# Free Space Optic Receiver with Strongly Overlapped Photodetectors' Field of View
## 1 Introduction
### Conditions Preserving Parallel Amplifier Advantages
Figure 5: Total surface screening depending on holder orientation and the \(\varphi_{z}\) value. Assuming that all the sensitive areas of the photodetector are perpendicular to the direction of falling rays, \(SNR_{out}\) will be enhanced \(\left(6/\sqrt{6}\right)^{2}\) times, which is six times. Since the useful signal current and the sensitive area are proportional [1], the currents and surfaces may be swapped, which matches the curve analysis shown in Figure 5.  
The most advantageous situation occurs when the direction of falling rays is perpendicular to the truncated pyramid base. Each photodetector surface screening is then maximized. Assuming that the default sensitive area is equal to the unit, and the photodetectors and the pyramid base are at a \(45^{\circ}\) angle, then one photodetector's surface screening is \(1/\sqrt{2}\) and the six photodetectors' surface screening is \(6/\sqrt{2}=3\sqrt{2}=4.243\). This is the same value shown in Figure 5 for the angle \(\varphi_{z}=90^{\circ}\). If \(\varphi_{z}\) decreases, the total surface screening also decreases and reaches its minimum at \(\varphi_{z}=0^{\circ}\), being slightly over 1. However, the noise level surface equivalent stays the same, being equal to \(\sqrt{6}=2.449\). To preserve the parallel amplifier's advantage, the total surface screening must be higher than \(2.449\). This is fulfilled for angles greater than the minimum angle \(\varphi_{z}=32^{\circ}\). In other words, at angle \(\varphi_{z}=32^{\circ}\), the \(SNR_{out}\) value of all PDAs is equal to the \(SNR_{out}\) of only one PDA in use. The best \(SNR_{out}\) is achieved for the angle \(\varphi_{z}=90^{\circ}\), though the maximum value is only \(\left(3\sqrt{2}/\sqrt{6}\right)^{2}=3\) (or \(4.8\) dB). The curve breaks in Figure 5 correspond to the angles when the photodetectors in shadow become illuminated. The calculations above have solved one particular case for a receiver equipped with six photodetectors fixed to truncated pyramid walls where the truncated pyramid walls and base are at a \(45^{\circ}\) angle. However, this is only one case out of many others. The question then arises as to design of a FSO receiver's optimal holder for a particular application? For example, how many photodetectors is best for that application? Which pyramid wall and base angle is optimal for that application? No universal answers can be provided; therefore, our attention returns to Equation (7). The number of photodetectors collecting the useful optical signal \(s^{\prime}\) from the total number of photodetectors \(s\) can be easily separated. Equation (7) can be rearranged as follows:  
\[SNR_{out}=\frac{s^{\prime^{2}}p_{s}K_{21}^{2}}{s^{2}p_{n}K_{21}^{2}+sp_{nA}}~ {}~{}[/]. \tag{16}\]  
Environments possessing a uniform signal coverage, such as interiors, provide reflection of light signals and lead to the equality \(s^{\prime}=s\). Based on these conditions, an increasing \(s\) eliminates noise power \(p_{nA}\) (the PDA noise power) and enhances the signal to noise ratio \(SNR_{out}\). Environments without reflections might keep some photodetectors permanently in shadow. Since they remain without a useful signal power, the equality \(s^{\prime}=s\) is replaced by the inequality \(s^{\prime}<s\), and the noise power \(p_{n}\) from an optical channel determines the situation. If it is much lower than the PDA noise power, \(p_{n}<<p_{nA}\), Equation (16) becomes:  
\[SNR_{out}=\frac{s^{\prime^{2}}p_{s}K_{21}^{2}}{sp_{nA}}~{}~{}[/]. \tag{17}\]  
Even here, enhancement of the \(SNR_{out}\) may be expected because the numerator grows by the power of two of the illuminated photodetector's number \(s^{\prime}\). Enlargement of the photodetector is again beneficial. The only problem is the non-zero noise power \(p_{n}\) from the optical channel.