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Discrete semiconductor devices
Four components with four jobs. A transistor whose gate takes no current at all switches amps from a logic pin, a diode run backwards pins a supply rail steady, and two more turn light and magnetic field into numbers a circuit can read.
Builds on Current-voltage characteristics and Force on a moving charge.
IN THIS TOPIC
- Read a MOSFET's drain characteristic, and use the threshold voltage to say whether the channel is open.
- Explain how a logic-level gate voltage switches a load, and why the gate itself takes no steady current.
- Use a zener diode and its series resistor to hold an output steady against a wandering supply.
- Say what a photodiode and a Hall effect sensor each measure, and choose between a photodiode and an LDR.
WHAT YOU PROBABLY THINK
Switching a heavy load needs a heavy current into the transistor's control terminal.
A gate that takes no current
A MOSFET has three terminals. The load current flows between the drain and the source, and the gate decides whether it flows at all. The gate sits on a thin insulating layer of oxide, out of contact with the silicon beneath, so it behaves as a very small capacitor rather than as a way in. Put a voltage on it and its field reaches through the oxide and draws carriers into a conducting channel underneath.
Nothing at all happens until the gate-source voltage passes the threshold voltage . Below it there is no channel, and the drain current is zero however large the drain voltage. Past it the channel opens and widens, and the current climbs steeply, roughly with the square of the excess above threshold.
WORKED EXAMPLE
One volt over, two volts over, three
The MOSFET drawn above has = 2.0 V, and at = 3.0 V it passes 0.10 A. Predict the drain current at 4.0 V and at 5.0 V.
The excess above threshold is 1.0 V, 2.0 V and 3.0 V in turn, and the current follows its square.
Doubling the excess quadruples the current, giving 4 × 0.10 = 0.40 A. Three times the excess gives nine times the current, 9 × 0.10 = 0.90 A, the top of the drawn curve.
That doubling test is the quick check on any drain characteristic. A straight line through those points would be the wrong shape entirely.
Switching a load
That shape of characteristic makes a superb switch. Wire the load in series with the drain and drive the gate from a logic output. At 0 V the channel is shut, the drain-source resistance is megohms, and the load sees microvolts. At 5 V the channel is fully open, its resistance falls to a fraction of an ohm, and almost the whole supply lands on the load.
Work the on state as a potential divider. Lamp and channel carry 12/24.05 = 0.499 A between them, the lamp keeps 11.98 V, and the MOSFET dissipates = 0.012 W against nearly 6.0 W in the lamp. A component wasting a hundredth of a watt to control six of them needs no heat sink.
The opening claim dies here. The gate is insulated, so no steady current enters it. A logic output has only to charge a gate capacitance of a nanofarad or two, which it does in microseconds, and in between it supplies nothing at all.
Holding a rail steady
Run an ordinary diode backwards and it blocks until the reverse voltage grows large enough to destroy it. A zener diode is built to break down gently at a chosen voltage and to survive the experience indefinitely. Past breakdown its characteristic is nearly vertical, so the current through it can swing by tens of milliamps while its pd barely moves.
A component whose voltage refuses to change is what a supply rail wants. So the zener goes across the load, reverse biased, with a series resistor between it and the raw supply. The load gets , everything left over is dropped across the resistor, and when the supply wanders it is the resistor's share that changes.
The resistor sets the current, and the current is the whole of the design:
Too small a current and the diode never properly reaches breakdown. Too large and it overheats. In the figure the resistor carries 10.0 mA at 9 V in and 25.4 mA at 15 V in, and the zener swallows whatever the load leaves.
YOUR TURN
Sharing the current
The regulator above runs from 12 V. Find the current in the 390 Ω resistor, then the current in the zener when the load draws 8.0 mA, and the power the zener dissipates.
Show the working
The resistor drops 12 − 5.1 = 6.9 V, so it carries 6.9/390 = 17.7 mA.
The load takes 8.0 mA of that, leaving 17.7 − 8.0 = 9.7 mA in the zener, and a power of 5.1 × 9.7 × 10−3 = 49 mW.
Ask the load for more than 17.7 mA and the zener current would have to run backwards, which it cannot. The output sags below 5.1 V and regulation is lost.
Light into current
A photodiode is a diode with a window, used reverse biased so that almost nothing flows in the dark. A photon absorbed in the depletion layer frees an electron-hole pair, and the strong field there sweeps the two apart before they can recombine. Every absorbed photon therefore adds to a small reverse current, the photocurrent, and doubling the illumination doubles the pairs freed each second.
Those flat lines carry two messages. Photocurrent is proportional to illumination, so the device is a linear light meter needing no calibration curve. And it hardly depends on the applied pd, so the diode behaves as a current source. Read it by dropping its current across a resistor: 40 μA through 50 kΩ gives 2.0 V, and 60 μA gives 3.0 V.
Against a light-dependent resistor the photodiode wins on speed and on linearity. Carriers in an LDR take milliseconds to build up and longer to disperse, so it cannot follow anything flickering faster than a few hundred hertz, and its resistance is nowhere near proportional to the light. A photodiode responds in nanoseconds, which puts one at the end of every optical fibre. Where the light changes slowly, the cheaper LDR is fine.
Measuring a field, and noticing a magnet
Send a current along a thin slab of semiconductor and put a magnetic field through its face. Each moving carrier feels a force at right angles to both, so carriers pile up along one edge and leave the opposite edge short of them. The separated charge builds a field that opposes further pile-up, and matters settle once the two forces balance. What remains is a steady Hall voltage across the faces.
Hold the current constant and that voltage is proportional to the flux density, so the slab is a direct-reading field meter. A Hall probe is exactly that, calibrated. Its reading is largest when the field runs perpendicular to the slab's face, so a probe is rotated for a maximum before the number is taken. The sensor above gives 2.4 mV in 0.20 T and 4.8 mV in 0.40 T, a sensitivity of 12 mV per tesla.
TRY IT UNSEEN
Counting the shaft round
That sensor is mounted beside a rotating shaft carrying one small magnet, which brings 0.35 T past the slab once per revolution. Find the peak output, and explain how the circuit measures the rotation rate.
Show the working
The sensitivity is 2.4/0.20 = 12 mV T−1, so 0.35 T gives 12 × 0.35 = 4.2 mV.
Away from the magnet the field is near zero and so is the output, so the sensor delivers one pulse per revolution. Counting pulses per second gives the rotation rate.
Proximity sensing is the same trick standing still. Bring a magnet close and the voltage appears, take it away and it vanishes, with no contacts to wear out.
THE EXAM BIT
- Threshold answers want the word channel. Below no channel exists, so no drain current flows however large the drain voltage, and above it the current rises with the square of the excess.
- For a switching calculation put the channel resistance in series with the load and divide. Quote the load's pd, and the power lost in the MOSFET as if the question asks.
- The insulated gate is worth two marks whenever a MOSFET is compared with anything else. No steady gate current flows, so the driver supplies only the charge the gate capacitance needs to change state.
- Zener answers begin with the resistor's drop, supply minus , then Ohm's law. The zener current is the resistor current minus the load current, and regulation holds only while that difference stays positive.
- Write reverse biased and proportional to intensity into every photodiode answer. A comparison with an LDR is won on response time and on linearity, not on cost alone.
CHECK YOURSELF
A MOSFET with a threshold voltage of 2.0 V switches a 24 Ω lamp across a 12 V supply, and its channel resistance is 0.05 Ω with the gate at 5 V. State the drain current with the gate at 1.5 V. Find the lamp's pd and the power wasted in the MOSFET with the gate at 5 V. Explain how a logic output rated at 1 mA can drive it.
Show a hint
Below the threshold nothing flows. Above it, treat the open channel as a small resistance in series with the lamp.
Show the answer
At 1.5 V the gate is below the threshold, so no channel forms and the drain current is zero.
With the gate at 5 V the series resistance is 24.05 Ω, so the current is 12/24.05 = 0.499 A and the lamp holds 0.499 × 24 = 11.98 V.
The MOSFET dissipates = 0.4992 × 0.05 = 0.012 W, against 6.0 W in the lamp.
The gate is insulated from the channel, so it draws no steady current. The logic output supplies only the brief charging current of the gate capacitance, well within 1 mA.
No channel below the threshold voltage; above it the drain current climbs with the square of the excess.
A MOSFET gate is insulated, so it takes no steady current and a logic pin can switch amps.
A zener holds its breakdown voltage across the load while the series resistor drops whatever is left.
Photocurrent follows the light; the Hall voltage follows the flux density.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Read a MOSFET's drain characteristic, and use the threshold voltage to say whether the channel is open.
- Explain how a logic-level gate voltage switches a load, and why the gate itself takes no steady current.
- Use a zener diode and its series resistor to hold an output steady against a wandering supply.
- Say what a photodiode and a Hall effect sensor each measure, and choose between a photodiode and an LDR.
Open the full revision checklist to track your progress across the whole unit.