4904_B2008_Quiz5_solns

4904_B2008_Quiz5_solns - Name SOLUTI OH S A 42.50;l ECEBox#...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

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

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

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

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Name SOLUTI OH S A 42.50;l ECEBox# _ B 3941; Average . g Problem Score Pomts mgwiijis 1 L 18 2 Ali 32 ECE4904 B2008 Semiconductor Devices Quiz 5 This is a closed book quiz! You are allowed one 8—1/2” X 11” sheet (both sides) of notes. Note: Potentially useful reference charts are given on pages 7 and 8 (tear off to use if you like). Show all your work. Partial credit may be given. Ifyou think you need something that you can't remember, write down what you need and what you'd do if you remembered it. Unless otherwise indicated, you may assume Si in equilibrium at T=300K. Look for the simple, straightforward way to solve the problem for the level of accuracy required. Don't get entangled in unnecessary algebra. As in real life, some problems may give you more information than you need. Don't assume that all information must be used! It's your job to decide what's relevant to the solution. You will have 25 minutes to complete this exam. There are 2 problems on a total of 8 pages. The figure on the opposite page shows the four major components of current flow in a pnp transistor. It is given (as shown in the figure) that the total emitter current is IE = 816uA, and the total collector current is Ic = 792uA. It is also given that the emitter injection efficiency is y=0.99. In the box provided on the opposite page, draw the correctly oriented circuit symbol for the pnp transistor. [3] For each of the four current components shown in the figure, indicate in the table on the opposite page whether the current flow is electron current or hole current, and indicate the magnitude of the current flow (in uA). [6] Determine (accuracy i10%) the base current 13 IB = 24 MA KCL ‘, 13 1 IE “1,; 3 gléyuA’ 792/uA ’ ZimA [3] d) Determine (accuracy i0.005) the base transport factor 0LT [3] on: 0.95 : $2-: 792 :0193 I 1 (50 8 Determine (i10%) the common emitter DC current gain Bdc l3] Bdc : 3 3 :: .ES. 3 E I I 3 24 802 * ' : smug oawfi-f-{A =17 1.1=(O,959)8’léMA = 808/«A _ MM FOR13 : 13 “f ?’ SiEMA FOR: 14 f 144 + 3/AA 3" W 12 'DJ l6] pnco = é) [3] [4] [6] [4] [5] 2. The figure on the opposite page shows an n+pn transistor: _Assume each region is uniformly doped (nondegenerate doping) with dopant densrties and other parameters as indicated in the table above the figure. a) Determine (accuracy i10%) the equilibrium majority (InnEO, ppBO,_and nnco) and minority (ano, anO, and pnco) carrier concentrations 1n the em1tter, base and collector regions. NUMERICAL RESULTS REQUIRED!!! j Emitter FROM ’ . . WM i» Mammy flew; Minority WP Tn; ' L b) Using dashed lines, sketch and label the equilibrium majority and minority carrier concentrations on the figure. For parts (c - g), the transistor is operated with a reverse bias on the collector-base junction and a forward bias of VBE = +0.6V on the emitter-base junction. c) What is the name of this operating condition? (Circle one) CUTOFF INVERTED ( ACTIVE SATURATION (1) Using solid lines, sketch the steady—state minority carrier concentrations on the figure. e) Using the boxes on the figure, indicate numerical values for the minority carrier concentrations at the edges of the depletion regions. (Note: accuracy ilO% or ::1E+3 /cm’, whichever is larger) LAW OF THE JUMCTTGM f) Given a junction area of A=4.0E-4 cmz, determine the resulting collector current 1C for the applied bias of VBE = +0.6V. (Accuracy i10%) [C = 75“ “A : A D 5.131% éea'l9)(uta~q)@o.7) fig Cl “Edy ’ . ' 1.3514 g) It is observed that an increase in the collector voltage (increasing the base—collector m7 reverse bias) causes a small increase in the collector current. What causes this dependence of collector current on collector voltage? Note: You must do more than give the name of this effect — explain what is happening that changes the current! C (. EXPLAIN!: BASE; WDTH MODULATTON ' [4] INCREASE w VCB REVERSE SEAS ‘ => lNCIQEASE' IN {1-8 0633!, iQEaroM MOW l :7 DECREASE; w BASE wtom as; mctiezxse: w cim/dx w BASE -i> WCREASEJ IQ ( l l 1 F _T Emitter Base Collector Doping N13}; 2 E+17 NAB = NBC 2 up}; = ting = upc : 461 / ,,,,,,,,,,,,,, N Diffusion DpE = 8.55 D“B = 30.2 (1:) DpC = 11.9 Coefficient Minority Carrier TpE = 1.0 E-7 13,113 = 1.0 E-6 1pc = 1.0 E-6 Lifetime MinorityCarrier LpE=9.3 13—4 x’inst.5E—3 Diffusion Length { Base Width W = 1.5E-4 \. Boxes for (c): E+iZ l5.8 E ‘r I3J‘ O O l (2 B 2%? (ea/0,0259) ’ ‘ _ “ ' Ppao 2.13 + Z WM L “B : Mina/am meme CONCENTRATmM 5 ~ LiNEAe ...
View Full Document

This note was uploaded on 01/24/2009 for the course ECE 4904 taught by Professor Mcneill during the Fall '08 term at WPI.

Page1 / 5

4904_B2008_Quiz5_solns - Name SOLUTI OH S A 42.50;l ECEBox#...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online