Question

Q1(a) [5 Marks The diagram of Fig Ql(a) below describes how a Bipolar Junction Transistor (BJT) may be represented as a two-port network. With the aid of this diagram, derive expressions for the small signal parameters: hie, hoe, hre and hfe. 01- Port 1 V1 Two-Port Network V2 Port 2 Fig Ql(a)

Answer 1

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Two-port Network The figure shows how a Bipolar Junction Transistor (BJT) may be represented as a two port network. we shall now represent the two-port networks with the help of h-parameters. From the knowledge of two port networks, we can express the variables V & Iz as a function of & & I with the help of a parameters. vo f (,,V2) I= f(I, 12) o vce A BJT in CE (common emitter) mode can be modeled by a two port network with m. vbe, 1,5%. V. Vce, Tric are small wantities. where V1 CS Scanned with CamScannee' berce

As discussed in the two-port network context, the hybrid on h- parameter model to describe the CE mode BJT can be given by neplnole) i = f(in veel Taking total derivative, we get & die dinge din toate dove on, droge = hie din + those da ce & die a hipe dig + hoe drece where hier koper haper hoe are bybrid parameters and are defined as a Input Ac impedance " hre- + Reverse transfer voltage ratio Once 0 & Porrw. , he O Forward transfer current gain m) het onder ► Output admittance " ou put odmstone hoe celi, 20 CS Scanned with CamScanner

If the variations in around the operating can be linearised as re be rece, i point, the į are small small signal model We a have b threuce ů = lipe y + hoelce CScanned with CamScanner

B hie in Given duta. hielke here - 10-4 life = 100 hoe = 12ms R = 2K2. 1) Ac current gain from the hybrid model, we can write - e be - vsa hie ist here ce o é = hçe ist hoe vce ----- CS$canned with me inscanner Ace - VL. LR. & iLav'c.

from equation ③ we have ů a lepe it hoe uce >> -ine uge io + hse il ru >in(19 hse Pl) (An je It hoe R. pulting the values, we get, 100 Ai - - 1 + (12x16 6x2*103) . A: 5-97.66 A: = -97.66 (Answers (Answer) . CSscanned with CamScanner

1) AC voltage gain:- from the equivalent circuit, we can write, - vsa hie it here uce --- - iez lige in + hoe vce. ----- Also, vee e Y and i į From equation , we get a insi, unge in + hoe die RL => in. - (1* We Re) in hee R IL - - 1 + hoe - lepe R AIR life where A - - Current gaun - It hoe ₂ (derived previc putting the value of ef in equation , we get e AIR CS Scanned with CamScanner

E t Voltage gain - Avevo heret hie AIRL AIRL As hore R + hie (Answer) putting the values, we get, - 97.66 2000 A = (-97.66*10-4, 2000)+ 1000 A: - 199.21 answers 1.) Input impedance :- we have, - vsa hie dot have ce i hage t hoe vce CSScanned with ca Also Vee av dizain o ---

from equation ③, wo get, į » » į z hoe in those y ia lepe if + hoe ik The one to : from equation o, we have, Vsa hie if + hore a hire is there į R : Wie is - kirpe hoge Rei (1 + hoe R.) . (we - bote - hoe CSscanned with CamScanner

Input impedance Zin - I hie - | here life RL (1 + use Re] (Answe pulting the values, we get Zin = 1000 - 10-4 x 100, 2000 1 + (12010> 2000) Zin 2 980.47 (Answers It is seen that the value of Zin is less than the value of hie. This happens due to loading effect . . .. Zin <hie. CSScanned with CamScanner