半导体器件原理 1.6 Carrier Actions in a PN Junction

About how minority and majority carriers move in a PN junction with external voltage applied, and the difference between short and long diodes.

Review

I_{D} = I_{0} (e^{\frac{q V_{A}}{k T}} - 1)

I_{0} = q (D_{n} \frac{n_{p 0}}{L_{n}} + D_{p} \frac{p_{n 0}}{L_{p}})

Mechanism for Minority Carriers

What’s the driving force for majority carriers?

When an electron move from N to P, it has two effects:
- Reduces concentration of electrons on the N side, causing diffusion
- Results charge imbalance, creating a positive charge behind, causing drift
The current density:
- $J_{n, diff} = q D_{n} \frac{d n}{d x}$
- $J_{n, drift} = q μ_{n} n E$ where $μ_{n}$ is the electron mobility (indicates how easily electrons move in the material)
On the N side, electrons are majority carriers, so $n$ is very large, while $d n / d x$ is small. Thus drift is the main driving force

In general:
- Diffusion is the main driving force to move minority carriers, because even a large $E$ cannot cause a large current with the limited number of carriers
- Drift is the main driving force to move majority carriers, because there are many carriers to move to balance out the effect of the charge imbalance

When electrons get pushed from N to P, recombination rate will increase with a tendency to restore the equilibrium
On average, an electron need to travel $L_{n}$ before recombination
When the diode is short, the neutral region length $W_{p}$ and $W_{n}$ may be smaller than $L_{n}$ and $L_{p}$
- $W_{p} < L_{n} and W_{n} < L_{p}$
- Note that the subscript of $L$ is different from that of $W$ , as the subscript of $L$ indicates the type of area, and the subscript of $W$ indicates the type of carrier
In this case, no combination can happen before an electron reaches the end of the diode, and the carrier distribution is a straight line
At the two ends, the diode contacts with metal, and the carrier concentration is determined by metal, as it has a large number of carriers
The metal will force carrier concentration to become $n_{p 0}$ and $p_{n 0}$ at the two ends
The current density becomes: $J_{n, diff} = q D_{n} \frac{n_{p 0}}{W_{p}} (e^{\frac{q V_{A}}{k T}} - 1)$ $J_{p, diff} = q D_{p} \frac{p_{n 0}}{W_{n}} (e^{\frac{q V_{A}}{k T}} - 1)$
The electron current on the P side must be supported by the continuous electron influx from the N side, which is driven by drift, thus the drift current on the N side equals the diffusion current on the P side
Similarly, the drift current of holes on the P side equals the diffusion current on the N side
The total current density is the combination of the two, and must be a constant across the diode

Why Carrier Distribution is a Straight Line

Within a short diode, no combination will happen, so the current flow must be a constant
$J_{n, diff} \propto \frac{d n}{d x}$
Thus, $\frac{d n}{d x}$ is a constant, and $n (x)$ is a straight line
Similar analysis also applies to holes

In a long diode, carriers can recombine before reaching the end of the diode
The carrier distribution is no longer a straight line
The diffusion current decreases as $x$ moves away from 0
The total current must be a constant, so the decrease in diffusion current must be compensated by an increase in drift current
The mechanism behind the increase in drift current:
- When an electron recombines with a hole, the carriers disappear, leving a negative charge behind (a hole is eliminated in the P side neutral region)
- The negative charge is quickly removed by the nearby holes
- Eventually, the missing hole must be compensated by an externally supplied hole coming from the battery connected to the P+ side
- This creates the drift current
- The current can be considered as:
  - By holes from the end of P side to the recombination point
  - By electrons from the recombination point to the end of N side
  - It’s like current changing lanes at the recombination point
  - Whenever a recombination happens, a drift current must be added to the left, but not the right, of the recombination point, thus the drift current increases as $x$ moves away from 0

Does Recombination Increase or Decrease Current?

Answer: Recombination increases the current.

Carriers face higher resistance when moving through low concentration areas
When electrons move on the N side, they face low resistance, and when they move to the P side, they face higher resistance
Recombination allows current to switch carriers from electrons to holes, which face lower resistance on the P side
It provides a mechanism for a higher current flow

By looking at the carrier distribution graph:
- If there is no recombination, the carrier distribution is a straight line, and $L_{n} \to \infty$ , $L_{p} \to \infty$ , so the gradient is small
- If there is recombination, $L_{n}$ and $L_{p}$ become smaller, so the gradient becomes larger
- A larger gradient means a larger current
- This additional current introduced by recombination is referred to as recombination current

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