98_exam2fall_sol

98_exam2fall_sol - 6.012 Fall 1998 Answers to Exam#2...

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6.012 Fall 1998 - Answers to Exam #2 Problem 1 a) Neglecting the depletion region, which we assumed you would do (and most of you did), the plot is as shown: p'(x) p'(1µm) x (µm) -1 0 1 2 b) Half of the injected minority carrier flux of MA flows to and across the junction and the other flows to the ohmic contact. The current I is minus the former multiplied by the charge per carrier; thus I = -qMA/2. c) Since many fewer minority carriers flow to the right, much more of the total injected minority carrier flux of MA must flow to the left and across the junction, meaning that the magnitude of I is now increased. In fact I is nearly -qMA. This "trick" is used in many photodiodes to increase the photocurrent and reduce undesirable recombination at ohmic contacts and device surfaces (which can look a bit like ohmic contacts in that they act as places of increased recombination). d) The charge on the gate is varies nearly linearly with the gate voltage in accumulation, which occurs for v GB < V FB in a MOS capacitor fabricated on a p-type substrate, and in inversion, which occurs for v GB > V T . In both cases the proportionality factor is the oxide capacitance, e ox A/t ox , which in this case is 3.5 x 10 -11 Coul/V, or 35 pF. e) With the switch closed, v GB is greater than V T so the interface is inverted and the inversion layer charge is -C ox (v GB - V T ), or -8.75 x 10 -11 Coul. f) The diode is reverse biased for v GB greater than zero, and for v GB greater than a few kT the current through the reverse biased diode is a constant I S . Thus it will look like a current source of this value for the v
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98_exam2fall_sol - 6.012 Fall 1998 Answers to Exam#2...

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