Question

Figure 1 shows an amplifier application using a JFET. Given that Vasq=-2.48 V. IDQ = 2.07 mA and you = 0.1 ms. 1) Determine the voltage gain at mid-range frequency, Awwid). ii) Determine fio, fic and fts for the network iii) Determine which value contributes to the low frequency response for the network. iv) Plot the asymptotes for low frequency response of the network. Use the semi-log graph paper given in APPENDIX II for your answer. v) Plot the actual low frequency response for the network. Use the same semi-log graph given in APPENDIX II. [20 Marks] [CO2, POI, C3] Voo = +18 V w Ro = 2.7 k 2 Cwi = 3 PF Cgd = 4 pF Cwo = 5 pF Cg. = 6 PF Cds - 1 pF Co Vo HE Сс 4.7 F loss = 6 mA Vp = -6 V 1 HF Rig 1 ΚΩ Ş RL 3.9 kg RG 10 M2 RS 1.2 ΚΩ Cs 10 PF w + TH

Answer 1

Given Vasa = -2.48V Isa : 2.07 MA Yos = Oolms ras 10 Ool ds das 10 kn Iogs: bmA Vp :- 6V » go: DID Trans conductances oras १ Je a loss vas' gm- ID = Ipss [I-Vas 010 1%) 2x6x103 6 ovas Up Im [ Im = 9x10-9 (0-586673 Im = 101733 may

Small signa? Ac -equivalent (capacitors are shout circuited] juriction capacitors are Open circuited a D Rsig + Vout Vgg guys RO كی S ने Pisig: 1x Im=101733 mA/ Ro- 2.7k dg 10kn R = 3.9kn Ro=10MM → Shill Roll R = R' = [10|19.4 || 3.1]1103 [1011 1.595u] Xo3 Re's 1037588 KN > Vgg [Rat Pisig Vs Vgg = Vgx Ra Rat Roig Ra

Yout= -9 m Ygs R Yout: ? k 9 ago Yout wy L-I mugs » Av mid Vout -Inilys Rex Ra - Vg Almid -Im R Ra Rat Risig Almid - 101733 X10 X 1037588 1 X 10 % 8x183 x 10x106 [loxldº + 13/X103 [Ra>> Rsig Avoid : -101733 x 1037588 Rathsign Ra Aumid :-1.61432 %

Low frequency ac- equivalent Imgs Roig www cu w Vys ککم Ro R Rg Cg Ca=luf Cg = 1045 CD: 4.7 uF frequency Woret ca (other capacitors are sol Ca Req - Resistance seen by ca Rsig wus Rsigt Ra 1 fue 91 Regla 2XX [R sigt Bella fua Pax [iovoºttoJx 141076

- fra 9XX 1000|x10x106 10 fra 21x10001 [fua = 001591 H3 7 Wedot Co [la and Cg and shoot circuíted] Co fes R QT Req Co Req Pollenst Re Req? nox 103/19:410° + 3.9 km 201259 kn + 3.9 kr [Req = 6:026 kr

feo 2KX 60026X10 X 4.7x10-6 103 foo QXx6.026x4.7 Ifo = 506190 HZ Worot Cg [ca and Co are shorted disconnected and supply given at Rg source is to find Req. X Rgig ods 20 vgs Ra Follo] Ix [Ex toy my }Rg Nx

→ at a KCL 1x + 9 mVgg I گے۔ at kel Ix 79 migs - Imgs I P = Ix x = Vgs Vgs + V x = 0 ... KUL Output equation. of -0 - x + das [ +9mx95) + Ex CrollRe] = Y = Odg I x + " do I'm [ugs] + 1 x [Roll Re] y = 1x [725 + Rolle] - Pas en Vi ده

Na [1+ 723 903 · 14 [75+ [role]] Ag+ [Roll Re Reg. He Ix Itras Im és T Resistance seen Reas by Cg Reall Rg Rage as This Rot It Rs zea Req Rg = Req -7 It Rs {to say rast Il R] Reqi 19x103 It lo8x10.3 leg

3 Regi odgt [Roll Re] 10x10 + [2111 397 118 200 Itras Im 3 It 10x10x 1-1733x103 10% (10 +1.59su] Req 11.5954x103 07 120733 1 +11.733 Rea 910.657 a 1200 Req 1 +1-3177 1200 It lo2x103 910-657 Req= 517.7481 fesa 2x 517 748 x 10x106 2T Req X Cg 105 22x517.748 fes = 30.7398 HZ

→ The frequency which 4 to 6 times greater than the other frequencies will be the lower Cut-off frequency fuco & fuca » fuero 30.7398 15. lower cut-off frequency:

(lv) low frequency response [Ayl indb odb/dec 20 las['10bd0323 29.5182 db rodb/dec roodby/dec fly fra fi > flg = 5.619443 fu, = 0.0159) H3 flg = 30.7398 H3 than less ful for frequencies fik 0.01591 H3 Slope of the plot is - 20 dblodec

fu= fra= 0.01591 H3 for frequencies lies between R 5.6194 0.01591 <f is 20db dec the plot is slope of lies between for frequencies Soblau a fa 30.7398 H2 -no doldec slope of the plot is no

(V) low response Woret dominant frequency frequency loblu32 | Arla fra 30.73943 1+ indb laul 9.578206 L 6.1728 db 30.739 HZ