THE RESONANT TUNNELLING DIODE OR RTD
The PN and Schottky diodes that we have discussed in chapter 3 and all the transistors we will discuss are based mainly on the drift-diffusion mechanism. The RTD, as the name signifies, is based on the tunnelling mechanism, and it cannot be analyzed by any classical method but only by such methods as the Landauer approach, so it is reasonable that we present it here.
The RTD involves the tunnelling of electrons through two barriers separated by a quantum well. A physical realization of such a structure is accomplished by the 5-layer sandwich of GaAs/AlxGcii-xAs/GaAs/AlxGai-xAs/GaAs. The two outer layers of GaAs are usually overdoped so as to act as contacts-leads, see figure 5.11. The compound AlxGax_xAs is an alloy of GaAs and AlAs. The latter has a band-gap of 2.3eV, whereas GaAs has a band- gap of 1.42eV. Therefore the conduction band offset Д£с (л:) = 60%AEg (x) acts as a barrier to flow from GaAs to AlxGaX-xAs. The stoichiometric index x varies between 0 and 0.43. The rather amazing feature of such a structure is that it exhibits negative differential resistance i.e. dl/dV < 0. This pair of semiconductors is not the only pair that exhibits resonant
FIGURE 5.11 (a) A five-layer sandwich of alternating GaAs and AlxGat_xA$ layers which constitutes
an RTD diode, and (b) the conduction band edge Ec of this system.
FIGURE 5.12 The experimental I - V characteristic of a III—IV system showing negative differential resistance. Figure reproduced from T. S. Moise et al J. Appl. Phys. 78, 6305 (1995).
tunnelling. Many pairs of III—IV and II—VI semiconductors exhibit resonant tunnelling. Figure 5.12 shows the experimental current-voltage characteristic of a III—V resonant tunnelling diode and one can clearly see regions in the characteristics where the current is decreasing while the voltage is increasing. A pair of II-VI semiconductors that exhibits resonant tunnelling is ZnMgO. The band diagram of any such structure is shown in figure 5.13. Layers II and IV constitute the barriers while layer III is the quantum well.
In the previous section, we have developed all the methodologies needed to analyze such a structure. In the two outer layers where the bands are flat, the eigenfunctions will be planewaves with an incoming plane wave and a reflected one in the left layer and only a transmitted one in the right one. On the other hand, in the layers where the potentials (or
FIGURE 5.13 The band structure diagram of the system in figure 5.11 under the application of a voltage difference.
FIGURE 5.14 Calculated transmission coefficient and experimentally determined quasi-bound eigenstate for the RTD with ZnO!Zn.xMgO alternating layers (calculation by the author).
conduction band edges) vary linearly with length, the wavefunctions will be combinations
of first and second kind Airy functions as described in detail in the previous section. Then
applying the continuity conditions for the wavefunction ЧК and its derivative —--at
the interfaces, we get a matrix equation relating the coefficients of the wavefunctions. We leave this as a rather long exercise for the student (see problem at the end of the chapter). The procedure is straightforward but the algebra is rather tedious. If such a methodology is applied to this double barrier problem, the transmission coefficient can be calculated with the mathematical apparatus given in section 5.5.
The transmission coefficient T(E) thus calculated for the system ZnO / Ztix-xMgxO / ZnO / Zti-xMgxO / ZnO with x = 0.2 and AEc = 0.6eV is shown in figure 5.14a. The experimentally determined band diagram for this case is shown in figure 5.14b. It can be seen that the transmission coefficient has a sharp maximum, equal to one, at roughly the energy that coincides with the experimental energy of a bound state in the ZnO well. Further calculations with much heavier effective masses for both materials but smaller A£c show four peaks with T = 1, again at energies where bound states appear in the well, see figures 5.15a and 5.15b. It appears therefore that the transmission coefficient has very sharp peaks at energies at which bound states exist in the well that is sandwiched between the two barriers. We can understand physically why this is so. Since the well is of a nanometric length, it can accommodate a small number of discreet states. On these states an electron can “momentarily step on” or more precisely resonate with.
FIGURE 5.15 Same as in figure 5.14 but with different material parameters; see text.
For a more accurate description of the currents in an RTD we need a 3-dimensional (3D) model. In the left lead, where bands are flat, we can write for the 3D wavefunctions
where л: denotes the direction along the RTD and r± denotes the 2-dimensional position vector perpendicular to x. kx and k± are the wavevectors along the respective directions. In the right lead of the RTD, where again the bands are flat, we can write for the wavefunction:
The potential energy in the effective mass equation is only a function of x, i.e. U = U(x). Hence this equation can be separated in longitudinal (x) and perpendicular (rx) components. The longitudinal x component will be identical to what we have described in the previous section 5.5, so that we can write
As already shown, T (Ex) has sharp peaks at the energies at which quasi-bound states appear in the well. (The prefix “quasi” is used because electrons do not stay indefinitely there but only for a limited time on their transport from one electrode to the other). We can therefore mathematically represent T(EX) in the vicinity of the ith quasibound-resonant state Erxes'' in the well by the following simple formula
where A£ is the width of the peak (at half maximum) and Q is a constant equal to 1 for a symmetric RTD and for an asymmetric RTD Q < 1.
Since there is an applied voltage V at the electrodes of the RTD, the energies of the electron in the bottom of the conduction band in the left and right electrode will differ by the potential energy eV. We therefore have
where L,R stand for the left and right reservoir or lead. We initially assume as in section 5.3 and section 5.5 that the conduction edge E$ is lowered by eV and Ec is unchanged. To use the Landauer formula we must now calculate NL and NR of equation 5.33, which are usually called supply functions. They are the number of electrons in the left and right lead reservoir respectively that have the energy Ex, the one parallel to the channel, fixed, i.e. they are the number of electrons in the leads incident on the barriers with a given normal to the barrier energy. Note that Ex, being along the channel, is also normal to the barrier. These functions can be calculated by employing our result for the density of states, DOS, of a 2-dimen- sional system. The leads of the RTD are not necessarily 2-dimensional systems since they are not necessarily of nanometric length, but the density of states of a 3-dimensional material with the energy in 1 dimension fixed, is mathematically the same as that of the DOS of a 2-dimensional system with a single energy in the remaining (confined) dimension.
Let the cross-section of the device be S. Then the required numbers in the leads are, using equation 5.12 with one term only in the summation and omitting the superscripts L, R for the moment,
where in 5.66 / stands for the Fermi-Dirac function (written more explicitly) and К is
Boltzmann’s constant. The above integral can be put into the form j dx / 1 + x so that Nl and Nr become (after a few manipulations)
Application of Landauer’s formula, equation 5.33 gives for the current density J = 11S
In using equation 5.68, care should be exercised in avoiding regions of the energies Ex at which the initial and final state of the electron are both occupied as shown in figure 5.16 and hence not contributing to the integral in 5.68. Such a case does not arise if eV = Ер -Ер » KT.
FIGURE 5.16 An arrangement of the bands in which transmission is not allowed because an electron goes from a filled state to a filled state also.
In fact if eV » KT only the current L —> R contributes to the total current. Assuming that the edge of the left conduction band is lifted by eV, compared to the unperturbed Fermi level and the right one remains unchanged (as shown in figure 5.16) and writing £/: = £f = £° at V = 0 equation 5.68 becomes for large V
The limit of T —> 0 (low temperatures) allows the following simplification Then equation 5.69 transforms into
where Ef = E°+eV.
If the Lorentzian in equation 5.63 is sharp around any of the resonances Erxes'' the integration in equation 5.71 can be performed analytically: any such integration gives the value at £(“’' of the remaining part of the integrand times the width of the Lorentzian AE. Hence we get for the current density near a maximum
where Q = 1 for a symmetric barrier and Q < 1 for an asymmetric one. So, very near the ith maximum at Erxe5J the current density first increases almost linearly as we approach £(“’' and then decreases linearly and then drops abruptly due to the Lorentzian shape of the transmission coefficient. At higher temperatures the approximation 5.70 is not valid so that a smooth curve is obtained as shown in figure 5.12.
5.1 Prove that the energy per unit area U/A in a 2-dimensional electron gas, when only the first sublevel is occupied is at T=0
where n is the number of electrons per unit area.
5.2 Use the calculated wavefunctions of problem 1.2 to express the transmission over a
potential step in matrix form as = T
where A,B,C,D are the coefficients of the incident and reflected waves from either side of the step and T is a matrix.