University Physics IV · topic UPIV-12.09
The p-n junction and semiconductor devices
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
The p–n junction and semiconductor devices: predict a finite change
A p–n junction is the boundary where p-type and n-type semiconductor meet: diffusion and drift cancel there at equilibrium, leaving a depletion region swept clear of mobile carriers and a built-in potential step fixed by the two doping levels. Use the supplied topic model Y_out = sqrt(1.29316*10^15*(10^(-pa)+10^(-pd))*max(0.0595263*(pa+pd-20)-v,0.02)), where Y_out is DEPLETION WIDTH (µm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. pa is p-side doping, log_10 N_a in cm^-3, dimensionless. Keep pd (n-side doping, log_10 N_d in cm^-3) = 16; v (Applied bias V, forward positive) = 0 V fixed. For this exercise use only 16.6062 ≤ pa ≤ 17.3938 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at pa = 16.7795 and 16.8897. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δpa across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
DEPLETION WIDTH against pa
Predictions from the supplied The p–n junction and semiconductor devices model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): DEPLETION WIDTH / µm
Horizontal axis (x): pa: p-side doping, log_10 N_a in cm^-3
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| pa: p-side doping, log_10 N_a in cm^-3 | DEPLETION WIDTH / µm |
|---|---|
| 16.60625 | 0.3479454429845339 |
| 16.803125 | 0.33773143669641964 |
| 17 | 0.3317787712534363 |
| 17.196875 | 0.3286967162221316 |
| 17.39375 | 0.32751356431561585 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of pa separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At pa = 16.7795, substitute into Y_out = sqrt(1.29316*10^15*(10^(-pa)+10^(-pd))*max(0.0595263*(pa+pd-20)-v,0.02)): Y₀ = 0.3387 µm.
- At pa = 16.8897, the same fixed-parameter model gives Y₁ = 0.33468 µm.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.00401912 µm; the input change is Δpa = 0.11025.
- Average sensitivity is ΔY_out/Δpa = -0.0364546 µm per unit input. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 0.3387 µm, Y₁ = 0.33468 µm; ΔY_out = -0.00401912 µm; average sensitivity = -0.0364546 µm per unit input.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
The p–n junction and semiconductor devices: infer an input with uncertainty
A p–n junction is the boundary where p-type and n-type semiconductor meet: diffusion and drift cancel there at equilibrium, leaving a depletion region swept clear of mobile carriers and a built-in potential step fixed by the two doping levels. Use the supplied topic model Y_out = sqrt(1.29316*10^15*(10^(-pa)+10^(-pd))*max(0.0595263*(pa+pd-20)-v,0.02)), where Y_out is DEPLETION WIDTH (µm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. pa is p-side doping, log_10 N_a in cm^-3, dimensionless. Keep pd (n-side doping, log_10 N_d in cm^-3) = 16; v (Applied bias V, forward positive) = 0 V fixed. For this exercise use only 16.6062 ≤ pa ≤ 17.3938 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.33013488 µm, with a bounded reading uncertainty ±0.0003932 µm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown pa by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
DEPLETION WIDTH against pa
Predictions from the supplied The p–n junction and semiconductor devices model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): DEPLETION WIDTH / µm
Horizontal axis (x): pa: p-side doping, log_10 N_a in cm^-3
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| pa: p-side doping, log_10 N_a in cm^-3 | DEPLETION WIDTH / µm |
|---|---|
| 16.60625 | 0.3479454429845339 |
| 16.803125 | 0.33773143669641964 |
| 17 | 0.3317787712534363 |
| 17.196875 | 0.3286967162221316 |
| 17.39375 | 0.32751356431561585 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(pa) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(16.6062) = 0.347945 µm and Y_out(17.3938) = 0.327514 µm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives pa_line = 16.6062 + (0.330135 − 0.347945)(17.3938 − 16.6062)/(0.327514 − 0.347945) = 17.2927. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives pa = 17.0866; substituting back gives Y_out = 0.33013488 µm.
- The interpolation error in the input is |pa_line − pa| = 0.206092. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.32974168 µm and 0.33052808 µm. Sort the roots even if the response decreases: 17.0633 ≤ pa ≤ 17.1122.
- The lower and upper input deviations are 0.023284 and 0.0255351. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer pa ≈ 17.0866; compatible interval [17.0633, 17.1122]. Endpoint interpolation gives 17.2927.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.