F2010A EEE 3308C Final Exam Solution

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

This preview shows pages 1–3. Sign up to view the full content.

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 =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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)
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online