Enhancement of the thermoelectric properties in bilayer graphene structures induced by Fano resonances

In Fig. 1a, b we illustrate the potential thermoelectric system primarily based on bilayer graphene barrier buildings. As a consultant case we’re displaying a double-barrier thermoelectric system. The warmth flows from the left to the proper lead, cold and hot aspect, respectively. Bilayer graphene is positioned on a substrate, which in flip is situated above a again gate. The potential limitations are generated by electrodes (gates) on high of bilayer graphene. The crystal construction is schematically proven in Fig. 1c. It corresponds to secure Bernal-stacked bilayer graphene. The related interactions between carbon atoms for our Hamiltonian are depicted as effectively, (gamma _{0}) and (gamma _{1}=t_{perp }). The potential limitations are illustrated in Fig. 1d. They correspond to gapless limitations, that’s, the potential power is identical in each graphene layers (V_1=V_2=V_0). It is usually vital to say that as graphene is supported by a substrate and usually encapsulated with dielectrics, we count on that phonon contribution to the thermal transport be negligible23,24,25. So, we’ll solely think about the contribution of electrons to the thermal transport.

Determine 1

(a) Prime and (b) aspect view of the schematic illustration of the potential thermoelectric system primarily based on bilayer graphene double limitations. The double barrier construction generated by the highest gates is sandwiched between cold and hot leads. (c) Crystal construction of Bernal-stacked bilayer graphene, with (gamma _{0}) and (gamma _{1}= t_{perp }) as the principle interactions between carbon atoms. (d) Band-edge profile of the conduction band of (a). Right here, (V_{0}=V_{1}=V_{2}) characterize the potential power on the bilayer graphene layers, respectively, additionally the potentials (V_{1}) and (V_{2}) are lower than (t_{perp }) and due to this fact the bands might be thought-about as paraboloids. (d_{B}) and (d_{w}) represents the widths of the barrier and effectively areas, respectively. (x_L) and (x_R) correspond to the left and proper ends of the nanostructure.

To review the thermoelectric properties of bilayer graphene buildings we’ll use the secure numerical technique referred to as the hybrid matrix technique40,56,57. On this drawback, we take care of a system of first-order unusual differential equations. To make use of switch matrix strategies, and particularly the so-called hybrid matrices, we have now made the mandatory changes, which then again are fairly apparent and are defined elsewhere40,41,57. To determine the hybrid matrix technique, we want firstly the efficient Dirac Hamiltonian that describes the cost carriers in bilayer graphene, specifically60,61,62:

$$start{aligned} mathcal {H}=left( start{array}{cccc} V_{1}&{}pi &{}t_{perp }&{}0 pi ^{*}&{}V_{1}&{}0&{}0 t_{perp }&{}0&{}V_{2}&{}pi ^{*} 0&{}0&{}pi &{}V_{2} finish{array} proper) , finish{aligned}$$


the place (pi =v_{F}(p_{x}+ip_{y})), (pi ^{*}=v_{F}(p_{x}-ip_{y})), (p_{x,y}=-ihbar partial _{x,y}) is the momentum operator and (v_{F}) is the Fermi velocity. This Hamiltonian is legitimate for energies lower than (gamma _{0}=3.09) eV60,61, it includes the interplay between graphene layers by way of the parameter (t_{perp }=390) meV61 and takes under consideration bandgap opening (E_{g}=V_{2}-V_{1}) when (V_1 ne V_2).

Now, it is very important point out that in our single and double barrier buildings there may be spatial homogeneity within the transverse path, that’s (q_{y}=k_{y}), then the propagation of Dirac electrons might be decreased to a one-dimensional drawback. So, we want secondly to unravel the one-dimensional eigenvalue equation (mathbf {H}cdot mathbf {F}(x)=Emathbf {F}(x)), which after primary algebra adopts the mathematical kind41

$$start{aligned} frac{dmathbf {F}(x)}{dx}+left( start{array}{cccc} q_{y}&{}ifrac{V_{1}-E}{hbar v_{F}}&{}0&{}0 ifrac{V_{1}-E}{hbar v_{F}}&{}-q_{y}&{}ifrac{t_{perp }}{hbar v_{F}}&{}0 0&{}0&{}-q_{y}&{}ifrac{V_{2}-E}{hbar v_{F}} ifrac{t_{perp }}{hbar v_{F}}&{}0&{}ifrac{V_{2}-E}{hbar v_{F}}&{}q_{y} finish{array}proper) cdot mathbf {F}(x)=mathbf {0}_{4times 1}. finish{aligned}$$


So, to discover a foundation, a operate of the shape (mathbf {F}(x)=mathbf {F}_{0}e^{iqx}) is proposed, that is substituted in Eq. (2) to reach at a secular equation that has the next eigenvalues as answer

$$start{aligned} q=pm sqrt{-q^{2}_{y}+frac{1}{2(hbar v_{F})^{2}}[(E-V_{1})^{2}+(E-V_{2})^{2}]pm frac{1}{2(hbar v_{F})^{2}}sqrt{[(E-V_{1})^{2}-(E-V_{2})^{2}]^{2}+4t^{2}_{perp }(E-V_{1})(E-V_{2})}}. finish{aligned}$$


We are able to see that there are 4 eigenvalues, this comes from the truth that the Hamiltonian is 4(occasions)4, then we have now 4 linear unbiased options given as

$$start{aligned} mathbf {F}(x)^{pm }_{j}=mathbf {F}_{0j}^{pm }e^{pm iq_{j}x} = (a_{j},b_{j}^{pm },c_{j},d_{j}^{pm })^{T}e^{pm iq_{j}x},quad j=1,2. finish{aligned}$$


the place the corresponding parts are

$$start{aligned} a_{j}= & {} 1; finish{aligned}$$


$$start{aligned} b_{j}^{pm }= & {} frac{hbar v_{F}(pm q_{j}-iq_{y})}{E-V_{1}};finish{aligned}$$


$$start{aligned} c_{j}= & {} frac{t_{perp }(E-V_{2})}{(E-V_{2})^{2}-(hbar v_{F})^{2}(q_{j}^{2}+q_{y}^{2})};finish{aligned}$$


$$start{aligned} d_{j}^{pm }= & {} frac{hbar v_{F}(pm q_{j}+iq_{y})}{E-V_{2}}c_{j}. finish{aligned}$$


A common answer might be expressed in matrix kind as follows

$$start{aligned} mathbf {F}(x) = left( start{array}{cccc} a_{1}e^{iq_{1}x}&{}a_{2}e^{iq_{2}x}&{}a_{1}e^{-iq_{1}x}&{}a_{2}e^{-iq_{2}x} b_{1}^{+}e^{iq_{1}x}&{}b_{2}^{+}e^{iq_{2}x}&{}b_{1}^{-}e^{-iq_{1}x}&{}b_{2}^{-}e^{-iq_{2}x} c_{1}e^{iq_{1}x}&{}c_{2}e^{iq_{2}x}&{}c_{1}e^{-iq_{1}x}&{}c_{2}e^{-iq_{2}x} d_{1}^{+}e^{iq_{1}x}&{}d_{2}^{+}e^{iq_{2}x}&{}d_{1}^{-}e^{-iq_{1}x}&{}d_{2}^{-}e^{-iq_{2}x} finish{array}proper) cdot left( start{array}{c} alpha _{1}^{+} alpha _{2}^{+} alpha _{1}^{-} alpha _{2}^{-} finish{array}proper) finish{aligned}$$


or equivalently by matrix blocks

$$start{aligned} mathbf {F}(x)=left( start{array}{c} mathbf {F}_{u}(x) mathbf {F}_{d}(x) finish{array}proper) =left( start{array}{cc} mathbf {U}^{+}(x) &{} mathbf {U}^{-}(x) mathbf {D}^{+}(x) &{} mathbf {D}^{-}(x) finish{array}proper) cdot left( start{array}{c} alpha ^{+} alpha ^{-} finish{array}proper) , finish{aligned}$$


the place (mathbf {F}_{u}(x)), (mathbf {F}_{d}(x)), (alpha ^{+}) and (alpha ^{-}) are a two-dimensional column vectors. (mathbf {U}^{pm }(x)) and (mathbf {D}^{pm }(x)) are the (2times 2) respective matrix blocks of the (4times 4) matrix in equation (9). We are able to outline the hybrid matrix as

$$start{aligned} left( start{array}{c} mathbf {F}_{u}(x_{L}) mathbf {F}_{d}(x_{R}) finish{array}proper) =mathbf {H}(x_{R},x_{L})cdot left( start{array}{c} mathbf {F}_{d}(x_{L}) mathbf {F}_{u}(x_{R}) finish{array}proper) . finish{aligned}$$


This equation relates the vectors (mathbf {F}_{u}(x)) and (mathbf {F}_{d}(x)) on the left ((x_{L})) and proper ((x_{R})) ends of the heterostructure, see Fig. 1d. The express expressions for the vectors (mathbf {F}_{u}(x)) and (mathbf {F}_{d}(x)) might be present in Ref. 41. Then from equations (10) and (11), the hybrid matrix might be written as

$$start{aligned} mathbf {H}(x_{R},x_{L})= & {} left( start{array}{cccc} a_{1}e^{iq_{1}x_{L}}&{}a_{2}e^{iq_{2}x_{L}}&{}a_{1}e^{-iq_{1}x_{L}}&{}a_{2}e^{-iq_{2}x_{L}} b_{1}^{+}e^{iq_{1}x_{L}}&{}b_{2}^{+}e^{iq_{2}x_{L}}&{}b_{1}^{-}e^{-iq_{1}x_{L}}&{}b_{2}^{-}e^{-iq_{2}x_{L}} c_{1}e^{iq_{1}(x_{R})}&{}c_{2}e^{iq_{2}(x_{R})}&{}c_{1}e^{-iq_{1}(x_{R})}&{}c_{2}e^{-iq_{2}(x_{R})} d_{1}^{+}e^{iq_{1}(x_{R})}&{}d_{2}^{+}e^{iq_{2}(x_{R})}&{}d_{1}^{-}e^{-iq_{1}(x_{R})}&{}d_{2}^{-}e^{-iq_{2}(x_{R})} finish{array}proper) nonumber &cdot left( start{array}{cccc} c_{1}e^{iq_{1}x_{L}}&{}c_{2}e^{iq_{2}x_{L}}&{}c_{1}e^{-iq_{1}x_{L}}&{}c_{2}e^{-iq_{2}x_{L}} d_{1}^{+}e^{iq_{1}x_{L}}&{}d_{2}^{+}e^{iq_{2}x_{L}}&{}d_{1}^{-}e^{-iq_{1}x_{L}}&{}d_{2}^{-}e^{-iq_{2}x_{L}} a_{1}e^{iq_{1}(x_{R})}&{}a_{2}e^{iq_{2}(x_{R})}&{}a_{1}e^{-iq_{1}(x_{R})}&{}a_{2}e^{-iq_{2}(x_{R})} b_{1}^{+}e^{iq_{1}(x_{R})}&{}b_{2}^{+}e^{iq_{2}(x_{R})}&{}b_{1}^{-}e^{-iq_{1}(x_{R})}&{}b_{2}^{-}e^{-iq_{2}(x_{R})} finish{array}proper) ^{-1}. finish{aligned}$$


Equation (12) is the hybrid matrix for a homogeneous area, in our case effectively or barrier. To acquire the hybrid matrix for a heterostructure it’s important to know the composition rule, for extra particulars see38,40,56. Then, to be able to know the transmission and transport properties of a heterostructure it’s essential to know the vectors (mathbf {F}_{u}(x)) and (mathbf {F}_{d}(x)) of Eq. (10) on the left ((x_{L})) and proper ((x_{R})) ends of the heterostructure and the entire hybrid matrix, see Ref. 41.

We assume {that a} wave (mathbf {F}^{+}_{01}e^{iq_{1}x}) touring from the left aspect hits the left finish of the barrier construction and leads to reflections (mathbf {F}^{-}_{01}e^{-iq_{1}x}) and (mathbf {F}^{-}_{02}e^{-iq_{2}x}) in that area, whereas on the proper finish of the barrier construction we have now solely transmitted waves (mathbf {F}^{+}_{01}e^{iq_{1}x}) and (mathbf {F}^{+}_{02}e^{iq_{2}x}), that is apparently a multi-channel drawback, however on each the left and proper aspect we have now non-propagating options. This makes us actually face an issue with a single channel. Then considering the above, equation (11) takes the shape

$$start{aligned} mathbf {M}_{1}+mathbf {M}_{2}cdot left( start{array}{c} r_{1} r_{2} t_{1} t_{2} finish{array}proper) = mathbf {H}(x_{R},x_{L})cdot mathbf {M}_{3}+mathbf {H}(x_{R},x_{L})cdot mathbf {M}_{4}cdot left( start{array}{c} r_{1} r_{2} t_{1} t_{2} finish{array}proper) , finish{aligned}$$


the place (r_{1}, r_{2}, t_{1}) and (t_{2}) are outlined when it comes to the coefficients (alpha ^{+}) and (alpha ^{-}) at every finish41. As well as, the matrices (mathbf {M}_{1},mathbf {M}_{2}, mathbf {M}_{3}) and (mathbf {M}_{4}) are given as

$$start{aligned} mathbf {M}_{1}= & {} left( start{array}{c} a_{1L} b^{+}_{1L} 0 0 finish{array} proper) ,quad mathbf {M}_{2}=left( start{array}{cccc} a_{1L}&{}a_{2L}&{}0&{}0 b^{-}_{1L}&{}b^{-}_{2L}&{}0&{}0 0&{}0&{}c_{1R}&{}c_{2R} 0&{}0&{}d^{+}_{1R}&{}c^{+}_{2R} finish{array} proper) , finish{aligned}$$


$$start{aligned} mathbf {M}_{3}= & {} left( start{array}{c} c_{1L} d^{+}_{1L} 0 0 finish{array} proper) ;quad mathbf {M}_{4}=left( start{array}{cccc} c_{1L}&{}c_{2L}&{}0&{}0 d^{-}_{1L}&{}d^{-}_{2L}&{}0&{}0 0&{}0&{}a_{1R}&{}a_{2R} 0&{}0&{}b^{+}_{1R}&{}b^{+}_{2R} finish{array} proper) . finish{aligned}$$


The subscripts L and R point out the exterior area the place the parts of the wavefunction amplitudes are calculated. Lastly, the transmission and reflection amplitudes might be written when it comes to the hybrid matrix in a extra compact kind as

$$start{aligned} left( start{array}{c} r_{1} r_{2} t_{1} t_{2} finish{array}proper) = left[ mathbf {M}_{2}-mathbf {H}(x_{R},x_{L})cdot mathbf {M}_{4}right] ^{-1} cdot left[ mathbf {H}(x_{R},x_{L})cdot mathbf {M}_{3}-mathbf {M}_{1}right] . finish{aligned}$$


Typically, we’re coping with an issue of two transmission channels. So, the transmittance can be given by

$$start{aligned} mathbb {T}(E,theta )=left| t_{1}proper| ^2+left| t_{2}proper| ^2, finish{aligned}$$


the place E and (theta) are the power and angle of incidence of the electrons within the bilayer graphene construction.

Nevertheless, if (E< t_{perp }) there is just one transmission channel, as might be seen from Eq. (3) on the semi-infinite areas (V_1=V_2=0). In such case, the transmittance is solely

$$start{aligned} mathbb {T}(E,theta )=left| t_{1}proper| ^2. finish{aligned}$$


Actually, in all our calculations the one transmission channel situation (E< t_{perp }) is fulfilled. Thus, the transmittance that we’ll use is decided by Eq. (18).

To calculate the transport and thermoelectric properties it’s needed to acquire the conductance operate, this may be accomplished utilizing the Landauer–Büttiker formalism58,59. Underneath this formalism, the conductance operate might be obtained by summing the transmittance over all angles of incidence

$$start{aligned} {G}(E)=G_{0}sqrt{E^{2}+t_{perp }E}int _{-frac{pi }{2}}^{frac{pi }{2}}mathbb {T}(E,theta )cos {theta } dtheta , finish{aligned}$$


the place (G_{0}=frac{2e^{2}L_{y}}{h^{2}v_{F}}) is the elemental conductance issue with e, (L_{y}) and h because the electron cost, the width of bilayer graphene sheet within the transverse y-coordinate and the Planck’s fixed, respectively. For our calculations, the width of the bilayer graphene is ready to (L_{y} = 200) nm63.

Within the case of thermoelectric properties, it’s essential to know the Seebeck coefficient, it may be accomplished with the assistance of Landauer–Büttiker formalism as effectively59,64. If we think about an efficient voltage (Delta mathcal {V}=mathcal {V}_1-mathcal {V}_2) and a temperature gradient (Delta T=T_1-T_2) between the supply and the drain in Fig. 1a, the linear-regime present of electrons within the system might be written as

$$start{aligned} I = G {Delta mathcal {V}} + G_{S}Delta T, finish{aligned}$$


the place the conductance is given by

$$start{aligned} G(mu )=int _{-infty }^{infty }G(E)left( -frac{partial f_{0}}{partial E}proper) dE, finish{aligned}$$


and the (G_{S}) coefficient by

$$start{aligned} G_{S}(mu )=int _{-infty }^{infty }G(E)frac{E-mu }{eT}left( -frac{partial f_{0}}{partial E}proper) dE, finish{aligned}$$


with T, (mu), and (f_{0}) as the typical temperature, the typical chemical potential and Fermi-Dirac distribution operate, respectively. Then the Seebeck coefficient is outlined because the voltage induced by a temperature gradient in open circuit situation ((I=0)), specifically,

$$start{aligned} S = -frac{{Delta mathcal {V}}}{Delta T} = frac{G_{S}}{G}. finish{aligned}$$


Furthermore, to characterize the thermoelectric efficiency of a tool we have to think about ZT, which is nothing aside from the quantity of electrical energy that may be generated from a temperature gradient. Particularly, ZT is given by the expression

$$start{aligned} ZT = frac{S^{2}GT}{G_{Ok}}. finish{aligned}$$


Right here, the amount (S^{2}G) is called the ability issue, for a fabric to be a very good thermoelectric, its energy issue must be as excessive as potential. (G_K) is the so-called digital thermal conductance at open circuit ((I=0)), which must be as little as potential for a excessive ZT. (G_{Ok}) is outlined by the thermal present

$$start{aligned} I_{Q} = frac{G_{P}}{G} I + G_{Ok}Delta T, finish{aligned}$$


and given as

$$start{aligned} G_{Ok} = -frac{I_{Q}}{Delta T} = G_{Q}-frac{G_{P}G_{S}}{G}, finish{aligned}$$


the place

$$start{aligned} G_{P}(mu )=int _{-infty }^{infty }G(E)frac{E-mu }{e}left( -frac{partial f_{0}}{partial E}proper) dE finish{aligned}$$



$$start{aligned} G_{Q}(mu )=int _{-infty }^{infty }G(E)frac{(E-mu )^{2}}{e^{2}T}left( -frac{partial f_{0}}{partial E}proper) dE finish{aligned}$$


are the Peltier transport coefficient and the thermal conductance at brief circuit (({Delta mathcal {V}}=0)), respectively. To finish the arithmetic of the thermoelectric properties, we want to point out that to compute the transport coefficients G, (G_{S}), (G_{P}), (G_{Ok}) and (G_{Q}) it’s enough to contemplate a number of (k_{B}T) round (mu) because of the broadening operate (left( -frac{partial f_{0}}{partial E}proper)) and never the whole power vary.

With the transport coefficients we will additionally compute straightforwardly the utmost energy54

$$start{aligned} P^{max}=frac{1}{4} S^2 G left( Delta Tright) ^2 finish{aligned}$$


and the effectivity at most energy54

$$start{aligned} eta (P^{max})=frac{eta _c}{2} frac{ZT}{2+ZT}. finish{aligned}$$


To finish this part, we’ll give the basics to find out the density of states (DOS) of single and double limitations. This calculation will give us the potential for know if the buildup of electron states is said to the development of the thermoelectric response. The DOS might be obtained straightforwardly by way of the system band construction. Within the case of one-dimensional periodic system the DOS per unit size and power is given as59

$$start{aligned} textual content {DOS}(E,theta ) = frac{1}{2pi }left| frac{partial q_{Bl}(E,theta )}{partial E}proper| , finish{aligned}$$


the place (q_{Bl}) is the Bloch wave vector related to the super-periodicity. Right here, we have now additionally remarked that in a one dimensional periodic system in 2D supplies there’s a dependence on the transverse wave vector or equivalently on the angle of incidence. Actually, we will receive the entire DOS by summing over all angles of incidence, specifically:

$$start{aligned} textual content {DOS}(E)=frac{L_{y}}{2pi }frac{sqrt{E^{2}+t_{perp }E}}{hbar v_{F}}int _{-frac{pi }{2}}^{frac{pi }{2}}textual content {DOS}(E,theta )cos {theta } dtheta . finish{aligned}$$


Up up to now, the basics to compute the DOS are fairly clear, nonetheless we’re not coping with periodic or superlattice buildings. To beat this element, we’ll think about single limitations as a periodic system with a unit-cell compose of a barrier and a effectively. Then, we’ll attempt to mimic single limitations by growing systematically the width of the effectively area. Within the case of double limitations, we’ll think about as unit-cell two limitations and two wells, and the width of the second effectively can be systematically elevated to imitate double limitations. By making use of the Bloch’s theorem to the talked about periodic methods of single and double limitations we will receive its corresponding band construction61. Particularly, from the Bloch situation

$$start{aligned} mathbf {F}(x+d_{uc})=e^{iq_{Bl}d_{uc}} mathbf {F}(x), finish{aligned}$$


we will derive the next transcendental equation

$$start{aligned} textual content {det}left( prod _{j=1}^{N}left( D_jcdot P_j^{-1}cdot D_j^{-1}proper) -e^{iq_{Bl}d_{uc}}Iright) =0, finish{aligned}$$


the place (d_{uc}) is the scale of the unit-cell, and (D_j) and (P_j) are matrices given when it comes to the coefficients and wave vectors of the wave capabilities of the barrier and effectively areas that compose the unit-cell, specifically:

$$start{aligned} D_{j}= & {} left( start{array}{cccc} a_{1,j} &{} a_{2,j} &{} a_{1,j} &{} a_{2,j} b^{+}_{1,j} &{} b^{+}_{2,j} &{} b^{-}_{1,j} &{} b^{-}_{2,j} c_{1,j} &{} c_{2,j} &{} c_{1,j} &{} c_{2,j} d^{+}_{1,j} &{} d^{+}_{2,j} &{} d^{-}_{1,j} &{} d^{-}_{2,j} finish{array} proper) ; finish{aligned}$$


$$start{aligned} P_{j}^{-1}= & {} left( start{array}{cccc} e^{iq_{1,j};d_j} &{} 0 &{} 0 &{} 0 0 &{} e^{iq_{2,j};d_j} &{} 0 &{}0 0 &{} 0 &{} e^{-iq_{1,j};d_j} &{} 0 0 &{} 0 &{} 0 &{} e^{-iq_{2,j};d_j} finish{array} proper) . finish{aligned}$$


Right here, (j=1,3) and (j=2,4) correspond to barrier and effectively areas respectively. Moreover, if (N = 2), we’re speaking a couple of unit cell fashioned by a barrier plus a effectively with (d_{uc} =d_B+d_W). For the case of two limitations (N = 4), the unit cell consists of two limitations and two wells with (d_{uc} = d_{B_1}+d_{W_1}+d_{B_2}+d_{W_2}). Lastly, with the dispersion relation (E=E(q_{Bl})), we will receive the DOS(E) straightforwardly by utilizing Eqs. (31) and (32).

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