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hw2soln

# hw2soln - 6 ID0=1mA when VD0=0.6V ID = IS exp(VD0/VT 2.717...

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6) I D0 =1mA when V D0 =0.6V I D = I S exp(V D0 /V T ) 2.717 I D0 = I S exp( (V D0 + Δ V) / V T ) (hopefully you recognize 2.717 as e) = I S exp(V D0 /V T ) exp( Δ V/V T ) = I D0 exp( Δ V/V T ) Therefore, exp( Δ V/V T ) = 2.717, and Δ V = V T ln(e) = V T So every time you increase the current through the diode by 2.717, you increase the voltage across the diode by V T . 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 I D0 = I D0 exp( Δ V/V T ), and therefore Δ V = V T ln(10) = 2.3 V T . . 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

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