# Solution to Exercise 5.1-2 Energy, Field strength and Photons

| You have a LED as a light source that emits a monochromatic light beam with wave length (in air) of l = 500 nm. The light is generated in a very small volume ("point source") and spreads out in a cone that illuminates a circle with radius 1 cm at a distance of 10 cm on some white paper. The LED has an over-all or plug efficiency of 50 % and is driven at 2V with 20 A.
|  | Question 1: How much power in W/m2 flows into the paper?
| Total power = UI = 40 W Light power = 50 % of total power = 20 W Light flux = 20 W / p r2 = 6,37 W/cm2 With eV instead of Ws = J and 1 J = 1 Ws = 6,24 · 1018 eV we have Light flux = 3,97 · 1019 eV / s · cm2.
|  | Question 2:How does that number compare with the light power coming form the sun at "AM 1" conditions (High noon, equation, no clouds)? You're supposed to know this basic number in some "simple number approximation".
| The sun at a cloudless day at high noon at the equator delivers about 1 kW/m2 = 1000 / 10000 W/cm2 = 0,1 W/cm2 Our LED thus delivers a very high (and unrealistic) intensity of 63,7 times more than the sun.
|  | Question 3: How many photons per second must hit the piece of paper if we discuss the energy flux now in the particle picture?
| A wavelength of 500 nm corresponds to a photon with energy hn = hc/l = {(4,1356 · 10–15) · (3 · 1017)} / 500 eVs · nms–1 · nm–1 = 2,48 eV. The necessary 3,97 · 1019 eV/scm2 thus corresponds to 3,97 · 1019 / 2,48 = 1,60 · 1019 photons/scm2 Since we need to illuminate an area of 3,14 cm2, we need 5,03 · 1019 photons/s
|  | Question 4: What kind of field strength would we have on the paper? Consider first that the light beam is fully coherent, next that the photons are completely uncorrelated.
| The energy flux in the light beam is given by <S> = ½E0H0 = (E0)2/(Zw) = 6,37 W/cm2
|  | For the electrical field strength in a fully coherent wave we have E0 = [Zw · 6,37 VW/Acm2]½ = [377 · 6,37 V2/cm2]½ = 49 V/cm That is a rather low field strength.
|  | A completely incoherent light consists of waves with all kinds of phases and all kinds of directions of the electrical field vectors. The total field strength then is a vector sum that tends to average to zero. We might assume that the number of independent "waves" equal the number of photons. That gives the field strength per wave = photon to EPh = 49 / 5,03 · 1019 V/cm = 9,74 · 10–19 V/cm. That is, of course, a rather meaningless number.
|  | If we consider that a photon delivers an energy of 2 eV to an area of about 1 µm2 within 1 ns, we would get EPh » 3,5 V/cm, which is more like it but still more or less nonsense.
|  | How about assuming that the energy WPh of a photon (= 2,5 eV, for example) is contained in a volume of l3 (1 µm)3, for example). We then have roughly WPh = (½e0 · E2)/l3 from the relation between energy density and field strength E. Going through the numbers we obtain EPh » 3.000 V/cm. That is a number one could live with.
|  | Question 5: What does the number of photons produced per second tell you about recombination rates, carrier densities, and current densities in the semiconductor?
| We need at least a recombination rate of R = 5,03 · 1019 s–1between electrons and holes that produces light. If we loose some of the light, the rate must be higher. The recombination ot the carriers take place in a device volume VDev given by lateral area F of the device times length lRecof the recombination zone; VDev = F · lRec.
|  | We can always express the specific recombination rate per cm–3 by R = 5,03 · 1019/ s · F · lRec = n/t with n = surplus carrier density, t = carrier life time » 1 ns.
|  | The necessary carrier density that must be supplied by the current is thus n = R t/ F · lRec. If we assume F = 10–4 cm2, t = 10–9 s; we have n = 5,03 · 1018 m–3 which looks reasonable.
| The total current I was 20 A. The current density j for a device with cross-sectional area F of 10–4 cm2cm2 is j = 2 · 105 A/cm2 which is a bit on the high side..
|  | The number Neof electrons (or holes "on the other side") that we inject is Ne = 20 (C/s) / 1,6 · 10–19 C = 1,25 · 1020 s–1. If half of this electrons recombine and produce light (we assumed an efficiency of 50 %), we have 6,25 · 1019 electrons per second available for this, a number that matches quite fortuitously with the required 5,03 · 1019 s–1. Or maybe, it's not that accidental?

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Questions and Answers about Details of Light Waves

Exercise 5.1.2 Energy, Field strength and Photons

© H. Föll (Advanced Materials B, part 1 - script)