Semiconductor depletion and charge collection
Separate depletion width, induced charge and the sources of an energy-resolution peak.
On this page
The question for today
What you’ll understand
- Compute a depletion voltage in a one-sided junction model.
- Relate pair creation and weighting potential to measured charge.
- Combine Fano fluctuations and input-referred electronic noise.
A semiconductor detector uses an electric field to collect charge produced by ionizing radiation. Its small mean pair-creation energy can provide many carriers without avalanche amplification, but those carriers are useful only if field geometry, trapping and readout preserve the signal. Reverse bias creates or enlarges a depleted sensitive region; it is not itself a statement that every deposited charge is fully measured. Understanding the detector requires separating electrostatics, transport, induced-current formation and the statistical width of a calibrated energy peak.
Deplete a low-doped region under explicit assumptions
Consider an abrupt one-sided planar junction with uniform effective ionized dopant density in the lightly doped bulk. The highly doped side contributes little depletion width. Solving Poisson’s equation gives a width proportional to the square root of built-in plus applied reverse-bias voltage. Use the dopant-density magnitude in that width formula, and use a consistent permittivity rather than a dimensionless relative dielectric constant alone. Assume the built-in field alone does not fully deplete the sensor. The approximation neglects complicated space-charge profiles and edge fields.
The width cannot exceed the physical sensor thickness. At full depletion, increasing reverse bias changes the field and collection speed rather than indefinitely increasing sensitive thickness. A parallel-plate capacitance estimate decreases as depleted width increases until that geometric limit. Breakdown, leakage, surface conditions and radiation damage impose further constraints. In particular, a fully depleted sensor can still lose carriers through trapping or recombination, so full depletion should not be equated with unit charge-collection efficiency.
Count pairs once and include both carrier motions
Deposited energy produces electron-hole pairs with a mean creation energy larger than the bandgap, because energy also goes into lattice excitations. Use a specified material and temperature for this mean energy. A fully collected pair supplies one elementary charge magnitude to a readout electrode after both carrier contributions are included, not two elementary charges merely because two carriers exist. The carrier paths and readout geometry determine how the signal develops in time.
The weighting potential is calculated with the readout electrode at unit potential, other electrodes at zero and no space charge. It differs from the actual bias potential driving carrier motion. The induced electrode-charge change from a signed moving carrier depends on its weighting-potential endpoints. Electrons and holes moving to opposite electrodes contribute complementary pieces for a simple fully collecting planar detector. Trapping, partial depletion, finite shaping time and segmented weighting fields can all alter the measured integral or pulse shape.
Distinguish intrinsic fluctuations from readout noise
At a fixed fully deposited energy, the charge-creation variance is often smaller than the Poisson value because electronic and lattice energy channels are constrained by energy conservation. A Fano factor describes that intrinsic variance in the stated material model. Independent electronic noise can be expressed as an equivalent input number of electrons and added in quadrature. This convention must be input referred; using an output noise without undoing gain gives the wrong energy width.
The resulting expression is a useful ideal peak-width budget, not a complete model for every spectrum. Incomplete collection can produce asymmetric low-energy tails rather than another Gaussian variance. Electronic noise also depends on detector capacitance, leakage current, temperature and shaping time. A Gaussian FWHM conversion applies to the approximately Gaussian core only. Measuring a broader peak therefore does not by itself disprove the Fano description; collection, calibration, electronic and source widths must be investigated separately.
Work it through
A pair count and a Gaussian core budget
Use Edep=36.5 keV, wi=3.65 eV per pair, F=0.10 and rms ENC=30 electrons. Assume complete collection and independent Gaussian electronics noise. Find mean pairs, charge magnitude and energy rms width; estimate Gaussian FWHM.
Converting energy to 36,500 eV gives 10,000 mean pairs. Their full collected charge magnitude is 10,000 times e, or about 1.602 fC using e=1.602 times 10^-19 C.
Intrinsic count variance is 0.10 times 10,000 = 1000. Adding electronic variance 30 squared gives total count variance 1900 and rms count width sqrt(1900), approximately 43.59 electrons.
Multiplying by 3.65 eV gives energy rms width about 159.1 eV. The Gaussian-core FWHM is approximately 374.7 eV. Counting two charges per pair would incorrectly double the signal calibration.
The mean is 10,000 pairs or 1.602 fC, with rms width about 159.1 eV and Gaussian FWHM about 374.7 eV.
The living notebook
Explore the model
Interactive lab · function
A depletion square-law mapping
Reset and set BOTH Coefficient a = 1 and Position x = 0.50; predict y = 0.25. Compare a fresh Reset run with a = 1 and x = 1; predict y = 1. If x is a scaled depletion width and y a scaled total junction voltage, this shows the ideal pre-full-depletion square law. The lab has no dopants, finite sensor thickness, weighting field, trapping or electronics noise.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA scaling relation is conditional on the electrostatic model and its finite thickness.
Pause & reason
Check your understanding
Your turn
Try it yourself
A stipulated one-sided junction has pre-limit width w=10 micrometers times sqrt(VR/volt), negligible built-in voltage and thickness 200 micrometers. Find width at 100 V and full-depletion voltage. Would 900 V make a 300-micrometer sensitive layer? How does ideal planar capacitance change from 100 V to full depletion?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Apply the square law only until the physical thickness is reached.
Compare with the solution
At 100 V the width is 100 micrometers, so the layer is only half depleted. Reaching 200 micrometers needs the square root of voltage to be 20, hence 400 V in this stipulated model. At 900 V the unconstrained formula would give 300 micrometers, but the actual width is capped at 200 micrometers; breakdown and field changes need separate evaluation. With fixed electrode area, doubling width from 100 to 200 micrometers halves the ideal capacitance. Neither calculation guarantees complete collection after trapping or finite-time shaping.
Keep the idea
Take it with you
- Depletion depth is bounded by the physical sensor.
- Weighting fields determine induced signals and differ from bias fields.
- Fano width and electronic noise are distinct parts of a resolution budget.
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