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Low Power Dissipation Semiconductor Display Device

IP.com Disclosure Number: IPCOM000078908D
Original Publication Date: 1973-Apr-01
Included in the Prior Art Database: 2005-Feb-26
Document File: 3 page(s) / 29K

Publishing Venue

IBM

Related People

Welber, B: AUTHOR [+2]

Abstract

Recent developments in molecular beam epitaxy make it practicable to deposit films of gallium phosphide or gallium aluminum arsenide on transparent substrates, with very high uniformity and thickness. A display device based on the alteration of the interference pattern in such films by the application of electric fields in selected portions of the substrate, utilizing the Franz-Keldish effect, can be fabricated. An important feature of this arrangement is that it does not require any electrical conduction or dissipation in its operation.

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Low Power Dissipation Semiconductor Display Device

Recent developments in molecular beam epitaxy make it practicable to deposit films of gallium phosphide or gallium aluminum arsenide on transparent substrates, with very high uniformity and thickness. A display device based on the alteration of the interference pattern in such films by the application of electric fields in selected portions of the substrate, utilizing the Franz-Keldish effect, can be fabricated. An important feature of this arrangement is that it does not require any electrical conduction or dissipation in its operation.

Considering a gallium aluminum arsenide or gallium phosphide film deposited on silicon dioxide, the three indices of refraction involved are. n(1) = 1 (air) n(2) =
3.5 (film) n(3) = 1.5 (SiO(2)).

At normal incidence, the variation of reflectivity versus film thickness passes through maximum and minimum values at quarter-wave intervals. These values calculated for the above indices are: R(min) = 0.04 R(max) = 0.61.

Corresponding to these, there will be a transmission maxima and minima given by: T(max) = 0.96 T(min) = 0.39.

The precise variation of T with film thickness governed by the expression T =
0.58 over 1.5-0.88 Sin/2/beta (1) where beta = 2 pi n(2)h over lambda Since R = 1-T R = 0.92 - 0.88 Sin/2/beta over 1.5 - 0.88 Sin/2/beta.

Assuming a film of approximately 20,000 angstroms thickness and that the wavelength lambda = 6,000 angstroms, then n(2)h - 70,000 angstroms and it would require a shift of delta n(2) in the index of delta n(2) over n(2) = 1,500 over 70,000 or approximately 2% to produce a full quarter wavelength shift in the interference pattern, thus changing R(max) to R(min) or vice versa. On the other and, if one were operating near R(min) and were to be satisfied with a change of R by a factor of 8 to 32%, a 0.5% shift in n(2) would suffice as can be readily verified from Equation (2). Such small shifts are achievable with the Franz- Keldish effect. It also follows, however, that variations in film thickness must be small compared to 0.5%, in order to preserve a uniform background against which to view the display. This fixes the requirements that the film thickness variation be less than +/- 100 angstroms.

Published data on GaP indicate th...