F2010A EEE 3308C Final Exam Solution

F2010A EEE 3308C Final Exam Solution - P roblem 1 (20 pts.)...

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Problem 1 (20 pts.) Short answer questions. Some device parameters are listed in Table 1. Diode MOSFET BJT n=1, V t =25 mV, I S =10 -14 A k n ’ = 4x10 -4 A/V 2 , W = 5 um, L= 1μm, |V T0 |= 0.7V, | λ |=0.02V -1 , | Φ F |= 0.3V, | γ |= 0.4 V 1/2 n=1, β =50, |V A |=100V, I S - 14 A, |V CEsat 0.3V a) What is one key difference between the I-V (or q-V or dI/dt – V) relations of Circuits I (R, L, C) components and E-Circuits I components (Diode, MOSFET, BJT)? b) List the 2 operating regions for a diode . c) If the current is measured to be 5mA, what is the ‘exact’ junction voltage? d) Write down a half-wave rectifier circuit using diodes . Label the input and the outputs. e) The rectifier example is a large-signal application of a diode. Another application of a diode is for small-signal inputs. Give the maximum small-signal voltage. R, L, C are linear while diode, MOSFET, and BJT are nonlinear. 1) forward, 2) reverse IS 10 14 A := Vt 25mV := n1 := ID 5mA := ID IS exp VD nV t 1 = VD n Vt ln ID IS 1 + := VD 0.673 V = vd < t 5 5mV =
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Darlington pair i) h) 1) cutoff, 2) saturation, 3) foward active, 4) reverse active VGS VT0 5 0.2V = vgs < VGS 1.7V = VGS VT0 2 IDsat knprime W L + := L1 u m := W5 u m := VT0 0.7V := knprime 4 10 4 A V 2 := VGS VT0 2 IDsat knprime W L = um 10 6 m := IDsat 1mA := IDsat 1 2 knprime W L VGS VT0 () 2 = since VSB=0 VT VT0 = g)
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This note was uploaded on 07/28/2011 for the course EEL 3308C taught by Professor Yoon during the Spring '11 term at University of Florida.

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F2010A EEE 3308C Final Exam Solution - P roblem 1 (20 pts.)...

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