Doping is the deliberate addition of a tiny amount of impurity atoms to silicon to control how many free carriers it has and what type they are. Adding phosphorus creates n-type silicon with extra free electrons; adding boron creates p-type silicon with mobile "holes". Every transistor, well, and contact on a CMOS chip is defined by where these two types of doping are placed.
12 min read | Beginner Level
What You'll Learn
- Why pure silicon is almost useless for building circuits
- How phosphorus creates n-type silicon and boron creates p-type silicon
- What a hole is and why it behaves like a positive carrier
- How doping level controls resistance, and where the proportion rule breaks down
Why Pure Silicon Is Not Enough
Pure (intrinsic) silicon at room temperature has only about 1010 free electrons per cm3 - roughly one for every 5 trillion silicon atoms. Every electron that breaks free also leaves behind one missing bond, so pure silicon has an equal number of electrons and holes.
That carrier count has two problems for circuit design:
- It is too small to carry useful current through a transistor.
- It is set by temperature alone, so the designer has no control over it.
Doping fixes both. It lets you choose the carrier count, typically anywhere from about 1014 to 1020 per cm3, and it lets you choose which carrier dominates.
N-Type Doping: Adding Extra Electrons
Silicon has four outer (valence) electrons, and each forms a bond with one of four neighbours. Phosphorus has five. When a phosphorus atom takes the place of a silicon atom in the crystal, four of its electrons form the usual bonds. The fifth has no bond to join.
That fifth electron is held very weakly - it needs only about 0.045 eV to break free, compared with 1.12 eV for a bonded silicon electron. At room temperature practically every one of them is already free. The phosphorus atom has donated an electron, so it is called a donor.
N-type doping, step by step 1. Replace one Si atom with P (P has 5 outer electrons) 2. 4 electrons bond with Si neighbours 3. 5th electron is loosely held (~0.045 eV to free it) 4. Room-temperature heat frees it -> 1 new free electron 5. P atom is left as a fixed +1 ion -> cannot move, carries no current Result: 1 free electron per phosphorus atom Other common donors: arsenic, antimony
In n-type silicon, electrons are the majority carriers and holes are the minority carriers.
P-Type Doping: Creating Holes
Boron has only three outer electrons. When it replaces a silicon atom, it can complete only three of the four bonds. The fourth bond is missing an electron. A neighbouring bonded electron can easily hop into that gap, which moves the gap to the neighbour's position. Then another electron hops in, and the gap moves again.
That moving gap is a hole. Because it represents a missing negative charge, it behaves exactly like a positive carrier: under an electric field it drifts in the opposite direction to electrons. Boron has accepted an electron from the crystal, so it is called an acceptor.
A helpful picture is a car park:
- P-type is a completely full car park with a few empty spaces. When a car moves forward into an empty space, the empty space moves backward. Watching the empty spaces is easier than tracking every car, so we treat the space itself as the thing that moves.
- N-type is a completely full car park with a few extra cars that have no space at all. Those extra cars drive freely on the road above the car park.
| Property | N-type | P-type |
|---|---|---|
| Typical dopant | Phosphorus, arsenic | Boron |
| Dopant outer electrons | 5 | 3 |
| Dopant name | Donor | Acceptor |
| Fixed ion left behind | Positive | Negative |
| Majority carrier | Electrons | Holes |
| Minority carrier | Holes | Electrons |
| Overall charge of the material | Neutral | Neutral |
Quick Check
Q: Is a piece of n-type silicon negatively charged?
Show Answer
No. Every free electron came from a phosphorus atom that is now a fixed positive ion. The positive ions and the free electrons cancel exactly, so the material is electrically neutral. "N-type" describes the majority carrier, not the net charge.
How Many Carriers? The Proportion Rule
Each dopant atom contributes one carrier, so the carrier count follows the dopant count directly. Dopant levels are tiny in relative terms - even heavy doping replaces only about 1 silicon atom in 500 - but the effect is enormous because pure silicon starts with so few carriers.
Proportion rule: dopant -> carriers -> resistance
more dopant atoms -> proportionally more majority carriers
more carriers -> lower resistance
Example: add 1e16 phosphorus atoms per cm^3
(about 1 P atom per 5 million Si atoms)
free electrons: 1e10 -> 1e16 (one million times more)
resistivity: ~230,000 -> ~0.5 ohm-cm
There is a second, less obvious rule. At a fixed temperature, the product of the electron count and the hole count stays constant. Push the electrons up and the holes fall by the same factor, because the extra electrons fill in holes when they meet them.
Majority up -> minority down (silicon, room temperature) electrons x holes = constant (about 1e20 per cm^6) pure Si: electrons 1e10 holes 1e10 n-type, 1e16 P: electrons 1e16 holes 1e4 p-type, 1e17 B: holes 1e17 electrons 1e3
Minority carriers are few, but they are not irrelevant. They are responsible for much of the leakage current wherever p-type and n-type regions meet.
Doping Levels in a CMOS Process
Chip designers use a shorthand for doping strength. A plain "n" or "p" means moderate doping; a "+" means heavy doping; a "-" means light doping.
| Region | Type | Approx. doping (per cm3) | Approx. resistivity (Ω·cm) | Purpose |
|---|---|---|---|---|
| Wafer substrate | p- | ~1015 | ~10 | Body for NMOS transistors |
| N-well | n | ~1017 | ~0.1 | Body for PMOS transistors |
| Channel implant | p or n | ~1017 - 1018 | - | Tunes the transistor's turn-on voltage |
| NMOS source/drain | n+ | ~1020 | ~0.001 | Low-resistance terminals for NMOS |
| PMOS source/drain | p+ | ~1020 | ~0.001 | Low-resistance terminals for PMOS |
Values vary widely between process nodes and foundries. Treat this table as a sense of scale: the body regions are lightly doped, and the terminals that must carry current are doped about a thousand to a hundred thousand times more heavily.
When a region receives both donors and acceptors, they cancel each other. Only the net difference counts. Fabs use this deliberately: an n-well is created by implanting phosphorus into a p-type wafer heavily enough to overwhelm the boron already there. This is called compensation or counter-doping.
Where Is Doping Control Used? Multi-Threshold Cell Libraries
Standard-cell libraries for phones and laptops usually offer the same logic gate in several "threshold voltage flavours", such as low-Vt, standard-Vt, and high-Vt. The versions differ mainly in channel doping (and, in newer nodes, gate materials). Low-Vt cells switch faster but leak more; high-Vt cells leak less but are slower. Synthesis and place-and-route tools swap between them to meet timing on critical paths while keeping standby power low elsewhere.
Electrons vs. Holes: Not Equal Carriers
Electrons and holes do not move equally easily. In silicon, electrons are roughly 2.5 to 3 times more mobile than holes. At the same doping level, n-type silicon therefore has about a third of the resistance of p-type silicon.
This one fact has a direct consequence in CMOS: a PMOS transistor, which conducts using holes, must be made wider than an NMOS transistor to deliver the same current. You will see this asymmetry every time you look at a standard-cell layout.
Where Is Doped Silicon Used as a Resistor?
Analog blocks on digital chips - bandgap references, bias circuits, and ESD protection networks - often need resistors. Instead of adding a separate material, designers use strips of doped silicon or doped polysilicon, choosing the doping level to get the resistance per square they need. Lightly doped strips give high resistance in a small area; heavily doped strips give low resistance.
Quick Check
Q: A region is implanted with 4 × 1017 boron atoms and 1 × 1017 phosphorus atoms per cm3. What type is it, and roughly how many majority carriers does it have?
Show Answer
P-type, because boron outnumbers phosphorus. The net doping is 4 × 1017 - 1 × 1017 = 3 × 1017 holes per cm3.
Common Mistakes
Mistake #1: Believing n-type silicon is negatively charged
Each free electron is balanced by a fixed positive donor ion, so n-type material is neutral. The same applies to p-type silicon, where each hole is balanced by a fixed negative acceptor ion. The "n" and "p" describe the majority carrier, not the net charge.
Mistake #2: Thinking a hole is a positive ion
The boron ion is negative and fixed in the crystal; it never moves. The hole is a missing electron in a bond that shifts from atom to atom. Confusing the two leads to wrong pictures of where charge sits and what carries current, especially once p-type and n-type regions meet.
Mistake #3: Assuming resistance falls in exact proportion to doping forever
The proportion rule works well at low and moderate doping. At high doping, carriers collide with the extra dopant ions more often and move more slowly. Going from 1014 to 1020 atoms per cm3 is a million-fold increase in dopant, but resistivity drops by only about 60,000×. Heavily doped regions still conduct very well, but not as well as a straight-line estimate predicts.
Mistake #4: Ignoring minority carriers
In n-type silicon doped at 1016, holes are outnumbered a trillion to one. They still matter: minority carriers are the source of the small reverse current in junctions, and their number grows quickly with temperature. Ignoring them hides a major source of leakage.
Practice Exercise
Challenge: Track the Carriers Through Two Implants
A region of silicon starts with 1 × 1017 boron atoms per cm3. Assume room temperature and that electrons × holes = 1020 per cm6.
- What type is it, and how many electrons and holes does it have?
- A second implant adds 3 × 1017 phosphorus atoms per cm3 to the same region. What type is it now, and how many electrons and holes does it have?
- Which state has lower resistance, and why?
Hint
Subtract the smaller dopant count from the larger to get the net doping. Then use the constant product to find the minority carrier count.
Solution
1. P-type. holes = 1e17 electrons = 1e20 / 1e17 = 1e3 2. Net doping = 3e17 P - 1e17 B = 2e17 donors -> n-type. electrons = 2e17 holes = 1e20 / 2e17 = 5e2 3. The n-type state has lower resistance: - it has twice as many majority carriers (2e17 vs 1e17) - its majority carriers are electrons, which are ~3x more mobile
This is exactly how an n-well is formed inside a p-type wafer: implant enough donors to overwhelm the acceptors already present.
Summary
In this tutorial, you learned:
- Pure silicon has too few carriers, and their number depends only on temperature
- Phosphorus (5 outer electrons) donates a free electron and creates n-type silicon
- Boron (3 outer electrons) creates a hole, a mobile missing electron that behaves as a positive carrier
- Doped silicon stays electrically neutral because every carrier is balanced by a fixed dopant ion
- More dopant gives proportionally more majority carriers and lower resistance, with diminishing returns at very high doping
- Raising the majority carrier count lowers the minority carrier count by the same factor
- Electrons are about 3× more mobile than holes, which is why PMOS devices are drawn wider
