The equivalent electrical circuit for the absorber is given below
In the above circuit Zg is purely resistive in nature that represents the solid graphite sheet of the absorber at the bottom. Z1, it represents the impedance of dielectric (SiO2) just above the bottom surface of absorber consisting of solid graphite sheet. Above this layer dielectric another annular graphite disc is placed that act as DRA unit 1. This unit of DRA is represented by series combination of R1 L1 and C1 connected in parallel to series combination of Z1 and Zg. Zd1 is impedance of DRA 1 similarly, Z2 represents the impedance of dielectric layer of SiO2 above DRA unit 1. Zd2 represents the impedance of DRA unit 2 consists of R2 L2 and C2 as shown in figure. Similarly, Zd3 and Zd4 represents the impedance of DRA unit 3 and unit 4 respectively. Zd3 consists of R3 L3 and C3 whereas Zd4 consists of R4 L4 and C4 passive elements. Zs is the source impedance. Z3 and Z4 are the impedance of dielectric layer consisting of SiO2 as shown in figure.
When the electrical excitation is given to the circuit, the input impedance Zin is inductive in nature at the input, so the maximum energy is stored as magnetic energy and very less energy is stored in electric field due to less capacitive nature of Zin As the energy flows towards the load, the impedance of the electrical network tends to become more and more capacitive and less inductive. Due to this the more power (energy) input gets stored in electric field due to the oscillation of electromagnetic energy between the inductive and capacitive element of the input impedance Zin. The resistive part of the input impedance act as energy dissipating elements. Thus, the electrical power which is travelling from the source to load undergoes change in its storing phenomenon from magnetic energy (power) to electrical energy stored in electric field. During this transaction of electrical power (energy), power loss occurs in the resistive part of the input impedance (Zin) that corresponds to absorption phenomenon of incident light within the absorber. As a result of energy transfer from inductive to capacitive element of Zin indicates the improvement of quality factor of equivalent RLC circuit, as an important condition of the working of absorber.
In the above equivalent absorber circuit applying the Transmission Line Concept (TLC) input impedance of the first stage \({z}_{i{n}_{1}}\)is calculated as given in Eq. (4)
$${z}_{i{n}_{1}}={z}_{c}\left[\frac{zg-J{z}_{c}{tan}\left({k}_{z}{h}_{1}\right)}{{z}_{c}-J{z}_{g}{tan}\left({k}_{z}{h}_{1}\right)}\right]$$
4
where \({z}_{c}\) is characteristics impedance of substrate, h1 is the height of dielectric Substrate, Zg is represented as impedance of bottom graphite sheet and \({k}_{z}\) is propagation constant. \({k}_{0}\) wave number in free space.
$${z}_{c}=\frac{\omega {\mu }_{0}}{{k}_{z}} ; {k}_{z}=\sqrt{{k}_{P}^{2}-{k}_{0}^{2}{\text{sin}}^{2}\theta };{k}_{P}=\omega \sqrt[.]{{\mu }_{0}{\epsilon }_{s}}$$
$${z}_{1}={z}_{2}={z}_{3}={z}_{4}={z}_{c}$$
$${z}_{{p}_{1}}={z}_{g} and {z}_{p\left(a+1\right)}=\left.{z}_{Da}|\right|{z}_{ina}$$
$${z}_{in\left(i+1\right)}={z}_{C}\left[\frac{{z}_{P\left(a+1\right)}-J{z}_{C}{tan}\left({k}_{z}{t}_{s}\right)}{{z}_{C}-J{z}_{P\left(a+1\right)}{tan}\left({k}_{z}{t}_{s}\right.}\right]$$
5
$$i=\text{1,2},3,\dots . and a=\text{1,2},3,\dots .$$
$${z}_{in}={z}_{s}+{z}_{P4}$$
6
$$\varGamma =\frac{{Real}\left({z}_{in}\right)-{z}_{0}}{{Real}\left({z}_{in}\right)+{z}_{0}}$$
7
$$A=1-{\left|{s}_{11}\right|}^{2}-{\left|{s}_{21}\right|}^{2}$$
8
Eq. (5) represents the generalize formula for input impedance at different stages while total input impedance is given by Eq. (6). The reflection coefficient and absorption are formulized in Eq. (7) & Eq. (8) respectively.
Different parameter of the above circuit can be evaluated by utilising other methods of network analysis. As per the above Equivalent absorber circuit, the circuit follows a ladder network pattern. The observed circuit input impedance in this manner is calculated as shown in Eq. (9).
$${\text{z}}_{\text{i}\text{n}}={\text{z}}_{\text{s}}+\frac{1}{{\text{y}}_{{\text{d}}_{4}+\frac{1}{{\text{z}}_{4}+\frac{1}{{\text{y}}_{{\text{d}}_{3}}+\frac{1}{{\text{z}}_{3}+\frac{1}{{\text{y}}_{{\text{d}}_{2}+\frac{1}{{\text{z}}_{2}+\frac{1}{{\text{y}}_{{\text{d}}_{1}}+\frac{1}{{\text{z}}_{1}+\frac{1}{{\text{y}}_{\text{g}}=\left(\frac{1}{\text{z}\text{g}}\right)}}}}}}}}}}}$$
9
\({\text{z}}_{\text{i}\text{n}}={\text{z}}_{\text{s}}+{\text{z}}_{{\text{p}}_{i}}\) , where \(i=\text{4,3},\text{2,1};\)
\({\text{z}}_{{\text{p}}_{\text{i}}}=\frac{1}{{\text{y}}_{{d}_{\text{i}}}+\frac{1}{{\text{z}}_{\text{i}}+{\text{z}}_{\text{p}\left(\text{i}-1\right)}}}\) , where \(\text{i}=\text{4,3},\text{2,1};\)
$$i=4, {z}_{{P}_{4}}=\frac{1}{{y}_{{d}_{4}}+\frac{1}{{z}_{4}+{z}_{{p}_{3}}}} , {z}_{{d}_{4}}={R}_{4}+J\left(\omega {L}_{4}-\frac{1}{\omega {c}_{4}}\right)=\frac{1}{{y}_{{d}_{4}}}$$
$$i=3, {z}_{{P}_{3}}=\frac{1}{{y}_{{d}_{3}}+\frac{1}{{z}_{3}+{z}_{{p}_{2}}}},{z}_{{d}_{3}}={R}_{3}+J\left(\omega {L}_{3}-\frac{1}{\omega {c}_{3}}\right)=\frac{1}{{y}_{{d}_{3}}}$$
$$i=2, {\text{z}}_{{\text{P}}_{2}}=\frac{1}{{\text{y}}_{{\text{d}}_{2}}+\frac{1}{{\text{z}}_{2}+{\text{z}}_{{\text{p}}_{1}}}}, {\text{z}}_{{\text{d}}_{2}}={\text{R}}_{2}+\text{J}\left({\omega }{\text{L}}_{2}-\frac{1}{{\omega }{\text{c}}_{2}}\right)=\frac{1}{{\text{y}}_{{\text{d}}_{2}}}$$
$$i=1, { \text{z}}_{{\text{P}}_{1}}=\frac{1}{{\text{y}}_{{\text{d}}_{1}}+\frac{1}{{\text{z}}_{1}+{\text{z}}_{{\text{p}}_{0}}}} ,{\text{z}}_{{\text{d}}_{1}}={\text{R}}_{1}+\text{J}\left({\omega }{\text{L}}_{1}-\frac{1}{{\omega }{\text{c}}_{1}}\right)=\frac{1}{{\text{y}}_{{\text{d}}_{1}}}$$
where \({\text{z}}_{{\text{p}}_{0}}=\text{Z}\text{g}\) is purely real function (it means only resistive part is present in the last graphite disk).
$${\text{z}}_{{\text{p}}_{1}}=\frac{1}{{\text{y}}_{{\text{d}}_{1}}+\frac{1}{{\text{z}}_{1}+\frac{1}{{\text{z}}_{\text{g}}}}}=\frac{1}{{\text{y}}_{{\text{d}}_{1}}+\frac{1}{{\text{z}}_{1}+{\text{z}}_{\text{g}}}}$$
\({\text{z}}_{{\text{d}}_{1}}={\text{R}}_{1}+\text{J}\left({\omega }{\text{L}}_{1}-\frac{1}{{\omega }{\text{c}}_{1}}\right)=\frac{1}{{\text{y}}_{{\text{d}}_{1}}}\) , \({\text{z}}_{{\text{d}}_{2}}={\text{R}}_{2}+\text{J}\left({\omega }{\text{L}}_{2}-\frac{1}{{\omega }{\text{c}}_{2}}\right)=\frac{1}{{\text{y}}_{{\text{d}}_{2}}}\)
\({\text{z}}_{{\text{d}}_{3}}={\text{R}}_{3}+\text{J}\left({\omega }{\text{L}}_{3}-\frac{1}{{\omega }{\text{c}}_{3}}\right)=\frac{1}{{\text{y}}_{{\text{d}}_{3}}}\) , \({\text{z}}_{{\text{d}}_{4}}={\text{R}}_{4}+\text{J}\left({\omega }{\text{L}}_{4}-\frac{1}{{\omega }{\text{c}}_{4}}\right)=\frac{1}{{\text{y}}_{{\text{d}}_{4}}}\)
\({\text{z}}_{{\text{p}}_{1}}=\frac{1}{{\text{y}}_{{\text{d}}_{1}}+\frac{1}{{\text{z}}_{1}+{\text{z}}_{\text{g}}}}\) , \({\text{z}}_{{\text{p}}_{1}}=\frac{1}{\frac{1}{{\text{R}}_{1}+\text{J}\left({\omega }{\text{L}}_{1}-\frac{1}{{\omega }{\text{c}}_{1}}\right)}+\frac{1}{{\text{z}}_{1}+{\text{z}}_{\text{g}}}}\), \({\text{z}}_{{\text{p}}_{1}}=\frac{1}{\frac{{\text{R}}_{1}-\dot{\text{J}}\left({\omega }{\text{L}}_{1}-\frac{1}{{\omega }{\text{C}}_{1}}\right)}{{\text{R}}_{1}^{2}+{\left({\omega }{\text{L}}_{1}-\frac{1}{{\omega }{\text{C}}_{1}}\right)}^{2}}+\frac{1}{{\text{z}}_{1}+{\text{z}}_{\text{g}}}}\)
\({\text{z}}_{{\text{P}}_{1}}=\frac{1}{\frac{{\text{R}}_{1}}{{\text{R}}_{1}^{2}+{\left({\omega }{\text{L}}_{1}-\frac{1}{{\omega }{\text{c}}_{1}}\right)}^{2}}+\frac{1}{{\text{z}}_{1}+{\text{z}}_{\text{g}}}-\text{J}\frac{\left({\omega }{\text{L}}_{1}-\frac{1}{{\omega }{\text{c}}_{1}}\right)}{{\text{R}}_{1}^{2}+{\left({\omega }{\text{L}}_{1}-\frac{1}{{\omega }{\text{c}}_{1}}\right)}^{2}}}\) , \({\text{z}}_{{\text{P}}_{1}}=\frac{1}{{\text{y}}_{{\text{p}}_{1}}};\)
$$\text{A}=\frac{{\text{R}}_{1}}{{\text{R}}_{1}^{2}+{\left({\omega }{\text{L}}_{1}-\frac{1}{{\omega }{\text{C}}_{1}}\right)}^{2}}+\frac{1}{{\text{z}}_{1}+{\text{z}}_{\text{g}}}, \text{B}=\frac{\left({\omega }{\text{L}}_{1}-\frac{1}{{\omega }{\text{C}}_{1}}\right)}{{\text{R}}_{1}^{2}+{\left({\omega }{\text{L}}_{1}-\frac{1}{{\omega }{\text{C}}_{1}}\right)}^{2}}$$
\({\text{z}}_{{\text{p}}_{1}}=\frac{1}{\text{A}-\text{j}\text{B}}=\frac{\text{A}+j\text{B}}{{\text{A}}^{2}+{\text{B}}^{2}}\) , \({\text{z}}_{{\text{P}}_{1}}=\frac{\text{A}}{{\text{A}}^{2}+{\text{B}}^{2}}+\dot{j}\frac{\text{B}}{{\text{A}}^{2}+{\text{B}}^{2}}\)
\({\text{z}}_{\text{p}1 \text{i}\text{m}\text{a}\text{i}\text{g}}=0\) , \(\frac{\text{B}}{{\text{A}}^{2}+{\text{B}}^{2}}=0, \text{B}=0;\)
$$\frac{\left({\omega }{\text{L}}_{1}-\frac{1}{{\omega }{\text{c}}_{1}}\right)}{{\text{R}}_{1}^{2}+{\left({\omega }{\text{L}}_{1}-\frac{1}{{\omega }{\text{c}}_{1}}\right)}^{2}}=0, {\omega }{\text{L}}_{1}-\frac{1}{{\omega }{\text{c}}_{1}}=0$$
\({{\omega }}_{{\text{R}}_{1}}=\frac{1}{\sqrt{{\text{L}}_{1}{\text{C}}_{1}}}, {\text{f}}_{{\text{R}}_{1}}=\frac{1}{2{\Pi }\sqrt{{\text{L}}_{1}{\text{C}}_{1}}}\) , \({{\omega }}_{{\text{R}}_{2}}=\frac{1}{\sqrt{{\text{L}}_{2}{\text{C}}_{2}}}, {\text{f}}_{{\text{R}}_{2}}=\frac{1}{2{\Pi }\sqrt{{\text{L}}_{2}{\text{C}}_{2}}}\)
\({{\omega }}_{{\text{R}}_{3}}=\frac{1}{\sqrt{{\text{L}}_{3}{\text{C}}_{3}}}, {\text{f}}_{{\text{R}}_{3}}=\frac{1}{2{\Pi }\sqrt{{\text{L}}_{3}{\text{C}}_{3}}}\) , \({{\omega }}_{{\text{R}}_{4}}=\frac{1}{\sqrt{{\text{L}}_{4}{\text{C}}_{4}}}, {\text{f}}_{{\text{R}}_{4}}=\frac{1}{2{\Pi }\sqrt{{\text{L}}_{4}{\text{C}}_{4}}}\)
$${\text{z}}_{\text{i}\text{n}}={\text{z}}_{\text{s}}+{\text{z}}_{\text{P}i}$$
Where the \({\text{z}}_{\text{P}i}\) is the recursive function after changing the i-value function recall itself.\({\text{z}}_{\text{P}i}=\frac{1}{{\text{y}}_{{d}_{i}}+\frac{1}{{\text{z}}_{i}+{\text{z}}_{\text{p}\left(i-1\right)}}}\)
$$i=4 {\text{z}}_{{\text{P}}_{4}}=\frac{1}{{\text{y}}_{{\text{d}}_{4}}+\frac{1}{{\text{z}}_{4}+{\text{z}}_{{\text{p}}_{3}}}}, i=3 {\text{z}}_{{\text{P}}_{3}}=\frac{1}{{\text{y}}_{{\text{d}}_{3}}+\frac{1}{{\text{z}}_{3}+{\text{z}}_{{\text{p}}_{2}}}}$$
\(i=2 {\text{z}}_{{\text{P}}_{2}}=\frac{1}{{\text{y}}_{{\text{d}}_{2}}+\frac{1}{{\text{z}}_{2}+{\text{z}}_{{\text{p}}_{1}}}}\) , \(i=1 {\text{z}}_{{\text{P}}_{1}}=\frac{1}{{\text{y}}_{{\text{d}}_{1}}+\frac{1}{{\text{z}}_{1}+{\text{z}}_{{\text{p}}_{0}}}}\)
$${\text{z}}_{{\text{P}}_{1}}=\frac{\text{A}}{{\text{A}}^{2}+{\text{B}}^{2}}+\text{j}\frac{\text{B}}{{\text{A}}^{2}+{\text{B}}^{2}}$$
10
ZP1 (10) is the input impedance of absorber’s first disk.
The outcome of absorber is achieved by the numerical analysis of ladder network and analyzed by network analysis and synthesis and concept of resonance. initially the sheet of rectangular graphite is placed which is purely resistive in nature. a dielectric is placed over it which is surrounded by a circular disc that is made up of graphite. This circular disc as inductive in nature with some resistive property. The disc has outer radius of 35 µm and inner radius of 8 µm. The capacitance of the disc can be calculated as (\(C=\frac{\epsilon A}{d}=\frac{\epsilon \pi {\left({r}_{01}-{r}_{i}\right)}^{2}}{2\pi \left({r}_{0},-{r}_{\dot{l}}\right)}=\frac{\epsilon }{2}\left({r}_{01}-{r}_{i}\right)\)) and the value obtained is 9.8 pF. Under this situation resonance will occurs, the resonating frequency of first resonance is 7.66 Thz. The inductance is calculated as \(L=\frac{1}{{\left(2\pi {f}_{R}\right)}^{2}C}\) that is obtained as 0.243 fh. Again, circular disc of reduced outer radius 30 µm but same inner radius is placed over the first circular disc and dielectric is kept between these plates, its capacitance is found to be 11.1 pF. Again resonance condition is obtained having resonating frequency of 8.55 Thz .in this situation calculated value of inductance is 0.267 fH. A conical shape is structured by placing another circular disc of graphite with less outer radius of 19 µm over second circular disc, but the inner radius remains unchanged. Dielectric is filled between these two plates so its capacitance will be 12.6 pF which results in occurrence of resonance having resonating frequency of 9.11Thz and the inductance L3 will be 0.267 fh. The topmost disc is made of Graphene which results is multi resonance condition having resonating frequency of 9.96 Thz and 10.513 Thz with capacitance 14.99 pF and 16.8 pF respectively. Corresponding to these frequencies, the value of inductance is 0.7685 fH and 0.267 fH .
The resistance of the disk can be calculated as \({R}_{i}=\rho \frac{L}{A}=\frac{l}{{\sigma }_{i}A}=\frac{2\pi \left({r}_{01}-{r}_{i}\right)}{\pi {\left({r}_{01}-{r}_{i}\right)}^{2}\pi }=\frac{2}{{\sigma }_{i}\left({r}_{01}-{r}_{i}\right)}\)Applying this, we get R1=0.08 𝞨, R2=0.09 𝞨, R3=0.117 𝞨 and R4=0.1818 𝞨.If the ray of light falls perpendicularly on the sheet, it reflects 80%, only 20% is absorbed to avoid this condition the light is given at an angle of 750 as it provides max absorption. due to this incident angle some deflection angle is created and light ray is trapped among the different discs. The ray which reaches at topmost disc loses its strength thus the maximum part of light is absorbed.