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2. Resonant Tunneling
The DBRTD's operation is based on two quantum phenomena: tunneling through a potential barrier, and resonance in a quantum well. Both are, in the lD case, easily treated in the transfer matrix approach. To use this approach, we suppose the electrons in the conduction band to be quasi-free particles of energy with m* the effective mass and the bottom of the conduction band (envelope function formalism [10]). Thus, a flux of incoming monoenergetic electrons can be described by a plane wave exp(ikz). In the following we take at the left spacer to be the zero of the energy scale (see Fig. · lb). Furthermore, we will use only one effective mass for the entire structure. The electronic potential is approximated by a piece-wise constant potential: this makes the model easier to handle, while it can be shown that no essential changes are introduced with regard to the resulting current-voltage features. [11].
Attributing to the first barrier (width . height potential drop across the barrier ) a reflection coefficient now means that an incoming wave of energy will be partly reflected by the barrier, resulting in a wave , and partly transmitted, leading to a wave , The barrier is completely described by the three functions and of wave number k or energy E. Analogously, the second barrier can be characterized by quantities and . To the two barriers together, seen as one complex barrier structure, can also be attributed a reflection coefficient R which can be expressed in and of the separate barriers. However, we cannot simply multiply (1- ) and (1- ) to find (1-R): the well between the barriers serves as a resonator, in which transmitted and reflected waves may interfere destructively or constructively. Taking into account this interference yields a total (1-R) given by:
(1)
where and
Since the energy dependence of 1-R is dominated by the positions of the maxima of 1-R are fairly well determined by the equation
(2)
This is the resonance condition and for every n we may fmd a resonance energy as a solution of (2). In the following we will consider one resonance energy (the lowest; n=0) only. From (1) we see that:
(3) corresponding to a peak-to-valley ratio of , which can be, for , as large as . The first maximum is thus a sharp peak, to which (if ) can be attributed a line width (full width at half maximum), which can be derived straightforwardly from (1):
(4)
From (3) we see that for the maximum of (1-R) equals unity. However, a good measure of the resonance's weight is the area under the peak rather than its height. Integrating over a from to , we find this area S to be:
(5)
One might conclude from (3) to (5) that only and are important, and that we can ignore the phase shifts and all together. However, and must be evaluated at , and since is determined by these phase shifts via (2), it follows that the phases play an essential role in this approach. If both and are close to unity, the peak is so sharp that, in integrations, we can approximate 1-R by a function,
(6)
From (6) we see that the double barrier filters out precisely the energy an energy channel is defined, through which the electrons with can pass to the well and through the whole structure. The resonance is not a truly bound state, since there is always a possibility for the particle to leave the well by tunneling through one of the two barriers (i.e. the wave function lea.ks out of the well). This implies a broadening of the resonance level. Although working with a time-independent SchrOdinger equation, we can use this to make an estimate of the life time , writing [12]:
(7)
where This is by a factor larger than the time needed to traverse the well. We will look at as a measure of the time that the electrons spend in the well (the so-called dwell time, see Section 6). Let us now calculate the densities in the well and the spacer layers on the left and the right side of the structure. Since the wave functions are plane waves, we have to consider only one z-value in each layer. In the left-hand spacer, there are an incoming and a reflected wave:
(8a)
while in the right-hand spacer, there is only a transmitted wave:
(8b)
In the case of the well, a little algebra leads to:
(8c)
Because of the energy filtering by the barrier structure, the electronic densities both in the well and at the end of the structure contain the factor 1-R, thus exhibiting the same resonant structure. However, this does not imply that the electrons in these regions a.11 have the same energy, since the E in (8) is only the part of the energy associated with the tunneling direction. To find the energy distribution in the well and at the end of the structure, we have to consider the problem in all three dimensions.
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