Semiconductor materials theory
Where silicon comes from, what it means to dope a crystal, and why a PN junction lets current flow in only one direction. From this come the diode, the power supply of any piece of equipment and, later on, the transistor and all integrated circuits.
01Conductors, insulators and semiconductors
What determines whether a material conducts or not is how free its valence electrons are — those in the outermost shell of the atom.
1 to 3 valence electrons, practically free. Copper, silver, aluminum. Resistivity ≈ 10−8 Ω·m.
4 valence electrons. Silicon and germanium. It neither conducts nor insulates: it depends on how it is treated. Resistivity ≈ 10−3 to 103 Ω·m.
5 to 8 electrons, tightly bound. Glass, mica, plastic. Resistivity ≈ 1012 Ω·m or more.
Pure (intrinsic) silicon forms a crystal in which each atom shares its four electrons with four neighbors: covalent bonds. With all the electrons occupied, at 0 K it would be a perfect insulator. At room temperature, thermal agitation breaks some bonds and frees a few carriers; that is why it barely conducts. It is also why semiconductors depend strongly on temperature, a fact that reappears in everything that follows.
When an electron breaks free it leaves a vacancy in the bond: a hole. A neighboring electron can jump in to fill it, and then the hole “moves” in the opposite direction. That is why a semiconductor has two types of carriers: electrons (negative charge) and holes (which behave like a positive charge). This does not happen in metals, and it is the key to everything else.
02Doping: N and P materials
Pure silicon is useless. What makes it useful is adding to it, in a controlled way, a tiny amount of impurities: roughly one foreign atom for every 10 million silicon atoms. This is called doping.
Pentavalent atoms are added (5 valence electrons): phosphorus, arsenic, antimony. Four electrons form bonds and one is left over, almost free.
- Majority carriers: electrons.
- Minority carriers: holes (from thermal agitation).
- The impurity is called a donor.
- The material remains electrically neutral.
Trivalent atoms are added (3 valence electrons): boron, gallium, indium. One electron is missing to complete the fourth bond: a hole is left.
- Majority carriers: holes.
- Minority carriers: electrons.
- The impurity is called an acceptor.
- It is also electrically neutral.
“N-type” does not mean the material is negatively charged, nor “P-type” positively charged. Both are neutral: for every extra free electron there is a nucleus with an extra positive charge that compensates it. The N and the P indicate which type of carrier conducts the current, not a net charge.
03The PN junction
If in the same crystal one region is doped P and the adjacent one N, something remarkable happens at the boundary. The excess electrons on the N side diffuse toward the P side and recombine with the holes. In doing so they leave behind fixed ions: positive on the N side, negative on the P side.
This region without carriers is called the depletion region, and the internal field that appears is the potential barrier. It is about 0.7 V in silicon and 0.3 V in germanium. It is responsible for all of the diode's behavior: as long as that barrier is not overcome from outside, no current flows.
04The diode: biasing and characteristic curve
The three diode models
Depending on the accuracy required, one or another is used. In Year 4 you work almost always with the second.
| Model | Forward | Reverse | When to use it |
|---|---|---|---|
| Ideal | Short circuit (0 V) | Open circuit | Quick analysis, when the source is tens of volts. |
| 0.7 V | 0.7 V source | Open circuit | The usual one. A good balance between simplicity and accuracy. |
| With rd | 0.7 V + dynamic resistance (a few Ω) | Very high resistance | Large currents or dissipation calculations. |
A 12 V source drives a silicon diode in series with a 1 kΩ resistor. Calculate the current and the power dissipated by the diode.
With the 0.7 V model, the voltage across the resistor is 12 − 0.7 = 11.3 V:
With the ideal model it would have given 12 mA: a 6 % error. With sources of 5 V or less, ignoring the 0.7 V already produces large errors and the ideal model stops being useful.
The diodes used in the workshop
| Diode | IF max. | VRRM | Use |
|---|---|---|---|
| 1N4148 | 200 mA | 100 V | Signal, fast switching. Glass package. |
| 1N4001 | 1 A | 50 V | Low-voltage rectification. |
| 1N4007 | 1 A | 1000 V | The general-purpose one. It costs the same as the 4001: it is best to stock only this one. |
| 1N5408 | 3 A | 1000 V | Higher-current power supplies. |
| 1N4733A | — | VZ = 5.1 V | 1 W Zener. |
| 1N5822 | 3 A | 40 V | Schottky: drops only 0.3 V. Switching power supplies. |
The painted band on the body always marks the cathode. It is the only way to identify it, and mixing it up is the most frequent mistake when building a power supply.
05Rectification
Rectifying means converting alternating current into pulsating direct current. It is the first stage of every power supply.
Half-wave
A single diode. It passes the positive half-cycle and blocks the negative one.
It uses half of the time, delivers little DC and is hard to filter. It is only used in very small chargers.
Full-wave with a center-tapped transformer
Two diodes and a secondary with a center tap. Each diode conducts during one half-cycle, so both half-cycles appear across the load with the same polarity.
Graetz bridge (bridge rectifier)
Four diodes and a plain secondary, without a center tap. It is the one used almost all the time.
| Rectifier | Diodes | VDC without filter | Ripple f | Total drop | PIV per diode |
|---|---|---|---|---|---|
| Half-wave | 1 | 0.318 Vp | 50 Hz | 0.7 V | Vp |
| Full-wave, center tap | 2 | 0.637 Vp | 100 Hz | 0.7 V | 2 Vp |
| Bridge | 4 | 0.637 Vp | 100 Hz | 1.4 V | Vp |
The bridge wins because it needs no center tap (a cheaper, better-utilized transformer) and because the PIV each diode must withstand is half that of the center-tap circuit. It pays with a drop of 1.4 V, which only matters in very low-voltage power supplies.
A “12 V” transformer delivers 12 V RMS. The peak is Vp = 12 × √2 = 17 V, and that is what ends up on the filter capacitor (minus 1.4 V from the bridge): about 15.6 V. Many projects fail because the capacitor is sized for 16 V when it actually receives almost 16 V. Always use a capacitor rated for at least 25 V.
06Capacitor filter and ripple
The rectifier output is direct but pulsating: it is good for charging a battery, not for powering a circuit. A large capacitor in parallel with the load charges at the peaks and delivers current in the valleys. The remaining undulation is called ripple (or ripple voltage).
Power supply with a 220/12 V RMS transformer, a bridge rectifier and a load drawing 500 mA. You want a ripple below 1 V peak to peak.
1. Peak voltage: Vp = 12 × 1.414 = 17 V.
2. Required capacitor, solving the ripple formula:
3. Choose the next higher standard value: 4700 µF (ripple 1.06 V, which is acceptable) or 6800 µF if you want some margin. Working voltage: 25 V at minimum.
4. Resulting DC: 17 − 1.4 − 0.5 ≈ 15.1 V.
5. Peak current in the diodes: as the capacitor charges, peaks of several amperes flow for about 1 ms. That is why a bridge for 500 mA of DC is chosen rated 2 A or more.
07Zener diode and regulation
An ordinary diode is destroyed if it is driven into reverse breakdown. The Zener is made to work precisely there: above its voltage VZ it conducts in reverse and keeps that voltage almost constant even as the current changes. That makes it a voltage reference.
12 V input, 1N4733A Zener (VZ = 5.1 V, Pmax = 1 W), 20 mA load. A minimum IZ of 10 mA is chosen so that the Zener stays well within the breakdown region.
Power in RS: (12 − 5.1)² / 220 = 0.22 W → a 1/2 W resistor.
Critical case: if the load is disconnected, everything goes through the Zener. IZ = 6.9 / 220 = 31 mA, and PZ = 5.1 × 31 mA = 0.16 W. That is well below a watt: the design tolerates having the load removed. Always check this case, because it is where Zeners burn out.
It is good for small currents (up to about 50 mA) and wastes energy: it draws the same with a load as without one. For more current a transistor is added as a follower (next topic) or an integrated regulator such as the 7805 or LM317 is used.
08Other diodes you will meet soon
| Type | What is special about it | Where it is used |
|---|---|---|
| LED | Emits light when conducting. Drop of 1.8 V (red) to 3.4 V (blue/white). Does not tolerate reverse voltage (5 V typical). | Indicators. Always with a series resistor. |
| Schottky | Drop of only 0.3 V and very fast switching. | Switching power supplies, reverse-polarity protection. |
| Varicap | Its reverse capacitance depends on the applied voltage. | Tuning of radios and TVs. |
| Photodiode | Its reverse current increases with light. | Sensors, remote controls, fiber optics. |
| Freewheeling diode | An ordinary diode used in antiparallel with an inductor. | Protects the transistor from the overvoltage a relay generates when it is switched off. |
Red LED (VF = 1.8 V, I = 15 mA) powered from 5 V — the resulting value is rounded to the nearest standard value:
Without a resistor, the LED conducts without limit and is destroyed in under a second. There are no exceptions to this rule.
09In the lab
With the multimeter on the diode range (symbol ▷|), measure in both directions. In forward bias it should read between 0.5 and 0.7 V (silicon) or 0.2 to 0.3 V (germanium/Schottky); in reverse, OL or out of range. If it reads low in both directions it is shorted; if it reads OL in both, it is open. Verify that the painted band matches the cathode (the black lead of the multimeter in forward bias).
Variable supply from 0 to 5 V, a 1N4007 diode in series with a 100 Ω resistor. Raise the voltage in steps of 0.1 V and note VD (with the voltmeter across the diode) and I (calculated from the drop across R). Plot I versus VD. The knee should appear at around 0.6 to 0.7 V, and it is clear that it is not a straight line: the diode does not obey Ohm's law.
220/12 V transformer, W04M bridge, 2200 µF/25 V capacitor, 470 Ω load. With the oscilloscope:
- View the sine wave at the secondary and measure Vp and the frequency.
- View the bridge output without the capacitor: both rectified half-cycles at 100 Hz.
- Connect the capacitor and measure the ripple with the oscilloscope on AC coupling and high sensitivity.
- Compare the measured ripple with the one calculated from Vr = I / (f·C).
- Double the load (235 Ω) and verify that the ripple doubles.
The transformer primary is at 220 V: wire it with the equipment unplugged, insulate it with heat-shrink tubing and never touch it while the supply is connected. Large electrolytic capacitors stay charged after disconnection: discharge them with a 1 kΩ resistor before handling the circuit board. And observe the polarity of the electrolytic: reversed, it explodes.
10Common mistakes
| Symptom | Usual cause |
|---|---|
| The power supply delivers nothing | A bridge diode backwards, or the primary fuse open. |
| The capacitor exploded or swelled | Reversed polarity, or insufficient working voltage (you forgot to multiply the RMS value by √2). |
| The DC is much lower than expected | The filter capacitor is missing: you are measuring the average value (0.637 Vp) instead of the peak. |
| 100 Hz hum in the powered equipment | Excessive ripple: capacitor too small for the current drawn. |
| The Zener heats up and burns out | The no-load case was not checked, where all the current goes through it. |
| The LED lasted a second | No current-limiting resistor, or connected in reverse at more than 5 V. |
| The bridge diodes burn out at power-on | The initial capacitor charging peak. Add a start-up resistor or choose a bridge with more current margin. |
11Self-assessment
Why does a semiconductor have exactly 4 valence electrons?
Because 4 is right halfway: it forms a crystal with complete covalent bonds (so it barely conducts when pure) but with very little energy a carrier is freed (so it conducts if doped or heated). With fewer electrons it would be a conductor and with more, an insulator.
Does an N-type material have a negative electric charge?
No. It is neutral: each donor atom contributes a free electron but also an extra proton in its nucleus. The N indicates that the majority carriers are electrons, not that there is a net charge.
What is the value of the potential barrier in silicon and what does it mean physically?
About 0.7 V. It is the potential generated by the depletion region, which is left with fixed ions of one sign on each side. For forward current to flow, the external source has to overcome that barrier; that is why the diode does not “turn on” until 0.7 V.
A 9 V RMS transformer feeds a bridge with a filter. What DC voltage is measured?
Vp = 9 × 1.414 = 12.7 V. Minus the 1.4 V of the two diodes that conduct: ≈ 11.3 V minus half the ripple. If the ripple is 1 V, about 10.8 V remain.
Why is bridge ripple easier to filter than half-wave rectifier ripple?
Because the ripple is at 100 Hz instead of 50 Hz: the capacitor has half the time to discharge between peaks. In the formula Vr = I/(f·C), when f doubles the ripple is halved with the same capacitor.
The load current of a power supply doubles. What happens to the ripple?
It doubles, because Vr is directly proportional to I. To keep the ripple you must also double the capacitor.
How is a Zener diode connected and why?
In reverse: the cathode (band) toward the positive. Because its job is to work in the breakdown region, where the voltage stays constant. Connected in forward bias it behaves like an ordinary 0.7 V diode and regulates nothing.
In a Zener regulator, what is the most demanding operating condition for the diode?
With the load disconnected and the maximum input voltage: all the current through RS flows through the Zener and its dissipation is maximum. That is the case to check against the diode's power rating.
What does a multimeter really measure on the “diode” range?
It injects a small, known current (typically 1 mA) and displays the resulting voltage drop. That is why a forward-biased silicon diode reads about 0.6 V: it is its VF at that current.