Rigorous Analysis of Shunt-Shunt
Feedback Circuit
• Nodal Analysis
• One Assumption… |g m |>>|g f|…( see below )
• Results on the final pages:
– Relationship to match Rs
– Gain expression (both as voltage gain and
trans-resistance with units “V/I”)
– Manipulations that put results into “classical”
form (ala step-by-step done in G ray & M eyer)
RF
RS
vs
vo
R in
R eq
+
vπ
-
gm v π
RL
Nodal − Equations :
(Vs − Vπ )gs − Vπ geq. − (Vπ − Vo )g f
(Vπ − Vo )g f − gmVπ − Vo gL = 0
rearranged :
Vπ (gs + geq. + g f ) − Vo g f = Vsgs
Vπ (gm − g f ) + Vo (g f + gL ) = 0
=0
Vo
is :
Vπ
Solving _ for _Vπ = f (Vs ) :
Vo  − gm + g f   −gm 

≈
=
Vπ  g f + g L   g f + gL 
 Vo 
use _  _in _ first _ equation :
 Vπ 
assu min g gm ⟩⟩ g f
 gm 
 Vπ = gsVs
Vπ (gs + geq. + g f ) + g f 
 gf + gL 
Vπ =
gs
 gf 
gs + geq. + g f + gm 

 g f + gL 
Vs
and...

 gf  
 geq. + g f + gm  g + g  
 f
L 
Vs
Vs − Vπ = 

 gf 
 gs + geq. + g f + gm 

 g f + gL  

solving _ for _ the _ input_ conductan ce :
iin = gs (Vs − Vπ )_ and...
gin =
Assumption
(I.e. that we want
“match” with
R s)
iin
Vs
which _in _ turn_ for _ a _matched _ condition :
g
gin = s =
2

 gf  
 geq. + g f + gm  g + g  
 f
L
= gs ⋅ 
 gf  
 gs + geq. + g f + gm 

 g f + gL  

therefore :

 gf 
 g f 
 = 2 ⋅  geq. + g f + gm 

gs + geq. + g f + gm 
 g f + gL 
 g f + gL  

which _ gives :
 gf 

gs = geq. + g f + gm 
 g f + gL 
solving _ for _
Vo
:
Vs




−
Vo  gm

=
Vs  g f + g L 
 gs + geq. + g f

Note: this equation is a
specific result of having
the Rin of amplifier with
feedback exactly equal
Rs (I.e. a matched load
condition)


gs
 gf  
+ gm 

 g f + gL  
Design Example
which _can _ be _ rewritten_ as :
− gm ⋅ gs
=
(g f + gL )(gs + geq. + g f )+ gm g f
this _ is_ exactly_ equivalent _ to _G & M (Eqn.8.56)
Coming Soon!
Vo
− gm ⋅ gs
=
Vs (g f + gL )(gs + geq. + g f ) + gm g f
Looking at equation once more, putting it
into the units which are consistent w ith
feedback theory (I.e. shunt-shunt… see
EE113 notes!)
which _can _ be _ rewritten_ in _"s tan dard"_ feedback _ form :
− gm
Actually, this is the
g f + gL )(gs + geq. + g f )
Vo
Vo
(
=
=
exact equivalent of the
Vs
gm g f
Is 1
+
Rs
G&M Eqn.8.56
(g f + gL )(gs + geq. + g f )
a
1+ | af |
where :
=
a=
− gm
(g f + gL )(gs + geq. + g f )
and
f = − gf
Which can be written in the
classical feedback equation
form...
These are the “a” and “f”
terms with natural units of:
V/I and I/V respectively
solving _the _ quadratic_ for _ g f :
gf =
− (gm + gL + geq − gs )±
(g
+ gL + geq − gs ) + 4(gs − geq )gL
2
m
2
and _the _ correct _(physical)_ solution_ is :
gf ≈
(g
(g
s
m
− geq )g L
+ gL + geq − gs )
Typical numbers based on Ic=5m A :
g m =0.193 (mho)
g eq =gp=0.0024 (mho)
and assuming…
g L =0.0005 (mho) ->R L =2K Ω
g s =0.05 (mho)
->R s =50 Ω
The above equations give:
g f =5.028 x 10^(-5) (mho) -> R f=19.99K Ω
and using this value in the gain expression:
Vo/Vs= -175
Sounds good but…
there are still potential issues
related to stability!