Microsoft Word - ECE162 HW 7 solutions

Microsoft Word - ECE162 HW 7 solutions - 5.3 Ge Photodiode...

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5.3 Ge Photodiode Consider a commercial Ge pn junction photodiode which has the responsivity shown in Figure 5Q3. Its photosensitive area is 0.008 mm 2 . It is used under a reverse bias of 10V when the dark current is 0.3 mA and the junction capacitance is 4 pF. The rise time of the photodiode is 0.5ns. a Calculate its quantum efficiency at 850, 1300 and 1550nm. b What is the light intensity at 1.55 mm that gives a photocurrent equal to the dark current? c What would be the effect of lowering the temperature on the responsivity curve? d Given that the dark current is in the range of microamperes, what would be the advantage in reducing the temperature? e Suppose that the photodiode is used with a 100 W resistance to sample the photocurrent. What limits the speed of response? 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.5 1 1.5 2 Wavelength(μm) The responsivity of a commercial Ge pn junction photodiode Responsivity(A/W) Figure 5Q3 Solution a At l = 850´10 -9 m, from the responsivity vs. wavelength curve we have R = 0.25 A/W. From the definitions of quantum efficiency (QE) h and responsivity we have, % 36.5 ) m 10 850 ( ) C 10 60218 . 1 ( ) A/W 25 . 0 ( ) ms 10 3 ( Js) 10 626 . 6 ( 9 19 1 8 34 = × × × × × × × = = - - - - λ η e hc R
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Similarly, we can calculate quantum efficiency at other wavelengths. The results are summarized in Table 5Q3. Table 5Q3 Wavelength, (nm) 850 1300 1550 Responsivity R , (A/W) 0.25 0.57 0.73 Quantum efficiency η , (%) 36.5 54.3 58.4 b Given, photocurrent I ph = I d = 0.3 μ A = 0.3 × 10 -6 A and area, A = 8 × 10 -9 m 2 , the incident optical power, P o = I ph / R = (0.3´10 -6 A)/(0.73 A W -1 ) = 4.1096 ´ 10 -7 W Light intensity, I o = P 0 / A = (4.1096 ´10 -7 W)/(8 × 10 -9 m 2 ) = 51.4 W m -2 or 5.14 mW cm -2 . c From semiconductor data under Selected Semiconductors, for most semiconductors dE g /dT is negative, E g increases with decreasing temperature. Stated differently, a vs l curve shifts towards shorter l with decreasing T . The change in a with T means that the amount of optical power absorbed in the depletion region and hence the quantum efficiency will change with temperature. The peak responsivity will shift to lower wavelengths with decreasing temperature. If maximum photogeneration requires a certain absorption depth and hence a certain a max , then the same a max will occur at a lower wavelength at lower temperatures. In Figure 5Q3, maximum responsivity corresponds to a max which occurs at l max at high T and at l ¢ max at lower T . λ Absortion coefficient = 1/ δ Low T High T max λ′ max α max The absorption coefficient depends on the temperature Figure 5Q3 d Dark current ( exp(- E g /kT )) will be drastically reduced if we decrease the temperature. Reduction of dark current improves SNR.
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e The RC time constant = 100 × (4 × 10 -12 ) = 0.4 ns. The RC time constant is comparable to the rise time, 0.5 ns. Therefore, the speed of response depends on both the rise time and RC time constant. (It is not simply 0.4 ns + 0.5 ns.)
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This note was uploaded on 12/02/2009 for the course ECE 000 taught by Professor O during the Spring '09 term at UCSB.

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Microsoft Word - ECE162 HW 7 solutions - 5.3 Ge Photodiode...

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