lecture19

lecture19 - 1 3.051J/20.340J Lecture 19 Drug Delivery:...

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1 3.051 J / 20 .340 J Lecture 19 Drug Delivery: Controlled Release What do we mean by “controlled” release? Control of: 1. delivery rate 2. site of release/activity Need for Control drug Traditional drug delivery concen. time from t 1 level Therapeutic range administration toxic t 2 longer period of dose efficacy = toxicity risk
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2 3.051 J / 20 .340 J drug Controlled drug delivery concen. toxic level time from administration Therapeutic range Types of Devices 1. Diffusion Controlled Delivery Devices Monolithic Devices Membrane Controlled Devices Osmotic Pressure Devices Swelling-Controlled Devices 2. Chemically Controlled Approaches Matrix Erosion Combined Erosion/Diffusion Drug Covalently Attached to Polymer Desorption of Adsorbed Drug 3. Electronic/Externally Controlled Devices
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3 3.051 J / 20 .340 J 1. Diffusion Controlled Devices a) Monolithic Devices Drug is released by diffusion out of a polymer matrix Release rate depends on initial drug concentration i) Case of C 0 < C s (drug concentration C 0 is below solubility limit in matrix C s ) Diffusion through matrix limits the release rate t 0 t 1 t 2 How can we control release rate? Rate control by choice of matrix : glassy matrix: D~10 -10 -10 -12 cm 2 /s rubbery matrix: D ~ 10 -6 -10 -7 cm 2 /s
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4 3.051 J / 20 .340 J Quantifying drug release Governed by Fick’s Laws. dC For 1D: The drug flux J is: J =− D dx C 2 C The change in drug concentration with time is: t = D x 2 We want to calculate: dM t /dt = release rate M t = amount released after time t Solve Fick’s 2 nd law with initial & boundary conditions. Example: For a 1D slab loaded at an initial concentration of C 0 , with drug concentration in solution resulting in constant surface concentration of C i . I.C.: C ( x ,0) = C 0 δ C = 0 B.C. 1: x x = 0, t C 0 B.C. 2: C( /2,t ) = C i Solve for C(x,t) (, dM t dC x t ) M t Adt dx D C i x = /2 A = cross-sectional area 0
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5 3.051 J / 20 .340 J The amount of drug released is given by the series solution: 8 Dn + 1 ) π ( 2 M t =− 2 2 2 2 t 1 2 M n = 0 ( 2 n + 1 ) e x p δ where: M = amount of drug released at long times (e.g., total amt of drug: M = C 0 A ) δ = slab thickness 1 M t /M 0 0.5 time Release rate (from derivative) : 1/ 2 dM t D short times: ~ t -1/2
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This note was uploaded on 11/11/2011 for the course BIO 2.797j taught by Professor Matthewlang during the Fall '06 term at MIT.

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lecture19 - 1 3.051J/20.340J Lecture 19 Drug Delivery:...

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