Biasing Analysis of NPN and PNP Transistors using PSpice Lite - Prof. Marc E. Herniter, Assignments of Electrical and Electronics Engineering

Download Biasing Analysis of NPN and PNP Transistors using PSpice Lite - Prof. Marc E. Herniter and more Assignments Electrical and Electronics Engineering in PDF only on Docsity! Problem 1 iso , Ne= Nae + Vag = Vee + TaRn @ YWvg (@ wo (1) Nec - Vee-Tarn ~ (P+!) Fa Re Salue For Tp Ze = Nee -Nee Re + (+8) Re Since oe = BTa Tes B(Vec- ve) i Ra + (i+) Re b) foc (e+) Re >> Re and Bo>\ The CKT will oe indePendent of B ed Since Ra isnot <é A+B) the Crrewie Will be dependent evr RB Semin = Bans (Vee Vee) _ So(i5-0.3) Tce ere Ret (4 Ban)Re — 2)S000 + (Si\lFP00) = lOm4 Zemay = Bmax (vec -Vae) _ 350 isv-o.av) - Sey Re + (b+ Bmax)Re SOO + (351) TPon ISemA \l > Ic is a Little dependent on B Problen 2 make Q thevenin eg ovr a ReaRe e Vee Ray = RRL @s7Klisk = \oiex Nags We Re _ sul sk = 1 42y Rae LvSk4+9,89K Nee Re =a 2 + | Low.k |.dw 2 [ Re =330 \f Frew Norte % te B) va Vee \ Rant (4p) Re For B= 350 Lez 3So ( F- 0.90) ——————__ = Sos LovaK + (350) (3200) Nee 2 CE Ne = Sey (Re+Re) = ISY— 2.07malo = 423v =D may + Swng s i 6AN 200A +3300) Nee P Na — Te (Reate) =~) — 2,43my (aaktie) = F.44U for B= $50 I, _ B(Vtn—V ae - 350 (4 rou _ 0,3~ J Rent Lite \R¢ F.open+ (351) (10002) = SZ 4rd Vig= Nee ~ k& {Re+Re) = \ev ~ Same (22k ak) = tNY 5 5 4 4 3 3 2 2 1 1 D D C C B B A A 2.942V 11.66V 0 + - V1 DC = 18 Model = Qx NPN BF=50 Qx Q1 2.884mA + R1 33k + R2 10k + R3 2.2k + R4 1k C:\WEBSITE\ROSE_CLASSES\ECE250\Homework\Spring_2004\HW6\problem 3-30-SCHEMATIC1-Bias.out.1 05/03/04 11:16:24 **** 05/03/04 11:15:39 ************** PSpice Lite (Mar 2000) ***************** ** Profile: "SCHEMATIC1-Bias" [ C:\WEBSITE\ROSE_CLASSES\ECE250\Homework\Spring_2004\HW6\problem 3-30-schematic1-bias.sim ] **** OPERATING POINT INFORMATION TEMPERATURE = 27.000 DEG C ****************************************************************************** **** BIPOLAR JUNCTION TRANSISTORS NAME Q_Q1 MODEL Qx IB 5.77E-05 IC 2.88E-03 VBE 8.02E-01 VBC -7.91E+00 VCE 8.71E+00 BETADC 5.00E+01 GM 1.12E-01 RPI 4.48E+02 RX 0.00E+00 RO 1.00E+12 CBE 0.00E+00 CBC 0.00E+00 CJS 0.00E+00 BETAAC 5.00E+01 CBX/CBX2 0.00E+00 FT/FT2 1.77E+18 JOB CONCLUDED TOTAL JOB TIME .05 Page: 1 reat mate 4 Thevenin owe lent OvX oF. Nice, Xa, And Ry Veg =10 +e Re Tczr $*S Ray verve % 5 Vee. oon Vee Va = Ve - se Re - Veg Ana tester dy = (Ro) Fa, So Ne = Na -Neg- (Qt\FQR- _ Nery Tees Va- Van _ | dee Veg =) Tah, ve. aN Raw } . Ray REAL milion aan opment eel Go\we Fx Te \) ct- Nan- Nee Ran * (\«8)Re tw % (ec ~ Vw, New) = Mv é WER =Oy % $e = pear pc Wot (®) ‘c FR -~ ee * “ De ke Ver Nae E Vez O _. -~ J - Ne- ER . J. -Ue SO -s_| 92 35 on 3.4 AS sO Vea VNu- te*e- Ve ton Vea = eet Fyn Gd, Zant 4.8 < ey Tae | A. 8T re 4 | SeliN _ Te 5 VE SA ew UL SN Ja, \ 7 s,\N < VY Nec |s, 5 5 4 4 3 3 2 2 1 1 D D C C B B A A 4.991V 5.802V 0 + - V1 DC = 10 Model = Qp PNP BF=50 + R1 20k + R2 15k + R3 1k Qp Q2-82.31uA 4.198mA C:\WEBSITE\ROSE_CLASSES\ECE250\Homework\Spring_2004\HW6\problem 3-32-SCHEMATIC1-Bias.out.1 05/03/04 14:19:37 **** 05/03/04 14:18:30 ************** PSpice Lite (Mar 2000) ***************** ** Profile: "SCHEMATIC1-Bias" [ C:\WEBSITE\ROSE_CLASSES\ECE250\Homework\Spri **** OPERATING POINT INFORMATION TEMPERATURE = 27.000 DEG C ****************************************************************************** **** BIPOLAR JUNCTION TRANSISTORS NAME Q_Q2 MODEL Qp IB -1.36E-05 IC -4.77E-03 VBE -8.15E-01 VBC 4.40E+00 VCE -5.22E+00 BETADC 3.50E+02 GM 1.84E-01 RPI 1.90E+03 RX 0.00E+00 RO 1.00E+12 CBE 0.00E+00 CBC 0.00E+00 CJS 0.00E+00 BETAAC 3.50E+02 CBX/CBX2 0.00E+00 FT/FT2 2.93E+18 JOB CONCLUDED TOTAL JOB TIME .11 Page: 1 P coblewn 3.52 a for O Dov Re ) — “iL 2 Ri, = Sookl| sook aoe = S4.7K qok Fok ||S00k Sook Vane Vo= Sv - So0l+ Swiw Gor lisoor) . = 1.4 + lov \ 4 SV Fo k. Il 500% 50k + cake) \ @ VUSW§ ~he - tWevensn egurwalen* ana sherxed No VV GO, Use Nave —~\hke Crccorv ALON Re- sok ® (v.—Vee) _ 00 (1azy - 0.79) Tes Oe S43 + (10v)( Scoo) Ran + (ue) Re = 297 mh alsok + SKY Neeo = Nee- 4ehhire) = Qov- LEM = FN ») QO is san Wren Nae Vee Ve = Wah Wwe Wave ‘tre C\Cous Nacesv Nic =Sw 2K Ry Re Vs Vo = \hesqt =0,20 —D —» Ow Vu ~~? te Tu + Re 3 —WwW ZX, = Neg-l-2vd O ~(~ | | ao Fe Cw) AW 2 gsun — — Ro, Qok 2oK  is deeply Barware A 5 5 4 4 3 3 2 2 1 1 D D C C B B A A 0 0 Vo 0 0 Model = Qx NPN BF=50 + R1 2000 + RC 2000 + - Vcc DC = 5 Dbreak D1 + - V2 DC = 2 Qx Q1Dbreak D2 + R2 20k + - Vin DC = 0