2013 nanoHUB-U Course: Principles of Electronic Nanobiosensors
2013 nanoHUB-U Course on
“Principles of Electronic Nanobiosensors”
HW1: Diffusion Limited Capture and Fractal Dimension of a Random Surface
Muhammad A. Alam
Network of Computational Nanotechnology
Discovery Park, Purdue University.
In Lectures 1-6, we have discussed several general topics related to nanobiosensor and the importance of
the diffusion geometry of the sensor in defining the settling time.
In Week 1, let us solve the following 5 problems as exercises (Problems 1.1, 1.3, 1.6, 1.8, 1.10).
Homework will not be graded, but once you have worked out a problem, you should look for the answer
among the multiple choices. Doing the homework before the solutions are posted will help you
understand the material better. We will discuss the solutions at the end of the week in a tutorial.
We will be using two softwares for this class: Octaview and BioSensorLab. Octaview is an
open-source, web-enabled version of MATLAB available through nanohub.org. See
http://nanohub.org/resources/octaview/supportingdocs for a video tutorial. The codes will be
provided as a part of the HW set. You will be asked to change parameters to see how the
solutions change as a result.
BiosensorLab is a software program to solve a wide variety of problems related to
nanobiosensors – again available through nanoHUB.org. The instructions to solve various
problems using BioSensorLab will also be available as a part of the HW set.
Part I: Density and Diffusion of Particles
Tutorial 1.0. For an intuitive understanding of particle diffusion in a fluidic background, visit
the website at http://www.falstad.com/gas/. On the right hand side, choose “Brownian Motion”
for the setup. You can change the parameters on the slider scale to see how the diffusion is
affected by the environment, such as temperature.
Problem 1.1: Calculating concentration. If you take a 1000 mg tablet of headache medicine,
what is the increase in the analyte concentration in blood? Assume, the molecular formula
of the medicine is C8H9NO2 and a typical volume of blood is 5 liters.
Solution:
The Molar weight is 151.16 gm, therefore 1000 mg contains 6.023e23/151.16~4e21 molecules. Per liter,
the number of molecules are 4e21/5~8e20. The molar concentration is 8e20/6.023e23=1.32 mM. It is a
relatively high concentration – this is why, we will see, headache medicine can work very quickly.
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2013 nanoHUB-U Course: Principles of Electronic Nanobiosensors
Problem 1.2: Diffusion coefficient of free particles. Calculate the diffusion coefficient of a
small protein molecule of radius 2nm. What is the diffusion coefficient for a cell of size ~ 2
um. Assume that the viscosity =0.01 Poise [g/cm.sec].
Solution:
1.1
10 1.1
10
/ .
/ .
Problem 1.3: Diffusion distance of molecules in water. Assume that the diffusion coefficient
of a protein is
.
/ and that of a cell is .
/ . How
far would the protein and the cell would travel in 10 minutes?
Solution:
6
= 629 micron for the protein, and 18 micron for the cell. A large molecule is
repeatedly scattered by the background and does not travel very far in a few minutes.
Problem 1.4: Diffusion Time of free molecules. After influencing receptors in the postsynaptic membrane, the neurotransmitters not degraded by enzymes diffuse back to the
pre-synaptic membrane, which lies 2x10-6 cm away, with D=5x10-6 cm2/s. How long does
this short journey take? (Taken from R Cotterill, Biophysics: An Introduction, Wiley.)
Solution:
0.8
10 ~1 micro-second.
The receptors are small; therefore, the response is relatively fast.
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2013 nanoHUB-U Course: Principles of Electronic Nanobiosensors
Part II: Diffusion flux towards sensors of various geometrical shapes
Problem 1.5: Exact solution across a 1D Membrane.
During the lectures in the class, we have discussed approximate analytical solution to a
variety of diffusion problems. Sometimes it may be important to solve the diffusion
equation exactly to compare the analytical results.
A membrane of thickness L separates two chambers kept at different concentration, namely, 0
and 0. The exact solution of the diffusion equation (by separation of variables) is given by the
following equation.
d
 D 2 
dt
dN
 kF ( N0  N ) s
dt
k F  ,  s  0
 x  2(1) n
  n 2 Dt  
n x
 ( x, t )   0   
sin
exp  

2


L
L
 L n 1 n



  n 2 Dt  
 D 0  
n
J (t )  

  1   2(1) exp  
2

L
 L   n 1


L
(a) Plot
,
for several different times.
(b) What is the steady-state concentration gradient? Interpret this result in terms of the
steady-state fluxes we have discussed in the class.
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2013 nanoHUB-U Course: Principles of Electronic Nanobiosensors
(c) Can you use the diffusion equivalent capacitance approach for this problem? Why or
why not?
(Ref: Sze, Semiconductor sensors, p. 432)
Solution:
(a) Plot the solution for several different times (when the exponent is 0, 1, 2, etc.)
L=100e-4; %%cm
x=linspace(0,L,100); %%cm
x_L=x/L;
D=1e-6; %%cm^2/Sec
t_vec=linspace(1,10,4); %%sec
rho_0=1e-6; %%Moles
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%rho
figure(1)
for m=1:length(t_vec)
t=t_vec(m);
rho_rho_0=0;
for n=1:10
rho_rho_0=rho_rho_0+2*(-1)^n/n/pi*sin(n*pi*x_L)*exp(-(n*pi)^2*D*t/L^2);
end
rho_rho_0=rho_rho_0+x_L;
rho=rho_rho_0*rho_0;
rho_mat(:,m)=rho;
plot(x*1e4,rho_mat(:,m))
hold on
end
title('rho vs. x');
xlabel('x [um]');
ylabel('rho [M]');
hold off
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%J
figure(2)
t_vec=linspace(1,10,100);
for m=1:length(t_vec)
t=t_vec(m);
J=0;
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2013 nanoHUB-U Course: Principles of Electronic Nanobiosensors
for n=1:10
J=J+2*(-1)^n*exp(-(n*pi)^2*D*t/L^2);
end
J=D*rho_0/L*(1+J);
J_vec(m)=J;
end
t_vec
J_vec
plot(t_vec,J_vec)
title('J vs. time');
xlabel('t [Sec]');
ylabel('J [Mcm/Sec]');
axis([0,20,0,3e-11])
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(b) What is the steady-state concentration gradient? Interpret this result in terms of the
steady-state fluxes we have discussed.
 ( x, ) 0
106 Moles


 104 Moles / cm
4
x
L 100 10 cm
 ( x, ) J ()  0
106 Moles



 104 Moles / cm
4
x
D
L 100 10 cm
(c) Can you use the diffusion equivalent approach for this problem? Why or why not?
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2013 nanoHUB-U Course: Principles of Electronic Nanobiosensors
Yes. In steady-state condition the structure will act like a parallel plate capacitor with
diffusion equivalent capacitance, CD,SS=D/L.
Problem 1.6: Steady state flux towards a disk sensor.
From the Wikipedia (http://en.wikipedia.org/wiki/Capacitance), find the capacitance of a
small disk embedded in an infinite media.
(a) Using the diffusion equivalent capacitance approach, calculate the steady-state diffusion
flux.
(b) Show using the formula for parallel capacitors that a small disk embedded in the half
plane (z>0), has half the capacitance.
(c) Compare the flux towards a spherical sensor vs. a disk sensor. Which flux is larger? Can
you explain physically the ratio of the fluxes?
Solution:
(a) The diffusion equivalent capacitance of a small disk embedded in an infinite media,
CD ,SS  8Da
D is the diffusion coefficient of the particles and is the radius of the circular disk.
The steady-state diffusion flux,
8 , (b) Since parallel capacitors add, therefore a disk embedded in a semi-infinite medium is half
that of the full medium, i.e.
4 .
, (c) The flux towards a spherical sensors is
large surface area for capture.
4
,
. The result is
times bigger, reflects
Problem 1.7: Diffusion-controlled rate for spherical Sensors
Compute the diffusion-controlled capture of a small molecules of density 1nM by a
spherical sensor of radius a=1nm and D= 10-5 cm2/sec.
Solution:
,
Now 1
4 ~1
4 1
5
1
9 6.023 10 Therefore, the total flux is
7
/1000
,
~8
6
1.256 1
~6.023 11
/
.
11
/ .
/
.
2013 nanoHUB-U Course: Principles of Electronic Nanobiosensors
Problem 1.8: Approximate and Exact Solutions.
Show that the exact and approximate solutions (given below) for time-dependent flux and
total integrated captured biomolecules by a spherical sensor are in good agreement with
each other.
Solution:
Exact:
J (t ) 
D 0 
a 
1 0 

a0 
6 Dt 

N total (t )  4 a02  J (t ')dt '  4 0 Da0 t  a0
0

t
6D

t

Approximate:
N total  t   N  t  A  Ct  0t
Ct 
a01 

4 D
6 Dt  a0

1
A  4 a02
(a) Derive the final expression for the approximate solution in the form given above.
(b) Compare the exact vs. approximate solutions by using the Octave code given below. How
would the results differ, if 6 were replaced by 2 for example? What does it say about the
importance of diffusion pre-factor?
(c) Given Ntotal, how would you calculate the flux toward the sensor?
(d) What is the time-exponent for the asymptotic response to the sensor surface?
Solution:
(a) Derive the final expression for approximate solution in the form given above.
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2013 nanoHUB-U Course: Principles of Electronic Nanobiosensors
The Diffusion equivalent capacitance of concentric spheres of radii a_0 and a_0+R is given
by
,
4
/
–
We can view the second cylinder as the edge of a receding diffusion front, i.e. ~√6
obtain the transient diffusion equivalent capacitance of the concentric spheres,
Ct 
a01 

4 D
6 Dt  a0

, to
1
The time evolution of particle aggregation with ρS set to zero,
→
4
(b) Compare the exact vs. approximate solutions. How would the results differ, if 6 was
replaced by 2 for example? What does it say about the importance of diffusion prefactor?
%% This program compares the exact and approximate solutions to a spherical
sensor.
%%J
prefac=6; %% Diffusion prefactor in 3D.
%%%%%%%%%%%
D=1e-6; %%diffusion coefficient of a small protein molecule, in cm^2/Sec
rho_0=1e-6; %%Moles
t_vec=linspace(0.3,100,10000);
a0=30e-7; %% radius of the sensor in cm.
J_exact=D*rho_0/a0*(1+a0./sqrt(prefac*D*t_vec));
plot(t_vec,J_exact);
title('J vs. time (J exact and J approx)');
xlabel('t [Sec]'); ylabel('J [Mcm/Sec]');
hold on
%%%%%%%%%%%
C_DSS=4*pi*D./((1/a0)-(1./(sqrt(prefac*D*t_vec)+a0)));
J_approx=C_DSS*rho_0/(4*pi*a0^2);
plot(t_vec,J_approx);
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2013 nanoHUB-U Course: Principles of Electronic Nanobiosensors
xlabel('t [Sec]'); ylabel('J [Mcm/Sec]');
legend('exact','approx')
hold off
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
figure(2)
prefac=6;
%%%%%%%%%%%
J_exact_prefac6=D*rho_0/a0*(1+a0./sqrt(prefac*D*t_vec));
plot(t_vec,J_exact_prefac6);
title('J vs. time with different prefac');
xlabel('t [Sec]'); ylabel('J [Mcm/Sec]');
hold on
%%%%%%%%%%%
prefac=2; %% Diffusion prefactor in 1D
J_exact_prefac2=D*rho_0/a0*(1+a0./sqrt(prefac*D*t_vec));
plot(t_vec,J_exact_prefac2);
xlabel('t [Sec]'); ylabel('J [Mcm/Sec]');
hold off
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%N_total
figure(3)
prefac=6;
N_t_exact=4*pi*rho_0*D*a0*(t_vec+(a0/sqrt(prefac*D)*sqrt(t_vec)));
plot(t_vec,N_t_exact);
title('Nt vs. time (Nt exact and Nt approx)');
xlabel('t [Sec]'); ylabel('N_{total} [a.u.]');
hold on
N_t_approx=C_DSS*rho_0.*t_vec;
plot(t_vec,N_t_approx);
xlabel('t [Sec]'); ylabel('N_{total} [a.u.]');
hold off
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%N_total
figure(4)
prefac=6;
N_t_exact_prefac6=4*pi*rho_0*D*a0*(t_vec+(a0/sqrt(prefac*D)*sqrt(t_vec)));
plot(t_vec,N_t_exact_prefac6);
title('Nt vs. time with different prefac');
xlabel('t [Sec]'); ylabel('N_{total} [a.u.]');
hold on
prefac=2;
N_t_exact_prefac2=4*pi*rho_0*D*a0*(t_vec+(a0/sqrt(prefac*D)*sqrt(t_vec)));
plot(t_vec,N_t_exact_prefac2);
xlabel('t [Sec]'); ylabel('N_{total} [a.u.]');
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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2013 nanoHUB-U Course: Principles of Electronic Nanobiosensors
-7
exact
approx
3.34
3.338
3.336
prefac=6
prefac=2
3.34
3.335
20 40 60 80 100
t [Sec]
x 10
20 40 60 80 100
t [Sec]
-15
x 10
Ntotal [a.u.]
3
2
1
-7
3.345
3.334
Ntotal [a.u.]
J [Mcm/Sec]
x 10
J [Mcm/Sec]
x 10
exact
approx
20 40 60 80 100
t [Sec]
-15
3
2
1
prefac=6
prefac=2
20 40 60 80 100
t [Sec]
D=1e‐6 cm2/Sec; ρ0=1e‐6 Moles; a0=30 nm The change is very small in both steady-state flux J and time evolution of particle
aggregation, Ntotal with the change of diffusion pre-factor from 6 to 2. There is no importance
of the diffusion pre-factor.
(c) Given Ntotal, how would you calculate the flux toward the sensor?
Flux toward the sensor,
/ .
(d) What is the time-exponent for the asymptotic response to the sensor surface?
Asymptotic response near → ∞
 D 0 0
 a t

0
D 0 
a 
1  0   4 D 0t 0
I (t  )  4 a02

a0 
6 Dt 
J (t  ) 
D 0 
a
1 0

a0 
6 Dt
The time-exponent is zero. The integrated flux is
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2013 nanoHUB-U Course: Principles of Electronic Nanobiosensors

N total (t )  4 0 Da0 t  a0

1
 4 0 Da0t
6D

t

With time-exponent=1. The results are identical to those from the approach based on
diffusion equivalent capacitance.
Problem 1.9: Steady State Diffusion Flux for system with complex geometry.
Consider a spherical source (radius a, surface density ) of biomolecules, which are being
captured by a planar sensor placed underneath at a distance d>a. Find the Maxwell’s
formula regarding electrical capacitance for this situation, make appropriate substitutions,
and calculate the steady-state flux from the spherical source to the planar sensor surface.
Solution:
The answer simply is


, where


sinh  ln  d / a    d / a   1 


 4 Da 
2
n 1
sinh  nln  d / a    d / a   1 



CD ,SS
,
2
Problem 1.10: nanoHUB Exercise using BiosensorLab.
Assume that you want to design a cylindrical nanowire (NW) sensor of 50 nm radius for the
purpose of capturing 20-bp (base pair) DNA molecules. The given device’s stability time is
limited to 10,000 seconds (about 3 hours). The surface conjugation parameters are not
known exactly, but assume for the time being typical values of kF = 3×106 M-1s-1 and kR = 1
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2013 nanoHUB-U Course: Principles of Electronic Nanobiosensors
s-1. The test fluid of 6 cc is injected to the sensor via pipette drop rather than continuous
external flow. What is the limit of detection for this biosensor?
Solution:




Log on using your nanoHUB ID on nanoHUB.
Launch the online tool BioSensor Lab (https://nanohub.org/resources/senstran) and select version
II
In the “Device parameter of Sensor” tab, select the option “Cylindrical nanowire Biosensor” and set
the nanowire radius of 50 nm (5×10-6cm). Keep length, oxide thickness, and doping density the same
as default values.
In the “Biological parameters” tab, set “Analyte Type” to be DNA. Set kf = 3×106 /(M×s) and kr =
1/s as indicated in the problem. In “Parameter of DNA” section, set DNA strand length (bp) to be 20.
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2013 nanoHUB-U Course: Principles of Electronic Nanobiosensors

We assume room temperature in this simulation, so set the Temperature in degree Kelvin to be 300 K
in the “Ambient Condition” tab.

Move on to the “Type of simulation” phase. In this problem, we are interested in the settling time. So
turn on the “settling time” switch. We will keep all the parameters in this option unchanged for the
time being.


Click on simulate.
Once the simulation is completed, select the graph “Settling time vs. analyte concentration” for
display.
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2013 nanoHUB-U Course: Principles of Electronic Nanobiosensors

You will be able to see that at the point with settling time = 10326.4 s, Analyte Concentration ≈ 8.5×
10-15 M.
Problems 1.11: Fractal Biosensors.
Consider two sensors with capture surfaces approximated by a Sierpinski carpet (DF=1.89)
and Sierpinski triangle (DF=1.58), respectively. Determine the number of molecules
captured by these sensors based on the diffusion limited transport theory discussed in the
class. Which one of this behaves like a planar sensor and which one as a cylindrical sensor?
Solution:
For a sensor defined by DF=1.89, the number of particles captured at time t is given by .
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2013 nanoHUB-U Course: Principles of Electronic Nanobiosensors
Since DF=2 defines a planar sensor, the response of a Sierpinski carpet will be similar to that of a
planar sensor. One can likewise calculate the response of a triangular sensors with
1.58.
.
.
The response is faster than a planar sensor, but not as fast as cylindrical sensors.
Problem 1.12: Diffusion flux with ellipsoidal Electrodes.
Calculate steady state flux between strip of size 2c and an ellipsoid defined by semi-major
axis a and semi-minor axis b. Recall that one can define an ellipse in elliptical coordinate
as follows
x2
y2

1
c 2 cosh 2    c 2sinh 2   
(Eq. 1)
where (tortuosity) is defined by either
≡ or
words, no sooner are , ,
defined, the is defined as well.
≡ . In other
If the steady state capacitance between the flat plate and the ellipsoidal surface
characterized by µ, is given by CD,SS= µD/π ~ πD/ln(r/a), and if the outer ellipsoidal cylinder
is kept at concentration ρ0,
(a) Find the steady-state flux to the sensor surface.
(b) Use the approximate relationship to define a time-dependent solution of the problem.
(c) How does the result compare to those related to concentric hemispherical surfaces?
(Ref: O. Benichou and R. Voituriez, PRL, 100, 168105, 2008.)
Solution:
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2013 nanoHUB-U Course: Principles of Electronic Nanobiosensors
(a) Find the steady-state flux to the sensor surface.
Fout  CD ,SS  0 
D

 0
(b) Use the approximate relationship to define a time-dependent solution of the problem.
N  t   CD ,SS 0t 
D
t
 0
(c) How does the result compare to those related to concentric hemi-cylindrical surfaces with
radius c and a, respectively?
N  t   CD ,SS  0t 
D
ln  a c 
0t
→ ln
The two results are comparable, because
16
, when
≫ .