"Pumping" Charges — Question 10

Section 27-1 · Physics pages 4330–4331

Question 10 Photon-driven charge separation has independent charge and energy limits An idealized photo-source receives photons at rate Phi, each with energy Eγ. A fraction f produces one successfully collected electron-hole pair each. Each collected pair delivers charge magnitude e through the load at voltage V. Neglect other energy inputs. low V high V positive charge non-electrostatic pump photons Schematic; labels, not drawing lengths, specify the data. (a) Find output current and output power. (b) Use energy conservation to bound V if all electrical work must come from the successfully collected photons, and compare current and voltage constraints.
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Question 10 - Solution Photon-driven charge separation has independent charge and energy limits low V high V positive charge non-electrostatic pump photons (a) Method and calculation I = efΦ. Pout = V I = efΦV, Puseful photons = fΦEγ. (b) Interpretation and further calculation efΦV ≤fΦEγ ⇒V ≤Eγ/e (fΦ > 0). I ∝fΦ, Vupper is independent of photon arrival rate in this bound. Physical check This is an energy upper bound, not a prediction of a real solar cell voltage or efficiency.

Question 10

Photon-driven charge separation has independent charge and energy limits

An idealized photo-source receives photons at rate Phi, each with energy EγE_\gamma. A fraction f produces one successfully collected electron-hole pair each. Each collected pair delivers charge magnitude e through the load at voltage V. Neglect other energy inputs.

Diagram shown in the original SVG worksheet.

  1. Find output current and output power.

  2. Use energy conservation to bound V if all electrical work must come from the successfully collected photons, and compare current and voltage constraints.

Question 10 - Solution

Photon-driven charge separation has independent charge and energy limits

Diagram shown in the original SVG worksheet.

(a) Method and calculation

I=efΦ.I=ef\Phi.Pout=VI=efΦV,Pusefulphotons=fΦEγ.P_{\mathrm{out}}=VI=ef\Phi V,\quad P_{\mathrm{useful\ photons}}=f\Phi E_\gamma.

(b) Interpretation and further calculation

efΦV≤fΦEγ⇒V≤Eγ/e(fΦ>0).ef\Phi V\le f\Phi E_\gamma\Rightarrow V\le E_\gamma/e\quad(f\Phi>0).I∝fΦ,Vupper is independent of photon arrival rate in this bound.I\propto f\Phi,\quad V_{\mathrm{upper}}\text{ is independent of photon arrival rate in this bound}.

Physical check

This is an energy upper bound, not a prediction of a real solar cell voltage or efficiency.