hw2soln - 6) ID0=1mA when VD0=0.6V ID = IS exp(VD0/VT)...

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Unformatted text preview: 6) ID0=1mA when VD0=0.6V ID = IS exp(VD0/VT) 2.717 ID0 = IS exp( (VD0+∆V) / VT) (hopefully you recognize 2.717 as e) = IS exp(VD0/VT) exp(∆V/VT) = ID0 exp(∆V/VT) Therefore, exp(∆V/VT) = 2.717, and ∆V = VT ln(e) = VT So every time you increase the current through the diode by 2.717, you increase the voltage across the diode by VT. So the answer is that at -40C, 25C, and +85C, the diode voltage increases are 20mV, 26mV, and 31mV – exactly the same as the answers to problem 1. 7) This time it’s 10 ID0= ID0 exp(∆V/VT), and therefore ∆V = VT ln(10) = 2.3 VT . . So the answers at -40C, 25C, and +85C are 46mV, 60mV, and 71mV. 9) a) Most of the electrons in silicon are tightly bound to the nucleus that they orbit, so they can’t contribute to the flow of electrons through the solid. Of the four electrons per atom that are in the outer shell, only a tiny fraction have the thermal energy necessary to escape that bond and move freely through the material. Those electrons, and the valence-band energy vacancies that they leave behind, are what contribute to conduction in a semiconductor. b) The electric field in the space charge region opposes the diffusion of carriers, and in equilibrium the diffusion and drift currents are equal and opposite. c) minority carriers: np and pn d) Recall that for every 60mV we increase the diode voltage, the diode current increases by a factor of 10. So from 0.7V to 1V, ∆V=300mV=5*60mV, so the current increases by five orders of magnitude, from 1mA to 100A. To go up another 500mV means more than 8 orders of magnitude increase in current, to more than 1010A, and at 2V another eight orders of magnitude, or more than 1018A. ...
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This note was uploaded on 08/30/2011 for the course EE 105 taught by Professor King-liu during the Fall '07 term at University of California, Berkeley.

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hw2soln - 6) ID0=1mA when VD0=0.6V ID = IS exp(VD0/VT)...

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