## 3 LPR Cell Design
### Transient Response  
The transient response of the LPR cell is analyzed using an equivalent circuit of Figure 12(a) for the LPR cell. The sub-threshold current \(I_{d}\) flowing through transistor M\({}_{1}\) is an exponential function of the gate-source voltage of M\({}_{1}\), \(V_{BI}\)-\(V_{CS}\), and it is given by:  
\[I_{d}=I_{d0}\exp\left\{\frac{1}{nV_{th}}(V_{th}-V_{CS}-V_{T})\right\} \tag{4}\]  
where \(V_{T}\) is threshold voltage, \(I_{d0}\) is the current if \(V_{BI}-V_{CS}=V_{T}\), \(n\) is an ideality factor, and \(V_{th}=k_{B}T\)/\(q\) is a thermal voltage. When the light is on, the drain current \(I_{d}\) increases up to the photocurrent \(I_{P}\) at \(t=T_{C}\)/2, where \(T_{C}\) is the cycle time of the light pulse. When the light is off, the drain current decreases, finally to \(I_{dM}\) at \(t=T_{C}\). The charge sensing node voltage at \(t=0\) is denoted by \(V_{CSM}\). Then the drain current can be written as:  
\[I_{d}=I_{dM}\exp\left\{\frac{-1}{nV_{th}}(V_{CS}-V_{CSM})\right\} \tag{5}\]  
From Figure 12(a), the drain current can be written as:  
Figure 11: Operation of LPR Cell. Figure 12: **(a)** Model of transient analysis of LPR cell, **(b)** Waveform of I\({}_{\text{ph}}\), i\({}_{\text{d}}\) and V\({}_{\text{CS}}\). \[I_{d}= \begin{bmatrix}C_{CS}\dfrac{dV_{CS}}{dt}+I_{p}&\left(0<t\leq T_{C}/2\right)\\ C_{CS}\dfrac{dV_{CS}}{dt}&\left(T_{C}/2<t\leq T_{C}\right)\end{bmatrix}  \tag{6}\]  
In Figure 12(a), the influence of the photodiode capacitance is ignored and the photo charge generated in the photodiode is instantly transferred to the charge sensing node. By solving Equation (6) with Equation (5) and initial conditions for turning light on and off, the drain currents during turning on, \(I_{d,on}\) and turning off, \(I_{d,off}\) are given by:  
\[I_{d,on}(t)=\frac{I_{p}}{1+\left(\frac{I_{P}}{I_{dM}}-1\right)\exp\left(- \frac{t}{\tau}\right)} \tag{7}\]  
\[I_{d,off}(t)=\frac{I_{p}}{1+\left(t-\frac{T_{C}}{2}\right)/\tau} \tag{8}\]  
where \(\tau\) is a time constant given by:  
\[\tau=\frac{nV_{sh}C_{CS}}{I_{p}} \tag{9}\]  
Since \(I_{d,\ off}(T_{C})=I_{dM}\), Equation (8) gives:  
\[\frac{I_{P}}{I_{dM}}=1+\frac{T_{C}}{2\tau}=1+\frac{1}{2\tau f} \tag{10}\]  
where \(f=1/T_{C}\) is the pulse frequency. From Equations (5), (7) and (8), for \(0<t\leq T_{C}/2\) :  
\[V_{CSM}-V_{CS}(t)=nV_{th}\ln\frac{I_{d,on}(t)}{I_{dM}}\] \[=nV_{th}\ln\frac{I_{p}}{I_{dM}}\frac{1}{1+\left(\frac{I_{p}}{I_{ dM}}-1\right)\exp(-\frac{t}{\tau})} \tag{11}\]  
and for \(T_{C}/2<t\leq T_{C}\) :  
\[V_{CSM}-V_{CS}(t)=nV_{th}\ln\frac{I_{d,off}(t)}{I_{dM}}\] \[=nV_{th}\ln\frac{I_{p}}{I_{dM}}\frac{1}{1+\left(t-T_{C}/2\right)/\tau} \tag{12}\]  
The signal amplitude, \(\Delta V_{CS}\) is then given by:  
\[\Delta V_{CS}=V_{CSM}-V_{CS}\left(T_{C}/2\right)\] \[=nV_{th}\ln\frac{I_{p}}{I_{dM}}\frac{1}{1+\left(\frac{I_{p}}{I_{ dM}}-1\right)\exp\left(-\frac{1}{2f\tau}\right)} \tag{13}\]  
where \(S_{{}_{L}}\) is the area of the lens, \(L\) is the distance to the LED and D is the diameter of the lens. The relationship between the area of the LED array \(S_{{}_{LED}}\) and the image size at the focal plane \(S_{{}_{F}}\) is given by:  
\[I_{d,off}(t)=\frac{I_{p}}{1+\left(t-\frac{T_{C}}{2}\right)/\tau} \tag{8}\]0  
where \(f_{{}_{I}}\) is the focal length. The optical power per unit area at the focal plane is given by:  
\[p_{{}_{0}}=\left\{ \begin{aligned} &\frac{FT_{{}_{L}}}{S_{{}_{F}}}& \quad\text{(if }\ S_{{}_{F}}>S_{{}_{P}}\text{)}\\ &\frac{FT_{{}_{L}}}{S_{{}_{P}}}&\quad\text{(if }\ S_{{}_{F}}\leq S_{{}_{P}}\text{ )}\end{aligned} \right. \tag{21}\]  
where \(T_{{}_{L}}\) is the transmittance of the lens and \(S_{{}_{P}}\)is the area of pixel. For \(S_{{}_{F}}>S_{{}_{P}}\), \(p_{{}_{0}}\)is given by:  
\[p_{{}_{0}}=\frac{\pi T_{{}_{L}}I}{4F_{{}_{N}}^{2}S_{{}_{LED}}} \tag{22}\]  
where \(F_{{}_{N}}(=f_{{}_{I}}/D)\) is the F number of the lens.Therefore \(p_{{}_{0}}\) becomes constant independent of the distance to the LED.