1
Extending Blackman's Formula to
Multiple-Transistor Feedback Circuits
Eugene Paperno

AbstractThe Blackman's formula for the impedance seen
from an arbitrary terminal of a single-transistor feedback circuit
is extended to the case of multiple-transistor circuits. To reach
this aim, we revisit the proof of the Blackman's formula for
single-transistor circuits, proof it in a similar way for doubletransistor circuits, and then extend for the case of multipletransistor circuits.
Index TermsBlackman's formula, feedback circuits,
impedance evaluation, single-, double-, triple-, multiple-transistor
circuits, return ratio.
I. INTRODUCTION
To examine the effect of negative feedback [1]-[17] on
terminal impedances it is important to describe the impedances
analytically as a function of the feedback partial gains, such as
return ratios. The impedance of an arbitrary terminal of a
single-transistor circuit can be described by the Blackman's
formula [3]:
Rt  Rt
1  RRsc
,
1  RRoc
(1)
where Rt is the closed-loop impedance seen by a test source vt
connected to an arbitrary terminal of a single-transistor circuit
[see Fig. 1(a)], R't is the open-loop terminal impedance, RRsc is
the return ratio for the short-circuited terminal, and RRoc is the
return ratio for the open-circuited terminal.
The aim of the present work is to extend the Blackman's
formula to the case of multiple-transistor circuits. To reach this
aim, we revisit the proof of the Blackman's formula for singletransistor circuits, proof it in a similar way for doubletransistor circuits, and then extend for the case of multipletransistor circuits.
s  vt Gv  s RRsc ,
where the input transmittance for the voltage test source
Gv 
s
,
vt
E. Paperno is with the Department of Electrical and Computer Engineering,
Ben-Gurion University of the Negev, P.O. Box 653, Beer-Sheva 84105, Israel
(e-mail: paperno@ee.bgu.ac.il).
(2)
and the return ratio for the short-circuited (vt=0) terminal
RRsc 
s
.
s
(3)
Note that the signals in (2) and (3) with the single and
double prime symbols corresponds to the case, where the only
active sources are vt and aOLs, respectively.
From (1), vt can be obtained as:
vt 
s (1  RRsc )
.
Gv
(4)
Let us now replace in Fig. 1(b) source vt with it, such that
it = vt/Rt, to keep the same conditions of the test source brunch.
Keeping the same brunch voltage and current leaves the signal
s. unchanged. As a result,
s  it Gi  s RRoc ,
(5)
where the input transmittance for the current test source
Gi 
s
,
it
(6)
and the return ratio for the open-circuited (it=0) terminal
II. SINGLE-TRANSISTOR FEEDBACK CIRCUITS
Let us consider a linear equivalent model of a generic
single-transistor circuit (see Fig. 1). To find the impedance of
an arbitrary terminal, we connect to it a voltage test source in
Fig. 1(a) and find by applying superposition the control signal
(1)
RRoc 
s
.
s
(7)
From (5), it can be obtained as:
it 
s (1  RRoc )
.
Gi
(8)
2
Hence, the terminal impedance
v
G 1  RRsc
Rt  t  i
.
it
Gv 1  RRoc
s
aOLs
Rt
(9)
RRsc
Gv
vt
(a)
Gv and Gi in (4) and (9) can be found from Figs. 1(c) and
(d):
Gv 
s
.
vt
(10)
s
aOLs
Rt
Gi 
s
v
for it  t ,
it
Rt
(11)
RRoc
Gi
(b)
where R't is the open-loop terminal impedance (for aOL=0).
Note that it=vt/R't in (11) helps keeping the same conditions
of the test source brunch for both vt and i't sources, as a result
s' in (10) and (11) has the same value.
Considering (9)(11), we obtain the Blackman's formula for
single-transistor circuits:
Rt  R't
1  RRsc
.
1  RRoc
(12)
it=
vt
Rt
s'
aOL=0
R' t
Gv
vt
(c)
III. DOUBLE-TRANSISTOR FEEDBACK CIRCUITS
Following similar approach, we extend in this section the
Blackman's formula for double-transistor circuits (see Fig. 2).
Figs. 2(a) and (b) suggests that the control signals of the
dependent sources
s'
R'
aOL=0
t
Gi
 s 1  vt G1v  s 1RR1sc  s 2 RR21sc

,

 s  v G  s RR  s RR
t 2v
2
2 sc
1
12sc
 2
(13)
 s 1  it G1i  s 1RR1oc  s 2 RR21oc

,

 s  i G  s RR  s RR
2
2oc
1
12oc
  2 t 2i
(14)
(d)
i't=
vt
R't
Fig. 1. Finding the impedance seen by a test source connected to an arbitrary
terminal of a generic single-transistor feedback circuit.
Equations (13) and (14) can be solved for vt and it:
Rt 
1  RR1sc  RR2 sc  RR1sc RR2 sc  RR12sc RR21sc
vt 
s 2 , (15)
G2 v  G2 v RR1sc  G1v RR12sc
G2i
G2 v
G1i
RR12oc
G2i
1   sc
,
G1v
1   oc

RR12sc
G2 v
1  RR1oc 
1  RR1sc
(17)
where
it 
1  RR1oc  RR2oc  RR1oc RR2oc  RR12oc RR21oc
s 2 , (16)
G2i  G2i RR1oc  G1i RR12oc
Dividing (15) by (16) gives
 sc  RR1sc  RR2 sc  RR1sc RR2 sc  RR12sc RR21sc
.
(18)
 oc  RR1oc  RR2oc  RR1oc RR2oc  RR12oc RR21oc
The first term in (17) represents the open loop impedance
seen by vt:
3
G2i
s
v
v
  2 v t  s  Rt ,
G2 v
i s s 2 v
it
s1
(19)
Rt
where i't= vt /R't [see Fig. 2(d)].
To find the second term in (17), let us consider Fig. 3, where
vt=1, it=1/R*t, s1=1, and s2=0, and, hence:
RR1sc
G1v
vt
RR12sc
(a)
RR21sc
G2v
 s*1  G1v  RR1sc

,

 *
*
 s 1  it G1i  RR1oc
s1
aOL1s1
Rt
(21)
RR1oc
G1i
RR12oc
(b)
it=
RR21oc
G2i
vt
Rt



RR2oc
s2
.
 RR1sc 
aOL2s2
s2
From (20) and (21),
G
G1i
G
RR12oc  G 2 v  1v  1i
G 2i
 G 2 v G 2i
RR2sc
(20)
 s*2  G2 v  RR12sc

.

 *
*
 s 2  it G2i  RR12oc
RR1oc 
aOL1s1
s'1
(22)
R'
G1i
RR12sc
G 2i
aOL2s2
aOL1=0
t
G1v
vt
(c)
Considering that in (22) in accordance with (19),
G2v
G1v G1i
G
G

 1v  2 v  Rt ,
G2 v G2i
G1i G2i
(23)
s'2
(22) can be rewritten as follows:
s'1
aOL2=0
aOL1=0
'
RR1oc 
G1i
G
RR12oc  RR1sc  1v RR12sc .
G 2i
G2 v
Rt
(24)
G1i
(d)
Considering (17), (19), and (24), Rt can eventually be
obtained as:
Rt  Rt
1   sc
.
1   oc
i't=
vt
R't
G2i
(25)
s'2
aOL2=0
Fig. 2. Finding the impedance seen by a test source connected to an arbitrary
terminal of a generic double -transistor feedback circuit.
4
s*1
IV. MULTIPLE-TRANSISTOR FEEDBACK CIRCUITS
aOL11
R*t
Following the approach given in the previous sections, the
terminal impedance can be obtained in accordance with (25)
for multiple-transistor feedback circuits. For example, for
triple-transistor circuits the short- and open-circuit return
ratios in (25) are as follows:
RR1sc
G1v
vt=1
RR12sc
(a)
G2v
 sc  RR1sc  RR1sc RR23sc RR32sc
 RR2 sc  RR2 sc RR13sc RR31sc
 RR3sc  RR3sc RR12sc RR21sc
,
 RR1sc RR2 sc  RR1sc RR3sc  RR2 sc RR3sc
(26)
 RR12sc RR21sc  RR13sc RR31sc  RR23sc RR32sc
s*2
aOL2=0
s*1
aOL11
R*t
 RR1sc RR2 sc RR3sc  RR12sc RR23sc RR31sc
 RR13sc RR21sc RR32sc
RR1oc
G1i
RR12oc
(b)
 oc  RR1oc  RR1oc RR23oc RR32oc
 RR2oc  RR2oc RR13oc RR31oc
i*t=
 RR3oc  RR3oc RR12oc RR21oc
.
 RR1oc RR2oc  RR1oc RR3oc  RR2oc RR3oc
G2i
1
R*t
(27)
s*2
 RR12oc RR21oc  RR13oc RR31oc  RR23oc RR32oc
 RR1oc RR2oc RR3oc  RR12oc RR23oc RR31oc
aOL2=0
Fig. 3. A generic double-transistor feedback circuit with vt=1, i*t=1/R*t, s1=1,
and s2=0.
 RR13oc RR21oc RR32oc
APPENDIX
DOUBLE-TRANSISTOR EXAMPLE CIRCUIT
Let us solve a double-transistor circuit (see Fig. A.1) for the
impedance seen by the input source. From Fig. A. 1(c),
Rt  RB  hie1 || [ R f  RE 2 || ( ro 2  RC 2 ) || ( hie2  ro1 )] , (A.1)
RR1oc  

From Fig. A.1(d),
RR1sc  


ib1
vs

vt 0
h fe1ro1
ro1  hie2  RE 2 || ( ro 2  RC 2 ) || ( R f  RB || hie1 )
RE 2
(A.2)
RE 2 || ( ro 2  RC 2 )
RB
|| ( ro 2  RC 2 )  R f  RB || hie1 RB  hie1
RR12sc  
ib2
vs
ro1  hie2  RE 2 || ( ro 2  RC 2 ) || ( R f  hie1 )
RE 2
h fe1ro1
ro1  hie2  RE 2 || ( ro 2  RC 2 ) || ( R f  RB || hie1 )
(A.3)
,
(A.4)
.
(A.5)
RE 2 || ( ro 2  RC 2 )
|| ( ro 2  RC 2 )  R f  hie1
ib2
vs
it 0
h fe1ro1
ro1  hie2  RE 2 || ( ro 2  RC 2 ) || ( R f  hie1 )
From Fig. A.1(e),
vt 0
it 0
h fe1ro1
RR12oc  

,

,
ib1
vs
5
RR2 sc  
ib2
vs
VCC
IC1
vt  0
RC2
vO
h fe2 ro 2

ro 2  RC 2  RE 2 || ( R f  RB || hie1 ) || ( hie2  ro1 )
,
Q2
RB
(A.6)
Q1
vt

Rf
RE 2 || ( R f  RB || hie1 )
RE2
(a)
RE 2 || ( R f  RB || hie1 )  hie2  ro1
Rt
RR21sc
i 
  b1
vs
RC2
RB
vt 0
ib1
hfe1ib1
hie1
ro1 ib2
hfe2ib2
hie2
ro2
vt
h fe2 ro 2


ro 2  RC 2  RE 2 || ( R f  RB || hie1 ) || ( hie2  ro1 )
, (A.7)
Rf
RE 2 || ( hie2  ro1 )
RB
|| ( hie2  ro1 )  R f  RB || hie1 RB  hie1
RE 2
RE2
R't
RC2
RB
RR2oc  
i'b1
it  0
ro1
Rf
.
i"b1

ro1
hfe2=0
hie2
Rf
ro 2  RC 2  RE 2 || ( R f  hie1 ) || ( hie2  ro1 )
.
ro2
it=0
it 0
h fe2 ro 2
RE 2
i"b2
hfe1ib1
hie1
vt=0
(c)
RC2
RB
RE 2 || ( R f  hie1 )  hie2  ro1

ro2
RE2
(A.8)
RE 2 || ( R f  hie1 )
ib1
vs
hfe2=0
hie2
vt
ro 2  RC 2  RE 2 || ( R f  hie1 ) || ( hie2  ro1 )
RR21oc  
i'b2
hfe1=0
hie1
h fe2 ro 2


ib2
vs
(b)
RE2
(d)
(A.9)
RE 2 || ( hie2  ro1 )
|| ( hie2  ro1 )  R f  hie1
By substituting (A.1)‒(A.9) into (18) and (25), one can
obtain the impedance seen by the input source.
RC2
RB
i'''b1
hie1
vt=0
i'''b2
hfe1=0
ro1
hfe2ib2
hie2
ro2
it=0
Rf
RE2
(e)
Fig. A.1. Finding the input impedance of a double-transistor circuit. (a).
Example circuit. (b) Original equivalent small-signal circuit. (c) The "prime"
circuit, where the independent source is the only active one. (d) The "doubleprime" circuit, where hfe1ib1 is the only active source. (e) The "triple-prime"
circuit, where hfe2ib2 is the only active source. Note that the dependent sources
in (d) and (e) are controlled by the corresponding signals of the original
circuit (b).
6
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