Question

(b) Refer to Figure Q5.2, given VBE = 0.6V, ro infinit re V.2, given VBE = 0.6V, ro infinity and thermal voltage of 26 mV, draw the ac equivalent circuit and calculate: (i) Input impedance, Zin (ii) Output impedance, Zout (iii) Gain, Av. (13 Marks) (PO3/C03/C4) Ic(mA) - lg = 400 UA RB w taylo = 300 UA 30++ - +/Vcc is --V-18 = 200 A Vin Bpc = 100 VBB 3.7 V Figure Q5.2

Answer 1

Approach:

From the given circuit and Ic-Vce Characteristics

Calculate:

• Vcc
• Ic and Vce at Q point
• RB
• Rc

Then calculate gm, rpi

Finally Draw AC Model and calculate the required quantities...

Note that since Ic-Vce graph is given hence Rb and Rc values are not given

Solulion Grireni • VBE =0.6 O pobrt From the Ic vs VCE graph at o Ico: J 30m Vceo= 3-4 Volb B= 160 30mt 26 mv = 86. 672 154 Gm the DC eguivalnt kt: • Ds laving Yas 1or the impue loop gives KVL in Figliy RB Ž RC VAB- B RB - VBEON=0 we hare 8= Ie_30yA YBB 30MA RA - 0.6=0 3.7 - 3.7-0-6 = 10-33 K-2 or RBa 30X1O RB=10-33 K-2

Also KYL in the olp loop gives Yec- Ico Rc-VCEO=0 > 10- 30x103 RC- 3+4 =0 RC= 10-3.tf 30X 103 = 220 s2 Rc= 2202 Thas we have all the Values reguired fr AC Aralipi RB- 10-33 IC 86-6752 Rc= 220-2 wenent draw the AC Feguivalent cep: DC Seur ces redaced to teso- Vin (86-672 Zout () Kine Vin - Vin Tin RBt TR= 10-417 K2

(ii) To CRlialate Lout det Vin eo Vin-o > VT=0 GnVe = oper cet' The Result ant Ckt s KB Zoat = Rr= 2202 (ii) from Figz- Vo= - Gm Vg RC (: Vettage Dinicion) Now VE= Vin · TE A Vo helomes Vo= - Jm · Vim TE . RC or Vin I54 X 86-67 X 220 -- 10417

konswers : O žin = 10-417K2 6 Zaut = 220 2 3 tye -2- //