ELE46703 TEST #1 Take-Home Solutions Prof. Guvench... Problem 1 : Light Emitting Diode (60 pts.) Consider a GaAs pn junction which has the following properties. N a 10 16 cm -3 (p-side), N d 10 19 cm -3 (n-side), B 7.21 10-16 m 3 s -1, cross sectional area A 1 mm 2. a. What is the diode current components due to diffusion I (P-side), I (N-side) in the P and N neutral regions at 300 K when the forward voltage across the diode is 1.1 V? b. What is the built-in potential? Calculate the thickness of the space charge (depletion) regions across the junction at V0 and at V1.1V? Calculate at 1.1V the recombination currents (in the space charge layers), I r(in Wp) and I r(in Wn). c. At this bias point, how many photons, in each region of the diode, are generated per unit time? Label these as I λ (P-side), I λ (N-side), I λ (in Wp) and I λ (in Wn). If only 5% of these can be transmitted out, what is the total optical power output, PλoutTOTAL in watts and external power efficiency, η Pλout TOTAL/of this LED? Note that Pλout 5% of (photon energy. I λ TOTAL) (20% extra) Bonus* Repeat the above calculations for a range of 0.8 < V < 1.2 V and plot on semi-log scale, All components of I vs V including ITOTAL, All components of Iλ vs V including ITOTAL, All components of Pλout vs V including PλoutTOTAL. Note: All calculations below are done by using MKS units. Data given/found are written/converted to MKS before use. Make a note of it. GaAs: kt 0.0259 A 1 10 6 m2 1.6 10 19 1.8 10 12 /m3 B 7.21 10 16 m3/s V 1 1.1 ε o 8.85 10 12 E G 1.42 ev ε GaAs 13.2 µm 10 6 ns 10 9 N++ Side P Side 10 19 10 6 m-3 10 16 10 6 m-3 2 p no 2 n po
µ nn 2400 10 4 µ pn 160 10 4 µ np 7000 10 4 µ pp 310 10 4 D pn kt µ pn D np kt µ np 1 1 τ pn τ np BN D BN A L pn I pn D pn τ pn L np D np τ np A D pn p no exp V 1 A D np n po 1 I np exp V 1 L pn kt L np kt I difftotal I pn + I np 1 1 10 25 m-3 1 10 22 m-3 p no 0.324 n po 324 µ nn 0.24 µ pn 0.016 µ np 0.7 µ pp 0.031 D pn 4.144 10 4 D np 0.018 τ pn 1.387 10 10 τ np 1.387 10 7 L pn µm 0.24 L np µm 50.145 I pn 2.496 10 4 I np 0.052 I difftotal 0.052 Problem 1.b Recombination Currents in The Space Charge Layers From the Class notes or the textbook: V 0.0, 1.1.. 1.1 V Bi kt ln V( V) V Bi V 2
V N ( V) V( V) + V P ( V) V( V) + W N ( V) 2 ε o ε GaAs V N ( V) N D W P ( V) 2 ε o ε GaAs V P ( V) N A I rwn ( V) A W N ( V) exp 2 τ pn V 2kT I rwp ( V) A W P ( V) exp 2 τ np V 2kT I rtotal ( V) I rwn ( V) + I rwp ( V) I pdiffn ( V) A D pn p no exp V L pn kt 1 I ndiffp ( V) A D np n po exp V L np kt 1 I DiffTotal ( V) I pdiffn ( V) + I ndiffp ( V) I λdiffn ( V) I pdiffn ( V) I λdiffp ( V) I ndiffp ( V) I λrwn ( V) I rwn ( V) I λrwp ( V) I rwp ( V) I Total ( V) I rtotal ( V) + I DiffTotal ( V) ( ) P λouttotal ( V) 0.05 E G + kt I Total ( V) η Ext ( V) P λouttotal ( V) VI Total ( V) Results: V Bi 1.341 For: V V( V) 0 1.1 1.341 0.241 V N ( V) W N ( V) I rwn ( V) V P ( V) W P ( V) I rwp ( V)
1.34 10-3 2.41 10-4 4.423 10-10 1.876 10-10 4.592 10-13 3.251 10-4 1.34 0.241 4.423 10-7 1.876 10-7 4.592 10-13 3.251 10-4 Note that most of the actios on the lowly doped P side I pdiffn ( V) I ndiffp ( V) 0 2.496 10-4 0 0.052 I λdiffn ( V) I λrwn ( V) I λdiffp ( V) I λrwp ( V) 0 1.56 10 15 2.87 10 6 2.032 10 15 0 3.263 10 17 2.87 10 6 2.032 10 15 I Total ( V) I DiffTotal ( V) I rtotal ( V) 9.185 10-13 0.053 0 0.052 9.185 10-13 6.501 10-4 P λouttotal ( V) 6.64 10-14 3.84 10-3 η Ext ( V ) BONUS Plots From the Class notes or the textbook: V 0.4, 0.41.. 1.3 V Bi kt ln 2 V( V) V Bi V
V N ( V) V( V) + V P ( V) V( V) + W N ( V) 2 ε o ε GaAs V N ( V) N D W P ( V) 2 ε o ε GaAs V P ( V) N A I rwn ( V) A W N ( V) exp 2 τ pn V 2kT I rwp ( V) A W P ( V) exp 2 τ np V 2kT I rtotal ( V) I rwn ( V) + I rwp ( V) I pdiffn ( V) A D pn p no exp V L pn kt 1 I ndiffp ( V) A D np n po exp V L np kt 1 I DiffTotal ( V) I pdiffn ( V) + I ndiffp ( V) I λdiffn ( V) I pdiffn ( V) I λdiffp ( V) I ndiffp ( V) I λrwn ( V) I rwn ( V) I λrwp ( V) I rwp ( V) I Total ( V) I rtotal ( V) + I DiffTotal ( V) ( ) P λouttotal ( V) 0.05 E G + kt I Total ( V) η Ext ( V) P λouttotal ( V) VI Total ( V)
I Total ( V) I DiffTotal ( V) I rtotal ( V) I ndiffp ( V) I pdiffn ( V) I rwn ( V) I rwp ( V) 1. 10 3 100 10 1 0.1 0.01 1. 10 3 1. 1. 10 4 1. 10 5 1. 10 6 1. 10 7 1. 10 8 1. 10 9 1. 10 10 1. 10 11 1. 10 12 1. 10 13 1. 10 14 1. 10 15 10 16 0.4 0.6 0.8 1 1.2 1.4 V Note that all photons generated, Iλ, and all optical power output values are simply proportional to the electric current component plotted above, the plots generated for those would be identical to the plots given above with a difference of scaling factor. 0.2 0.15 η Ext ( V) 0.1 0.05 0.4 0.6 0.8 1 1.2 1.4 V Conclusions:
1. Due to the one-sided structure of this junction, the extent of the space charge is mostly into the lightly doped P-side. 2. The space-charge recombination currents are proportional to the space-charge layer thickness and inversely proportional to the recombination times, τ ~ 1/(B.N). Although relative components of the recombination currents are expected to be favored by the wider space charge in the lowly doped P-side, Ir ~ W/τ Β.Ν.W, N.W product dependency (the total charge on each side of the junction which is eual) makes both currents to be the same. 3. At low bias (~0.8V) the recombination total dominates over the diffusion total. Later the diffusion total takes over, at 1.1V 52mA passing through the diode is 99% due diffusion. 4. Due to the one-sided nature of this junction, the diffusion (injection) components are one sided, i.e., the diffusion current component corresponding to injection from highly doped side (N+) to the lowly doped side (P) overwhelmingly dominates. 5. Therefore, the LED glow will be coming predominantly from th P-side of the junction. 6. The external efficiency decreases with the bias voltage is increased (as the light output increases).