Math Bridge: Photodiode Responsivity

A field team faces an unresolved physical question: How does one microamp reveal the light power? They must answer it before changing photodiode wavelength in nanometres on the real device. Predict the direction first.

See the relationship before changing it

Set the input, apply the rule, then read the result with its unit.

The retained audit below checks several chapter fixtures. This added model holds every other chapter fixture fixed, so the numeric fixture does not switch without explanation.

Derive the baseline in four named moves

1
Name the input. The chapter baseline for photodiode wavelength in nanometres is 850.
2
Name the relationship. E=hc/λ; Rmax=q/E; R=ηRmax; P=I/R
3
Substitute the chapter fixture. Set photodiode wavelength in nanometres to 850. The page ledger gives photon energy as 1.46 eV.
4
Read the result. Keep eV beside the value. Use it only inside the technical boundary on this page.

Predict, then change photodiode wavelength in nanometres

Try Predict the direction of photon energy. Move one control, calculate, then check your prediction.

Observe Longer wavelength lowers energy per photon and raises the ideal current per watt, provided the material can still absorb those photons. Reset the control to 850 and compare photon energy.

Explain Only photodiode wavelength in nanometres moves here. The other chapter fixtures remain fixed.

Check yourself

What should you do before trusting a moved-control result?

Answer: Predict its direction, apply the shown relationship, keep the units, and reset to the worked baseline.

What does this small model leave out?

Answer: Only photodiode wavelength in nanometres moves. Field effects named in the page's technical boundary stay fixed.

1. A watt arrives as photons

Optical power is energy per second. Light at a chosen wavelength packages that energy into photons, so the first job is to find the energy in one photon.

Phoebe: Longer-wavelength photons each carry less energy, so the same watt contains more of them.

2. Find one photon's energy

Convert wavelength850 nm = 850 × 10⁻⁹ m.

Substitute into E = hc/λUse h = 6.626 × 10⁻³⁴ J·s and c = 3.00 × 10⁸ m/s.

Ephoton = hc/λ = 2.34 × 10⁻¹⁹ J = 1.46 eV

3. Count useful electrons

Apply quantum efficiencyOnly the fraction η of photons creates collected charge.

Divide current by optical powerR = I/P = ηq/Ephoton = ηqλ/(hc).

Rmax = 0.685 A/W; η = 0.8 ⇒ R = 0.548 A/W

4. Try the wavelength

E=hc/λ; Rmax=q/E; R=ηRmax; P=I/R

TryMove wavelength while the chapter's η = 0.8 and 1 µA current stay fixed.

ObserveAt 850 nm, one photon carries 2.34 × 10⁻¹⁹ J and the 80%-efficient diode gives 0.548 A/W. A 1 µA current therefore traces back to 1.82 µW.

ExplainLonger wavelength lowers energy per photon and raises the ideal current per watt, provided the material can still absorb those photons.

Technical boundaries.

The light-power estimate keeps quantum efficiency at 0.8 while wavelength changes. It excludes wavelength-dependent absorption, reflection, dark current, shot noise, junction capacitance, amplifier noise, saturation, temperature, and optical geometry; actual responsivity needs the photodiode's data sheet.

wavelength-dependent absorption: Needs separate evidence
reflection: Needs separate evidence
dark current: Needs separate evidence
shot noise: Needs separate evidence
junction capacitance: Needs separate evidence
amplifier noise: Needs separate evidence
saturation: Needs separate evidence
temperature: Needs separate evidence
optical geometry: Needs separate evidence
Real responsivity must come from the actual photodiode datasheet: Needs separate evidence

Use field evidence or a deeper model before release.

5. Invert the chapter's current

Poptical = Iphoto/R = 1 × 10⁻⁶/0.548 = 1.82 µW

This is the optical power at the diode, not LED output power and not total light in the room. Geometry and reflection decide how much source power reaches the detector.

6. Connect to the voltage readout

The chapter's transimpedance stage turns the photocurrent into a voltage. Responsivity explains the optical-to-current step; feedback resistance and amplifier limits govern the current-to-voltage step.

7. Check yourself

Why does 850 nm light have about 1.46 eV per photon?

Answer: Substitute 850 × 10⁻⁹ m into E = hc/λ, then divide joules by the electron charge.

Why is real responsivity below 0.685 A/W here?

Answer: The chapter uses η = 0.8, so only 80% of arriving photons produce collected charge.

Does 1.82 µW state the LED's emitted power?

Answer: No. It is the inferred power reaching the photodiode under the stated responsivity.

Honesty boundary.

850 nm wavelength: Incident-light wavelength
η ≈ 0.8: Assumed photon-to-collected-electron efficiency
0.548 A/W responsivity: Photodiode current per incident optical watt
1 µA current: Measured photocurrent
1.82 µW result reproduce the chapter: Inferred incident optical power
The constants h: Planck constant
c: Speed of light
q are physical constants: Elementary electric charge

This is a one-target responsivity runway, not a complete optical link or amplifier design; Under the Hood retains those limits.