EE 321 Analog Electronics, Fall 2011Homework #8 sol ution5.103. An npn BJT with grounded emitter is operated with VBE= 0.700 V, atwhich the collector current is 1 mA. A 10 kΩ resistor connects the collector to+15 V supply. What is the resulting collector voltage VC? Now, if a signal appliedto the base raised vBEto 705 mV, find the resulting total collector current iCand total collector voltage vCusing the exponential iC-vBErelationship. For thissituation, what are vbeand vc? Calculate the voltage gan vc/vbe. Compare withthe value obtained using the small-signal approximation, that is −gmRC.The o utput voltage is VC= VCC− ICRC= 15 − 1 × 10 = 5 V. Next we rais the base voltage.First we want to find the value of ICfrom the bias point, andiC= ISevBEVTIS=iCevBEVT=1 × 10−3e0.725×10−3= 6.91 × 10−16ANext for vBE= 0.705 V:iC= ISevBEVT= 6.91 × 10−16× e0.70525×10−3= 1.22 mAvC= VCC− iCRC= 15 − 1.22 × 10 = 2.8 VFor this situation vbe= 5 × 10−3, and vc= 2.8 − 5 = −2.2 V, and the gain is vc/vbe= −440.The gain using the linear approximation isAvo= −ICRCVT= −1 × 1025 × 10−3= 4005.104. A transitor with β = 120 is biased to operate at a DC collector current of1.2 mA. Find the values of gm, rπ, and re. Repeat for a bias current of 120 µA.gm=ICVT=1.225= 0.048 Ω−1rπ=βgm=1200.048= 2500 Ωre=αgm=ββ + 11gm=12012110.048= 20.7 ΩIf instead the transistor is biased at IC= 120 µA, we getgm=ICVT=0.1225= 0.0048 Ω−1rπ=βgm=1200.0048= 25 kΩ1re=αgm=ββ + 11gm=12012110.0048= 207 Ω5.110. The following table summarizes some of the basic att ributes of a number ofBJTs of different types, operating as amplifiers under various conditions. Providethe missing entries.I will just provide the explicit equations for the first column, (a ):β =α1 − αIE=ICαIB=ICβgm=ICVTre=αgmrπ=βgmTra nsistor a b c d e f gα 1.000 0.99 0.98 1 0.99 0.9 0.984β ∞ 100 50 ∞ 100 9 62.8IC(mA) 1.00 0.99 1.00 1 0.25 4.5 69.1IE(mA) 1.00 1.00 1.02 1 0.253 5 70.2IB(mA) 0 0.0099 0.02 0 0.0025 0.5 1.10gm(mA/V) 40 39.6 40 40 9.9 180 700re(Ω) 25 25 24.5 25 100 5 1.41rπ(Ω) ∞ 2530 1250 ∞ 10.1K 50 89.75.114. A biased B J T operates as a grounded-emitter amplifier between a signalsource, with a source resistance of 10 kΩ, connected to the base and a 10 kΩ2load connected as a collector resistance RC. In the corresponding model, gmis40 mA/V and rπis 2.5 kΩ. Draw the complete amplifier model using the hybrid-π BJT equivalent circuit. Calculate the overall voltage gain vc/vs. What is thevalue of BJT β implied by the values of the model parameters? To what valuemust β be increased to double the overall voltage gain?RpRsRcvsvcvo= −gmRCvi= −gmRCrπRs+ rπvs= −0.04 × 10 × 103×2.510 + 2.5= −80The value of β can be found fromrπ=βgmorβ = rπgm= 2.5 × 103× 0.04 = 100To double the gain we would want to double t he factorrπRs+ rπIt is currently equal to2.510 + 2.5= 0.2To double it we would need to change rπ:rπRs+ rπ= 0.4rπ=0.40.6Rs= 0.67Rs= 6.7 kΩThe factor increase in β is the same as the factor increase in rπ:βnew= βoldrπ newrπold= 100 ×6.72.5=
View Full Document
Unlocking...