double carbon nanotube antenna as a detector of modulated terahertz radiation

Double carbon nanotube antenna as a detector of modulated terahertz

radiationV. Semenenko1, V. Leiman1, A. Arsenin1, Yu. Stebunov1, and V. Ryzhii2

1Laboratory of Nanooptics and Femtosecond Electronics,

Moscow Institute of Physics and Technology, Dolgoprudny 141700, Russia2Computational Nanoelectronics Laboratory,

University of Aizu, Aizuwakamatsu, Fukushima 965-8580, Japan

Outline

• Introduction• Double carbon nanotube antenna• Analytical solution• Results• Conclusion

IntroductionSome schemes of detectors of modulated terahertz radiation:

0~ 1 um, ~ 1 um, ~ 50 nm, ~ 1 5 nmL l z a

V. Ryzhii, et al // Appl Phys Lett, 2007 V. Leiman, et al // J. Appl Phys, 2008

IntroductionDetector scheme on the basis of single-walled carbon nanotubes:

Yu. Stebunov et al // Appl. Phys. Ex., 2011

Output signal is the current induced by the variable capacitance:

… but there is other scheme:0I C V

INTRODUCTION

• Small mass• Lower electron collision frequency (in comparison with

2DEG)

• High contact resistance (about 6.5 kΩ)

Main advantages of using metallic SWCNTs as mechanical and plasma resonators:

… and main disadvantage:

Double carbon nanotube antennaMathematical model.Plasma resonator:1) Hydrodynamic model for electron transport in metallic SWCNTs:

2

e

= 0

= ,

S

Fz

S

uen

t z

v eu u E

en z m

2) Maxwell equations,3) Boundary conditions on thenanotubes surface:

4 , 0.n n n zE E

Mechanical resonator:dynamic Euler-Bernoully equation for beam deflection:

2 4b

m2 4

/2/2

= ,

0, 0.

l

l l

z Lz L

J Fx x x

t t M z Mx

xz

and their ends:

Mechanical resonatorDynamic Euler-Bernoully equation:

2 4b

m2 4= ,l

l l

J Fx x x

t t M z M

Consider the case when the linear force ( , ),lF z t can be

represented in the following view: ( , ) ( ) ( ).l lF z t g z f t

We will find the solution in the form of expansion in a series:

1

, ,j jj

x z t x t z

where j z of orthonormal functions

That are eigenfunctions of the homogeneous equation

4

bm 4

0,jj

l

Ji z

M z

/2/2

0, 0.jj z L

z Lz

( , ) Re( ( ) )i tj jx z t z e

Mechanical resonatorExpressions for the eigenmodes:

/22

/2

1= / , = ,

L

j j j j j

L

z y z N N y z dzL

S

cos / 2= cos cosh , = , 2 1, 0.. ,

cosh / 2j j j j

Ly z z j n n

L

sin / 2= sin sinh , = , 2 , 1.. ,

sinh / 2j

j j j j jj

Ly z z j n n

L

4m 4 b 1

= , æ = , cos = .æ cosh

j j

j jl j

i JL

M L

Reducing to the lumped oscillator jSet of functions forms a full basis, so we can expand the function ( )g z

into a series of them: 0.5

=1 0.5

= , = .j j j jj

zg z z g z z d

L

Partial sums of the series are shown in the figure for 1 sin

.2

z

Lg z

Lumped mechanical oscillator2 4

bm2 4

( )= ( ),l

l l

J f tx x xg z

t t M z M

So, substituting

1

, j jj

x z t x t z

=1

= j jj

g z z

we get:

2m ,j l

j j j jl

f tx x x

M

2

b2

1= , cos = .

coshj

j jl j

J

M L

Next we will consider only the oscillation of the general mode (j=1),and the dynamic equation for the lumped oscillator will be:

2m m 1 1 m 1, , ,l

l

f tx x x x x

M

Plasma resonatorIn the case Ez=0

Solution of electrodynamic equations for the nanotubes of infinite length gives:

Dispersion equation:

2 2 2 2 2ED

22ED 0 0

= , ,

4= , = ,

F

S

i v k v v v

e n av K ka K kd

m

Sign “-” corresponds to the anti-symmetric mode, that carries a signal in the double line. In this case

0 ( ) ln , 1,= ln ,d

x x xa

K

i.e. phase velocity doesn’t depend on k.

one can obtain an equation for the forced plasma oscillation in the Fourier space:

2

,

= 0,

= ,

k S k

k k z kS

i ik en u

v eu i ik E

en m

Here is implied that only symmetric mode can be exited in the system, i.e.

0 0 .= K ka K kd

0,zE When

Plasma resonator

Then it possible to make analytically reverse Fourier transformation of the previous equations system:

2 2

= 0

= ,ED z

I

t zI

I v v Et z

Boundary conditions:

/2 /2

, 0, , 0.z L z L

I z t I z t

Solution of the system:2 2

, ,2 2 2

1 sin cos= , = 1 ,

cos cos2 2

ED EDz z

v kz i v kzE I E

L Lk v k vk k

is the linear charge of the two nanotubes= 4 a

= 4 SI a en u is the electric current in them

Because of

We consider that ( ) const, / .k k L

/ ,L c the external electric fieldcan be considered as uniform near the system,thus, we can put ( , ) ( )z zE z t E t

Lumped plasma resonator2

,2

1 sin= ,

cos2

EDz

v kzE

Lk v k

e2

( ) .i

kv

Remind the expression for in

z-space:

The frequencies of plasma resonances are 1 2 , 0.. ;n

vn n

L

For the case e 0 one can put:2

,2

= ,cos

2

sinzEDEL

v

zvL

k L

and introduce a new value2

,ED0, 2

= ,cos

2

zEL vLv k

Lumped plasma resonatorIn the vicinity of the general resonance 1( ~ ) the approximate

relation is valid:20

2 20 e

1 4.

cos / 2kL i

Graphs are built for the oscillations quality factor 0 e/ 10.Q

Lumped plasma resonatorSo, in Fourier space we have the relation:

2,ED

0, 2=

cos2

zEL vLv k

22 2 2ED

, 0 0 e2 2

4/z

L vE i

v

or2 2

2 2 0 ED0, e 0 ,2 2

4( ) ,z

L vi i E

v

…that corresponds to the following relation in the t-space:

2

2 20 e 0 e 0 e2 2

4= ,ED

z

L vE t

v

and

The tz-dependence of ρ will be: 0( , ) ( ) sin .z

z t tL

Interaction between the resonators

1) Plasma resonator affects the mechanical one:

/2

2m m 1

/2

1, 0.2189.

Ll

l L

f tx x x g z z dz

M L

The linear force acting on one nanotube from another:22

20 ( )( , )( ) ( ), ( ) , ( ) sin

2 2l l l

tz t zF f t g z f t g z

d d L

Thus, we obtain:2

2 0m m .

2 l

x x xM d

Dynamic equation for the mechanical resonator

Interaction between the resonators1) Mechanical resonator affects the plasma one:

Deformation of the plasma resonator causes its eigenfrequency shift.

In the case of non-deformed nanotubes we had:

2 2

= 0

= ,ED z

I

t zI

I v v Et z

/2 /2

, 0, , 0.z L z L

I z t I z t

2 2 2ED

0 0 co .nst

,

=

Fv v v

K ka K kd

When the NTs are deformed,

0 0 1( ) ( 2 ( ))( ) ka K k d x zz K

For small oscillation amplitude x d

1 0 0

2( ), ( )( )

xz ka K kd

dz K

Wavevector also depends on coordinate:

e2

2 2 2ED, ( ),F

ik z

v zv v v z

Interaction between the resonators

The solution in the case of non-deformed NTs was:

2 2

, ,2 2 2

1 sin cos= , = 1 ,

cos cos2 2

ED EDz z

v kz i v kzE I E

L Lk v k vk k

To get the solution for the deformed NTs, we should substitute , / 2kz kL

In all the trigonometric functions with:/2

0 0

( ) , ( ) .z L

k d k d

Then the resonant condition will be:/2

0

( ) ,2

L

k d n

or, for general frequency e 0 we will have the relation:

/22

e12

e 0 /2

1= 1 , = 0.831.

L'ED

L

v xd

v d L

Equations for coupled resonators

22

m m

2 22 2ED ED

e e e2 2 2

=2

41 2 = ,

l

z

x x xM d

v x L vE t

v d v

22

m m

2e

2

( )(1 )

qy y y

SMR y U t

q q qL d L

The same structure of the equations as in the case of those describing capacitance transducers:

Results

Consider the modulated incoming signal: 0= 1 cos cos ,zE t E m t t

For the small oscillations x d we can put e econst .'

12 0 ,x d / 0.3x d Maximum :x and e e/ 1 0.035'

Amplitude of the nanotubes mechanical oscillation.

So, we get:

22 2 20

e e e e e2 20

4= Re , = / ,

/ 2 / 2 .

ED

i t i ti t

E L vt G f t G i

v

f t e m e m e

And for the mechanical oscillation:

4

2 22 2 2 20 ED

e m m m m m4 2m

4= Re , = / .i t

l

E Lm vx t G G e G i

M d v

Detection of mech. oscillationOutput signal: out 0

CV V

C

2 2 14 , .

4ln /l l l

xC C C

d d a

Thus,42 2

max 20out e m4 2 2

m 0

32= .l ED

l

C E Lm vV Q Q

M d v

If consider this device as a detector of modulated THz radiation,its responsivity will be given as out in/ .VR V P

e me m

e m

, .Q Q

in m e rad .P P P P

/2 2 3 2

8 5 max 2e 0 EDm e rad e e e e2 4 2

0/2

4/ ~ 10 , / ~ 10 , , = .

L

ED L

E L vP P P P P I z t dz Q

v v

So,

2

max 60 EDe m2 2

m

8= 10 V/W,l

Vl

C Vm vR Q Q

M d v v

for 3m m 00.5um, 1nm, 20nm, 10 , 0.2GHz, 1V, 0.1,L a d Q V m

7 8 8F ED e e8.7 10 cm/s, 2.2 10 cm/s, 6.1 10 cm/s, 6.1THz, 74,1.v v v Q

Parametric instability threshold

e e 2e e

1= Re , , = , = .

1 /i tG s e G s

s is Q

22 2ED

m m 2 2

2 2e e e

=

1 = cos ,l

v

M d v

t

2 2ED ED

02 2 2

4= 2 , ,

v x L vE

v d v

Now consider the monochromatic incoming signal: 0= cos ,zE t E t2

2m m

2 22 2ED ED

e e e 02 2 2

=2

41 2 = cos ,

l

x x xM d

v x L vE t

v d v

so,

In the assumption x d we can get:

and 22

e2( ),

= .2

G t st

… but actually, 22

e2( ),

= .2

G t st

Parametric instability threshold

22

22 EDm m e2 2

= , , = , , = , .2l

vK F t s K F s G s

M d v

1e, , , ~ ,

FF t s F s

So, for the mechanical oscillations we have the following equation:

Using the expansion of F into the series of the degrees,

we get: 2m m = , ,

FK K F s

sign 1.F

1m e 0,

FK

Parametric instability thresholdSo, when there is a self-excitation of the

mechanical oscillations in the system.

3e

3max ,

8

FQ

so we can estimate

624 22 e 0th 4 2

e m

= .3

l

ED

M v dE

nQ Q v L

For the parameters stated before we obtain: th ~ 1kV/cm,E

that is equivalent to the incident radiation intensity2

th ~ 1kW/cm .I

In comparison, the maximum amplitude of the modulated signal (with modulation depth m=0.1), under which the nanotubes begin touch one another (2x1(0)=d) is estimated as: that corresponds to the intensity of

max ~ 50V/cm,E2

max ~ 0.2mW/cm .I

Conclusion• The model describing combined plasma and

mechanical oscillations in the system of the two parallel metallic carbon nanotubes is developed.

• Proposed scheme detecting modulated THz radiation featured a remarkably high responsivity (about 106 V/W).

• For the self-excitation of the mechanical modes in the system, quite a high preamplification (by a factor about 10-100) of the incoming signal is needed.

Thank you very much for your attention