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Infrared and millimeter wave absorber structures for thermal detectors
Infrared
Phys.
Vol.
30, No.
6, pp. 46J478,
0020-089 I/W $3.00 + 0.00 Copyright G 1990 Petgamon Press plc
1990
Primed inGreatBritain. Al1rights reserved
INFRARED AND MILLIMETER WAVE ABSORBER STRUCTURES FOR THERMAL DETECTORS ARMAND HADNI and XAVIERGERBAUX Laboratoire lnfrarouge Lointain, Universite de Nancy 1. B.P. 239, 54506 Nancy, France (Received
13 November
1989)
Abstract-fn a classical (i.e. Type I) pyroelectric detector, a crystal plate is coated with two metal electrodes. In order to achieve an absorbing power as close as possible to 100%. different structures have been used in the past: (i) a metal-black coating on the front electrode and (ii) a very thin front electrode having a square resistance matching the impedance of vacuum, with the back electrode making a quarter wavelength structure. However. the quarter wavelength structure becomes inefficient when the absorption length becomes smaller than the plate thickness. The simpler solution is to use a transparent antireflective layer, so that the whole radiant energy would enter the pyroelectric plate and be absorbed. This can be a perfect solution when the double path through the layer matches the phase shift on reflection at the pyroelectric material, and a nearly perfect one for a broad band centered at that wavelength. Transformation of radiant power into heat occurs directly within the pyroelectric. It is shown that a number of semiconductors can be used to make such antireflective layers, and practical solutions are feasible for many pyroelectrics. The advantages of this simple solution over i./4 structures stuck to the pyroelectric plate, as proposed by Parsons et al., are a negligible added heat capacity as the antireflective coating can generally have enough electrical conductivity to be used as an electrode, and a negligible lateral heat conductivity. When such a solution is not feasible (i.e. reflectivity cannot be cancelled completely), we can still use a very thin metal electrode. It is shown that the electrical conductivity may be 2 orders of magnitude smaller than for the bulk. This leads to small indexes in the IR (e.g. n, cz k, LX6). very convenient to give the structure a null reflectivity, when the granular metal layer is covered by a suitable, experimentally available coating (index of refraction nr = &). Additionally, the determination of the wavelengths and values of the absorption maxima give useful info~ation on the complex refractive index of the coated crystal at wavelengths where reflectivity and transmission m~surements are often inaccurate or impossible (T _ 0).
1. INTRODUCTION
A thermal absorber of neariy 100% efficiency for a large range of wavelengths is needed for thermal detectors (e.g. pyroelectric detectors), and its thermal mass must be low compared to the pyroelectric slab. Several solutions have been proposed: (i) metal-black coatings e.g. gold-blacks (ii) very thin metal coatings (iii) quarter-wavelength structures. 1.1. Metal-black
coatings
They show a significant absorption in the visible and NIR when their square resistance R, is sufficiently low. For instance we have measured”’ a 20% transmission at R = 1 pm for R, = IO9R i.e. for a thick enough layer of bismuth black deposited on a glass plate. A 92-9876 absorption efficiency has been claimed for platinium black absorbers in the 8-14pm band, with thicknesses of several pm.(2) They have been widely used in the past to blacken thermopiles, bolometers and pyroelectric detectors.‘3d’ 1.2. Very thin metal /ayer coatings They have been promoted by Golay to absorb the MIR in his famous pneumatic detector”) which is still the most sensitive thermal detector at room temperature. The absorber is made of 500 A 465
466
ARMAND HADNI and XAVIER GERBAUX
thick aluminium layer evaporated membrane 100 A thick to get:
(at a low residual
gas pressure,
e.g. IO -4 mm) over a collodion
Z, being the free space impedance of vacuum, Z, = a = 377 R. It is well known that such an adjustement of R, should give a maximum absorption at long wavelengths. However the aluminium film thickness used is 500 times larger7 than would be expected from the conductivity value cB of bulk aluminium (or, = 0.58 x IO’R- ’ rn-‘, hence e = I/(u,~) = 2/(az0); e rr IO “I m), and we must take go = IO’ Q ’ m ’ for the conductivity of evaporated aluminium, the decrease being due more to the granular structure of such films than to oxidization since it is also observed with gold which is often used to replace aluminium in such devices. The heat capacity is that of a water film 200 8, thick, hence c’ = 0.4 calg/cm3, i.e. slightly smaller than for bulk aluminium (cb = 0.52 calg/cm’ . K). and probably smaller still if we subtract the contribution of the nitrocellulose membrane (-50 8, thick water film), hence c’ = 0.3 calgjcm K). ‘Anyhow the specific heat of evaporated aluminium has the same order of magnitude as the bulk, and also the specific mass. Only the electric conductivity is lower by 2 or 3 orders of magnitude. Taking g,, = IO5 R .-I m ‘, it is possible to get the complex conductivity u = dR -@, at any angular frequency 0:
with g, = N,[e’/(m *a,)], the collision or relaxation, or scattering angular frequency and ilo = (g,)/(2nc ) the collision frequency, e.g. in cm- ‘. In the case of granular aluminium. assuming 3 free-electrons per atom, and crO= 1O’Q ’ m ‘, N’, = 6.23 x 10” m ‘, m* = lO-3o kg, we get: go = I .5 x IO” rad ss’: collision frequency go is larger than plasma frequency wP = 2 x lOI rad s ’ (co: = N,[e’/(m * fz,nt)]}, i.e. plasma oscillations are overdamped in the thin layer of evaporated aluminium. It occurs because co has been divided by 500 and thus go multiplied by this factor, while w,, was not modified. Now:
nk
=2!L 2q)o
n’-k’=c’
-go
go c(‘(g:, + 02) =
(y -
[~“/(cogo)l I + (w2!g;)
Assuming t ;_ = 2, we can get n and k vs frequency 1’ < vg x 0.001 = 8.105 x IO-’ = 800cm-‘,
v in cm-’
(Fig.
I). It is seen that
for
(1) tGolay”l p. 360, has written: “the ratio of the actual thickness of 270 n metallic layers deposited on collodion (I 1 50 8, thick) by evaporation in vacuum to the thickness computed from the specific resistivity for the same resistance is many hundred fold. Thus the caloritic inertia of a radiation absorbing aluminium film is of the order of 200 8, water film.” In a more basic paper, Reynolds er al.“’ have said: “In general. thin metallic films exhibit wide variations in properties with method of preparation and have much larger resistivities than would be predicted from theory.” They have been successful to decrease the ratio of thin film resistivity to bulk resistivity to less than three by using a fast rate of evaporation (e.g. 100 A/s) and low residual gas pressure (e.g. 10e5 mm) in the case of silver and copper.
Infrared and millimeter wave absorber structures
1500
IOW
0
467
2ow
(cm-‘)
Y
Fig. 1. Refractive index of a thin-layer of evaporated aluminium, calculated assuming a d.c. conductivity rrO= 105R-’ m-r (500 times less than for the bulk), and a collision frequency g,, = I.5 x IO” rad s-l (500 times higher). In the range of frequencies considered (v < 2000 cm-‘). we have o
is independent of g,, . That is because n* -k* has a small, nearly constant value, while nk is increasing to infinity when o 40. For instance, for granular aluminium, a,,/(cOgO) = 0.07, O<(Z)/(1+$)<0.07 and 2 > n* - k* > 1.93, i.e. n* - k* * 2, now (nk &, =
lo5 (2 x lo-“) x 1.5 x 1.5 x lo’7=o.03Y
and thus nk > 30, for o G gO/lOOO. From n and k, taking into account all possible internal reflections, we calculate transmission T, reflectivity R and absorption power A vs film thickness t, i.e. T(n, k, 1) (I - R*)*[l + (k*/n*)]T* T(n’ k’t)=(l
-R*T*)*+aR*T*
sinz(cp + $)’
(2)
with R*=(n
tg*
=
- t)*+k* (n + l)*+k*’
2k n*+k*-
;
1
(2’)
&2ko; C
is the wavelength in vacuum, and I// is the phase advance for reflection at the metal/air interface. 1.2.1. Low frequencies. Figure 2 gives T, R and A for a low frequency v = 100 cm-‘. It is seen that with t = 500 A, A is reaching a maximum whose value is 50%. This is the classical Wolterdorff’s result.‘9’ t = 500 A corresponding to R, = Z,,/2 for the Al layer considered above. For thicker films, it is seen on Fig. 2 that A is smaller than 50% because R is increasing from the 25% value for z = 500 A, up to lOO%, and transmission Tis decreasing to zero. This does not mean that R could not be decreased with some antireflection coating, and T increased substantiously, but such a two layer system is another problem (see section 3.4.2). It is also seen that for thinner films A is smaller than 50%, this time because T is larger than 25%. i
ARMAND HADNI and XAVIER GERBAUX
468
Fig. 2. Transmission T, reflectivity R and absorption power A for a free-standing aluminium film vs thickness I. For I = 500 A, we have T = R = 25%. and A = 50%. The calculations have been made with Y = IOOcmbut they lead to the same curves in the whole IR (see Fig. 3).
1.2.2. High frequencies. Figure 3 gives the absorption power A vs v for the thickness t = 500 8, which gives a maximum value A = 50% at low frequencies. It is seen that A keeps this constant value up to lO”cm-‘. i.e. in the whole IR, and that explains the wide use of such films as IR absorbers. For still higher frequencies, i.e. v coming closer to the relaxation frequency, A is no longer a constant, even for f = 500 A, which is not surprising since the approximation n z k is no longer applicable. 1.3. Quarter-wavelength structures in vacuum It is well known”” that for very long wavelengths the thin metal film absorption can be raised from the above demonstrated 50% value to lOO%, by backing the film with a perfect reflector at a i./4 optical distance and taking a film thickness twice as small, i.e. t = 250 A, giving R, = Z. = 377 Q/o. In vacuum the device will behave as a perfect anti-radar system for A, but what happens when a dielectric slab is put between the two metal layers?
100 -
1oooA 60 500
2
A
60 -
a
250 A 40 -
20 I I 100
10'
I IO2 Y
lcm-’
I
I
I
103
10'
10'
1
Fig. 3. Absorption A vs I* for a free-standing aluminium film 500 A thick, shows a quite constant value 155,OOOcm~‘) (v, A = 50% up to 104 cm-’ where it increases as Y moves closer to the plasma frequent and to the collision frequency (vn = 796.178 cm-‘). Data are also given for I = 250 x and =I = 1000 A: the tolerance is large for the optimal thickness.
2.
PYROELECTRIC
Infrared
and millimeter
PLATE
COVERED
wave absorber
WITH
structures
TWO
METAL
469
ELECTRODES
Let us consider the case of sodium nitrite at 300 K with two damned oscillators which are IR active when the fR electric field is parallel to the polar axis; v, = 148 cm-‘; AC, = 2; S, = 0.12 and v2= 230 cm-‘; AC,= 0.55; 6, = 0.17. Assuming c3 = 1.8, the optical constants n and k are calculated vs v (Fig. 4), and the bulk reflectivity R *, and also the transmission T of a crystal plate t = 4 pm (Fig. 5), using equations (2) and (2’). Let us deposit a very thin aluminium film (thickness t = fM) on the front face, and a thick perfect reflector on the rear face. Figure 6 gives the calculated reflectivity power R vs v for a pyroelectric crystal plate, thickness r,, = 100 pm, coated with a thin film r, = 170 A, corresponding roughly to R, = 2,. The calculation is a classical one using boundary conditions. The spectral range includes a low frequency interval from 1 to 60 cm-’ where k is small, and a high frequency range from 60 to 240cm-’ where k is high.
2. I. Small absorption frequency range 2.1.1. Possibility of perfect absorbers for a series of frequencies. From 0 to about 60 cm-‘, k is very small and we observe in Fig. 6, a series of fringes whose minima at v,, v2 and vXare close to zero, with wavelengths %,, i, and I,, such that n, t, = 2, /4; n,r, = 3(i,/4); n,r,, = 5(&/4), hence Y!= 1/(4n, to); v2= 3/(4n,f,); v3= 5/(4~~f*). In that spectral region, k 5 0; n = Ct and the minima are roughly equidistant. For these frequencies R = 0, T = 0, hence A = 100%. The quarter wavelength structure is working as nicely with a nearly transparent medium, as it works in vacuum. We have the perfect absorbers we needed. The metal layers are thin, have a small heat capacity, and anyhow are needed to make electrodes on both sides of the crystal plate. 2.1.2. Some problem.~. There are two problems: (i) the frequency range Av,,, where A > 70% is small: Av,,, = 8 cm-‘, for the above example. (ii) for very tow frequencies, i.e. v 4 v,, R is close to lOO%, and A is negligible. 2.1.X Some soi~tions.
(9 The frequency range where A > 70% for instance, is proportional
to the free spectral interval Av,,, = 1/(2nt,) i.e. inversally proportional to tO. For instance, if we divide I, = 100 pm by a factor 100, Av,,, = 80 cm ’ at a frequency v, = 1000 cm-’ suitable for most IR imaging applications. The problem is that such a very thin film is difficult to make except by evaporation, or sputtering.(“)
0 Y
Fig. 4. NaNO,
km-‘)
refractive index ti = n -jk vs v. for the electric field of the IR wave parallel to the polar axis. assuming the two oscillators parameters listed in the text.
ARMAND HADNI and XAVIER GERBACX
470
\
\T
(t=4pm) \
\
/
\
‘\
60 2 40 -
20 -
I
I
I
0
50
100
I 150
I
I
I
zoo
250
300
Y lcm-‘) Fig. 5. Bulk reflectivity power R*, and
transmission power 7‘ for a NaNOz calculated from Fig. 4.
(ii) For very low frequencies, A z 0 because all parts of distance from the perfect reflector if it is measured in of the stationnary waves is vanishingly small in the We have shown”*’ that the perfect reflector at the to be replaced by a second very thin metal layer, increased up to 50%.
crystal
plate
I = 4 pm vs Y.
the crystal plate are at a negligible wavelength-units. The electric field crystal, and so the absorption. back-face of the crystal plate has and the absorption coefficient is
2.2. High absorption frequency runge 2.2. I. Di&ulties. For v > 60 cm-‘, k > 0. I: we observe an increasing damping in the channelled spectrum and finally fringes can no longer be seen in the reststrahlen band. At resonance, the reflectivity power is 0.52 i.e. roughly the same one given by a bare single crystal bulk and the absorption power is limited to 0.48. The quarter-wavelength structure is no longer efficient by the simple fact that the absorption length is so short now that the rear-face of the crystal plate cannot see the impinging radiation. It is also clear from this simulated experience that a thin metal-layer cannot cancel reststrahlen because E, c E,,, while EI z 0 (Fig. 7). In conclusion a pyroelectric plate covered with two thin metal electrodes is a good absorber device except in parts of the spectrum where the pyroelectric absorption index is especially high.
60
0
so
loo
150
200
250
300
Y (cm-‘)
Fig 6. A thin layer (! = 170 A) of granular alummium evaporated on a NaNO, single crystal I = 100 pm can reduce the reflectivity close to zero for a series of frequencies in the low frequency region where k is small, but the reststrahlen is hardly modified (a slight reduction in the frequency range lSC~190 cm ‘. a slight increase from 190 to 230cm. ‘). Calculations are made from Figs I and 4.
Infrared
and millimeter
wave absorber
\
\
471
structures
3
Fig. 7. A thin layer of a semiconductor (medium no. 2) is deposited on a bulk single crystal (medium no. 3) to modify the gross reflectivity power. The electric field of the impinging wave is E, at the l-2 interface, E, for the first reflected wave, and E2 for the second one which has crossed twice semiconductor (SC) layer no. 2.
This is a serious problem with such pyroelectrics as PZLT, LiTaO, etc. . . which have indeed a high absorption index around 16pm, where most IR imaging devices have still to be sensitive. 2.2.2. Drastic solution. The best known solution in our knowledge is the one proposed by Parsons and Pedder in 1988:“3’ stick the pyroelectric plate to a quarter-wavelength absorber made of a nearly transparent polyimide 1/4 sheet, metal coated on each side. On the pyroelectric side the coating is a nearly perfect reflector of titanium, and on the front-face it is a nichrome film 3770/O. The drawback is the increase of thermal mass: the absorber structure has a thermal capacity of 2.4 J/m’. K. For comparison, typical pyroelectric ceramics has a thermal capacity of 2.6 J/m’. K per micron of thickness. We shall now propose a new solution.
3. ANTIREFLECTION
COATING
ON
A HIGHLY
ABSORBING
DIELECTRIC
3. I. Advantage Is it possible to make a perfect antireflection coating on a highly absorbing dielectric? If it were, we should have A = I - R = loo%, since T = 0, and hence we should have made the perfect absorber. 3.2. Feasibility, the cue
of SrTiO,,
as an example
Let us, for instance, consider the case of SrTiO, in the FIR where the absorption index is fairly high (Fig. 8, from Ref. (14)) e.g. n3 = 18, and k, = 7, at 300 K for v = 6Ocm-’ and R* = 0.81, and let us coat it with a thin layer (r2 = 10 pm) of easily available germanium (e.g. p0 = 50 R. cm; i.e. c,, = 2 R ’ m ‘), with nz = 4, and k, z 0, (Fig. 7). In germanium, the band gap is large compared to kT(E, = 10,000 cm-‘. kT(300 K) = 208 cm-‘), and the small d.c. conductivity in Ge is due to impurity atoms, i.e. each one bringing one electron value (a” = 2 R-’ m- ‘) is so small that to the conductivity band. The conductivity n2 - k2 z c x, k $ n, and n = fi = Cre, and k = [ao/(2~oon)]‘~2. With N, = lp el./m3, (i.e. lOI eI./cm’, i.e. lOI impurity atoms/cm’), we get g,, = (N,e’)/(m*a,) = 1.5 x lOI rad ss’ (or r0 = 80 cm -I), and wp = [(N,,e2)/(m*c,nS)]“2 = 1.4 x IO” rad s-‘. The important fact is that germanium has a k small enough, around 60 cm-’ (k = 4.7 x 10.-2, or K = 35 cm-‘, or 6 = 0.3 mm), that it is possible to make a i./4 plate nearly transparent in the IR (r,;, = 10 pm at v = 60 cm-‘). It thus seems possible to cancel El, by E2 in Fig. 7. Let us note that the thickness t, of a Ge plate able to match the free impedance of vacuum is r, = l/(aoZ,,) = 1.3 mm for microwaves, but not at v = 60 cm- ’ which is too close to v0 = 80 cm-‘, the relaxation frequency.
472
ARMAND
HAVNI and XAVIER GERBACX
-/
Fig. 8. Optical
II (300Kl
constants
n and k of SrTiO,
at 300 K (-)
and at 93 K
(
), from Tinkham.“”
Figure 9 gives the calculated reflectivity power R in vacuum (i.e. n, = I) vs v taking into account all multiple reflections in the three layers structure (Fig. 7). It is seen that at 59.3 cm -I, R goes very close to zero and A = 100%. If we accept A 2 70%, the band-width at this frequency is seen to be Av = 30 cm-‘. It is thus very easy to make SrTiO, a perfect absorber around 60 cm -’ with a thin coating of germanium. An additional advantage is that such a layer can be used as the front electrode of the pyroelectric detector since the resistivity of the dielectric is many orders of magnitude higher than the one of the purest germanium at room temperature. For application at IOOOcm- in the same conditions, the thickness of the germanium layer ought to be divided by 1000/60 = 16.7. then I, = 0.6 pm and Av,,,?~ = 500 cm ‘. 3.3. Position
of the reflectMy
minima
The optical-length retardation in a double pass through the germanium layer has to compensate for the phase shift advance I++at the Ge/SrTiO, reflection. Taking into account all multiple reflections in the germanium layer, the electric field E’ of the resultant reflected wave is calculated:
Ec
=
r
E
0 [
+ 42~21P exM+ I2 I - r,,p exp[j($
1’
- cp)l
(3)
- cp]
20 -
0
20
40
60
60
100 Y
120
140
160
160
I 200
(cm-‘)
Fig. 9. Reflectivity of a bulk SrTiO, single crystal covered with a thin layer of Gc (12 = 10 gm) at 300 K (--_). and at 93 K ( .- -). It is seen that at 300 K. the Ge layer is efficient to reduce R to zero at 6Ocm-‘. and to 16% at 175 cm ‘. At 93 K the Ge layer is less efficient, and R is only reduced to 20%. The minima have frequencies slightly smaller than at 300 K.
Infrared and millimeter wave absorber structures
473
with r,*
=
-
r2,
=
4 4
3.3.1. Phase shift.
+
nz
t,2
;
=
4
-. n,
24
12,
2n2
=-.
+n,’
n,
+n2
cp = 4nn, tv,
(4)
is the phase shift retardation due to a double pass through the germanium layer, and I/I the phase advance at the reflection from 2 to 3, given by:
‘%n2
tg$ = n: -n:
-k:’
with n: - (n: + k:)
cos * =
P[h+nd2+kil
It is clear that tg $ and cos II/, have the same sign, the one of n: - (n: +
k:);
(i) n: < n: + k:, e.g. n2 < n3, we can choose a solution with z2‘< &, < IC,
*=*o,
(7)
(ii) n: > n: + k:, we can choose a solution rjO with 0+.
(8)
Then, more generally, $ = $,, + 2K’7t, (9) with K’ = 0; 1; 2; 3; etc. . . 3.3.2. Reflectivity factor modulus. p is the modulus of the reflectivity factor from 2 to 3: P = P23 =
(n:-n:-k:)2+4k:n: [(n2+n3)‘+k:12
(8’) 3.3.3. Reflectivity power. Reflectivity power is minimum when a double path compensates exactly the phase shift at reflection. The intensity of the gross reflected wave is written: I = EC. ET, and putting a = + - cp; we have: ~12~12t21~ 1
now R = I/E:,
exp(ia)
+
-rzlp exp(ia)
r12+t12~21
1 -r2,
P
exp(-@)
+
P exp(-ja)
(112
11-~2,
P
12,
PI2
exp(.h)l[l -~2,
P
ev(-&)I
1 ’
hence: R
=r2
I2
+(t1212,~)2+2r12r12t2,pbNlL - cp)-r2,pl I +r$p’-2r2,p .cos(ll/ -cp) ’
(9)
Numerically, for SrTiO, at T = 300 K, for v = 60cm-‘: n, = 18; k, = 7, and with a germanium coating in air (n, = 1, n, = 4). we have: r ?, = -r12 - 0.60;
(,, = 0.40;
r,, = 1.60;
r,,t,, = 0.64;
p23 = 0.68.
Hence R = 0.36 +
0.402 - 0.520x 1.166-0.816x’
(10)
with x = cos(I(I - cp). - 0.28 dR/dx = 02: R is a decreasing function of x;
forx=-l,R=0.83;
forx=O,R=0.70andforx=+l,
R=0.02.
474
and XAVIER GERRAL’X
ARMAND HAIN
In conclusion,
to find the frequencies cos(@ -cp)=l;J/ cp =+
-2K”n:
of the minima
--cp =2K”n cp =+,,+2Kn
of R, we have to take .Y = 1. i.e
with K”=O.
with K=O.
_+t. &2....: fl.
k2 . . . . .
(11)
and 4nn,r
11= t+b,, i- 2Kn.
(1 I’)
Equation (I I ‘) gives values of 11which make R a minimum. which is very close to zero in this example. The significance of equation (I 1) is that the electric field Ez (Fig. 7) is finally in phase with the electric field E, of the incident wave. i.e. E: is in phase opposition with E, _ It is also the same for all these waves E;. E,. E., . . add together and can destroy E, if the final sum has the E,.E.,.... same amplitude as E, 3.3.4. Vuriution of‘ rhe _fieyuenc.,* v,,, of fhc reflecti1:ii.b. minima. H.ith the optical constants of‘ the absorbing dielectric no. 3. (9 For SrTiO, at T = 93 K. and around 60 cm ‘. k, = 20, n, = 18; t+b= 2.92 rad. ($ = 167’ ), and assuming I, = IO pm. we get \I,,, = 58 cm ’ with K = 0. in good accordance with the computer calculation reported in Fig. 9. The phase shift advance is smaller than 3.14 rad. the value expected if dielectrics no. 3. were transparent (k! = 0). (ii) For SrTiO, at T= 30OK; II,= 18:/~~=7,$ =2.98rad.$ = 171 . and keeping ’ in good accordance with Fig. 9. 1,= IOpm, weget v,=59.3cm (iii) For li, = 0, i.e. a transparent dielectric we have Ic/ = 3.14 rad. and still keeping f2 = IO pm, we get \trn = 62.5 cm ‘. The Cc plate in this case is exactly a quarter-wavelength plate (2ncj, = i-/2). In conclusion, the germanium coating I~ = IO jtrn thick deposited on medium no. 3 gives a series of minima in the reflectivity spectrum. whose positions are very sensitive to the optical constants of medium no. 3. For instance the first reflectivity minimum (k = 0) has a frequency which is decreased from 62.5 to 58 cm ‘. i.c. AV = 4.5 cm ‘, or AV/Y = 8% when n3 is increased from 1 to 18. and k, from 0 to 20. For the second reflectivity minimum, the frequency is decreased from 187.5 ‘. and Av.‘Y = 12X, when medium no. 3 is changed from vacuum to 165cm ‘. i.e. Av = 22.5cm into SrTiO, at T = 93 K. A measurement of such shifts should give a new relation between n3 and k,. Such a data in more accurate than a classical reflectivity power measurement because of the high accuracy of frequency measurements in Fourier Transform Spectroscopy. The value of the reflectivity minimum should give a second, but less accurate relation, linking n3 and k, through 11~.
3.4. I. An equarion Ricing nijor tllruniirc~~ec.tion Iu.ver. To get a null reflectivity at the minimum, and so A = loo%, i.e. to achieve a perfect absorber. WC must have E’ = 0. (t,,t,, p)i( I - r,,p) = r’,;
and thus, p3(n,. I+, k,) = (n? - n, ).‘(n: +- n,). which is seen directly T+R=I. In short to get a null reflectivity. we must have pJi = r?, . n, and k3 being given, to a perfect absorber. n2 = 4 gives effectively [(n, + n?)’ + ki] = [(n: -
since T = !,:I~, . R = ri,, and
(12)
and assuming n, = I, equation (12) will give the value of 11~which leads As an example we have seen that for II! = 18. k! = 7, germanium, with A = 99%. More generally, equation (12) is written: [(n: - n3)’ + k!]/ I),“(n: + l)]’ and we get:
Infrared
and millimeter
wave absorber
structures
47s
3.4.2. Immediate applications.
(9 For k, = 0, we have the classical result: n: = n,; 6Ocm-‘, we have to make (ii) In the case of SrTiO, at 300 K, around n: = 18 + (49/17) = 20.9, and n, = 4.5 is indeed close to the refractive index of germanium. At 93 K, n: = 18 + (400/17) = 41.5, and n2 = 6.5 is now significantly higher than the refractive index of germanium and reflectivity is not completely cancelled by a germanium coating (see Fig. 9, dashed line). (iii) For a metal in the FIR, n3 z k,, and a thin layer with n: z 2n, i.e. n, = 6
(14)
ought to give an efficient anti-reflection coating on the metal, and hence a high absorption power A = 1 - R, if the metal layer is thick enough to be sure that T = 0.
Let us try germanium again (i.e. nz = 4) for such a coating, we then must have n, = k, = 8 for the metal if we want the germanium layer to be efficient against reflection. That is exactly the case of granular-aluminium for a large range of frequencies around 500 cm ’ (see Fig. I). In that case, tg # = 64/( 16 - 128) = -0.57, and I(/,,= -3O”+ 180” = 150; $0= 2.62rad. Now K =47rkv = 5 x 104cm-‘, and the absorption length 6 = l/K = 2 x 10 ’ cm, 6 = 0.2 pm: if the aluminium layer is thicker than some tenths of a micrometer, and covered with a germanium layer with a similar thickness, we can expect the absorption power to be very close to 100% on very broad bands. Such a composite coating has several obvious advantages: small heat capacity, small lateral heat conductivity, good electrical conductivity needed for an electrode in a pyroelectric detector and larger frequency range where the power is kept close to 100%. These possibilities depend on the accuracy of calculations in Fig. 1. If n3 was smaller, we could use silicon for coating (nz = 3), but if n, was much larger than 8 it would be difficult to find an efficient material for the coating. Figure 10 gives the reflectivity R of a 30 pm thick granular Al, coated with a 0.92 pm thick Ge layer. The reflectivity is zero around 560 cm-‘, and 5% at 1800 cm-‘. (iv) To get n: > 0, and thus a real number for n,, we must have n3 + [k$(n, - l)] > 0. Difficulties appear only when n3 c I, then: n,(n, - 1) + k: < 0, or approximately : .(14’Such a relation may be difficult to fulfill in some -n,+k:n,>k parts of the spectrum, and an antireflection coating might appear impossible in this spectral range. This is the case of LiTaO, around 600 cm-‘: n3 = 0.1, k, = 1. In fact this is a very special case. We are in a part of the spectrum where R* 2~0, A = loo%, and any coating can only increase R* and decrease A. 3.4.3. Value of the reflectivity factor minimum for a given coating, on a given material. If we use a given material (e.g. Ge) for coating, equation (3) will give the factor of reflection for the minima (i.e. cp = IL,,- 2Krr): EC r=%=r,,+-.
T.P 1 -
(3’)
r2,p
For LiTaO, at 300 K, around 600 cm .-I, p = 0.81, and r = 0.42 is far from zero: a germanium layer cannot be used efficiently to make LiTaO, a good absorber in this part of the spectrum. 3.5. The case of NaN02 as a second example We have seen in section 2.2 the difficulty to reduce the NaNO, reststrahlen reflectivity around 150 cm-’ and 275 cm-’ with a thin metal layer deposited on it. Let us try a coating with germanium which has a smaller refractive index and a negligible absorption index in the MIR, if pure enough. 3.5.1. For the v, = 160cm -’ reststrahlen. For that frequency, it is seen in Fig. 4 that n3 = 1.3 and k, = 2.7, hence pzj = 0.63, very close to rzI = 0.60: an interference minimum in the reflectivity, located at 160 cm-’ will be close to zero (R z 0).
ARNAIW HADNI and XAVIER
476
GERBAUX
tf (cm-‘) Fig. IO. Reflectivity R% of a granular Al film t = 30pm thick, coated with a Ge film I =0.92pm. showing a decrease of retlectivity to zero at S50cm-‘, and to 6% at 18OOcm-‘.
To get the germanium coating thickness which leads to a reflection minimum at 160cm ‘, we have to calculate the phase shift advance at the reflection from Ge to NaNO,: tg $ = (8 x 2.7)/[16 - (1.69 + 7.29)] = 3.08; cos $ > 0, hence + = 1.26 rad i 21Y’n; ($ = 72” _+ 360” x K). We thus have to compensate that advance by a double pass through the plate such as: 4nnz fv, = Il/(,+
the compensation is possible t, = 1.5 pm + 7.81 pm x K; K = 0:
for t, =
the
values
1.5 pm;
2Kn
of
K which
(II’) make
I,&~ + 2&r > 0, hence:
K = 1: I, = 9.3 pm.
In Fig. 11 R shows a minimum at v = 160 cm-’ for a germanium coating thickness t, = I .45 pm and f, = 9.3 pm, and the minima values are very close to zero. For the thinnest coating, the reflectivity minimum at 160cm‘-’ is broad, and the integrated reflectivity from 0 to I7Ocm ’ is especially small. Figure 12 shows R vs thickness t. For I = 0, R = 56%, and for t, = 1.5 pm + 7.8 pm x K, R 2: 0. 3.5.2. For the v2 = 27.5~1 -’ resrstrahlen. In Fig. 4 n3 zk3 = 0.5, then pz3 =0.79. and tjz7 = 0.254 rad. To get a phase shift retardation Ic/which is suitable to compensate exactly for II/. 100
r
Y
(cm-‘1
Fig. II. Reflectivity of a bulk NaNOz single crystal covered with a thin layer of Ge in the case of two thicknesses, I* = I .45 pm (- - -). and I~ = 9.30 pm (-) at 300 K. The antireflective coating is perfect at v = l6Ocm-” for both thicknesses. Another reflectivity minimum is observed at Y = 275cm-’ for rz = 9.3 pm.
Infrared and millimeter wave absorber structures
t (pm) Fig. 12. Reflectivity of a bulk NaNO, single crystal coated with Ge, at v = 160 cm-‘. vs coating thickness 1, showing 4 minima close to zero.
we have to make the germanium coating thickness t = 0.18 pm + K x 4.5 pm; K = 0 :r= 0.18 pm; K= 1: f =4.68pm; K=2: t =9.18pm. It is seen in Fig. 11 that we have indeed a minimum at v = 275 cm-’ for t = 9.3 pm and the reflectivity is low (R = 10%) but not negligible. Figure 13 shows R vst.For t = 0, R = R* = 20%, and for t,=0.18~m;4.68~m; 9.18pmetc. .., R = 10%. 4.
CONCLUSIONS
Metal-black coatings can give a 92% absorption in the 8-14pm band, but they have to be several pm thick, hence a thermal capacity of the same order of magnitude as a 3 pm thick pyroelectric ceramics, and a lateral heat conductivity which is a serious drawback especially for pyroelectric arrays. Moreover their reflectivity increases rapidly for longer wavelengths, and they are not reproducible. Very thin metal layers, in contrast, have a heat capacity and lateral thermal-conductivity which are both negligible, and in a number of cases the absorption power can be higher than 50% and even close to 100%. For free standing layers, it has been shown that A is kept constant at a value of 50% for frequencies up to 10,000 cm-‘. The thickness has to be chosen to give R, = Z,,/2 = 188 R. loo-
60 2 a 40-
I
0
I
I 10
20 t
I 30
(pm)
Fig. 13. Reflectivity of a bulk NaNO, single crystal coated with Ge, at v = 275 cm-‘, vs coating thickness I, showing 7 minima with the same value R, = 12%. The first one is corresponding to a very thin layer, fI = 0.3 pm.
ARMAND
478
HAD~‘Iand XAVIERGERLIAUX
For layers deposited on a transparent material, the absorption power A can be slightly higher than 50% for a given number of wavelengths for which the transparent material plays the role of a more or less antireflective layer. A can even be increased up to 100% if the transparent material is backed by a thick metal layer to give a quarter-wavelength structure for a number of radiations, and if the thin metal layer thickness is two times less than above to make R, = Z, = 377 R. The bandwidth can cover the whole 7-14 pm band if the transparent pyroelectric plate is thick enough (e.g. I pm thick for a PZLT ceramics). In the reststrahlen regions, the reflectivity of the pyroelectric is generally not decreased by the thin metal layer deposited on it. We have shown that it has to be replaced by a thin semiconductive layer which has a very small absorption index. and a refractive index which is nearly independent of wavelength (e.g. germanium). It will act as an antireflective layer, and for a well adapted material, we get R = 0, hence A = 100%. The semiconductor thickness can be very small, and A kept to values close to 100% on a broad frequency range. There are marked advantages in using the nearly i.!4 germanium coating instead of the metal-coated A/4 polyimide film used by Parson et al.:
(9 With the germanium
coating, absorption occurs directly in the high absorption dielectric plate at a small depth beyond the germanium coating. In Parson’s device absorption is localized in the nichrome film 377R/O. layer optimum thickness of coatings at IO pm is around 0.2 pm (ii) The germanium while the polyimide film is 1.4 pm thick and has to be metal-coated on both sides. Moreover cCie= 0.074 calg/g . K = 0.3 I J/g . K is slightly smaller than cpoly,,,,= 0.40 calg/ g. K. However pcir = 5.46 g/cm’, hence c&, = 1.70 J/cm3 K has the same order of magnitude as c&,,,,. Finally, c&~(, ,,,” = 0.50 J/m’. K. and we had seen in eqn (5): (.;ly,m.I 4pm.coatcd = 2.4 Jim’ K. The small heat capacity of the germanium device is a noticeable advantage over the Parsons’ device. However the reduction ration (i.e. 5) is not the effective reduction in heat capacity for thick detectors. The true gain is
With a I pm thick pyroelectric
ceramics, G=
taken
above
as an example:
2.4 + 2.6 = , 6 0.5 + 2.6
’ ’
This is quite appreciable. For much thicker pyroelectric ceramics, the gain will be less but the first advantage (production of heat in the pyroelectric directly), will still have to be considered. coating 0.2 pm thick, has a lateral thermal conductivity which is at least (iii) A germanium 12 times smaller than the 1.4pm thick polyimide film metal-coated on both faces. REFERENCES P. Strimer. X. Gerbaux. A. Hadni and T. Souel. Infrured fhyr. 21, 37 (1981). D. B. Belts. F. J. J. Clarke, L. J. Cox and J. A. Larkin, J. fhys. 18, 689 (1985) J. Lecomte, Handhuch der Ph.&. Springer, Berlin (1960). S.-I. Takahashi. Infrared Phys. 13, 301 (1973). S. Kasukabe. S. Yatsurya and R. Uyeda. J. C‘r.wa/ Growh 24, 315 (1974). Y. Hatanaka, S. Okamoto and R. Nishida, Jap. J. oppl. PhJs. 49, 563 (1980). M. J. E. Golay. Rev. Sri. Insrrum. 20, 816 (1949); l&347 (i947): 18, 357 (1947) F. W. Revnolds and G. R. Stilwell. Phvs. Rer. 88. 418 (1952). W. Wolteidorf, Z. Phys. 91, 230 (1934). G. Bruhat. Cours de Physique, Electricit& Masson, Paris (1956). K. Nakamura, T. Ishigaki, A. Kaneko. S. Takahashi. J. Nishida, Y. Wakabayashi and H. Nakamura. fnr. J. m/rared millimeter Waves 10, 907 (I 989). 12. J. Claude1 and A. Hadni. Elecrro-OprrcsiLaser Inrernatiotud. C‘onferenw Proceedings (Edited by H. G. Jerrard), p. 149. IPC Science and Technology Press’ (1976). 13. A. D. Parsons and D. J. Pedder. J. Var. Sci. Technol. A6. 1686 (1988) 14. M. Tinkham. Science 145, 240 (1964). 1. 2. 3. 4. 5. 6. 7. 8. 9. IO. I I.
Phys.
Vol.
30, No.
6, pp. 46J478,
0020-089 I/W $3.00 + 0.00 Copyright G 1990 Petgamon Press plc
1990
Primed inGreatBritain. Al1rights reserved
INFRARED AND MILLIMETER WAVE ABSORBER STRUCTURES FOR THERMAL DETECTORS ARMAND HADNI and XAVIERGERBAUX Laboratoire lnfrarouge Lointain, Universite de Nancy 1. B.P. 239, 54506 Nancy, France (Received
13 November
1989)
Abstract-fn a classical (i.e. Type I) pyroelectric detector, a crystal plate is coated with two metal electrodes. In order to achieve an absorbing power as close as possible to 100%. different structures have been used in the past: (i) a metal-black coating on the front electrode and (ii) a very thin front electrode having a square resistance matching the impedance of vacuum, with the back electrode making a quarter wavelength structure. However. the quarter wavelength structure becomes inefficient when the absorption length becomes smaller than the plate thickness. The simpler solution is to use a transparent antireflective layer, so that the whole radiant energy would enter the pyroelectric plate and be absorbed. This can be a perfect solution when the double path through the layer matches the phase shift on reflection at the pyroelectric material, and a nearly perfect one for a broad band centered at that wavelength. Transformation of radiant power into heat occurs directly within the pyroelectric. It is shown that a number of semiconductors can be used to make such antireflective layers, and practical solutions are feasible for many pyroelectrics. The advantages of this simple solution over i./4 structures stuck to the pyroelectric plate, as proposed by Parsons et al., are a negligible added heat capacity as the antireflective coating can generally have enough electrical conductivity to be used as an electrode, and a negligible lateral heat conductivity. When such a solution is not feasible (i.e. reflectivity cannot be cancelled completely), we can still use a very thin metal electrode. It is shown that the electrical conductivity may be 2 orders of magnitude smaller than for the bulk. This leads to small indexes in the IR (e.g. n, cz k, LX6). very convenient to give the structure a null reflectivity, when the granular metal layer is covered by a suitable, experimentally available coating (index of refraction nr = &). Additionally, the determination of the wavelengths and values of the absorption maxima give useful info~ation on the complex refractive index of the coated crystal at wavelengths where reflectivity and transmission m~surements are often inaccurate or impossible (T _ 0).
1. INTRODUCTION
A thermal absorber of neariy 100% efficiency for a large range of wavelengths is needed for thermal detectors (e.g. pyroelectric detectors), and its thermal mass must be low compared to the pyroelectric slab. Several solutions have been proposed: (i) metal-black coatings e.g. gold-blacks (ii) very thin metal coatings (iii) quarter-wavelength structures. 1.1. Metal-black
coatings
They show a significant absorption in the visible and NIR when their square resistance R, is sufficiently low. For instance we have measured”’ a 20% transmission at R = 1 pm for R, = IO9R i.e. for a thick enough layer of bismuth black deposited on a glass plate. A 92-9876 absorption efficiency has been claimed for platinium black absorbers in the 8-14pm band, with thicknesses of several pm.(2) They have been widely used in the past to blacken thermopiles, bolometers and pyroelectric detectors.‘3d’ 1.2. Very thin metal /ayer coatings They have been promoted by Golay to absorb the MIR in his famous pneumatic detector”) which is still the most sensitive thermal detector at room temperature. The absorber is made of 500 A 465
466
ARMAND HADNI and XAVIER GERBAUX
thick aluminium layer evaporated membrane 100 A thick to get:
(at a low residual
gas pressure,
e.g. IO -4 mm) over a collodion
Z, being the free space impedance of vacuum, Z, = a = 377 R. It is well known that such an adjustement of R, should give a maximum absorption at long wavelengths. However the aluminium film thickness used is 500 times larger7 than would be expected from the conductivity value cB of bulk aluminium (or, = 0.58 x IO’R- ’ rn-‘, hence e = I/(u,~) = 2/(az0); e rr IO “I m), and we must take go = IO’ Q ’ m ’ for the conductivity of evaporated aluminium, the decrease being due more to the granular structure of such films than to oxidization since it is also observed with gold which is often used to replace aluminium in such devices. The heat capacity is that of a water film 200 8, thick, hence c’ = 0.4 calg/cm3, i.e. slightly smaller than for bulk aluminium (cb = 0.52 calg/cm’ . K). and probably smaller still if we subtract the contribution of the nitrocellulose membrane (-50 8, thick water film), hence c’ = 0.3 calgjcm K). ‘Anyhow the specific heat of evaporated aluminium has the same order of magnitude as the bulk, and also the specific mass. Only the electric conductivity is lower by 2 or 3 orders of magnitude. Taking g,, = IO5 R .-I m ‘, it is possible to get the complex conductivity u = dR -@, at any angular frequency 0:
with g, = N,[e’/(m *a,)], the collision or relaxation, or scattering angular frequency and ilo = (g,)/(2nc ) the collision frequency, e.g. in cm- ‘. In the case of granular aluminium. assuming 3 free-electrons per atom, and crO= 1O’Q ’ m ‘, N’, = 6.23 x 10” m ‘, m* = lO-3o kg, we get: go = I .5 x IO” rad ss’: collision frequency go is larger than plasma frequency wP = 2 x lOI rad s ’ (co: = N,[e’/(m * fz,nt)]}, i.e. plasma oscillations are overdamped in the thin layer of evaporated aluminium. It occurs because co has been divided by 500 and thus go multiplied by this factor, while w,, was not modified. Now:
nk
=2!L 2q)o
n’-k’=c’
-go
go c(‘(g:, + 02) =
(y -
[~“/(cogo)l I + (w2!g;)
Assuming t ;_ = 2, we can get n and k vs frequency 1’ < vg x 0.001 = 8.105 x IO-’ = 800cm-‘,
v in cm-’
(Fig.
I). It is seen that
for
(1) tGolay”l p. 360, has written: “the ratio of the actual thickness of 270 n metallic layers deposited on collodion (I 1 50 8, thick) by evaporation in vacuum to the thickness computed from the specific resistivity for the same resistance is many hundred fold. Thus the caloritic inertia of a radiation absorbing aluminium film is of the order of 200 8, water film.” In a more basic paper, Reynolds er al.“’ have said: “In general. thin metallic films exhibit wide variations in properties with method of preparation and have much larger resistivities than would be predicted from theory.” They have been successful to decrease the ratio of thin film resistivity to bulk resistivity to less than three by using a fast rate of evaporation (e.g. 100 A/s) and low residual gas pressure (e.g. 10e5 mm) in the case of silver and copper.
Infrared and millimeter wave absorber structures
1500
IOW
0
467
2ow
(cm-‘)
Y
Fig. 1. Refractive index of a thin-layer of evaporated aluminium, calculated assuming a d.c. conductivity rrO= 105R-’ m-r (500 times less than for the bulk), and a collision frequency g,, = I.5 x IO” rad s-l (500 times higher). In the range of frequencies considered (v < 2000 cm-‘). we have o
is independent of g,, . That is because n* -k* has a small, nearly constant value, while nk is increasing to infinity when o 40. For instance, for granular aluminium, a,,/(cOgO) = 0.07, O<(Z)/(1+$)<0.07 and 2 > n* - k* > 1.93, i.e. n* - k* * 2, now (nk &, =
lo5 (2 x lo-“) x 1.5 x 1.5 x lo’7=o.03Y
and thus nk > 30, for o G gO/lOOO. From n and k, taking into account all possible internal reflections, we calculate transmission T, reflectivity R and absorption power A vs film thickness t, i.e. T(n, k, 1) (I - R*)*[l + (k*/n*)]T* T(n’ k’t)=(l
-R*T*)*+aR*T*
sinz(cp + $)’
(2)
with R*=(n
tg*
=
- t)*+k* (n + l)*+k*’
2k n*+k*-
;
1
(2’)
&2ko; C
is the wavelength in vacuum, and I// is the phase advance for reflection at the metal/air interface. 1.2.1. Low frequencies. Figure 2 gives T, R and A for a low frequency v = 100 cm-‘. It is seen that with t = 500 A, A is reaching a maximum whose value is 50%. This is the classical Wolterdorff’s result.‘9’ t = 500 A corresponding to R, = Z,,/2 for the Al layer considered above. For thicker films, it is seen on Fig. 2 that A is smaller than 50% because R is increasing from the 25% value for z = 500 A, up to lOO%, and transmission Tis decreasing to zero. This does not mean that R could not be decreased with some antireflection coating, and T increased substantiously, but such a two layer system is another problem (see section 3.4.2). It is also seen that for thinner films A is smaller than 50%, this time because T is larger than 25%. i
ARMAND HADNI and XAVIER GERBAUX
468
Fig. 2. Transmission T, reflectivity R and absorption power A for a free-standing aluminium film vs thickness I. For I = 500 A, we have T = R = 25%. and A = 50%. The calculations have been made with Y = IOOcmbut they lead to the same curves in the whole IR (see Fig. 3).
1.2.2. High frequencies. Figure 3 gives the absorption power A vs v for the thickness t = 500 8, which gives a maximum value A = 50% at low frequencies. It is seen that A keeps this constant value up to lO”cm-‘. i.e. in the whole IR, and that explains the wide use of such films as IR absorbers. For still higher frequencies, i.e. v coming closer to the relaxation frequency, A is no longer a constant, even for f = 500 A, which is not surprising since the approximation n z k is no longer applicable. 1.3. Quarter-wavelength structures in vacuum It is well known”” that for very long wavelengths the thin metal film absorption can be raised from the above demonstrated 50% value to lOO%, by backing the film with a perfect reflector at a i./4 optical distance and taking a film thickness twice as small, i.e. t = 250 A, giving R, = Z. = 377 Q/o. In vacuum the device will behave as a perfect anti-radar system for A, but what happens when a dielectric slab is put between the two metal layers?
100 -
1oooA 60 500
2
A
60 -
a
250 A 40 -
20 I I 100
10'
I IO2 Y
lcm-’
I
I
I
103
10'
10'
1
Fig. 3. Absorption A vs I* for a free-standing aluminium film 500 A thick, shows a quite constant value 155,OOOcm~‘) (v, A = 50% up to 104 cm-’ where it increases as Y moves closer to the plasma frequent and to the collision frequency (vn = 796.178 cm-‘). Data are also given for I = 250 x and =I = 1000 A: the tolerance is large for the optimal thickness.
2.
PYROELECTRIC
Infrared
and millimeter
PLATE
COVERED
wave absorber
WITH
structures
TWO
METAL
469
ELECTRODES
Let us consider the case of sodium nitrite at 300 K with two damned oscillators which are IR active when the fR electric field is parallel to the polar axis; v, = 148 cm-‘; AC, = 2; S, = 0.12 and v2= 230 cm-‘; AC,= 0.55; 6, = 0.17. Assuming c3 = 1.8, the optical constants n and k are calculated vs v (Fig. 4), and the bulk reflectivity R *, and also the transmission T of a crystal plate t = 4 pm (Fig. 5), using equations (2) and (2’). Let us deposit a very thin aluminium film (thickness t = fM) on the front face, and a thick perfect reflector on the rear face. Figure 6 gives the calculated reflectivity power R vs v for a pyroelectric crystal plate, thickness r,, = 100 pm, coated with a thin film r, = 170 A, corresponding roughly to R, = 2,. The calculation is a classical one using boundary conditions. The spectral range includes a low frequency interval from 1 to 60 cm-’ where k is small, and a high frequency range from 60 to 240cm-’ where k is high.
2. I. Small absorption frequency range 2.1.1. Possibility of perfect absorbers for a series of frequencies. From 0 to about 60 cm-‘, k is very small and we observe in Fig. 6, a series of fringes whose minima at v,, v2 and vXare close to zero, with wavelengths %,, i, and I,, such that n, t, = 2, /4; n,r, = 3(i,/4); n,r,, = 5(&/4), hence Y!= 1/(4n, to); v2= 3/(4n,f,); v3= 5/(4~~f*). In that spectral region, k 5 0; n = Ct and the minima are roughly equidistant. For these frequencies R = 0, T = 0, hence A = 100%. The quarter wavelength structure is working as nicely with a nearly transparent medium, as it works in vacuum. We have the perfect absorbers we needed. The metal layers are thin, have a small heat capacity, and anyhow are needed to make electrodes on both sides of the crystal plate. 2.1.2. Some problem.~. There are two problems: (i) the frequency range Av,,, where A > 70% is small: Av,,, = 8 cm-‘, for the above example. (ii) for very tow frequencies, i.e. v 4 v,, R is close to lOO%, and A is negligible. 2.1.X Some soi~tions.
(9 The frequency range where A > 70% for instance, is proportional
to the free spectral interval Av,,, = 1/(2nt,) i.e. inversally proportional to tO. For instance, if we divide I, = 100 pm by a factor 100, Av,,, = 80 cm ’ at a frequency v, = 1000 cm-’ suitable for most IR imaging applications. The problem is that such a very thin film is difficult to make except by evaporation, or sputtering.(“)
0 Y
Fig. 4. NaNO,
km-‘)
refractive index ti = n -jk vs v. for the electric field of the IR wave parallel to the polar axis. assuming the two oscillators parameters listed in the text.
ARMAND HADNI and XAVIER GERBACX
470
\
\T
(t=4pm) \
\
/
\
‘\
60 2 40 -
20 -
I
I
I
0
50
100
I 150
I
I
I
zoo
250
300
Y lcm-‘) Fig. 5. Bulk reflectivity power R*, and
transmission power 7‘ for a NaNOz calculated from Fig. 4.
(ii) For very low frequencies, A z 0 because all parts of distance from the perfect reflector if it is measured in of the stationnary waves is vanishingly small in the We have shown”*’ that the perfect reflector at the to be replaced by a second very thin metal layer, increased up to 50%.
crystal
plate
I = 4 pm vs Y.
the crystal plate are at a negligible wavelength-units. The electric field crystal, and so the absorption. back-face of the crystal plate has and the absorption coefficient is
2.2. High absorption frequency runge 2.2. I. Di&ulties. For v > 60 cm-‘, k > 0. I: we observe an increasing damping in the channelled spectrum and finally fringes can no longer be seen in the reststrahlen band. At resonance, the reflectivity power is 0.52 i.e. roughly the same one given by a bare single crystal bulk and the absorption power is limited to 0.48. The quarter-wavelength structure is no longer efficient by the simple fact that the absorption length is so short now that the rear-face of the crystal plate cannot see the impinging radiation. It is also clear from this simulated experience that a thin metal-layer cannot cancel reststrahlen because E, c E,,, while EI z 0 (Fig. 7). In conclusion a pyroelectric plate covered with two thin metal electrodes is a good absorber device except in parts of the spectrum where the pyroelectric absorption index is especially high.
60
0
so
loo
150
200
250
300
Y (cm-‘)
Fig 6. A thin layer (! = 170 A) of granular alummium evaporated on a NaNO, single crystal I = 100 pm can reduce the reflectivity close to zero for a series of frequencies in the low frequency region where k is small, but the reststrahlen is hardly modified (a slight reduction in the frequency range lSC~190 cm ‘. a slight increase from 190 to 230cm. ‘). Calculations are made from Figs I and 4.
Infrared
and millimeter
wave absorber
\
\
471
structures
3
Fig. 7. A thin layer of a semiconductor (medium no. 2) is deposited on a bulk single crystal (medium no. 3) to modify the gross reflectivity power. The electric field of the impinging wave is E, at the l-2 interface, E, for the first reflected wave, and E2 for the second one which has crossed twice semiconductor (SC) layer no. 2.
This is a serious problem with such pyroelectrics as PZLT, LiTaO, etc. . . which have indeed a high absorption index around 16pm, where most IR imaging devices have still to be sensitive. 2.2.2. Drastic solution. The best known solution in our knowledge is the one proposed by Parsons and Pedder in 1988:“3’ stick the pyroelectric plate to a quarter-wavelength absorber made of a nearly transparent polyimide 1/4 sheet, metal coated on each side. On the pyroelectric side the coating is a nearly perfect reflector of titanium, and on the front-face it is a nichrome film 3770/O. The drawback is the increase of thermal mass: the absorber structure has a thermal capacity of 2.4 J/m’. K. For comparison, typical pyroelectric ceramics has a thermal capacity of 2.6 J/m’. K per micron of thickness. We shall now propose a new solution.
3. ANTIREFLECTION
COATING
ON
A HIGHLY
ABSORBING
DIELECTRIC
3. I. Advantage Is it possible to make a perfect antireflection coating on a highly absorbing dielectric? If it were, we should have A = I - R = loo%, since T = 0, and hence we should have made the perfect absorber. 3.2. Feasibility, the cue
of SrTiO,,
as an example
Let us, for instance, consider the case of SrTiO, in the FIR where the absorption index is fairly high (Fig. 8, from Ref. (14)) e.g. n3 = 18, and k, = 7, at 300 K for v = 6Ocm-’ and R* = 0.81, and let us coat it with a thin layer (r2 = 10 pm) of easily available germanium (e.g. p0 = 50 R. cm; i.e. c,, = 2 R ’ m ‘), with nz = 4, and k, z 0, (Fig. 7). In germanium, the band gap is large compared to kT(E, = 10,000 cm-‘. kT(300 K) = 208 cm-‘), and the small d.c. conductivity in Ge is due to impurity atoms, i.e. each one bringing one electron value (a” = 2 R-’ m- ‘) is so small that to the conductivity band. The conductivity n2 - k2 z c x, k $ n, and n = fi = Cre, and k = [ao/(2~oon)]‘~2. With N, = lp el./m3, (i.e. lOI eI./cm’, i.e. lOI impurity atoms/cm’), we get g,, = (N,e’)/(m*a,) = 1.5 x lOI rad ss’ (or r0 = 80 cm -I), and wp = [(N,,e2)/(m*c,nS)]“2 = 1.4 x IO” rad s-‘. The important fact is that germanium has a k small enough, around 60 cm-’ (k = 4.7 x 10.-2, or K = 35 cm-‘, or 6 = 0.3 mm), that it is possible to make a i./4 plate nearly transparent in the IR (r,;, = 10 pm at v = 60 cm-‘). It thus seems possible to cancel El, by E2 in Fig. 7. Let us note that the thickness t, of a Ge plate able to match the free impedance of vacuum is r, = l/(aoZ,,) = 1.3 mm for microwaves, but not at v = 60 cm- ’ which is too close to v0 = 80 cm-‘, the relaxation frequency.
472
ARMAND
HAVNI and XAVIER GERBACX
-/
Fig. 8. Optical
II (300Kl
constants
n and k of SrTiO,
at 300 K (-)
and at 93 K
(
), from Tinkham.“”
Figure 9 gives the calculated reflectivity power R in vacuum (i.e. n, = I) vs v taking into account all multiple reflections in the three layers structure (Fig. 7). It is seen that at 59.3 cm -I, R goes very close to zero and A = 100%. If we accept A 2 70%, the band-width at this frequency is seen to be Av = 30 cm-‘. It is thus very easy to make SrTiO, a perfect absorber around 60 cm -’ with a thin coating of germanium. An additional advantage is that such a layer can be used as the front electrode of the pyroelectric detector since the resistivity of the dielectric is many orders of magnitude higher than the one of the purest germanium at room temperature. For application at IOOOcm- in the same conditions, the thickness of the germanium layer ought to be divided by 1000/60 = 16.7. then I, = 0.6 pm and Av,,,?~ = 500 cm ‘. 3.3. Position
of the reflectMy
minima
The optical-length retardation in a double pass through the germanium layer has to compensate for the phase shift advance I++at the Ge/SrTiO, reflection. Taking into account all multiple reflections in the germanium layer, the electric field E’ of the resultant reflected wave is calculated:
Ec
=
r
E
0 [
+ 42~21P exM+ I2 I - r,,p exp[j($
1’
- cp)l
(3)
- cp]
20 -
0
20
40
60
60
100 Y
120
140
160
160
I 200
(cm-‘)
Fig. 9. Reflectivity of a bulk SrTiO, single crystal covered with a thin layer of Gc (12 = 10 gm) at 300 K (--_). and at 93 K ( .- -). It is seen that at 300 K. the Ge layer is efficient to reduce R to zero at 6Ocm-‘. and to 16% at 175 cm ‘. At 93 K the Ge layer is less efficient, and R is only reduced to 20%. The minima have frequencies slightly smaller than at 300 K.
Infrared and millimeter wave absorber structures
473
with r,*
=
-
r2,
=
4 4
3.3.1. Phase shift.
+
nz
t,2
;
=
4
-. n,
24
12,
2n2
=-.
+n,’
n,
+n2
cp = 4nn, tv,
(4)
is the phase shift retardation due to a double pass through the germanium layer, and I/I the phase advance at the reflection from 2 to 3, given by:
‘%n2
tg$ = n: -n:
-k:’
with n: - (n: + k:)
cos * =
P[h+nd2+kil
It is clear that tg $ and cos II/, have the same sign, the one of n: - (n: +
k:);
(i) n: < n: + k:, e.g. n2 < n3, we can choose a solution with z2‘< &, < IC,
*=*o,
(7)
(ii) n: > n: + k:, we can choose a solution rjO with 0+.
(8)
Then, more generally, $ = $,, + 2K’7t, (9) with K’ = 0; 1; 2; 3; etc. . . 3.3.2. Reflectivity factor modulus. p is the modulus of the reflectivity factor from 2 to 3: P = P23 =
(n:-n:-k:)2+4k:n: [(n2+n3)‘+k:12
(8’) 3.3.3. Reflectivity power. Reflectivity power is minimum when a double path compensates exactly the phase shift at reflection. The intensity of the gross reflected wave is written: I = EC. ET, and putting a = + - cp; we have: ~12~12t21~ 1
now R = I/E:,
exp(ia)
+
-rzlp exp(ia)
r12+t12~21
1 -r2,
P
exp(-@)
+
P exp(-ja)
(112
11-~2,
P
12,
PI2
exp(.h)l[l -~2,
P
ev(-&)I
1 ’
hence: R
=r2
I2
+(t1212,~)2+2r12r12t2,pbNlL - cp)-r2,pl I +r$p’-2r2,p .cos(ll/ -cp) ’
(9)
Numerically, for SrTiO, at T = 300 K, for v = 60cm-‘: n, = 18; k, = 7, and with a germanium coating in air (n, = 1, n, = 4). we have: r ?, = -r12 - 0.60;
(,, = 0.40;
r,, = 1.60;
r,,t,, = 0.64;
p23 = 0.68.
Hence R = 0.36 +
0.402 - 0.520x 1.166-0.816x’
(10)
with x = cos(I(I - cp). - 0.28 dR/dx = 02: R is a decreasing function of x;
forx=-l,R=0.83;
forx=O,R=0.70andforx=+l,
R=0.02.
474
and XAVIER GERRAL’X
ARMAND HAIN
In conclusion,
to find the frequencies cos(@ -cp)=l;J/ cp =+
-2K”n:
of the minima
--cp =2K”n cp =+,,+2Kn
of R, we have to take .Y = 1. i.e
with K”=O.
with K=O.
_+t. &2....: fl.
k2 . . . . .
(11)
and 4nn,r
11= t+b,, i- 2Kn.
(1 I’)
Equation (I I ‘) gives values of 11which make R a minimum. which is very close to zero in this example. The significance of equation (I 1) is that the electric field Ez (Fig. 7) is finally in phase with the electric field E, of the incident wave. i.e. E: is in phase opposition with E, _ It is also the same for all these waves E;. E,. E., . . add together and can destroy E, if the final sum has the E,.E.,.... same amplitude as E, 3.3.4. Vuriution of‘ rhe _fieyuenc.,* v,,, of fhc reflecti1:ii.b. minima. H.ith the optical constants of‘ the absorbing dielectric no. 3. (9 For SrTiO, at T = 93 K. and around 60 cm ‘. k, = 20, n, = 18; t+b= 2.92 rad. ($ = 167’ ), and assuming I, = IO pm. we get \I,,, = 58 cm ’ with K = 0. in good accordance with the computer calculation reported in Fig. 9. The phase shift advance is smaller than 3.14 rad. the value expected if dielectrics no. 3. were transparent (k! = 0). (ii) For SrTiO, at T= 30OK; II,= 18:/~~=7,$ =2.98rad.$ = 171 . and keeping ’ in good accordance with Fig. 9. 1,= IOpm, weget v,=59.3cm (iii) For li, = 0, i.e. a transparent dielectric we have Ic/ = 3.14 rad. and still keeping f2 = IO pm, we get \trn = 62.5 cm ‘. The Cc plate in this case is exactly a quarter-wavelength plate (2ncj, = i-/2). In conclusion, the germanium coating I~ = IO jtrn thick deposited on medium no. 3 gives a series of minima in the reflectivity spectrum. whose positions are very sensitive to the optical constants of medium no. 3. For instance the first reflectivity minimum (k = 0) has a frequency which is decreased from 62.5 to 58 cm ‘. i.c. AV = 4.5 cm ‘, or AV/Y = 8% when n3 is increased from 1 to 18. and k, from 0 to 20. For the second reflectivity minimum, the frequency is decreased from 187.5 ‘. and Av.‘Y = 12X, when medium no. 3 is changed from vacuum to 165cm ‘. i.e. Av = 22.5cm into SrTiO, at T = 93 K. A measurement of such shifts should give a new relation between n3 and k,. Such a data in more accurate than a classical reflectivity power measurement because of the high accuracy of frequency measurements in Fourier Transform Spectroscopy. The value of the reflectivity minimum should give a second, but less accurate relation, linking n3 and k, through 11~.
3.4. I. An equarion Ricing nijor tllruniirc~~ec.tion Iu.ver. To get a null reflectivity at the minimum, and so A = loo%, i.e. to achieve a perfect absorber. WC must have E’ = 0. (t,,t,, p)i( I - r,,p) = r’,;
and thus, p3(n,. I+, k,) = (n? - n, ).‘(n: +- n,). which is seen directly T+R=I. In short to get a null reflectivity. we must have pJi = r?, . n, and k3 being given, to a perfect absorber. n2 = 4 gives effectively [(n, + n?)’ + ki] = [(n: -
since T = !,:I~, . R = ri,, and
(12)
and assuming n, = I, equation (12) will give the value of 11~which leads As an example we have seen that for II! = 18. k! = 7, germanium, with A = 99%. More generally, equation (12) is written: [(n: - n3)’ + k!]/ I),“(n: + l)]’ and we get:
Infrared
and millimeter
wave absorber
structures
47s
3.4.2. Immediate applications.
(9 For k, = 0, we have the classical result: n: = n,; 6Ocm-‘, we have to make (ii) In the case of SrTiO, at 300 K, around n: = 18 + (49/17) = 20.9, and n, = 4.5 is indeed close to the refractive index of germanium. At 93 K, n: = 18 + (400/17) = 41.5, and n2 = 6.5 is now significantly higher than the refractive index of germanium and reflectivity is not completely cancelled by a germanium coating (see Fig. 9, dashed line). (iii) For a metal in the FIR, n3 z k,, and a thin layer with n: z 2n, i.e. n, = 6
(14)
ought to give an efficient anti-reflection coating on the metal, and hence a high absorption power A = 1 - R, if the metal layer is thick enough to be sure that T = 0.
Let us try germanium again (i.e. nz = 4) for such a coating, we then must have n, = k, = 8 for the metal if we want the germanium layer to be efficient against reflection. That is exactly the case of granular-aluminium for a large range of frequencies around 500 cm ’ (see Fig. I). In that case, tg # = 64/( 16 - 128) = -0.57, and I(/,,= -3O”+ 180” = 150; $0= 2.62rad. Now K =47rkv = 5 x 104cm-‘, and the absorption length 6 = l/K = 2 x 10 ’ cm, 6 = 0.2 pm: if the aluminium layer is thicker than some tenths of a micrometer, and covered with a germanium layer with a similar thickness, we can expect the absorption power to be very close to 100% on very broad bands. Such a composite coating has several obvious advantages: small heat capacity, small lateral heat conductivity, good electrical conductivity needed for an electrode in a pyroelectric detector and larger frequency range where the power is kept close to 100%. These possibilities depend on the accuracy of calculations in Fig. 1. If n3 was smaller, we could use silicon for coating (nz = 3), but if n, was much larger than 8 it would be difficult to find an efficient material for the coating. Figure 10 gives the reflectivity R of a 30 pm thick granular Al, coated with a 0.92 pm thick Ge layer. The reflectivity is zero around 560 cm-‘, and 5% at 1800 cm-‘. (iv) To get n: > 0, and thus a real number for n,, we must have n3 + [k$(n, - l)] > 0. Difficulties appear only when n3 c I, then: n,(n, - 1) + k: < 0, or approximately : .(14’Such a relation may be difficult to fulfill in some -n,+k:
T.P 1 -
(3’)
r2,p
For LiTaO, at 300 K, around 600 cm .-I, p = 0.81, and r = 0.42 is far from zero: a germanium layer cannot be used efficiently to make LiTaO, a good absorber in this part of the spectrum. 3.5. The case of NaN02 as a second example We have seen in section 2.2 the difficulty to reduce the NaNO, reststrahlen reflectivity around 150 cm-’ and 275 cm-’ with a thin metal layer deposited on it. Let us try a coating with germanium which has a smaller refractive index and a negligible absorption index in the MIR, if pure enough. 3.5.1. For the v, = 160cm -’ reststrahlen. For that frequency, it is seen in Fig. 4 that n3 = 1.3 and k, = 2.7, hence pzj = 0.63, very close to rzI = 0.60: an interference minimum in the reflectivity, located at 160 cm-’ will be close to zero (R z 0).
ARNAIW HADNI and XAVIER
476
GERBAUX
tf (cm-‘) Fig. IO. Reflectivity R% of a granular Al film t = 30pm thick, coated with a Ge film I =0.92pm. showing a decrease of retlectivity to zero at S50cm-‘, and to 6% at 18OOcm-‘.
To get the germanium coating thickness which leads to a reflection minimum at 160cm ‘, we have to calculate the phase shift advance at the reflection from Ge to NaNO,: tg $ = (8 x 2.7)/[16 - (1.69 + 7.29)] = 3.08; cos $ > 0, hence + = 1.26 rad i 21Y’n; ($ = 72” _+ 360” x K). We thus have to compensate that advance by a double pass through the plate such as: 4nnz fv, = Il/(,+
the compensation is possible t, = 1.5 pm + 7.81 pm x K; K = 0:
for t, =
the
values
1.5 pm;
2Kn
of
K which
(II’) make
I,&~ + 2&r > 0, hence:
K = 1: I, = 9.3 pm.
In Fig. 11 R shows a minimum at v = 160 cm-’ for a germanium coating thickness t, = I .45 pm and f, = 9.3 pm, and the minima values are very close to zero. For the thinnest coating, the reflectivity minimum at 160cm‘-’ is broad, and the integrated reflectivity from 0 to I7Ocm ’ is especially small. Figure 12 shows R vs thickness t. For I = 0, R = 56%, and for t, = 1.5 pm + 7.8 pm x K, R 2: 0. 3.5.2. For the v2 = 27.5~1 -’ resrstrahlen. In Fig. 4 n3 zk3 = 0.5, then pz3 =0.79. and tjz7 = 0.254 rad. To get a phase shift retardation Ic/which is suitable to compensate exactly for II/. 100
r
Y
(cm-‘1
Fig. II. Reflectivity of a bulk NaNOz single crystal covered with a thin layer of Ge in the case of two thicknesses, I* = I .45 pm (- - -). and I~ = 9.30 pm (-) at 300 K. The antireflective coating is perfect at v = l6Ocm-” for both thicknesses. Another reflectivity minimum is observed at Y = 275cm-’ for rz = 9.3 pm.
Infrared and millimeter wave absorber structures
t (pm) Fig. 12. Reflectivity of a bulk NaNO, single crystal coated with Ge, at v = 160 cm-‘. vs coating thickness 1, showing 4 minima close to zero.
we have to make the germanium coating thickness t = 0.18 pm + K x 4.5 pm; K = 0 :r= 0.18 pm; K= 1: f =4.68pm; K=2: t =9.18pm. It is seen in Fig. 11 that we have indeed a minimum at v = 275 cm-’ for t = 9.3 pm and the reflectivity is low (R = 10%) but not negligible. Figure 13 shows R vst.For t = 0, R = R* = 20%, and for t,=0.18~m;4.68~m; 9.18pmetc. .., R = 10%. 4.
CONCLUSIONS
Metal-black coatings can give a 92% absorption in the 8-14pm band, but they have to be several pm thick, hence a thermal capacity of the same order of magnitude as a 3 pm thick pyroelectric ceramics, and a lateral heat conductivity which is a serious drawback especially for pyroelectric arrays. Moreover their reflectivity increases rapidly for longer wavelengths, and they are not reproducible. Very thin metal layers, in contrast, have a heat capacity and lateral thermal-conductivity which are both negligible, and in a number of cases the absorption power can be higher than 50% and even close to 100%. For free standing layers, it has been shown that A is kept constant at a value of 50% for frequencies up to 10,000 cm-‘. The thickness has to be chosen to give R, = Z,,/2 = 188 R. loo-
60 2 a 40-
I
0
I
I 10
20 t
I 30
(pm)
Fig. 13. Reflectivity of a bulk NaNO, single crystal coated with Ge, at v = 275 cm-‘, vs coating thickness I, showing 7 minima with the same value R, = 12%. The first one is corresponding to a very thin layer, fI = 0.3 pm.
ARMAND
478
HAD~‘Iand XAVIERGERLIAUX
For layers deposited on a transparent material, the absorption power A can be slightly higher than 50% for a given number of wavelengths for which the transparent material plays the role of a more or less antireflective layer. A can even be increased up to 100% if the transparent material is backed by a thick metal layer to give a quarter-wavelength structure for a number of radiations, and if the thin metal layer thickness is two times less than above to make R, = Z, = 377 R. The bandwidth can cover the whole 7-14 pm band if the transparent pyroelectric plate is thick enough (e.g. I pm thick for a PZLT ceramics). In the reststrahlen regions, the reflectivity of the pyroelectric is generally not decreased by the thin metal layer deposited on it. We have shown that it has to be replaced by a thin semiconductive layer which has a very small absorption index. and a refractive index which is nearly independent of wavelength (e.g. germanium). It will act as an antireflective layer, and for a well adapted material, we get R = 0, hence A = 100%. The semiconductor thickness can be very small, and A kept to values close to 100% on a broad frequency range. There are marked advantages in using the nearly i.!4 germanium coating instead of the metal-coated A/4 polyimide film used by Parson et al.:
(9 With the germanium
coating, absorption occurs directly in the high absorption dielectric plate at a small depth beyond the germanium coating. In Parson’s device absorption is localized in the nichrome film 377R/O. layer optimum thickness of coatings at IO pm is around 0.2 pm (ii) The germanium while the polyimide film is 1.4 pm thick and has to be metal-coated on both sides. Moreover cCie= 0.074 calg/g . K = 0.3 I J/g . K is slightly smaller than cpoly,,,,= 0.40 calg/ g. K. However pcir = 5.46 g/cm’, hence c&, = 1.70 J/cm3 K has the same order of magnitude as c&,,,,. Finally, c&~(, ,,,” = 0.50 J/m’. K. and we had seen in eqn (5): (.;ly,m.I 4pm.coatcd = 2.4 Jim’ K. The small heat capacity of the germanium device is a noticeable advantage over the Parsons’ device. However the reduction ration (i.e. 5) is not the effective reduction in heat capacity for thick detectors. The true gain is
With a I pm thick pyroelectric
ceramics, G=
taken
above
as an example:
2.4 + 2.6 = , 6 0.5 + 2.6
’ ’
This is quite appreciable. For much thicker pyroelectric ceramics, the gain will be less but the first advantage (production of heat in the pyroelectric directly), will still have to be considered. coating 0.2 pm thick, has a lateral thermal conductivity which is at least (iii) A germanium 12 times smaller than the 1.4pm thick polyimide film metal-coated on both faces. REFERENCES P. Strimer. X. Gerbaux. A. Hadni and T. Souel. Infrured fhyr. 21, 37 (1981). D. B. Belts. F. J. J. Clarke, L. J. Cox and J. A. Larkin, J. fhys. 18, 689 (1985) J. Lecomte, Handhuch der Ph.&. Springer, Berlin (1960). S.-I. Takahashi. Infrared Phys. 13, 301 (1973). S. Kasukabe. S. Yatsurya and R. Uyeda. J. C‘r.wa/ Growh 24, 315 (1974). Y. Hatanaka, S. Okamoto and R. Nishida, Jap. J. oppl. PhJs. 49, 563 (1980). M. J. E. Golay. Rev. Sri. Insrrum. 20, 816 (1949); l&347 (i947): 18, 357 (1947) F. W. Revnolds and G. R. Stilwell. Phvs. Rer. 88. 418 (1952). W. Wolteidorf, Z. Phys. 91, 230 (1934). G. Bruhat. Cours de Physique, Electricit& Masson, Paris (1956). K. Nakamura, T. Ishigaki, A. Kaneko. S. Takahashi. J. Nishida, Y. Wakabayashi and H. Nakamura. fnr. J. m/rared millimeter Waves 10, 907 (I 989). 12. J. Claude1 and A. Hadni. Elecrro-OprrcsiLaser Inrernatiotud. C‘onferenw Proceedings (Edited by H. G. Jerrard), p. 149. IPC Science and Technology Press’ (1976). 13. A. D. Parsons and D. J. Pedder. J. Var. Sci. Technol. A6. 1686 (1988) 14. M. Tinkham. Science 145, 240 (1964). 1. 2. 3. 4. 5. 6. 7. 8. 9. IO. I I.