Lecture 7 - Knowledge Check

Advanced Assessment: Haynes-Shockley, Recombination Kinetics & PN Junction Electrostatics.

Progress Question 1 of 20
1

Haynes-Shockley Parameter Extraction

In a Haynes-Shockley experiment conducted on an n-type Silicon bar, the distance between the injection and collection contacts is $d = 1.2 \text{ cm}$. A steady voltage of $V_1 = 24 \text{ V}$ is applied across the total bar length of $L = 2 \text{ cm}$. The peak of the excess hole pulse arrives at the collection contact at exactly $t_0 = 250 \text{ } \mu\text{s}$. The temporal width of the pulse (measured at $1/e$ of its peak amplitude) is $\Delta t = 35 \text{ } \mu\text{s}$. Calculate the hole diffusion coefficient $D_p$ in $\text{cm}^2/\text{s}$.

Step-by-Step Derivation

Correct Answer: B.
  1. Determine the electric field: $E_0 = \frac{V_1}{L} = \frac{24 \text{ V}}{2 \text{ cm}} = 12 \text{ V/cm}$.
  2. Extract the drift mobility: $\mu_p = \frac{d}{E_0 t_0} = \frac{1.2}{12 \times 250 \times 10^{-6}} = \frac{1.2}{0.003} = 400 \text{ cm}^2/\text{Vs}$.
  3. Utilize the pulse spread equation: $D_p = \frac{(\mu_p E_0)^2 (\Delta t)^2}{16 t_0}$.
  4. Substitute values: $D_p = \frac{(400 \times 12)^2 (35 \times 10^{-6})^2}{16 \times 250 \times 10^{-6}}$.
  5. Compute numerator: $(4800)^2 \times (35 \times 10^{-6})^2 = 23,040,000 \times 1.225 \times 10^{-9} = 0.028224$.
  6. Compute denominator: $16 \times 0.00025 = 0.004$.
  7. Final division: $D_p = \frac{0.028224}{0.004} = 7.056 \text{ cm}^2/\text{s} \approx 7.06 \text{ cm}^2/\text{s}$.
2

Quasi-Fermi Energy Splitting

An n-type silicon sample ($n_i = 10^{10} \text{ cm}^{-3}$) is uniformly doped with a donor concentration of $N_d = 10^{16} \text{ cm}^{-3}$. It is optically illuminated, injecting steady-state excess carriers such that $\delta n = \delta p = 10^{14} \text{ cm}^{-3}$. Assuming complete thermalization at $T = 300\text{ K}$ ($kT = 0.0259\text{ eV}$), calculate the total energy separation between the electron and hole quasi-Fermi levels, $E_{Fn} - E_{Fp}$.

Derivation

Correct Answer: C.
  1. Find the total electron concentration: $n = n_0 + \delta n \approx 10^{16} + 10^{14} = 1.01 \times 10^{16} \text{ cm}^{-3}$.
  2. Find the total hole concentration: $p_0 = \frac{n_i^2}{n_0} = \frac{10^{20}}{10^{16}} = 10^4 \text{ cm}^{-3}$. Then $p = p_0 + \delta p = 10^4 + 10^{14} \approx 10^{14} \text{ cm}^{-3}$.
  3. Calculate electron Quasi-Fermi level relative to $E_{Fi}$:
    $E_{Fn} - E_{Fi} = kT \ln(n/n_i) = 0.0259 \ln(1.01 \times 10^{16} / 10^{10}) = 0.0259 \times \ln(1.01 \times 10^6) \approx 0.358 \text{ eV}$.
  4. Calculate hole Quasi-Fermi level relative to $E_{Fi}$:
    $E_{Fi} - E_{Fp} = kT \ln(p/n_i) = 0.0259 \ln(10^{14} / 10^{10}) = 0.0259 \times \ln(10^4) = 0.0259 \times 9.210 \approx 0.239 \text{ eV}$.
  5. Total separation: $E_{Fn} - E_{Fp} = (E_{Fn} - E_{Fi}) + (E_{Fi} - E_{Fp}) = 0.358 + 0.239 = 0.597 \text{ eV}$.
3

SRH Recombination under High Injection

The full SRH recombination rate equation is $R = \frac{C_n C_p N_t (np - n_i^2)}{C_n(n + n') + C_p(p + p')}$. Assume a single midgap trap level ($n' = p' = n_i$) in a p-type semiconductor. Under the specific condition of extreme high-level injection where $\delta n = \delta p \gg p_0 \gg n_i$, what does the mathematical formulation for the excess carrier lifetime $\tau$ cleanly converge to?

Derivation

Correct Answer: D.
  1. Under extreme high injection, total concentrations are entirely dominated by the excess carriers: $n \approx \delta n$ and $p \approx \delta n$.
  2. The numerator simplifies: $C_n C_p N_t ((\delta n)^2 - n_i^2) \approx C_n C_p N_t (\delta n)^2$.
  3. The denominator simplifies: $C_n(\delta n) + C_p(\delta n) = \delta n (C_n + C_p)$. (Dropping $n'$ and $p'$).
  4. The recombination rate becomes: $R \approx \frac{C_n C_p N_t (\delta n)^2}{\delta n (C_n + C_p)} = \delta n \frac{C_n C_p N_t}{C_n + C_p}$.
  5. Lifetime is defined as $\tau = \frac{\delta n}{R}$. Thus, $\tau = \frac{C_n + C_p}{C_n C_p N_t} = \frac{1}{C_p N_t} + \frac{1}{C_n N_t}$.
  6. Since $\tau_{p0} = \frac{1}{C_p N_t}$ and $\tau_{n0} = \frac{1}{C_n N_t}$, the lifetime cleanly converges to the sum of the fundamental lifetimes: $\tau = \tau_{n0} + \tau_{p0}$.
4

PN Junction Doping Back-Calculation

In an ideal Silicon PN step junction at $300\text{ K}$, precision metrology indicates a built-in potential barrier of exactly $V_{bi} = 0.750\text{ V}$. The donor doping concentration on the n-side is known to be $N_d = 5 \times 10^{15}\text{ cm}^{-3}$. Given $n_i = 10^{10}\text{ cm}^{-3}$ and $V_t = 0.0259\text{ V}$, calculate the acceptor doping concentration $N_a$ on the p-side.

Derivation

Correct Answer: A.
  1. The built-in potential is governed by: $V_{bi} = V_t \ln\left(\frac{N_a N_d}{n_i^2}\right)$.
  2. Isolate the logarithmic term: $\frac{0.750}{0.0259} = 28.9575$.
  3. Exponentiate both sides: $\frac{N_a N_d}{n_i^2} = e^{28.9575} \approx 3.768 \times 10^{12}$.
  4. Isolate $N_a$: $N_a = \frac{3.768 \times 10^{12} \times n_i^2}{N_d}$.
  5. Substitute values: $N_a = \frac{3.768 \times 10^{12} \times 10^{20}}{5 \times 10^{15}} = \frac{3.768 \times 10^{32}}{5 \times 10^{15}} = 0.7536 \times 10^{17}$.
  6. Convert to standard scientific notation: $7.536 \times 10^{16} \text{ cm}^{-3} \approx 7.54 \times 10^{16} \text{ cm}^{-3}$.
5

Peak Field to Depletion Width Ratio

In a step PN junction under thermal equilibrium, what is the exact algebraic representation of the ratio of the peak electric field $E_{max}$ to the total depletion width $W$?

Derivation

Correct Answer: C.
  1. The peak electric field is $E_{max} = \frac{e N_d x_n}{\varepsilon_s}$.
  2. The total width is $W = x_n + x_p$. We apply charge neutrality $N_a x_p = N_d x_n$ to express $x_p$ in terms of $x_n$: $x_p = x_n \frac{N_d}{N_a}$.
  3. Substitute $x_p$ into $W$: $W = x_n + x_n \frac{N_d}{N_a} = x_n \left(\frac{N_a + N_d}{N_a}\right)$.
  4. Isolate $x_n$ as a fraction of $W$: $x_n = W \left(\frac{N_a}{N_a + N_d}\right)$.
  5. Substitute $x_n$ back into the $E_{max}$ equation: $E_{max} = \frac{e N_d}{\varepsilon_s} \left[ W \left(\frac{N_a}{N_a + N_d}\right) \right]$.
  6. Divide by $W$ to find the ratio: $\frac{E_{max}}{W} = \frac{e}{\varepsilon_s} \frac{N_a N_d}{N_a + N_d}$.
6

Intrinsic Fermi Level Alignment

In the depletion region of a forward-biased PN junction, the Quasi-Fermi levels for electrons ($E_{Fn}$) and holes ($E_{Fp}$) split by an amount $eV_A$. At what precise spatial location $x$ does the intrinsic Fermi level $E_{Fi}$ exactly intersect the geometric average of the two quasi-Fermi levels, meaning $E_{Fi} = \frac{E_{Fn} + E_{Fp}}{2}$?

Explanation

Correct Answer: B. The carrier concentrations are defined by the distance between the respective quasi-Fermi level and the intrinsic level: $n = n_i e^{(E_{Fn}-E_{Fi})/kT}$ and $p = n_i e^{(E_{Fi}-E_{Fp})/kT}$. For $E_{Fi}$ to sit exactly halfway between the quasi-Fermi levels, the energy distances must be equal: $E_{Fn} - E_{Fi} = E_{Fi} - E_{Fp}$. If these exponential arguments are equal, the resulting concentrations must be equal ($n = p$). Because the junction doping is generally asymmetrical, the location where $n=p$ is not at $x=0$, but rather shifted toward the more lightly doped side.
7

Haynes-Shockley Lifetime Extraction

The area $S$ under a Haynes-Shockley voltage readout curve is proportional to the number of surviving minority carriers, scaling as $S \propto \exp(-t_0/\tau)$, where $t_0$ is the peak arrival time. Two sequential experimental runs are performed by merely varying the applied electric field $E_0$.

Run 1: Arrival $t_{01} = 200\text{ }\mu\text{s}$, measured Area $S_1 = 150$.
Run 2: Arrival $t_{02} = 100\text{ }\mu\text{s}$, measured Area $S_2 = 407.7$.

Determine the minority carrier recombination lifetime $\tau$.

Derivation

Correct Answer: B.
  1. Write the proportional equations: $S_1 = K e^{-t_{01}/\tau}$ and $S_2 = K e^{-t_{02}/\tau}$.
  2. Set up a ratio to eliminate the unknown proportionality constant $K$: $\frac{S_2}{S_1} = \frac{e^{-t_{02}/\tau}}{e^{-t_{01}/\tau}} = e^{(t_{01} - t_{02})/\tau}$.
  3. Substitute the known experimental values: $\frac{407.7}{150} = e^{(200 - 100)/\tau} \implies 2.718 = e^{100/\tau}$.
  4. Recognize that $2.718 \approx e^1$. Therefore, we take the natural log of both sides: $\ln(2.718) \approx 1$.
  5. Solve the exponent: $1 = \frac{100}{\tau} \implies \tau = 100\text{ }\mu\text{s}$.
8

Depletion Voltage Division

The total built-in potential barrier $V_{bi}$ is the sum of the electrostatic potential drop across the n-side depletion region ($\Delta V_n$) and the p-side depletion region ($\Delta V_p$). By integrating the respective triangular electric field distributions across each half of the space charge region, what is the exact algebraic ratio of $\Delta V_n$ to $\Delta V_p$?

Derivation

Correct Answer: D.
  1. The potential drop across the n-side is $\Delta V_n = \frac{e N_d x_n^2}{2 \varepsilon_s}$.
  2. The potential drop across the p-side is $\Delta V_p = \frac{e N_a x_p^2}{2 \varepsilon_s}$.
  3. The raw ratio is $\frac{\Delta V_n}{\Delta V_p} = \frac{N_d x_n^2}{N_a x_p^2} = \frac{N_d}{N_a} \left(\frac{x_n}{x_p}\right)^2$.
  4. Utilize the charge neutrality condition $N_a x_p = N_d x_n$ to isolate the ratio of widths: $\frac{x_n}{x_p} = \frac{N_a}{N_d}$.
  5. Substitute this width ratio back into the voltage ratio equation: $\frac{\Delta V_n}{\Delta V_p} = \frac{N_d}{N_a} \left(\frac{N_a}{N_d}\right)^2 = \frac{N_a}{N_d}$.
  6. Insight: The voltage drop is heavily concentrated in the more lightly doped side.
9

SRH Trap Density Perturbation

The simplified SRH recombination rate for an n-type material under low-level injection is $R = \delta p / \tau_{p0}$. During an irradiation event, the physical trap density $N_t$ is abruptly quadrupled ($4 \times N_t$). However, the external optical source generation rate $g'$ is held perfectly constant (meaning $g' = R$ in steady state). What happens to the steady-state excess hole concentration $\delta p$ as a result of the quadrupled trap density?

Derivation

Correct Answer: A.
  1. The minority carrier lifetime is inversely proportional to the trap density: $\tau_{p0} = \frac{1}{C_p N_t}$. If $N_t$ is quadrupled, the new lifetime is $\tau_{p0}^{new} = \tau_{p0} / 4$.
  2. In steady state, the generation rate perfectly matches the recombination rate: $g' = R = \delta p / \tau_{p0}$.
  3. Because the optical source is held constant, $g'$ (and therefore $R$) must remain constant regardless of internal material changes.
  4. Equating the two states: $\frac{\delta p_{old}}{\tau_{p0}} = \frac{\delta p_{new}}{\tau_{p0}^{new}}$.
  5. Substitute the new lifetime: $\frac{\delta p_{old}}{\tau_{p0}} = \frac{\delta p_{new}}{\tau_{p0}/4} \implies \delta p_{new} = \delta p_{old} / 4$.
  6. Because the traps pull holes out 4 times faster, the standing pool of holes is drained to 25% of its original size before reaching the dynamic equilibrium state that matches the constant influx.
10

Ambipolar Transport Coefficient

In a Haynes-Shockley experiment, an ambipolar pulse drifts and spreads. The spreading is governed by the effective ambipolar diffusion coefficient $D' = \frac{\mu_n n D_p + \mu_p p D_n}{\mu_n n + \mu_p p}$. Consider a near-intrinsic semiconductor sample where electrons are exactly twice as abundant as holes ($n = 2p$) and electron mobility is exactly three times hole mobility ($\mu_n = 3\mu_p$). Assuming the Einstein relation inherently holds true ($D_n = 3D_p$), evaluate the effective ambipolar diffusion coefficient $D'$ as a fractional multiple of the fundamental hole diffusion coefficient $D_p$.

Derivation

Correct Answer: C.
1. Start with the core equation: $D' = \frac{\mu_n n D_p + \mu_p p D_n}{\mu_n n + \mu_p p}$.
2. Substitute the given ratios: $\mu_n = 3\mu_p$, $n = 2p$, and $D_n = 3D_p$.
3. Evaluate the Numerator: $(3\mu_p)(2p)D_p + (\mu_p)(p)(3D_p)$.
    $= 6 \mu_p p D_p + 3 \mu_p p D_p = 9 \mu_p p D_p$.
4. Evaluate the Denominator: $(3\mu_p)(2p) + (\mu_p)(p)$.
    $= 6 \mu_p p + \mu_p p = 7 \mu_p p$.
5. Divide the terms: $D' = \frac{9 \mu_p p D_p}{7 \mu_p p}$.
6. Cancel the common $\mu_p p$ terms to find the final ratio: $D' = \frac{9}{7} D_p \approx 1.286 D_p$.
11

Haynes-Shockley Variable Scaling

In a Haynes-Shockley experiment, the applied electric field $E_0$ is suddenly reduced to exactly half its original value ($E_0 / 2$) for a second run on the identical sample. How does the newly measured temporal pulse width $\Delta t'$ (measured at $1/e$ of the peak amplitude) geometrically scale compared to the original $\Delta t$?

Derivation

Correct Answer: C.
  1. The peak arrival time is $t_0 = d / (\mu_p E_0)$. Halving $E_0$ strictly doubles the arrival time: $t_0' = 2t_0$.
  2. The pulse spread equation is $D_p = \frac{(\mu_p E_0)^2 (\Delta t)^2}{16 t_0}$.
  3. Rearrange for $\Delta t$: $\Delta t = \frac{\sqrt{16 D_p t_0}}{\mu_p E_0}$.
  4. Substitute the new values ($t_0' = 2t_0$ and $E_0' = E_0/2$): $\Delta t' = \frac{\sqrt{16 D_p (2t_0)}}{\mu_p (E_0 / 2)}$.
  5. Pull out the constants: $\Delta t' = 2 \sqrt{2} \left[ \frac{\sqrt{16 D_p t_0}}{\mu_p E_0} \right] = 2\sqrt{2} \Delta t$.
12

Ambipolar Mobility Collapse (High Injection)

The ambipolar mobility is given by $\mu' = \frac{\mu_n \mu_p (p - n)}{\mu_n n + \mu_p p}$. Consider a lightly doped p-type semiconductor subjected to an extraordinarily intense laser pulse, plunging it into extreme high-level injection where $\delta n = \delta p \gg p_0$. What physical behavior does the ambipolar packet exhibit under an applied external electric field?

Explanation

Correct Answer: C. Under extreme high-level injection, the injected excess carrier concentrations completely overwhelm the background doping, meaning $n \approx p \approx \delta n$. Substituting $n = p$ into the numerator of the ambipolar mobility equation yields $(p - n) = 0$. Therefore, $\mu' = 0$. The packet mathematically loses its ability to drift as a cohesive unit because the internal Coulomb field perfectly counteracts the external field. It will only diffuse radially outward.
13

SRH Recombination Rate Optimization

The SRH recombination rate is governed by $R = \frac{C_n C_p N_t (np - n_i^2)}{C_n(n + n') + C_p(p + p')}$. To mathematically maximize the non-radiative recombination rate in a manufactured device (e.g., to "kill" minority carrier lifetime for fast switching), at what precise energy level within the bandgap should the dopant traps ($E_t$) ideally be introduced?

Derivation

Correct Answer: C.
  1. The terms $n'$ and $p'$ represent the carrier concentrations that would exist if the Fermi level were located exactly at the trap energy level $E_t$.
  2. Mathematically, $n' = n_i e^{(E_t - E_{Fi})/kT}$ and $p' = n_i e^{(E_{Fi} - E_t)/kT}$.
  3. To maximize the recombination rate $R$, we must completely minimize the denominator.
  4. The sum $(n' + p')$ reaches its absolute mathematical minimum when $E_t = E_{Fi}$, which forces $n' = p' = n_i$.
  5. Therefore, midgap traps are the most highly effective recombination centers.
14

Quasi-Fermi Level Thermal Sensitivity

An n-type semiconductor ($N_d = 10^{16} \text{ cm}^{-3}$) is illuminated to maintain a steady-state excess carrier concentration $\delta n = \delta p = 10^{13} \text{ cm}^{-3}$. If the ambient temperature is increased significantly while maintaining the exact same optical illumination intensity, what happens to the total energy separation between the quasi-Fermi levels ($E_{Fn} - E_{Fp}$)?

Explanation

Correct Answer: B. The separation is $E_{Fn} - E_{Fp} = kT \ln\left(\frac{np}{n_i^2}\right)$. While the $kT$ multiplier implies a linear increase, the denominator $n_i^2$ inside the logarithm scales exponentially with temperature ($n_i^2 \propto T^3 e^{-E_g/kT}$). This massive exponential growth of $n_i^2$ completely dominates the logarithm, causing the $\ln$ term to shrink rapidly. Physically, at high temperatures, the massive thermal generation of intrinsic carriers drowns out the optical excess carriers, pushing the system closer to thermal equilibrium where $E_{Fn} = E_{Fp}$.
15

PN Junction Field Scaling

Consider a perfectly symmetric Silicon PN junction ($N_a = N_d = N_0$). If an entirely new junction is fabricated with exactly 100 times the doping on both sides ($N_a = N_d = 100 N_0$), how does the maximum electric field $E_{max}$ approximately scale at thermal equilibrium? (Assume the slight logarithmic change in $V_{bi}$ is negligible).

Derivation

Correct Answer: B.
1. The peak electric field is $E_{max} = \frac{e N_d x_n}{\varepsilon_s}$.
2. In a symmetric junction, $x_n = W/2$.
3. The total depletion width is $W = \sqrt{\frac{2\varepsilon_s V_{bi}}{e} \frac{N_a+N_d}{N_a N_d}} = \sqrt{\frac{4\varepsilon_s V_{bi}}{e N_0}}$.
4. Thus $E_{max} = \frac{e N_0}{2\varepsilon_s} \sqrt{\frac{4\varepsilon_s V_{bi}}{e N_0}} = \sqrt{\frac{e N_0 V_{bi}}{\varepsilon_s}}$.
5. If the uniform doping $N_0$ is scaled by 100, $E_{max}$ scales by $\sqrt{100} = 10$.
(The slight logarithmic increase in $V_{bi}$ will make the true factor slightly higher than 10, but 10 is the overwhelmingly dominant scaling factor).
16

Asymmetrical Depletion Dynamics

In a highly asymmetrical $p^+n$ junction step diode where $N_a \gg N_d$, which of the following statements strictly defines the electrostatic characteristics of the depletion region?

Explanation

Correct Answer: D. Charge neutrality physically dictates that $N_a x_p = N_d x_n$. Since $N_a \gg N_d$, it mathematically requires that $x_n \gg x_p$. Thus, the total depletion width $W = x_n + x_p \approx x_n$. The space charge region must expand deeply into the lightly doped side to uncover enough fixed charge to balance the heavy charge tightly packed into the microscopic thin layer on the highly doped side.
17

Lifetime Asymmetry in SRH Recombination

In the simplified SRH model under low-level injection, the excess carrier lifetime in a p-type material is $\tau_{n0} = \frac{1}{C_n N_t}$, while in an n-type material it is $\tau_{p0} = \frac{1}{C_p N_t}$. If a specific physical trap defect has an electron capture cross-section that is strictly 10 times larger than its hole capture cross-section ($C_n = 10 C_p$), what is the ratio of the minority hole lifetime in an n-type sample to the minority electron lifetime in a p-type sample ($\tau_{p0} / \tau_{n0}$) assuming identical trap densities?

Derivation

Correct Answer: C.
  1. The lifetimes are directly inversely proportional to their respective capture probabilities.
  2. Set up the ratio: $\frac{\tau_{p0}}{\tau_{n0}} = \frac{1 / (C_p N_t)}{1 / (C_n N_t)} = \frac{C_n}{C_p}$.
  3. Since the problem states $C_n = 10 C_p$, the ratio evaluates to exactly 10.
  4. Physically, the trap is highly effective at snagging electrons, meaning electrons die quickly in p-type material. It is poor at catching holes, meaning holes survive much longer in n-type material.
18

Intrinsic SRH Recombination

Using the full SRH formulation for an absolutely intrinsic semiconductor in strict thermal equilibrium ($n = p = n_i$), what does the net mathematical recombination rate $R$ strictly evaluate to?

Explanation

Correct Answer: B. The numerator of the full SRH equation is strictly $C_n C_p N_t (np - n_i^2)$. In thermal equilibrium, the mass action law strictly holds, meaning $np = n_i^2$. Therefore, the entire numerator is perfectly zero. In thermal equilibrium, spontaneous generation exactly balances structural recombination, yielding a net recombination rate $R$ of identically zero.
19

Ambipolar Diffusion vs Electric Field

The ambipolar diffusion coefficient $D'$ mathematically dictates the thermal spreading of an injected excess carrier packet. Does the absolute magnitude of a large applied external electric field $E_0$ mathematically alter the numerical value of the ambipolar diffusion coefficient $D'$?

Explanation

Correct Answer: C. The mathematical formulation for $D' = \frac{\mu_n n D_p + \mu_p p D_n}{\mu_n n + \mu_p p}$ is entirely independent of the applied electric field $E_0$. The external field only dictates the ambipolar drift velocity (via $\mu'$). The geometric spreading (variance) of the packet as it travels down the bar is governed purely by $D'$ and time $t$, completely irrespective of how fast the packet is being dragged by $E_0$.
20

PN Junction Neutral Bulk Profile

In deriving the electrostatics of the PN junction, we apply the approximation that the space charge region abruptly ends at $+x_n$ and $-x_p$. What is the mathematical charge density $\rho(x)$ assumed to be at any $x > x_n$ (deep into the heavily doped n-type neutral bulk)?

Explanation

Correct Answer: C. Outside the depletion region (for $x > x_n$ and $x < -x_p$), the semiconductor is assumed to be in a perfect state of macroscopic charge neutrality. The fixed positive donor ions are perfectly neutralized by the mobile sea of majority carrier electrons ($n_0 \approx N_d$). Therefore, the net charge density $\rho(x)$ is exactly zero, which mathematically requires the electric field to remain completely flat (at zero) deep within the neutral bulk regions.