.

Accepted Answer

The resistor drops 15 − 5.1 = 9.9 V, so I = 9.9/390 = 25.4 mA (1). The load takes 12 mA of that, leaving 25.4 − 12 = 13.4 mA in the zener (1). P = V_ZI = 5.1 × 13.4 × 10⁻³ = 68 mW (1). Only the leftover 9.9 V appears across the resistor, so dividing the whole 15 V by 390 Ω overstates the current.

Question 10

A reverse-biased photodiode delivers a photocurrent of 24 μA at a certain illumination, and its current is passed through a 75 kΩ load resistor. Calculate the pd across the resistor, and the pd when the illumination is trebled.

Accepted Answer

V = IR = 24 × 10⁻⁶ × 75 × 10³ = 1.8 V (1). Photocurrent is proportional to illumination, so three times the light gives 3 × 24 = 72 μA (1) and V = 5.4 V (2 s.f.) (1). The diode is not a resistor. Its flat characteristics make it behave as a current source, so trebling the illumination trebles the current rather than altering a resistance.

Question 11

A Hall effect sensor carries a constant current and gives an output of 2.4 mV when the flux density through its face is 0.20 T. Calculate its sensitivity in mV T⁻¹ and its output in a field of 0.55 T.

Accepted Answer

Sensitivity = 2.4/0.20 = 12 mV T⁻¹ (1), the Hall voltage being proportional to B at constant current (1). Output = 12 × 0.55 = 6.6 mV (1). The sensitivity is fixed only while the sensor current is held constant, and the division must be taken as voltage over flux density: reversing it gives 0.083 T mV⁻¹.

Question 12

A 5.1 V zener diode is fed through a 390 Ω series resistor from a supply that can fall as low as 9.0 V. Calculate the resistor current at 9.0 V and hence state the largest load current the regulator can supply.

Accepted Answer

The resistor drops 9.0 − 5.1 = 3.9 V, so I = 3.9/390 = 10.0 mA (1). Every milliamp the load takes comes out of that current, so the largest load current is 10.0 mA (1), and at that point the zener current has fallen to zero (1). Ask for more and the zener current would have to run backwards, which it cannot: the output sags below 5.1 V and regulation is lost. Take the load current at the lowest supply voltage, which is the worst case for the zener current.

Question 13

A MOSFET with a threshold voltage of 2.0 V passes 0.10 A when the gate-source voltage is 3.0 V. Without calculating k, deduce the drain current at a gate-source voltage of 6.0 V.

Accepted Answer

The excess above threshold goes from 1.0 V to 4.0 V, a factor of 4, and the current follows its square, a factor of 16 (1). I_D = 16 × 0.10 = 1.6 A (1). Doubling the gate voltage does not give four times the current: the square law acts on the excess above the threshold, not on V_GS itself.

Question 14

A 5.1 V zener regulator must supply a load drawing up to 15 mA from a supply that wanders between 9.0 V and 15.0 V. The zener needs at least 5.0 mA to stay in breakdown and is rated at 400 mW. Calculate the largest series resistance that meets the specification, choose the preferred value 180 Ω, and check the design at both ends of the supply range, including the power rating needed for the resistor.

Accepted Answer

Worst case for the resistance is the lowest supply, where its drop is only 9.0 − 5.1 = 3.9 V (1). It must still carry the load's 15 mA plus the zener's 5.0 mA, that is 20 mA, so R ≤ 3.9/(20 × 10⁻³) = 195 Ω (1). With R = 180 Ω the resistor carries 21.7 mA at 9.0 V in, comfortably above the 20 mA needed (1). At 15.0 V in the resistor carries 9.9/180 = 55 mA (1), and with the load disconnected all of it goes through the zener, giving P = 5.1 × 55 × 10⁻³ = 280 mW (1), inside the 400 mW rating (1). The resistor itself dissipates 9.9 × 55 × 10⁻³ = 0.54 W, so a 1 W resistor is needed (1).

Question 15

A logic output that can supply 4.0 mA drives the gate of a MOSFET. The gate behaves as a 1.5 nF capacitor and must be taken to 5.0 V to switch the transistor on. Estimate the switching time, and state the current the logic output supplies once the MOSFET is on.

Accepted Answer

Charge needed: Q = CV = 1.5 × 10⁻⁹ × 5.0 = 7.5 × 10⁻⁹ C (1). At a steady 4.0 mA, t = Q/I = 7.5 × 10⁻⁹/(4.0 × 10⁻³) (1) = 1.88 × 10⁻⁶ s, about 1.9 μs (1). Once the gate is charged the current falls to zero, because the oxide insulates the gate from the channel (1).

Question 16

A Hall sensor giving 2.4 mV in a flux density of 0.20 T is mounted beside a rotating shaft carrying four small magnets, each bringing a peak flux density of 0.45 T past the slab. A counter registers 120 pulses in 2.0 s. Calculate the peak output voltage and the rotation rate of the shaft in revolutions per minute.

Accepted Answer

Sensitivity = 2.4/0.20 = 12 mV T⁻¹, so the peak output is 12 × 0.45 = 5.4 mV (1). Pulse rate = 120/2.0 = 60 pulses per second (1). Four magnets give four pulses per revolution, so the shaft turns 60/4 = 15 times a second (1), that is 15 × 60 = 900 revolutions per minute (1). Count the magnets before dividing: assuming one pulse per revolution would give 60 rev s⁻¹.

Question 17

A photodiode gives a photocurrent of 20 μA per unit of illumination. It is to produce 2.5 V across a load resistor at an illumination of 3 units. Calculate the resistance required, state the pd obtained with the nearest preferred value of 43 kΩ, and explain why a light-dependent resistor could not replace the photodiode in a fibre-optic receiver working at 1 MHz.

Accepted Answer

At 3 units the photocurrent is 3 × 20 = 60 μA (1). R = V/I = 2.5/(6.0 × 10⁻⁵) = 41.7 kΩ (1). With 43 kΩ the pd is 6.0 × 10⁻⁵ × 43 × 10³ = 2.58 V (1). An LDR takes milliseconds for its carriers to build up and disperse, so at 1 MHz, a microsecond per bit, it could not follow the light at all, while a photodiode responds in nanoseconds (1). R = V/I: dividing current by voltage instead gives 24 μA V⁻¹, which is a conductance rather than a resistance.

Question 18

A MOSFET with a threshold voltage of 2.0 V switches a lamp. A student argues that a gate voltage of 2.1 V must be enough, since it exceeds the threshold. Explain why the gate is instead driven to the logic-level voltage of 5 V named on the data sheet.

Accepted Answer

The threshold is where conduction begins: just above it the channel is barely formed and its resistance is still high (1). The guaranteed low on-resistance arrives only at the higher gate voltage the data sheet names, where the channel is fully enhanced (1). With a partly formed channel the MOSFET would keep a large share of the supply pd and dissipate significant power itself, instead of delivering almost the whole supply to the lamp (1).