Chapter 15: Semiconductors and the Diode

More tightly than in a conductor, but less than in an insulator

A semiconductor lies between the two: its outer electrons are held more firmly than a conductor’s but less firmly than an insulator’s, so heat can free some of them. The close option gets the order backwards.

Much like an insulator, as few electrons are free

In very pure silicon the outer electrons are tied up in the bonds of the crystal, so at room temperature it behaves much like an insulator. Doping or heating is needed to give it free charge carriers.

Some outer electrons gain enough energy to break free, so it conducts

Heat gives the outer electrons extra energy, and some break away from their atoms, so a current can flow when a voltage is applied. Heat loosens the electrons rather than binding them tighter: a hot semiconductor conducts better.

Germanium

Early semiconductor devices were mostly germanium; most are now silicon, with materials such as gallium arsenide kept for special purposes. Aluminium is a metal and glass an insulator, neither a semiconductor.

Four

A silicon atom’s 14 electrons sit in three layers, 2, 8 and 4, and only the four outer ones are moderately bound and shared with neighbouring atoms. Eight is the full second layer, not the outer one.

Adding tiny amounts of another element to pure silicon

Silicon is first refined until it is very pure, then small amounts of another element are added on purpose: that addition is doping. Refining comes before it and is not doping itself.

A doping agent

The element added on purpose, such as boron or phosphorus, is the doping agent, and it becomes part of the crystal. A catalyst only speeds up a chemical reaction and is not used up or built in.

P-type semiconductor

Boron and aluminium atoms have only three outer electrons, so each leaves one bond short of an electron, a hole, and the silicon becomes P-type. Elements with five outer electrons, such as phosphorus, make N-type.

Phosphorus or arsenic

Phosphorus and arsenic have five outer electrons; four bond with the neighbouring silicon atoms and the fifth is left free, giving N-type silicon. Boron and aluminium, with three, give P-type instead.

Free electrons, loose in the crystal

Each doping atom with five outer electrons leaves one electron free to move, and these negative electrons carry the current, hence N for negative. Holes, acting as positive charges, are the carriers in P-type material.

Holes, gaps where electrons are missing

Each doping atom with three outer electrons leaves a hole, a missing electron, which behaves like a free positive charge, hence P for positive. Free electrons are the carriers in N-type material.

From the positive terminal towards the negative

Electrons step into holes from the negative side, so each hole moves the other way, from the positive terminal towards the negative, as a positive charge would. It is the electrons that drift towards the positive terminal.

A place where an electron should be but is missing

A hole is a gap in the bonds of the crystal where an electron ought to be; it acts like a positive charge. A free electron, not bound into the crystal, is the other kind of carrier, in N-type material.

A neighbouring electron moves into it, leaving a new gap behind

An electron beside the hole moves in to fill it and leaves its own place empty; repeated, this makes the hole appear to travel. A hole is only a gap, so it cannot move unless electrons move.

The depletion layer

Free electrons from the N side fill holes on the P side, leaving a thin region, about 1 µm, with almost no free carriers: the depletion layer or depletion region. The PN junction is the boundary it forms at.

The P side negative and the N side positive

Electrons cross from the N side to fill holes on the P side, so the P side gains negative charge and the N side, having lost electrons, is left positive. P and N name the carriers, not this charge.

None: it is electrically neutral

Each free electron came with a doping atom whose nucleus carries one extra positive charge, so the material is neutral overall. The N in N-type refers to its negative charge carriers, not to any net charge.

About 1 µm

The depletion layer is very thin, only about 1 µm, a thousandth of a millimetre; 1 mm would be a thousand times too thick. Being so thin, it gives a reverse-biased junction a noticeable capacitance.

Positive to the P-type side, negative to the N-type side

Forward bias puts positive on the P side and negative on the N side, pushing carriers into the depletion layer so that current flows once the forward bias voltage is exceeded. The other way round is reverse bias.

It lets a current pass through it

Forward bias, P side positive, pushes carriers into the depletion layer, so once the forward bias voltage is exceeded current flows readily. Blocking all but a tiny leakage current is what a reverse-biased junction does.

Only a tiny leakage current, usually under 1 µA

Reverse bias widens the depletion layer, so after a brief initial pulse only the reverse leakage current flows, typically under 1 µA; 1 mA would be a thousand times too much. Large currents flow only under forward bias.

20 MΩ

R = V ÷ I = 10 ÷ 0,000 000 5 = 20 000 000 Ω = 20 MΩ, so a reverse-biased diode is practically an open circuit. Reading µA as mA gives the 20 kΩ option.

It becomes wider

Reverse bias pulls more electrons into the P side and draws free electrons out of the N side, deepening the depletion layer until the junction voltage matches the applied voltage. Forward bias does the opposite.

Its forward bias voltage

A silicon junction needs a forward bias voltage of about 500 to 800 mV, typically 600 mV, before it conducts, so 650 mV fits. Reverse breakdown comes at far higher voltages.

Lower: about 0,2 V against about 0,6 V

A germanium junction conducts from about 0,2 V (the manual says 100 to 200 mV; 0,2 to 0,3 V is usual in practice) and a silicon one from about 0,6 V. The close option swaps the two materials.

3,6 mA

The diode keeps its forward voltage, leaving 6 − 0,6 = 5,4 V across the resistor, so I = 5,4 ÷ 1 500 = 3,6 mA. Ignoring the diode’s drop gives 6 ÷ 1 500 = 4 mA.

1,8 V

Each conducting diode holds about its forward voltage, and voltages in series add: 3 × 0,6 = 1,8 V. The 0,6 V option counts only one of the three diodes.

It stays at about VF, hardly changing

Once a diode conducts, the current rises steeply with very little change in voltage, so the voltage across it can be taken as VF at any current. Ten times the voltage is how a resistor behaves (Ohm’s law).

A junction diode

One P-type and one N-type region meeting at a single PN junction make a junction diode, anode on the P side. A bipolar junction transistor has two junctions (NPN or PNP), the subject of chapter 17.

The anode

The anode is joined to the P region and the cathode to the N region, so making the anode positive forward-biases the diode. A diode has just these two terminals.

The anode must be more positive

A diode conducts when its anode is more positive than its cathode by at least the forward bias voltage. With the anode more negative the diode is reverse-biased and passes only a tiny leakage current.

The way the arrow points: anode to cathode

Current flows into the anode and out of the cathode, the way the arrow points. Electrons, being negative, drift the other way, which is where the close option comes from: the symbol shows conventional current.

Rectifying AC into DC

Letting current through one way only turns alternating current into a one-way current: rectification, the job of every rectifier circuit. Regulating a voltage is the Zener diode’s job, and relies on its reverse breakdown instead.

A half-wave rectifier

A single diode conducts only while the input makes its anode positive, so the load gets one half of each cycle and nothing in the other: a half-wave rectifier. A full-wave rectifier uses both halves.

8,4 V

The output follows the input, less the diode’s forward voltage: 9 − 0,6 = 8,4 V. Forgetting the drop gives 9 V, and a half-wave rectifier passes half of each cycle, not half the voltage.

What remains of the original AC, riding on the DC output

The capacitor cannot hold the output quite steady, so remnants of the original AC ride on the DC: that rise and fall, at the supply frequency or twice it, is ripple. Brief switching spikes are a different kind of noise.

A lowpass filter, passing DC and attenuating the ripple

A capacitor across the output is a one-element lowpass filter: its reactance is low to the AC ripple, which it shunts away, and infinite to DC, which passes to the load. A highpass filter would keep the ripple.

A low hum in the sound

Ripple at 50 Hz or 100 Hz gets into the audio and is heard as hum, which is why ripple is called hum in audio systems. Hiss is random noise, not a tone from the mains.

It charges at each peak and feeds the load between peaks

While the diode conducts near each peak the capacitor charges; while the diode is off it discharges into the load, filling the gaps. The diode, now reverse-biased, stops the charge draining back into the supply.

60 Hz

A half-wave rectifier recharges the capacitor once per AC cycle, so the ripple is at the supply frequency, 60 Hz here (50 Hz in South Africa). Twice the supply frequency, 120 Hz, comes from full-wave rectification.

It rectifies the AC, turning it into DC

The diode bridge is a full-wave rectifier: it steers both half cycles of the AC the same way through the load, giving pulsating DC. Smoothing out the ripple is the job of the capacitor that follows it.

A bridge rectifier

Four diodes arranged as a bridge rectify both half cycles: the full-wave bridge rectifier. A centre-tap rectifier is a full-wave circuit with only two diodes and a centre-tapped transformer winding (chapter 16).

It gives no output for more than half of every cycle

For the negative half cycle, plus the moments when the input is below VF, the output is zero, so the capacitor must carry the load through long gaps. Its ripple is at the supply frequency, not twice it.

It turns them into positive output pulses

On negative half cycles the other pair of diodes conducts, D2 and D4 in the manual’s circuit, so current still flows the same way through the load and the output stays positive. Blocking them is half-wave behaviour.

800 Hz

A full-wave rectifier gives an output pulse on every half cycle, two per cycle, so the ripple is at twice the supply frequency: 2 × 400 = 800 Hz. Ripple at 400 Hz would come from a half-wave rectifier.

8 ms

Full-wave ripple is at 2 × 60 = 120 Hz, so the time between peaks is 1 ÷ 120 = 8,33 ms, nearest 8 ms. Using 60 Hz, as for a half-wave rectifier, gives 16,7 ms, hence 17 ms.

The current passes through two diodes in series

On each half cycle the current passes through two conducting diodes in series, D1 and D3 or D2 and D4, each dropping about 0,6 V: 2 × 0,6 = 1,2 V. A silicon diode drops only about 0,6 V.

8,8 V

Two diodes conduct in series on each half cycle, so the output peak is 10 − 2 × 0,6 = 8,8 V. Taking off a single 0,6 V gives 9,4 V, the half-wave result.

25 V

AC voltages are given as RMS, and a sine wave peaks at √2 times its RMS value: 1,414 × 18 = 25,5 V, nearest 25 V. Taking the 18 V as the peak is the RMS-for-peak slip.

It can be much smaller

The capacitor is recharged twice per cycle, so it feeds the load for less than half as long between charges and can be much smaller, not larger. It is still needed, as the output still pulsates.

Where the cathodes of two diodes meet

Current leaves each conducting diode by its cathode, so the positive output is where the D1 and D4 cathodes join and the negative where the D2 and D3 anodes join. The AC goes in at the other two corners.

A Zener diode

A Zener diode is designed so that its reverse breakdown voltage can be exceeded safely, and it is used that way. A varicap is also reverse-biased, but kept below breakdown, where it acts as a capacitor.

The reverse voltage at which it breaks down

VZ is the Zener voltage, the carefully controlled reverse breakdown voltage, at which this diode works, holding about 6,8 V. It is not a damage limit: a Zener is made to exceed its breakdown voltage safely.

Permanent damage to the diode itself

Reverse breakdown will usually ruin an ordinary diode for good, so its VR must never be exceeded. Breaking down with no harm done is what a Zener diode is specially made for.

Holding a DC voltage steady

Reverse-biased through a series resistor, a Zener keeps the voltage across it near VZ whatever the input voltage and load current, so it is a voltage regulator. The load current is free to change.

The Zener takes more current, so the resistor drops more voltage

A rise in input tries to lift the output above VZ, so the Zener conducts harder; the extra current through the series resistor drops the extra voltage, and the output stays near VZ. Less current would let it rise.

To limit the Zener diode’s current

Once in breakdown the Zener conducts strongly, so without a resistor the current would destroy it; the resistor limits the current and takes up the difference between input and VZ. The Zener itself sets the voltage.

230 Ω

The resistor drops the difference, 12 − 5,1 = 6,9 V, so R = 6,9 ÷ 0,03 = 230 Ω. Using the whole 12 V gives 400 Ω, forgetting that the Zener holds 5,1 V of it.

180 Ω

R = (13,8 − 9,1) ÷ 0,025 = 4,7 ÷ 0,025 = 188 Ω, so 180 Ω is nearest. Dividing the whole 13,8 V by 25 mA gives 552 Ω, hence the 560 Ω option.

15 mA

The resistor’s current divides between the Zener and the load (Kirchhoff’s current law), so IZ = 40 − 25 = 15 mA. Adding them to 65 mA would need more current than the resistor carries.

A non-conducting gap separates two conducting regions

The P and N regions conduct, while the depletion layer between them has no free carriers, so the junction is like two plates with a thin insulator between. The leads add a little stray capacitance, but not this effect.

It becomes smaller

More reverse bias widens the depletion layer, like moving a capacitor’s plates further apart, so the capacitance falls. That is how a DC voltage tunes a varicap: more bias, less capacitance.

A varactor diode

Varicap (variable capacitance) and varactor are two names for the same diode. A varistor is a different part altogether: a resistor whose resistance changes with the voltage across it.

Varying an oscillator’s frequency with a DC voltage

A varicap’s capacitance follows its reverse bias, so in an oscillator’s tuned circuit a DC tuning voltage sets the frequency. The varicap changes the tuning, not the output level.

It rises to twice its value

f = 1 ÷ (2π√(LC)), so the frequency varies as one over the square root of C: a quarter of the capacitance gives √4 = 2 times the frequency. Four times forgets the square root.

20 pF

In series, 1 ÷ C = 1 ÷ 100 + 1 ÷ 25 = 0,05, so C = 20 pF, less than the smaller of the two. Adding them to 125 pF is the rule for capacitors in parallel.

It passes the DC bias but blocks the RF signal

L1 has a high reactance at the signal frequency, so RF cannot leak into the bias supply, which would add losses and lower the Q, while the DC bias passes freely. Passing RF and blocking DC is what C1 does.

To stop L2 from shorting out the DC bias

L2 has almost no resistance to DC, so without C1 it would short the bias; C1 blocks the DC but passes RF, its capacitance in series with the varicap’s. Keeping RF out of the bias supply is L1’s job.

Making the control voltage X positive

X positive with respect to the chassis forward-biases D1, so DC flows through L1, D1 and R1 and the diode’s low resistance passes the signal. X negative reverse-biases D1, which then blocks the signal.

So that the diode does not rectify the signal

A large signal’s peaks would overcome the control voltage and flip the diode’s bias for part of each cycle, rectifying and distorting the signal. The risk is distortion, not damage: a small signal keeps the diode on or off.

It passes the DC control current but blocks the RF signal

L1 has a high reactance at the signal frequency, so the RF cannot leak into the control circuit, while the DC control current flows through it freely. The capacitors do the opposite: they pass RF and block DC.

They pass the RF signal but block DC

The capacitors keep the DC control current out of whatever is connected at A and B, while their low reactance lets the RF through. Passing DC and blocking RF is what the inductor L1 does.

A low capacitance when switched off

Capacitance across a diode that is switched off would let RF leak through it, which would be disastrous in a switch, so PIN diodes have low capacitance. High capacitance is what a varicap wants, not a switch.

Intrinsic: pure silicon between the P and N regions

A PIN diode has a layer of pure, undoped (intrinsic) silicon between its P and N regions, hence P-I-N. The middle layer is semiconductor, not an insulating oxide.

Faster switching and no inherent limit on their life

PIN diode switches have no inherent lifetime limit, surviving a very high number of operations, and they switch much faster than mechanical relays. It is the relay that wears out.

Energy the current releases in the semiconductor

With suitably chosen semiconductor material and doping, the energy change at the junction is given off directly as electromagnetic energy, which can be light. It is not glowing from heat, as a lamp filament does.

More efficient, and they degrade very slowly

LEDs give far more light per watt than incandescent or fluorescent lights, and are reliable and slow to degrade, so an LED replacement gives more light for less power. The close option has the efficiency backwards.

It is much higher

Most LEDs drop about 2,0 V when conducting, against about 0,6 V for an ordinary silicon diode, so their forward voltage is much higher. Every LED is a diode and has a forward voltage.

500 Ω

The resistor drops what the LED does not: 12 − 2 = 10 V, so R = 10 ÷ 0,02 = 500 Ω. Dividing the whole 12 V by 20 mA gives 600 Ω, ignoring the LED’s forward voltage.

1 kΩ

R = (13,8 − 2,0) ÷ 0,012 = 11,8 ÷ 0,012 = 983 Ω, so 1 kΩ is nearest. Forgetting the LED’s 2,0 V gives 13,8 ÷ 0,012 = 1 150 Ω, hence the 1,2 kΩ option.

RF noise from their switching power supplies

Many cheap LED lamps have switching power supplies that put out strong RF noise, and in towns this can make parts of the HF spectrum unusable. Light itself does not interfere with radio reception.