Investigation of Exciton Diffusion and Charge Transfer Dynamics in Nano Phase-Separated P3HT:PCBM blend films by Utrafast Time Resolved Fluorescence

Transcription

1 This journal is (c) The Royal Society of Chemistry Investigation of Exciton Diffusion and Charge Transfer Dynamics in Nano Phase-Separated P3HT:PCBM blend films by Utrafast Time Resolved Fluorescence Hai Wang a,b, Hai-Yu Wang* a, Bing-Rong Gao a, Lei Wang a,b, Zhiyong Yang a,b, Xiao-Bo Du a,b, Qi-Dai Chen a, Jun-feng Song a, Hong-Bo Sun* a,b a State Key Laboratory on Integrated Optoelectronics, College of Electronic Science and Engineering, Jilin University, 699 Qianjin Street, Changchun 3, China. b College of Physics, Jilin University, 9 Jiefang Road, Changchun 33, China Corresponding authors: haiyu_wang@jlu.edu.cn, hbsun@jlu.edu.cn Electronic Supplementary Information

2 This journal is (c) The Royal Society of Chemistry Excitation density calculation: The power of the excitation pulse of 65nm, at levels of, 5,, 5,.5 nj/pulse, was adjusted by the neutral density filters and used to excite samples. The diameter of the excitation spot was measured by two ways, one was to place a photo sensitive paper at the sample position and the size of the exposed area under excitation pulse was measured with a microscope. The other way was the knife-edge measurement, by moving a sharp knife edge slowly going across the pulse at the sample position, and recording the transmitted power vs. the position of the knife edge to determine the profile of the spot. The results of the two measurements agree well, giving a spot diameter of 3 μ m. For pristine P3HT film, the film thickness was measured to be about nm by Stylus profiler (MBIOS TECHNOLOGY INC XP-), the excitation volume is calculated to 9 3 be.65 cm. The incident photons/pulse is calculated by dividing the incident power/pulse 7 by one photon energy of 65nm, which is3.7 J. nj/pulse corresponds to an incident light intensity of 6.3 photons/pulse. The absorbance of the file at 65nm is about.34, corresponding to a 54.3% absorption of the incident photons, resulting in a total absorption of 3.4 photons. The total number of the absorbed photons divide by the excitation volume 9 3 results in the final excitation density:.9 / cm. The blend film excitations are calculated the same way. Solution of the rate equation: The rate equation can be given as: d a CT = 4πRD a + R N 4πRCTDN CT + R πdt πdt dm t Let M t =,we have dn ( t) = N,substituting it into the above equation, we get dm t 4 R CT a πrct DNCT M t 4πRa D R = πdt πdt We can set = + 4π RCT DNCT, B = 4 π DRCT NCT, C = 4π Ra D, D π D = 4 Ra, dm B D = + M + C+ t t,let B D as D, and P t = + t, D Qt = C+,we have t dm = P t M + Q t Let M t = exp P t,substituting it into, we get

3 d ( ) = exp P t Q t d So, N t = Q t exp P t Supplementary Material (ESI) for Nanoscale This journal is (c) The Royal Society of Chemistry = Q t exp P t Set x = t, D B N t = C+ exp + t t ( ) D = C + exp t B t t D = C + exp ( x Bx) xdx x = + ( Cx D) exp ( x Bx) dx C BC ( ) ( ) = x + B + D exp x Bx dx C BC = ( x + B) exp ( x Bx) + D exp ( x Bx) dx C BC = exp ( x Bx) d ( x Bx) D exp ( x Bx) dx + C BC B B = exp ( x Bx) + D exp x + + dx Set B u = x+,we get C BC B N t = exp ( x Bx) + D exp exp u du π ( ) = exp cx dx erf cx 4c C BC B π B N t = exp ( x Bx) + D exp erf x + + E 4 const, 3

4 This journal is (c) The Royal Society of Chemistry C BC B π B B M t = exp ( x Bx) + D exp erf x Econst exp t C BC B π B = exp ( x Bx) + D exp erf x + + Econst exp ( ( x + B) dx) 4 C BC B π B = exp ( x Bx) + D exp erf x Econst exp ( x Bx) C BC B π B = + D exp exp ( x + Bx) erf x Econst exp( x Bx) C BC B π B = + D exp exp ( t + B t ) erf t + + Econst exp ( t + B t ) 4 So we can give = C BC B B + D exp exp t + B t erf t + + E exp t + B t 4 ccording to initial condition,n=n when t=,so, π const C BC B π B Econst = + D exp erf N 4 and N t = C BC B B C BC B B + D exp exp t + B t erf t D exp erf exp t + B t 4 N 4 Finally, it can be simplified as following form: π π t Nexp γt β t N t = γ t π βγ β t β β + exp γt β t β exp erf γt erf γ + + γ γ γ γ γ Where γ = 4π RCT DNCT, γ = 4π RDN a, β = πdr N, 4 CT CT π D β = 4 RN a. For the pristine P3HT film, N CT =, so the solution for exciton-exciton annihilate only can be expressed as: = ( ) N exp t / D a + 4π RDN a exp t/ + 4π RNerf t/ 4

5 This journal is (c) The Royal Society of Chemistry For the blend film, if the excitation power is low enough, the exciton-exciton annihilate is absent, so for the charge transfer only we have Nt = N exp ( + 4 πrct DNCT ) t 8 πdrct NCT t 5