Final Exam Solution, Spring Semester 2011 ECSE 6962, Light Emitting Diodes and Solid State Lighting (Prof. E. F. Schubert)

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1 Final Exam Solution, Spring Semester 2011 ECSE 6962, Light Emitting Diodes and Solid State Lighting (Prof. E. F. Schubert) Note: Put your name on your paper. Show your work. Underline all results. Always show units. All materials are allowed (textbook, other books, manuscripts, excerpts, homework, calculators, etc.). Use ball point pen, no pencil. 1. Consider a tri chromatic white light source consisting of three LEDs emitting a monochromatic light at 450 nm, 525 nm, and 625 nm. Each of the emission line has an optical power of 1 W. Assume that the electrical to optical power conversion efficiency of the blue emitter is 60%, or the green emitter is 25% and the red emitter is 75%. (a) Calculate the luminous efficacy of radiation of this light source (lumens per optical watt). (b) Calculate the luminous efficiency of the source of this light source (lumens per electrical watt). Next, assume that one of the three LEDs fails to operate. (c) For which LED failing will the luminous efficiency increase most? Explain. (d) For which LED failing will the luminous efficiency decrease most? Explain. Next, assume that all LEDs are working again and that the external quantum efficiency (EQE) of the three LEDs is equal to their electrical to optical power conversion efficiency. Next we connect the three LEDs in series, the current through each of them is 0.35 A. (e) Given these assumptions, are there any resistive electrical losses in the LEDs? (f) What are the forward voltages of the three LEDs? (g) What is the total electrical power consumed by the light source? (h) What is the total optical power emitted? (i) What is the total luminous flux (luminous power) emitted? (j) What is the luminous efficiency of the source? Solution: (a) From Fig in the LED textbook, we read the eye sensitivity V() for 450 nm, 525 nm, and 625 nm wavelength light as 0.05, 0.8, and 0.3 respectively. Eye sensitivities of all wavelengths are referenced to the 555 nm light eye sensitivity. Using Eq. 16.3, the total luminous flux for this light source is Φ 683 lm W lm W The total optical power of this light source is 3 W. As a result, the luminous efficacy is lm Luminous efficacy lm/w. 3 W (b) Assume that the electrical to optical power conversion efficiency of the blue emitter is 60%, the green emitter is 25%, and the red emitter is 75%. The electrical power consumed by the three emitters are 1

2 1 W W W 7 W Using Eq. 16.4, the luminous efficiency is lm luminous efficiency lm/w. 7 W (c) Consider each case when one of the three LEDs fails to operate, the subscript indicates which LED fails to operate. The total power consumed when one LED fails to operate is 1 W W 5.33 W, W W 3 W, W W 5.67 W, 0.25 The total luminous flux from the remaining two LEDs is Φ 683 lm W lm W Φ 683 lm W lm W Φ 683 lm W lm W Therefore, the luminous efficiency is lm luminous efficiency lm/w 5.33 W lm luminous efficiency lm/w 3 W lm luminous efficiency lm/w 5.67 W From part (b), we know the luminous efficiency increases the most, when the blue LED fails to operate. (d) From part (b) and (c), we learned that the luminous efficiency decreases the most when the green LED fails to operate. (e) Given these assumptions, there are no resistive losses in the LEDs. From Eq. 5.3 of the LED textbook, the external quantum efficiency is defined as / /, where P is the optical power emitted into free space, I is the injection current, and is the frequency of the emitted light. The electrical to optical power conversion efficiency, from Eq. 5.4, is defined as:, 2

3 Using the assumption given in the question, we have, i. e.,. The forward voltage of an LED can be expressed by, where R S is the series resistance of the LED. With the assumption given in the question, the term must be equal to zero. So, no resistive electrical losses exist in these LEDs. (f) From part (e), the voltage of the LED is determined by. Using the relation /, we obtain. So, the voltage across blue, green, and red LEDs are V 2.76 V V 2.37 V V 1.99 V (g) From part (e), the total voltage across the three LEDs connected in series is 7.12 V. Given the current through the three LEDs is 0.35 A, the total electrical power consumed by the three LEDs is W (h) For each LED, we assume the external quantum efficiency (EQE) of the three LEDs is equal to their electrical to optical power conversion efficiency, so, W (i) From Eq. 16.1, Φ 683 lm W 450 nm 525 nm 625 nm, where V(450 nm), V(525 nm) and V(625 nm) are the eye sensitivities at wavelength 450 nm, 525 nm, and 625 nm respectively. Consequently, 3

4 lm Φ 683 W W W W lm. (j) From part (g) and part (i), using Eq. 16.4, we obtain the luminous efficiency of the source as luminous efficiency lm W 96.3 lm/w. 2. Semiconductor LED chips are commonly encapsulated Assume that the refractive index of an encapsulant encapsulant air boundary can be calculated as: d by an opti is n 1 = 1.6. cally transpa The Fresnel arent encapsulant. reflection at the. (1) (a) What is the reason that an encapsulant is being used in LEDs? (b) Calculate the Fresnel reflection at the encapsulant air boundary. (c) To reduce Fresnel reflection at the encapsulant aion the first encapsulant (graded refractive index encapsulation) ). The second encapsulant has a refractive index of n 2. Sketch the device having boundary, one can employ a second encapsulant forming a thin layer the two encapsulants. (d) Assume that the reflectivity at the two interfaces (i.e. encapsulant 1 to encapsulant 2 and encapsulant 2 to air) add up (R = R 1 + R 2 ). Give a formula for the total reflectance as a function of n 2. (e) Show that n 2 has an optimum value, that is, a value at which the Fresnel reflection is minimized. Solution: (a) An encapsulant helps to (i) enhance the light extraction efficiency of an LED, and (ii) provide a protection of LED chips from mechanical forces and adverse environmental conditions (e.g., humidity). (b) Using the Equation (1), provided in the question, we calculate (c) The following figure shows a device having two encapsulants %. 4

5 (d) Using Equation (1), and the formula for the total reflectance, we have, where, and, where. Note that. So, neglecting interference effects, the total reflectance as a function of n 2 is (e) The minimum reflectance occurs when the first derivative of R(n 2 ) is zero, d 0. d Taking the derivative of R(n 2 ), we obtain d 4 d 4. To find the minimum, we require that the derivative be zero. Note that when,. the derivative indeed becomes zero. You may be familiar with this result since it is a common result used in quarter wavelength anti reflection coatings. When the second encapsulant has a refractive index of, then the total reflectance is minimized. The minimized total reflectance is It is generally difficult to measure the temperature of an LED with an Hg thermometer. (a) Explain why. (b) The forward voltage measurement is a common method to measure the junction temperature. Assume that at 300 K an LED has, at a current of 1 ma, a forward voltage, V f, of 1.45 V. Assume further that the change in V f is 2 mv for a temperature change of 1 kelvin. If the V f is 1.40 V, what is the junction temperature? (c) Assume that two of these diodes are wired in a series circuit configuration and are injected with 1 ma. What is the measured V f of the circuit at 300 K and what is it at 400 K? Draw the circuit diagram including the volt meter. (d) Assume that two of these diodes are wired in a parallel circuit configuration, each being injected with 1 ma. What is the measured V f of the circuit at 300 K and what is it at 400 K? Draw the circuit diagram including the volt meter. 5

6 Solution: (a) Typical LEDs are very small, so is very difficult to insert an Hg thermometer inside an LED chip to accurately measure the LED temperature. Also, the large Hg thermometer would cool the LED thereby influencing the measurement. (b) Assuming linear dependence of junctionn temperaturee on temperature, with d 2 mv/k, d then, d d, where T 0 is the initial temperature, i.e. 300 K, and 1.45 V. So, when 1.44 V, d 325 K. d (c) Voltage acrosss two diodes connected in series is the sum of the voltage across each diode. At 300 K, the measured V f of the circuit is 300 K 2.9 V. At 400 K, from part (b), the voltage across each diodee is 400 K 300 K d 100 K 1.25 V. d so, the measured V f of the circuit at 400 K is 400 K 2.5 V. A possible schematic circuit diagram is shown below.. (d) Voltage acrosss two diodes connected in parallel is also the voltage across each diode, since the two diodes connected in parallel are exactly the same, at a current of 1 ma in each diode, the voltage across each diode is also the same. So, the measured V f of the circuit is the same as the voltage across each diode. 6

7 At 300 K, the measured V f of the circuit is 3000 K 300 K 1.45V. At 400 K, from part (c), 400 K 1.25 V. so, the measured V f of the circuit at 400 K is 4000 K 400 K 1.25V. A possible schematic circuit diagram is shown in the following figure. 4. Calculate the (i) critical angle for total internal reflectionn and (ii) the solid angle of the light escape cone divided by the solid angle of a sphere ( 4 steradians) for a (a) AlGaInP LED chip with a refractive index of 3.5. (b) GaN LED chip with a refractive index of 2.5. (c) Polymer LED chip with a refractive index of 1.5. (d) What can we learn from the results calculated above? Solution: (a) From Eq. 5.18b, for AlGaInP LED chip with refractive index of 3.5, the critical angle for total internal reflection is arcsin radians. 3.5 Using Eq. 5.22, we obtain the fraction of light that escapes from the semiconductor 1 2 1cos 2.08% %. (b) For GaN LED chip with refractive index of 2.5, the critical angle forr total internal reflection is arcsin radians. 2.5 Using Eq. 5.22, we obtain the fraction of light that escapes from the semiconductor 1 2 1cos 4.17% %. (c) For polymer LED chip with refractive index of 1.5, thee critical angle for total internal reflection is 7

8 arcsin radians. 1.5 Using Eq. 5.22, we obtain the fraction of light that escapes from the semiconductor 1 2 1cos 12.73%. (d) LED chips made from lower index materials help to improve the light extraction efficiency. 5. LEDs employing p n heterojunctions show advantages over LEDs with p n homojunctions: (a) Through which mechanism do LEDs with heterojunctions improve the performance? (b) Give two carrier loss mechanisms (carriers not being injected into the active region) existing in a p n heterojunction LED. (c) What is the difference between double heterostructures LEDs and a single quantum well LEDs? Which device structure may perform better under what circumstances? Explain why. Solution: (a) LEDs employing heterostructures improve the LED efficiency by confining carriers to a small active region. In such a small active region, the number of defects, that can cause non radiative recombination, is limited. Heterostructures prevent the diffusion of free carriers over long distances where the number of defects, that the carriers could interact with, would be much larger. (b) Mechanisms include: (i) Carriers may escape from the active region into the barrier layers by thermal emission. (ii) Carriers may not be captured by the active region and ballistically fly over the active region (particularly if the active region is very thin). (iii) At high current density levels, carriers may overflow the active region and escape into the confinement regions. (c) A double heterostructures LED has a relatively thick quantum well. A single quantum well LED has a relatively thin quantum well. At low current injection density, the efficiency of LED is determined by non radiative recombination. At high current injection density, the efficiency of LED is determined by current overflow. At low current injection density, single quantum well LED may perform better, because thinner quantum well (QW) contains less defects compared to thicker QW, so non radiative recombination is less severe in single QW LED. At high current injection density, double heterostructures LED may perform better, because the current overflow in thicker QW is less likely than in a thinner QW. 8