14-1. Biasing the Photodiode For a particular value of R, this equation is solved numerically for i, and the efficiency (cid:3) sc = i2R/Pin is calculated. By varying R, the graph shown in Fig. 14-4 is obtained. The optimum efficiency of 8.94% is obtained for R = 2.5(cid:8). If (cid:5)= 2, Eq. (14-4) becomes iR (cid:1) 0.0506 i = (6 108) exp(cid:2) (cid:3) 0.129 and the optimum efficiency increases to 17.9% at R = 5(cid:8). A more detailed model of solar cell efficiency would take into account the variation with wavelength of the optical power from the sun and the fraction of this light that is absorbed by the silicon. In practice, solar cells based on crystalline silicon can have efficiencies as high as 24% in the laboratory, with ~ 15% being typical in commercial devices. Thin films of amorphous silicon (atoms not ordered periodically) are inexpensive to manufacture but have lower efficiencies, typically 13% in the laboratory and 57% in commercial devices.

14-2. OUTPUT SATURATION In the case of the solar cell just discussed, the primary goal is to convert as much optical power as possible into electrical power. When the photodiode is used as a light detector, however, it is generally more important that the detector output be linear with the incident light power. In this section, we examine the linearity of photodiode detector circuits using the two types of biasing modes. Photovoltaic Mode Consider first the photovoltaic bias mode shown in Fig. 14-2a. When R is very large (open-circuit condition), the load line is nearly horizontal, and the operating point is close to the Vd axis where i (cid:7) 0. In that case, Eq. (14-3) becomes Figure 14-4 Variation of solar cell efficiency with load resistance for Example 14-1. Area of the cell is 4 cm2. Optimum efficiency is higher and occurs at a higher load resistance when the diode ideality factor ((cid:5)) is 2 rather than 1. c14.qxd 2/22/2006 3:23 PM Page 254 254

Chapter 14 Photodiode Detectors eVd(cid:1) (cid:5)kBT 0 (cid:5) i0(cid:1)exp(cid:2) (cid:3) 1(cid:4) i(cid:2) with i(cid:2) given by Eq. (14-2). Solving for the diode voltage gives (cid:5)kBT (cid:1) e i(cid:2)(cid:1) i0 ln (cid:2)1 + (cid:3) Vd = (14-6) If the induced photocurrent is much greater than the dark current (i(cid:2) (cid:7) i0), this becomes Pine(cid:3) abs(cid:1) i0h(cid:4) (cid:5) (cid:5)VT ln (cid:2) (cid:3) (open circuit, i(cid:2) (cid:7) i0) (14-7) Vd where kBT (cid:1) e (cid:6) (voltage equivalent of temperature) (14-8) VT The diode voltage, therefore, varies logarithmically with the incident power for i(cid:2) (cid:7) i0. For i(cid:2) (cid:9) i0, however, Pine(cid:3) abs(cid:1) i0h(cid:4) i(cid:2)(cid:1) i0 (cid:5) (cid:5)VT = (cid:5)VT (14-9) Vd Defining the quantity (cid:5)kBT (cid:1) ei0 (cid:5)VT(cid:1) i0 Rsh = = (shunt resistance) (14-10) the diode voltage can be written as Pine(cid:3) abs(cid:1) h(cid:4)

(cid:5) Rshi(cid:2) = Rsh(cid:2) (cid:3) (open circuit, i(cid:2) (cid:9) i0) (14-11) Vd This result suggests that the photodiode can be modeled as an ideal current source connected in parallel with a resistor Rsh, as depicted in Fig. 14-5. Since i = 0 for an open cirFigure 14-5 When i(cid:2) (cid:9) i0, the photodiode can be modeled as an ideal current source in parallel with a shunt resistance Rsh. c14.qxd 2/22/2006 3:23 PM Page 255 255