Hmwk #3 (Sample Midterm) Solutions PDF (1)

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EE121B Winter 2006 Midterm Name: EE 121B Midterm 8 February 2006 Read the question and the possible answers before attempting any calculations. Write your name on the front of each page (tests may be separated for grading purposes). This test is closed book. You may use a calculator and a single 8.5x11 sheet (back and front) of formulas. This can include the sheet that I provided to you earlier with your own formulas on the back. Section 1: Multiple choice (45%) Circle the correct answer. Any problem with multiple answers circled will be marked wrong! (45 pts.) There is no partial credit on the multiple choice questions. Some of the answers are obvious by looking. Look and think before calculating! 1. (5 pts.) The junction capacitance of a pn junction in equilibrium is 3.5 fF (fF = 10" F) and the contact potential (built-in voltage) is 0.5 V. What is the junction capacitance under a reverse bias of -1.5 V? a. 7F (b 175 ) c.10.5fF d.35fF QJ& must be smaller, only b.is sialle— e = 8B JCu(9” | Wit g;_:.; Jr=2 w i T (15) w(o) 2. (5 pts.) An npn transistor has an emitter injection efficiency 0f 0.990, a base transport factor of 0.980 and a minority carrier lifetime of 0.1 psec in the base. What is the transit time of holes through the base? X anse2? 503 e essie Zljed e\ ¥= 0.99 0y =058 %, =0.Aps . e ~ Cn - 100ns ol% I-01-.02=697 P B= ooz =325 2 Y33 OR. Nr =098 tuke $= .00 H B 093-50 2 Yy ’Zg'-‘LD;Q’\:ZHS 3. (5 pts.) The Ebers-Moll equations of a specific npn bxpo]ar transistor are/tglven o by: 1072, —9.8x[07" ) et 9.8x107° * 1.47x107 ?{”—1 What is the approximate common emitter current gain in normal active mode? b.2 c.98 d. 980 Vie /KT i Io_.zea be/kT £ Eeg 0 3 u= 08¢ - .00 o3 e‘hz/g«l € B 9.gxlo . jomg ey oot 1 i ;o

EE121B Winter 2006 Midterm 1.0E+00 s — 1.0E-01 1.0E-02 " 1.0E-03 1.0E-04 1.0E05 le, Ib (Amperes) 1.0E-06 T 1.0E-07 1.0E-08 1.0E-09 1.0E-10 02 03 04 05 06 07 08 08 Ves (Volts) 4. (5 pts.) The Gummel plot shown is for a silicon npn transistor with base doping of N,=10". What is the approximate maximum value of the common emitter current gain (). a. 10 b. 1000 c.1 3.1 ""SP“"."‘ (see Sf‘PLX 5. (5 pts.) An ideal transistor is biased in common emitter mode. If the base current is 100pA, the common emitter current gain is 100, and the load resistance is 50 ©, what is the approximate small signal voltage gain ? a. 02 b. 100 d. 20,000 58 jaouh, [z 100 RiESA b = BI"Q 2 Ve = =c = — 7.‘:‘. = 8mRL V- R Vr : - (100)ioopN)(02) 26mV ‘UCLA Henry Samueli SEAS

EE121B Winter 2006 Midterm Name: et 12 g e 10 lg=50 pA oc = U I = 40 pA 8 B H R, s ¢ 30uA Iy I * = * Ve 4 Vag : i E \ = 0 | 0 05 1 1.5 2 25 Vce (Volts) = IS0 6. (5 pts.) Referring to the common-emitter Ic-Vce characteristic above. The 15 v - 15D transistor with the above characteristic is biased from a 1.5V supply (Vcc = f = 1.5V) with a 150 Q load. Vg is chosen such that the base current is 30 pA. What is the approximate collector-emitter voltage (Vcg) across the transistor? = 104 a 0.1V b. 1.5V c. 0.8V o 7. (5 pts.) The transistor with the above characteristic is biased from a 1.5V supply (Vee = 1.5V) with a 150 Q load. This time Vg is chosen such that the base current is 50 pA. The transistor is biased in 2 a. Cutoff b. Normal ( c. Saturation ) d. Reverse Active Forward Active

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EE121B Winter 2006 Midterm 1.0E+01 g = 1.0E+00 1.0E-01 1.0E-02 1.0E-03 1.0E-04 1.0E-05 1.0E-06 1.0E-07 1.0E-08 1.0E-09 0.00 0.50 1.00 1.50 Ve (Volts) 8. (5pts.) In the transistor with measured Gummel plot as shown find the base series resistance Rg in the model? 0.35V #10 a 4 b. 10 d. 100 oy EZQ Ic, Is (Amps) n p n N, =1x10% /em* | N, =1x10'/cm’ | N,=1x10"/cm’ (¢— 0.5 pm —| 9. (5 pts.) Refer to the symmetrical transistor above. We desire to operate this transistor in normal active mode with Vgg = 0.5V. What is the approximate maximum reverse bias voltage that can be applied |-Vpc| before a punch through condition is reached? b.5V-10V ¢ 10V-100V d.>100V Loolc up Zero H‘J A&f\e’ﬂw\ fe nrp :)u\—\c"\‘-\n W(Na=10%) .33 Tgprere ER junchon (unknowrn fid L;,.;) A 0.33 UCLA Henry Samueli SEAS RSsums- Vo 202V Jo1 = — —_— 0.5° 0.7 +1°Veel [~vge) OV

EE121B Winter 2006 Midterm Name: Section 2: Problems (55%) Show your work. Full credit for the correct answer with work shown. Sensible answers (correct order of magnitude) get partial credit. Generous partial credit for an incorrect answer with the correct ideas if clear and brief. Negative credit for irrelevant or incorrect equations (never less than zero but can negate some positive credit). 1. (25pts.) (18 pts.) a) (15) Derive an expression for the base transit time in an npn transistor in normal active mode. Your final expression should be in terms of the neutral base width and other constants. Start with equations given on the formula sheet and assume that the emitter injection efficiency is unity. ’L‘ L 5. - 0 ( (3° ~ < _°(_T PR Frow sheet T | 2 s -0y 3 § w , ’Ys = IE = %y (l‘d~r>:_."_.. Tn 'é(—l:.) Ighore ( ) s 3 we o i Aonorinsdor ' ne 7y wWS W) 2ISe - 2hn There are many ways 4o do this ¢ Many ot valid exprerisas Ls]"""-'*\ R +le bave M-'m'mm) b) (5) Given a silicon npn transistor with base width of 0.5 pm and a doping of Ng = 10", calculate the approximate transit time. -y )’- -3 = -3 2 (osxig™) _ oa2sxi0 em . [Oal0 em = < = 5 _kiT( 2(0.025) l4o0 s 2800 cm'/s R 7"0—‘7 :EJ?- nsec v L{OPSQC 28 (357) ¢) (5) If this is the dominant delay time (and we can neglect all others) calculate the unity current gain cutoff frequency (f;) of the transistor. 9 o L - 23x0 Y 27 y GH= (4us) a0 R

EE121B Winter 2006 Midterm N, = 1077 N, = 4x1017 N,g = 1017 2. (30 pts.) Three silicon npn bipolar junction transistors differ only in the design of their base. #1 is the “nominal” transistor, #2 differs from the nominal by having a base doping four times as high as #1 and #3 differs from the nominal by having a base twice as wide as #1,. (assume Wy/L, is small (<< 1) in all three transistors. Ignore depletion effects in the base...assume Wy, or 2W,, is the neutral base width.) a) (12 pts.) On the plot supplied below carefully draw the quantitative excess electron concentration (and label clearly #1, #2, #3) in the base for the three ‘transistors in normal active mode (assume Vg is the same for all three cases with exp[qVee/kT] = 101 and Vi negative with exp[qVpc/kT] << 1). Label the el vertical axis with numerical quantities for your excess electron concentrations. " Aﬂi =Npy (EKP - ,> = 100 npe 2. lO Moo (#1)= 2 =N (#3) = 2000 gl #3)= 200000 27‘)0 #2)= 7= o leo{. 2) .‘ xio” [=el Mz (ﬂz) = 50,000 b) (8 pts.) Given that the nominal transistor has an emitter injection efficiency of 0.990 calculate the emitter injection efficiency of transistor #2 and transistor #3 (to 3 decimal places). Assume py(10'7) = p,(4x10"7). el s 2 A NN o) ¥= ks ! 1+fi;‘:’ ﬂ; VTN = Naz _ = 0.04 x1-A g 1-b, Aac LS U4, 5, Ml 2bc TR (Continued on next page) {3 s g Y =099 ¥,F 1-004 ¥, % 1008 UCLA Henry Samueli SEAS JE = 0960 = 6980 g [w\ore et oM RIS ([ 02N SAESg 0-780‘1)

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EE121B Winter 2006 Midterm Name: c) (5 pts.) Given that the base transport factor of the nominal transistor is 0.990 calculate the base transport factor of transistor #2 and transistor #3 (assume that L, and 1, are the same for all three transistors). ' a2 7/ ot = 35S 2o-E) 7.2 0\1\‘1 o\e(eual.s on WI So *2 is Sar= = 6990 7 dT: J_(’\*i;/>1- a4 (2_‘{"- >7. b8 S v $ SJ 2 \Ln 2 L F3e 3 = Yxo0) D(—,‘? = D,O"’O = oY \o{.,, = plyy = 0.990 ATT:T%;_I d) (5 pts.) Calculate the approximate common emitter current gain (B1, Bz, B3) for all three transistors (label answers clearly, 2pt for the general expression and 1pt for each labeled answer). If you can’t get a numerical answer or expression, 1pt for ordering them from largest to smallest B...like a>b>c. o, = ¥, o, = 0.990 -0.990 % 0.98 48 i » pito BE S 0.95 ; ¢ D388 10060 B 0.9Y X3 ~ ).'30173 5 g,> B 2P3 From inspection Yo sholld ser fi,>fia l31>(g3 bir chan -)J\L:-\«.-)L.. e u* Ba vibs it \ QA\:. L\c.*n\— 5