Transcript
Hybrid Model
https://in.linkedin.com/in/rakeshmandiyarakesh.yadav1211@gmail.com
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At low frequencies, we analyze transistor using h-parameter. But for high frequency analysis the h-parameter model is not suitable, because :-
(1) The value of h-parameters are not constant at high frequencies.
(2)At high frequency h-parameters becomes very complex in nature.
• Hybrid- model is used for high frequency analysis of the transistor.
• This model gives a reasonable compromise between accuracy and simplicity to do high frequency analysis of the transistor.
• So we can summaries the merits of hybrid- model as; it is simple and accurate.
• The value of all parameters are constant with high frequency and all the resistive components of this model can be obtained from low frequency h-parameters.
In the derivation of trans conductance gm and diffusion capacitance Cde (≈Ce), it is assumed that Vbe changes so slowly with time that in the base region, concentration of minority carrier holes is along straight line from Je to Jc.
Thus the collector current Ie remains equal to the emitter current IE and the base current is extremely small compared with Ic.
VALIDITY OF HYBRID-∏ MODEL
Hence under dynamic condition, the hybrid-π model is valid only when the rate of change of Vbe with time is so small that the base current increment Ib is small in comparison with the collector current increment Ie.
Giacoletto proved that the network elements in the hybrid model are frequency invariant provided that
Since 2πf W2/6 DB<<1 But from previous sections we have Ce and Cde
Ce = gm/2πfT
Cde = W2gm/2DB
Thus we can reduce the previous equatio
f<<6πfT/2π Orf<<3fT
Thus we conclude that the hybrid-π model is valid for
frequency up to about fT/3 .
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This model gives reasonable compromise b/w accuracy and simplicity to do high frewuency analysis of transistor.
Figure shows the hybrid- model for transistor in CE configuration. For this model, All parameters (resistances and capacitances) are assumed to be independent of frequency. But they may vary with the Q point or operating point.
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The parameter gm is called trans conductance (or mutual conductance). The trans conductance represents the small change in collector current about the operating point produced by the small changes in base emitter voltage .
This effect of change in collector current due to small change in base-emitter voltage, accounts for the current generator gm Vb’c at output and it is called as trans conductance. gm=ΔIc/ΔVb’e at a constant Vce.
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Capacitance Cc :-The collector-junction capacitance :-
Cc = Cbe (The output capacitance measured at CB
junction with the input open, i.e. Ie = 0 and is specified by the manufactures as Cob.) Since in the active region the collector,junction is reverse biased, then Cc is a transition capacitance, and hence varies as Vcb where n is 1/2 or 1/3 for an abrupt or gradual junction, respectively.
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Hence the charge stored in the base region (QB) is given by :-
QB =P'(0) *(A*W*q) / 2
Where W = base width
A = cross-sectional area of base AW = volume of base q = electronic charge Pʹ(0) = Injected charge concentration Pʹ(0)/2 = Average concentration of carrier in base region QB = p’(0)*(v*q)/2
Thus , Aq = 2QB/P'(0)*W ………….(1)Current density (J) = -qDp(dP/dx) Where Dp = diffusion constant dp / dx = Change in concentration w.r.t. distance Diffusion current :-
I = J.A = - AqDp (dP'/dx) = +AqDBP'(0)/W (where , DB= diffusion constant for minority carrier)
Thus, Aq = I*W / DB*P'(0) ……………………(2)
Equating both the equations :-
2QB/P'(0)*W = I*W/DB*P(O) QB = I*W2/2DB ………………………....(3) Finally,The static diffusion capacitance Cde=dQB/dV From above equation:- Cde = dI*W2/ 2DB*Dv
= [W2/ re *2DB]Where re = emitter junction incremental resistance and is given by :- re=dV/dl = VT/IE
Thus , Cde=[W2/2DB].[IE/VT] Cde=gm.[W2/2DB] = Ce
This indicates that the Cde is proportional to the IE and inversely proportional with VT. Cde is almost independent of temperature.
Experimentally, Ce is determined from a measurement of the frequency flat which the CE short-circuit current gain drops to unity ; Ce =gm/2πft
The equation for the trans conductance can be derived as follows :Let us consider a P-N-P transistor in the CE Configuration with Vee bias in the collector circuit a shown in Fig.
T1 !P NPR5 1V 1 5TP 5TP 6
rb'b
Vbe VE
B
C
B'
E
VCC
The trans conductance is defined as the ratio of change in the collector current due to small changes in the voltage Vb’e across the emitter junction.We know that, the Collector current in active region is given as For a pnp transistor Ic=Ico- IE
Also , Isc = gm Vb’e =Ic or gm = dIc / dVb’e …….(1)
δIc= αδIE
Substituting values of δIC in eq (1) we get :- gm= αdIE/ dVb’e
The emitter diode resistance is given as :- 1/re = dIE/dV
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For P-N-P transistor Ic is negative and for an N-P-N transistor Ic is positive, but in the foregoing analysis (with VE = +VBE) we get
Hence, for either type of transistor gm is positive.
gm = α (IC -ICO) / VT
= α IC / VT
Substituting value of VT we get,
gm = 11600IC/T
Substituting/ the value of re in we get,
gm= αIE/VT = α ( Ico-Ic )/VT
Equation shows that trans conductance gm is inversely proportional to temperature. At room temperature, 300k gm=11600Ic/300
= Ic/26*10-3
=Ic /26This is for, Ic=1.3mA, gm=0.05m mho while for, Ic=2.6mA, gm=0.1m mhoThese values are much larger than the trans conductance obtained with FET’s
As discussed earlier, the high frequency analysis takes into account the capacitive effect of PN Junction in transistor.
We can not assume at high frequency that transistor responds instantly to changes of input voltage or current because the mechanism of the transport of charge carrier from emitter to collector is essentially one of diffusion.
The increase of hfe with temperature has been determined experimentally, where as the increase with |VCE| is due to the decrease of the base width and the reduction in recombination which increase the transistor ‘Alpha’.
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For calculating the resistance we are conidering the Hybrid ∏-model in terms of low frequency H parameter aiso neglacting the capacitance effect.
First consider the hybrid- model for CE Configuration:-
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We know that hie is the input resistance (in h-parameter model) when the output terminals i.e. collector terminals are shorted. Under this condition in Hybrid - model, rb'c comes in parallel with rb'e. But since rb'c »rb'e
rb’e || rb’c ≈ rb’e
:. With hybrid- model input resistance with output shorted is rbb' + rb'e.
hie = rbb’ + rb’e
rbb’ = hie – rb’e
In h-parameter model, hre is defined as reverse voltage gain: hre = Vb’e/Vce = rb’e /(rb’e + rb’c)
By applying voltage divider formula hre rb’e + hre.rb’c = rb’e
rb’e(1-hre) = hre.rb’c
Since ; hre <<1 rb’e = hre.rb’c
rb’c = rb’e / hre
Feedback Conductancegb’c = hre*gb’e
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Ic= Vce / rce + gm. Vb’e + Vce /(rb’c + rb’e )Vb’e = hre. Vce
Ic= Vce / rce + gm.hre. Vce + Vce /(rb’c + rb’e )gb’c = hre.hfe
gm = gb’e.hfe
hoe = gce + gb’c.hfe .gb’c/gb’e +gb’c
hoe = gce + gb’c(1+ hfe)hfe>>1gce = hoe – gb’c.hfe
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Looking at the above fig.we can write IL= -gmVb’e And Ii= Vb’e [ gb’e + jw (Cc+ Ce) ] Current gain under short circuit condition is AI = IL/ Ii
, = -gmVb’e / Vb’e [gb’e + jw (Cc+Ce) ] AI = -hfe/[1 + j2f(Cc+Ce)/ gb’e] ………..Eq.(1) AI =- hfe/[1 + jf/fβ]
Where fβ =gb’e/ 2Π(Cc+Ce) As we know gb’e= gm/hfe
where fβ = gm/ hfe 2Π(Cc+Ce) …………….. Eq.(2)
β- Cut-Off Frequency(fβ)
fβ is defined as the frequency at which CE short circuit current gain falls (1/ 2)1/2 (or 0.707 or falls by 3 dB ) of its low frequency value i.e. hfe.
So te value of hfe can be neglacted
because hfe >>1
The value of fβ can be
found out using expressionfβ= gm/2(Cc+Ce)
The frequency range up to fβ forms the 3 dB band width or simply the band width of the circuit.The Parameter(fT) :
Frequency fT is defined as the frequency at which CE Short circuit current gain becomes unity.
at f= fT ,
AI becomes : AI = 1= hfe/[1 +( fT/fβ)2 ] … eq.(3) The ratio fT/ fβ, is quite large as compared to (fT/ fβ>>1)) hence Equation (1) becomes 1= hfe/ fT/fβ substituting this value of
fβ= gm/ hfe2f(Cc+Ce) therefore
fT = gm/2f(Cc+Ce) …. Eq.(4)
since Ce>>Cc we can write So we can neglacted the value of Cc
fT = gm/ 2f Ce ……. Eq.(5)
fT as CE short circuit current gain bandwidth product:- We know that,
fT = hfe*fβ
That is fT is the product of low frequency gain (hfe) and 3dB bandwidth of short circuited CE stage (fβ) . Hence fT represents the CE short circuit current gain-bandwidth product.
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α-Cut-Off Frequency, fα We have already seen β-cut-off frequency for transistor operating in CE configuration. a cut off frequency is a similar frequency for transistor operating in CB configuration.fα can be defined as the frequency at which the transistor's short circuit CB current gain drops 3 dB or (1/2) times from its value at low frequency hfb. The expression for fα is given as:
fα = hfe*gb'e / 2Π*Ce ………….. Eq.(6)
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As K = Vce/Vb’e
= ILRL/ Vb’e
= (-gmVb’e)RL/Vb’e
= -gmRL
So ; 1-K=1+ gmRL
Thus, AI = IL/ Ii
AI = -gmVb’e / Vb’e [gb’e + jwCeq ]
As we know gb’e= gm/hfe
Thus , Ai = -gb’ehfe/[gb’e+ jwCeq ] Ai = -hfe/[1 + j2Πf(Ceq)/ gb’e] ………..Eq.(1)
Ai= - hfe/[1 + jf/fH]
fH= gb’e/ 2 Πceq
=1/ 2Πrb’e[Ce+Cc(1 +gmRL)]
The simplified equivalent circuit is at RL= 0 :-
fH=1/ 2Πrb’e(Ce+Cc)= f
So the current gain = Ai = hfe / [ 1+(f /fh)2]1/2
= hfe / (2)1/2,
= 0.707 hfe
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Rs’=Rs+rbb’=1/Gs’By applying KCL at B’ & C, we get : -Gs’Vs=[Gs’+gb’e+s(Ce+Cc)]Vb’e-sCcVo..(1)
0=(gm-sCc)Vb’e+(1/RL+sCc)Vo……(2)
We know that the voltage gain is
By solving the equation 2 and find the value of And put I in equation 1
Avs=Vo/Vs
Vb’e
Vb’e = (1/RL+sCc)Vo /(sCc-gm)
Vo/Vs= -Gs’RL(gm-sCc)/ s2CeCcRL +s[Ce+Cc+CcRL
(gm+gb’e+Gs’)]+Gs’+gb’e……eqn.(3)
By solving the quadratic equation We get Avs=Vo/Vs=K1(s-s0)/[(s-s1)(s-s2)…eqn.(4) Where K1=Gs’/Ce, s0=gm/Cc
s1=Complex value
s2 =Complex value
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fH =1/ 2ΠReqCeq
Req =(Rs+rbb’)||rb’e
Req=Rs rbb’rb’e/Rs+rbb’+rb’e
hie=rbb’+rb’e
Put the value of Req=(Rs+rbb’)rb’e/(Rs+hie)
fH = (Rs+hie)/2ΠC(Rs+rbb’)rb’e
Avs= Vo/Vs
=-gmVb’eRL/Is(Rs+rbb’+rb’e)
Avs =-gmVb’eRL/Is(Rs+hie)
Vb’e=Ib’erb’e
Avs =-gm(Ib’erb’e)RL/Is(Rs+hie)
Ib=Is
Avs =-gmrb’eRL/Rs+hie
since fH = (Rs+hie)/2ΠC(Rs+rbb’)rb’e
therefore|Avsfh|
=-gmrb’eRL/Rs+hie(Rs+hie)/2ΠC(Rs+rbb’)rb’e
|Avsfh|=-gmRL/2ΠC(Rs+rbb’)
|Avsfh|=gmRL/2ΠC(Rs+rbb’)
|Avsfh|=gmRL/2ΠCeq(Rs+rbb’) Ceq=Ce+Cc(1+gmRL)
|Avsfh|=gmRL/2ΠCe+Cc(1+gmRL) (Rs+rbb’)
divide both numerator & denominator by 2Π (Ce+Cc)|Avsfh|=gmRL/2Π(Ce+Cc)/[2Π[Ce+Cc(1+gmRL) ](Rs+rbb’)]/2Π(Ce+Cc)
|Avsfh|=gmRL/2Π(Ce+Cc)/[(1+CcgmRL) /(Ce+Cc) ](Rs+rbb’)
fT = gm/2Π(Ce+Cc)
|Avsfh|= fT RL/(1+ 2ΠfTCcRL)(Rs+rbb’)
A+
Is
Rs
1k
rbb' 1k
R1
1k
C1
1u
gmVb'e
RL
1k
rb'e
1k
(b) Applying current dividing rule we getVs/(Rs+rbb’+rb’e) =Vb’e/rb’e
IsRsrb’e=(Rs+rbb’+rb’e)Vb’e
Is=(Rs+hie)Vb’e/Rsrb’e
AIS=-gmVb’e/Vb’e (Rs+hie/Rsrb’e)
=-gmrb’eRs/Rs+hie
fH=(Rs+hie)/2ΠC(Rs+rbb’)rb’e
fT=gm/2Π(Ce+Cc)
|AIsfh|= gmrb’eRs/Rs+hieX(Rs+hie)/2ΠC(Rs+rbb’)rb’e
|AIsfh|=(-gm/2ΠC)X(Rs/Rs+rb’e)
C=Ce+Cc(1+gmRL)
|AIsfh|= -gmRs/2Π(Ce+Cc)
|AIsfh|=gmRs/2Π(Ce+Cc)/(2ΠCe+Cc(1+gmRL))(Rs+rbb’)/2Π(Ce+Cc) fT=gm/2Π(Ce+Cc)|AIsfh|=fTRs /(1+2ΠfTCcRL)(Rs+rbb’)
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We know that low freq. gain of emitter follower is almost unity. Therefore A=1;1-K=0;K-1/K=0, thus we can ignore Ce(1-K),gb’e(K-1/K)and Ce(K-1/K). The simplified ckt is: -
Voltage gain A = Vo/Vi = Ve/Vi….............(1)
Ve=IZ
=I/(1/RL+jwCL)
=gmVb'e/((1+jwCLRL)/RL)
=gmRLVb'e/(1+jwCLRL)
=gmRL(Vi-Ve)/(1+jwCLRL)Ve = gmRLVi/(1+jwCLRL)– gmRLVe /(1+jwCLRL ) Ve+VegmRL/(1+jwCLRL) =gmRLVi/(1+jwCLRL) (Ve[(1+jwCLRL +gmRL)]) / (1 +jwCLRL)=gmRLVi/(1+jwCLRL)
Therefore Av=Vo/Vi gmRL
1+gmRL+jwCLRL
Upper cut-off frequency fH Av = gmRL/(1+gmRL)(1/1+[jwCLRL/(1+gmRL)])
gmRL
1+gmRL(1+jf/fH)
Thus , Av =AVL/(1+jf/fH)
Where , AVL=gmRL/1+gmRL<1
fH=1+gmRL/2ΠCLRL
If f = fH ThenAv = gm RL
√2 (1+gm RL)Now, since the input impedance between terminals B' and C is very large in comparison with (rb'b + Rs). Hence A, also represents the over all voltage gain Avs:-
Avs = Ve / Vi
Avs= Ao/1+jf/fH
Numerical
Ques1. The following low frequency parameters are known for given transistor at Ic =10mA,Vce =10Vand at room temperature, hie=500Ω,hoe=4*10-5 A/V , hfe=100,hre=10-4 .At same operating point ft = 50Mhz,Cob=3PF.Compute values of all hybrid parameters.
Ans Given:- Ic =10mA,Vce =10V,hie=500Ω, hoe=4*10-5 A/V , hfe=100, hre=10-4 ft =50Mhz,Cob=3PF
To Find:- All values of hybrid parametersFormula Used:-VT=T/11600,gm =Ic /VT ,rb’e=hfe /gm, rbb’=hie-rb’e , rb’c = rb’e /hre, gce= hoe-(1+ hfe)gb’c ,rce =1/ gce ,Ce = gm /2ΠfT
gm=Ic /VT
=10x10-3 /0.026 =0.3846A/V 2 rb’e=hfe /gm,
=100/0.3846 =260Ω
3 rb’c = rb’e /hre
=260/10-4
= 260x10+4 Ω
5 gce= hoe-(1+ hfe)gb’c
=4x10-5 –(1+100)/260x104
=0.12x10-5 A/V
6 rce=8.33x105 Ω
7 Ce=gm /2ΠfT
=0.3846/2x3.14x30x106
8 Ce=2.04PF
4 rbb’=hie-rb’e
=500-260 =240 Ω
Solution:- 1 gm =Ic /VT VT=T/11600 =2.98/11600 VT =0.026V
Ques2.Given following transistor measurements made at Ic=5mA,Vce =10V & at room temperature hfe =100 Ω, hie=600Ω,Aie =10 at 10 Mhz, Cc =3PF.Find fß , ft ,Ce ,r b’e & r
b’b.
Ans Given :-
Ic=5mA,Vce =10V, hfe =100 Ω, hie=600Ω,Aie =10 at 10 Mhz, Cc =3PF
To Find:- fß , ft ,Ce ,r b’e & r b’b.
Solution:- 1 ft=Aie xf =10x10 =100Mhz 2 fß= ft / hfe
= 100/100 = 1Mhz 3 gm=Ic/VT
=5/26 =0.0192A/V
4 rb’e = hfe/gm
=100/0.192
=520Ω 5 rb’b = hie - rb’e
=100-520
=-420 Ω
6 Ce=gm /2Π fT
=0.192/2X3.14X100X106
=30.5PF
Ques3.The hybrid parameters of a transistor used in circuit as shown in fig. gm=50mA/V, rb’e=1KΩ, rb’c=4MΩ, rb’c=80KΩ, Cc=3PF, Ce=100PF, rbb’=100Ω. +Vcc
Find:- (1)Upper 3dB frequency of current gain. (2) the magnitude of voltage gain at Avs=Vo/Vs at frequency of part1.
Given:- gm=50mA/V, rb’e=1KΩ, rb’c=4MΩ, rce=80KΩ, Cc=3PF, Ce=100PF, rbb’=100Ω.
Solution:-
First draw its h-equivalent diagram
(1)fh=1/2ΠReq Ceq
Req=(Rs+rbb’)||rb’e
=(900+100)||1000
=500Ω Ceq=Ce+Cc(1+gmRL) =100X10-12 +(1+50X10-3X1X103)3X10-12
=253PF
fh=1/2ΠReq Ceq
=1/2X3.14X500X253X10-12
fh=1.26Mhz (2) Voltage Gain Avs=-gmRL
=-50X103X10-3 Avs =-50dB
Ques4 . Consider a single stage CE transistor amplifier with a load resistance shunted by load capacitance. Prove that (a)internal voltage gain K=Vce/Vb’e = gmRL/1+jw(Cc+CL)RL
(b) 3dB frequency is given as FH =1/2Π(Cc+CL)RL
Ans:-
Proof:- (a) K=Vce/Vb’e
Vce=I/G
I=gm Vb’e
G=1/RL+jw(CL+CC)
=>Vce= -gm Vb’e/ 1/RL+jw(CL+CC)
= -gm Vb’eRL /1+jw(CL+CC)
K=Vce/Vb’e
=>K= -gm Vb’eRL /(1+jw(CL+CC)) Vb’e
K= -gm RL /1+jw(CL+CC)
Hence Prooved
(b) C=Ce+Cc(1-K) K= -gm RL /1+jw(CL+CC) =>C=Ce+Cc(1+gm RL /1+jw(CL+CC)) K= -gm RL /1+j(w/fh) w=1/ RL(CL+CC) fh=w/2Π fh=1/2Π RL(CL+CC) Hence Prooved
Ques5.(a) From circut diagram assume |K|>>1 Prove that K= -gm Rc +jwCC RL/1+jwRLCC Why may be the term jwCC RL be neglected in the numerator but not in the denominator? (b) the Miller’s admittance in the output circuit is given byYo=jwCc(1-1/K)Prove that this represents a capacitance Co in parallel with a resistance Ro given by Co= Cc (1+gm RL)/gm RL, Ro= -gm /w2Cc
circuit diagram
Vce=I/G
I=gm Vb’e
G=1/RL+jwCC (K-1/K)
=>Vce=-gm Vb’e/ 1/RL+jwCC (K-1/K)
Vce = KVb’e (since K= Vce/Vb’e )
=>K= -gmVb’e / 1/RL+jwCC (K-1/K) Vb’e
-gm =K1/RL+jwCC (K-1/K)
-gm =K/RL+jwCC (K-1)
-gm =K(1/RL+jwCC)-jwCC
K=(-gm+jwCC)/(1/RL+jwCC)
K=(-gmRL+jwCCRL)/(1+jwCCRL)
Hence prooved
w=gm/Cc
=50X10-3/3X10-12
=16.7X108 radians wT=gm/Ce
= 50X10-3/100X10-12
=0.5X108 radiansgmRL= 50X10-3 X2X103
=100in numerator(gmRL+jwCCRL)
jwCCRL=j(16.7X108 X3X10-12X2X103)
=10.02 which is <<100 & can be neglected But In denominator(1+jwCCRL)
The value of jwCCRL(10.02)>>1 hence it can’t be
Neglected.(b) Yo=jwCc(1-1/K) K=(-gmRL)/(1+jwCCRL)(since jwCCRL is neglected in num.) Yo=jwCc(1-1+jwCCRL/-gmRL) =jwCc(-gmRL-1-jwCCRL/-gmRL) =jwCc(1+gmRL/gmRL)+(jwCCRL)(jwCC)/(gmRL) Yo =jwCc (1+gmRL/gmRL)-w2Cc
2/gm - (1) General Equation Yo=jwC0-1/R0
By comparing equation (1) with gen eqnC0= Ce(1+gmRL/gmRL), R0 =-gm/w2Cc
2 H.P
Ques6. Verify(a)|Avsfh|=gm /2ΠCRL /Rs +rbb’ =(fT/1+ 2ΠfTCcRL)( RL/Rs +rbb’) (b)|AIsfh|=(fT/1+ 2ΠfTCcRL)( RS/Rs +rbb’)
PROOF:- (a) fH =1/ 2ΠReqCeq
Req =(Rs+rbb’)||rb’e
Req=Rs rbb’rb’e/Rs+rbb’+rb’e
hie=rbb’+rb’e
Put the value of Req=(Rs+rbb’)rb’e/(Rs+hie)
fH = (Rs+hie)/2ΠC(Rs+rbb’)rb’e
Avs= Vo/Vs =-gmVb’eRL/Is(Rs+rbb’+rb’e)
Avs =-gmVb’eRL/Is(Rs+hie) Vb’e=Ib’erb’e
Avs =-gm(Ib’erb’e)RL/Is(Rs+hie) Ib=Is
Avs =-gmrb’eRL/Rs+hie
since fH = (Rs+hie)/2ΠC(Rs+rbb’)rb’e
therefore|Avsfh|=-gmrb’eRL/Rs+hie(Rs+hie)/2ΠC(Rs+rbb’)rb’e |Avsfh|=-gmRL/2ΠC(Rs+rbb’)|Avsfh|=gmRL/2ΠC(Rs+rbb’)Hence Prooved
|Avsfh|=gmRL/2ΠCeq(Rs+rbb’) Ceq=Ce+Cc(1+gmRL) |Avsfh|=gmRL/2ΠCe+Cc(1+gmRL) (Rs+rbb’)
divide both numerator & denominator by 2Π (Ce+Cc)|Avsfh|=gmRL/2Π(Ce+Cc)/[2Π[Ce+Cc(1+gmRL) ](Rs+rbb’)]/2Π(Ce+Cc)
|Avsfh|=gmRL/2Π(Ce+Cc)/[(1+CcgmRL) /(Ce+Cc) ](Rs+rbb’)
fT = gm/2Π(Ce+Cc)
|Avsfh|= fT RL/(1+ 2ΠfTCcRL)(Rs+rbb’)Hence Prooved
(b) Applying current dividing rule we getVs/(Rs+rbb’+rb’e) =Vb’e/rb’e
IsRsrb’e=(Rs+rbb’+rb’e)Vb’e
Is=(Rs+hie)Vb’e/Rsrb’e
AIS=-gmVb’e/Vb’e (Rs+hie/Rsrb’e)
=-gmrb’eRs/Rs+hie
fH=(Rs+hie)/2ΠC(Rs+rbb’)rb’e
fT=gm/2Π(Ce+Cc)
|AIsfh|= gmrb’eRs/Rs+hieX(Rs+hie)/2ΠC(Rs+rbb’)rb’e
|AIsfh|=(-gm/2ΠC)X(Rs/Rs+rb’e)
C=Ce+Cc(1+gmRL)
|AIsfh|= -gmRs/2Π(Ce+Cc)|AIsfh|=gmRs/2Π(Ce+Cc)/(2ΠCe+Cc(1+gmRL))(Rs+rbb’)/2Π(Ce+Cc) fT=gm/2Π(Ce+Cc)|AIsfh|=fTRs /(1+2ΠfTCcRL)(Rs+rbb’) Hence prooved
Ques7.Consider hybrid –Π circuit at low frequency so that Ce &Cc may be neglected .omit none of other elements in circuit. If RL=1/gL. Prove that(a) K=Vce/Vb’e =-gm+gb’c/ gce+gb’c +gL
(b)Using Miller’s theorm draw the equivalent circuit between C& E. Applying KCL to this network , show that above value of K is Obtained.(c) Using Miller’s theorm draw the equivalent circuit between B& E.prove that AI under RL is
AI=gL/ (gb’e+gb’c /K)-gb’c
(d)Using results of part (a) & (c) & relationship between hybrid h&Π parameters prove that AI=-hfe/(1+hoeRL)
Proof:-(a)K=Vce/Vb’e
Vce=IscZce
IL=Isc=Vb’egb’c -gmVb’e =Vb’e(gb’c -gm)
Zce=(rb’c||rce||RL)(Z seen between C&E) =1/ gb’c+gce+gL
Vce= Vb’e(gb’c -gm)/(gb’c+gce+gL) K=Vce/Vb’e
=>K=(gb’c -gm)/(gb’c+gce+gL)Hence proved
(b)Applying Miller’s theorm b/w C&E we getVce= -gmVb’e/gce+gL+(K-1/K)gb’c Vce= -gmVb’e K/K(gce+gL+gb’c)- gb’cVce/Vb’e= (-gmK)/K(gce+gL+gb’c)- gb’c K = (-gmK)/K(gce+gL+gb’c)- gb’c
10/21/2015
K = (-gmK)/K(gce+gL+gb’c)- gb’c
Taking common K both sides we get K(gce+gL+gb’c)- gb’c =-gm
K(gce+gL+gb’c)=-gm + gb’c
K= (-gm + gb’c) /(gce+gL+gb’c)
Hence proved(c) Applying Miller’s theorm b/w B &E we getAI=IL/Ii=gmVb’e/Ii
rbb' 1k
rb'e
1k
(rb'
c/1-
K)
1k
B
E
IL=VcegL
Vce=Vb’eK (since K=Vce/Vb’e)
=>IL=gLVb’eK
Also IL=K gLIi/(gb’e+(1-K)gb’c) IL/Ii=K gL/(gb’e+gb’c-Kgb’c)Dividing by K throughout we get IL/Ii =gL/(gb’e+gb’c /K)-gb’cHence proved.(d)As gm>>gb’c & gb’e>>gb’c
Thus K=-gm/(gce+gL+gb’c) (1)
AI=gLK/gb’e-Kgb’c (2)
from (1)& (2)
AI=gmgL/(gce+gL+gb’c)/gb’e+gb’c gm/(gce+gL+gb’c)
AI=-gmgL/[gb’e (gce+gL+gb’c) +gb’c gm] dividing both num.&denom. by gb’e gL we getAI=-gm/gb’e/[(gce+gL+gb’c)/gL]+[gb’c gm /gL]hfe=-gm/gb’e
gL= gm gb’c
AI=-hfe/1+(gce+gb’c+gmgb’c)/gL
AI = -hfe/(1+RLhoe )Hence proved
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