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Homework 2_soln - ECE 2100 Professor Alyosha Molnar...

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ECE 2100: Professor Alyosha Molnar Homework Assignment #2 ( Solution ) Due Feb 17, 2010 Areas covered: Node, mesh analysis. Thevenin, Norton equivalents. Maximum power transfer. Superposition. LEDs, Photodiodes, Photovoltaics. 1) Consider two LED lighting systems, one using amber LEDs ( λ = 600nm) and the other using blue ( λ = 450nm). a. What is the expected turn-on voltage of each? V TO = 1235/ λ , so for λ = 600nm, V TO = 2.06V for λ = 450nm, V TO = 2.74V b. In each case you put 3 diodes in series, and place a resistor in series with them. In both cases you want to limit current to 50mA when 12V is placed across the whole string (Current sets the number of photons generated). In each case, choose an approximate resistor value to accomplish this. For the amber LED, 3 in series will give a voltage of ~6.18V, so the resistor will see 5.82V, which with a current of 50mA, means R=V/I = 5.82V/.05A = 120 For the blue LED, 3 in series will give a voltage of ~8.22V, so the resistor will see 3.8 V, which with a current of 50mA, means R=V/I = 3.8V/.05A = 75 2) Modeling an optical isolator: Optical isolators are used to pass signals between parts of a system that are operating at different voltages (for example in hospital rooms where everything is isolated on a floating ground to avoid discharges). The basic circuit is shown in Fig. 3a: it uses a resistor in series with an LED, whose light couples to a photodiode. Specifically, light intensity is proportional to LED current (L = k 1 I LED ), and photodiode current is proportional to absorbed light (I PD = k 2 L). 0 < k 1 , k 2 <1. An I-V curve of the input port looks like that shown below (Fig 2b). I PD , when plotted vs I LED gives the curve below it (Fig 2d). a. For Vin > V TO, the turn-on voltage of the LED, this circuit can be modeled as shown. Based on the curves shown, find values for V TO and A I and Rs. The x-intercept of plot 1 is V TO = 2V , The slope of plot 2 is A I = 0.001 One over the slope of plot 1 (above V TO ) is 1V/10mA = 100 b. For, Vin = V TO + Δ Vin, what must R L be to ensure Vo = Δ Vin? I LED = (Vin – V TO )/R S for Vin > V TO So I LED = ( Δ Vin)/R S = ( Δ Vin)/ 100 I PD = A I ·I LED = 0.001( Δ Vin)/ 100 =( Δ Vin)/ 100k Vo = R L ·I PD = R L · ( Δ Vin)/ 100k We want Vo = Δ Vin, so R L =100k
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3) Solar Cell Problem. Here again we will use the diode equation to model the Silicon Solar Cell which we will test in the lab. For parameter values in the diode equation: choose I 0 = 20 μ A , η = 1 , and kT/q = 25.86 mV . In fact, you have four of these diodes in series, what, then, is the I(V) equation of the 4-stack? Restricting the model to the forward bias regime only, add a 16 series resistance and plot the solar cell’s IV curve (representing operation in the dark) on a linear scale. Generate this plot in excel or another math program (not by hand). Compute V as a function of I as in HW 1. I(V) is hard to solve here, but V(I) is easy: for one diode, V= kT/q·ln(I/Io) = 25.86mV·ln(I/20 μ A) For 4 diodes, V = 4(25.86mV·ln(I/20 μ A)) = 103.44mV ln(I/20 μ A), so I = 20 μ A[exp(V/103.44mV)-1] With the resistor, V =103.44mV ln(I/20 μ A)+I·16 a) b) c) d)
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-30.00 0.00 30.00 60.00 -0.50 0.00 0.50 1.00 1.50 V I, mA dark Light a. Plot on the same graph (as part a) the illuminated diode IV curve. Now assuming an independent bias generation
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