Download IV. Measuring Return Ratio at the Terminals of a Dependent Source

1
Simulating Return Ratio in Linear Feedback
Networks by Middlebrook's Method
© Eugene Paperno, 2012

Abstract—The proof of Middlebrook's formula for simulating
return ratio is revisited in order to extend it to linear networks
with multiple bilateral feedback loops. Instead of an idealized
feedback model, the new proof is based on a generic feedback
model. It is shown that Middlebrook's formula can be applied
with no approximations to any linear feedback network with a
single dependent source. An example circuit is simulated, and its
return ratios obtained by conventional and Middlebrook's
approaches are compared. The relative difference between the
two methods is less than 0.2 ppm. This small difference is caused
by the limited computing accuracy.
Index Terms—Current injection, generic feedback model,
idealized feedback model, linear feedback networks,
Middlebrook's method, multiple bilateral feedback loops, return
ratio, voltage injection.
I. INTRODUCTION
RATIO concept is a very powerful tool to analyze
R ETURN
linear feedback networks with a single dependent source
[1]-[22]. Such networks represent a very wide class of singletransistor or single operational amplifier electronic circuits.
Finding return ratio is very important for revealing the effect
of feedback on the network closed-loop gain, impedances, and
stability. Conventionally,
the return ratio is found by
suppressing all the independent network's sources, assigning a
fixed value to the dependent source, and calculating the signal
that returns to its controlling terminals.
This procedure is not suitable, though, for simulating
electronic circuits, where there is no access to circuit's
dependent sources, or for testing real electronic circuits.
An alternative approach was suggested by Middlebrook in
[8], [21] and consists in connecting test sources to the
accessible terminals of a transistor or operational amplifier.
According to Middlebrook's method, two partial return ratios
are measured first, one for a current injection and the other for
a voltage injection, and then they are translated into the
circuit's return ratio.
Unfortunately, the proof of Middlebrook's method is based
on an idealized feedback model, does not account for nonzero
This work was supported by Analog Devices, Inc.
Eugene 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: [email protected]).
reverse loop gain and for multiple feedback loops.
In this work, we revisit the proof of Middlebrook's method
and extend it to linear networks with multiple bilateral loops.
Instead of an idealized feedback model, our proof is based on
a generic feedback model.
II. RETURN RATIO AND THE CLOSED-LOOP GAINS
Let us consider in Fig. 1(a) a generic linear feedback
network with a single dependent source. It is important to note
that the generic network in Fig. 1 is not necessarily a singleloop feedback one and can include a number of bilateral
feedback paths connecting between the dependent source and
its controlling terminals.
The network's return ratio is defined as follows [see Figs.
1(b) and (c)]:
T 
s
ss  0
s
,
(1)
where ss is the input signal source, and s is the signal
controlling the dependent source.
In Figs. 1(b) and (c), where ss=0, we assume that the entire
network seen by the dependent sources can be replaced by an
equivalent impedance composed of Zo, representing the output
impedance of a transistor or operational amplifier, and Z,
representing the rest of the network's total equivalent
impedance.
Injecting either a test current in Figs. 2(a), (b) or a test
voltage in Figs. 2(c), (d), right at the output terminals of the
dependent source, allows calculating T in each test. The
closed-loop gains, iy/ix and vy/vx in Figs. 2(a), (c) can be found
as follows:
ACL 
so
A
 G OL  D ,
st
1 T
(2)
where so denotes either iy or vy output signals, st denotes either
it or vt test sources,
G
s
a 0
st
is the input transmission, s denotes either i, for the current
(3)
2
si
s
aOL si
aOLs
it
iy
ix
ss
(a)
Zo
ss0
(a)
Z
s
aOLs
ss0
si
T
aOL0
aOL0
Zo
it
ss0
iy
aOL0
ix
aOL0
(b)
Z
s
it
Zo
(b)
aOL(ro||Z)s
ss0

ss0
T
Gi a
OL
ss0
0
Z
Zo
si
aOL(Z||ro)s v
(b)
Z
vt
vy
vx
Fig. 1. Finding the return ratio for a generic linear feedback network: (a)
original network, (b) and (c) suppressing the signal source and replacing the
dependent source with an equivalent independent one. Zo denotes the output
impedance of a transistor or operational amplifier, and Z denotes the rest of
the total equivalent impedance seen by the dependent source when ss=0.
Zo
ss0
(c)
Z
sv
injection, or v , for the voltage injection,
D
so
a 0
st
,
aOL 0
aOL0
vt
(4)
is the direct transmission, so denotes either iy or vy output
signals, and
Gv
ss0
T
aOL(Zo||Z)
vy
aOL0
vx
aOL0
0
vt
Zo
(d)
Z
AOL 
so
st  0
s
.
(5)
is the open-loop gain.
Fig. 2. Finding the return ratio for a generic linear feedback network: (a) by
current injection and (c) by voltage injection, (b) finding input and direct
transmissions Gi, and Di, finding input and direct transmissions Gv and Dv.
III. MEASURING RETURN RATIO DIRECTLY AT THE
DEPENDENT SOURCE TERMINALS
If a test source can be connected directly to the dependent
source, as shown in Figs. 2(a) and (c), a single test is enough
to calculate the return ratio. According to (2),
iy
ix

Gi
AOLiy
 Diy
1 T
A
Gi OLix  Dix
1 T
T a OL
0
a OL 1  T

 T ,
T a OL

1
a OL 1  T

(6)
3
vy
vx

Gv
AOLvy
s*i
 Dvy
1 T
A
G v OLvx  Dvx
1 T
aOLs*i
Zo
,
a OL ( Z 0 || Z )
T
0
a OL ( Z 0 || Z )
1T

 T
a OL ( Z 0 || Z )
T

1
a OL ( Z 0 || Z )
1 T
i*y
it
(7)

i*x
ss0
(a)
Z
where AOLiy, AOLix , AOLvy, and AOLvx are the open-loop gains for
output signals iy, ix, vy, and vx, respectively; Gi and Gv are the
input transmissions from it and vt sources; Diy and Dix are the
direct transmissions from it source to output signals iy and ix;
Dvy and Dvx are the direct transmissions from vt source to
output signals vy and vx, and aOL is the dependent source gain.
Note that the only difference between calculating T in Fig.
1(b) and calculating Gi in Fig. 2(b) is in the active independent
source value: aOLs against it. Therefore, according to (5),
GiT/aOL in (6). Similarly, the active independent sources in
Fig. 1(c) and Fig. 2(d) are aOL(Zo||Z)s and vt, therefore,
GvT/[aOL(Zo||Z)] in (7).
Note also that suppressing the dependent sources in Figs.
2(b) and (d) zeroes Diy and Dvy and forces ixit and vxvt,
therefore, according to (4), Diy0 and Dix1 in (6), and Dvy0
and Dvx1 in (7).
Note as well, that suppressing the test sources in Figs. 2(a)
and (c) to calculate the open-loop gains in accordance to (5)
results in iy=ix in Fig. 2(a) and vy=vx in Fig. 2.(c). Therefore,
AOLiy= AOLix in (6), and AOLvy= AOLvx in (7).
IV. MEASURING RETURN RATIO AT THE TERMINALS OF A
DEPENDENT SOURCE LINKED TO ZO
If a test source can be connected only to the dependent
source linked to its output impedance Zo, as shown in Figs.
3(a) or (c), two tests are needed to calculate the return ratio. In
each test, we will find the circuit's partial return ratios, either Ti
for the current injection in Fig. 3(a), or Tv for the voltage
injection in Fig. 3(c), and then translate them into T.
According to Figs. 3(a) and (b),
s*i
aOL=0
aOL0
Zo
ss0
G*i
it
i*y
aOL0
i*x
aOL0
Zo
T
aOL Zo+Z
(b)
Z
s*v
aOLZos*v
Zo
vt
v*y
aOL0
v*x
aOL0
ss0
(c)
Z
s*v
aOL=0
aOL0
Zo
Ti 
i *y
 *
ix

Gi*
Gi*
*
AOLiy

Diy*
vt
1 T
*
AOLix
 Dix*
1 T
,
Zo
T
aOL Z o  Z

Zo
T

aOL Z o  Z

aOL
Z

1  T Zo  Z T (Zo  Z )  Z

aOL
Zo
Zo

1  T Zo  Z
(8)
ss0
G*v
Z
T
aOL Zo+Z
v*y
aOL0
v*x
aOL0
(d)
Z
Fig. 3. Finding partial return ratios in a generic linear feedback network: (a)
injection of the test current, (b) finding G*i, D*iy, and D*ix, (c) injection of
the test voltage, and (d) finding G*v, D* vy, and D* vx.
4
According to Figs. 3(c) and (d),
Tv  
v *y
v *x

*
AOLvy
0
RCa
5k
RFa

1 T
A*
*
Gv* OLvx  Dvx
1 T
Gv*
*
Dvy
Zo
Z aOL Z o

aOL Z o Z o  Z 1  T
Zo  Z
.

Z o aOL
T
Z


aOL Z o Z o  Z 1  T Z o  Z

CBCa 10p
5.9k
RBa
2.7k
Cbe3
100p
hiea
2.6k
(a)
Fa
roa
100k
T
F
0
(9)
GAIN = 100
CEa
10u
REa
1234567
0
CBCb 10p
T (Zo  Z )  Zo

Z
0
0
RCb
5k
RFb
5.9k
Solving (8) and (9) for T yields Middlebrook's formula
RB2
2.7k
CBEb
100p
T T 1
T  i v
.
Ti  Tv  2
(b)
hieb
2.6k
rob
100k
RR
ita
0Vac
0Vdc
(10)
0
100Aac
0Adc
CEb
10u
REb
1234567
0
0
V. CONCLUSION
The proof of Middlebrook's formula for simulating return
ratio is extended to linear networks with multiple bilateral
feedback loops. Instead of an idealized feedback model, the
proof is based on a generic feedback model. Thus, it is shown
that Middlebrook's formula can be accurately applied to a
much wider class of feedback networks.
CBCc 10p
0
RCc
5k
RFc
5.9k
RBc
2.7k
ACKNOWLEDGMENT
The author wishes to express his deepest gratitude to Prof.
Shmuel (Sam) Ben-Yaakov for very fruitful and inspiring
discussions.
CBEc
100p
hiec
2.6k
0Vac
0Vdc
iy
0Vac
0Vdc
ix
(c)
it
Fc
roc
100k
1Aac
0Adc
F
0
GAIN = 100
CEc
10u
REc
1234567
0
0
APPENDIX
To illustrate the accuracy of finding a return ratio by
applying Middlebrook's method, let us consider in Figs. 4 and
5 a SPICE simulation of a feedback network with two bilateral
feedback loops.
The original circuit is shown in Fig. 4(a). Fig. 4(b)
illustrates measuring return ratio by conventional method, Fig.
4(c) illustrates injection of the test current, and Fig. 4(d)
illustrates injection of the test voltage.
Fig. 5 compares between the return ratios found by
conventional and Middlebrook's methods. Note that the
relative difference is less than 0.2 ppm. This small difference
is caused by the limited computing accuracy.
CBCd 10p
0
RCd
5k
RFd
5.9k
vt
vy
vx
(d)
RBd
2.7k
Cbd
100p
hied
2.6k
Fd
rod
100k
F
0
1Vac
0Vdc
GAIN = 100
v xy
CEd
10u
REd
1234567
0
0
Fig. 4. Example circuit: (a) original circuit, (b) simulating return ratio by
replacing the dependent source with an equivalent independent one, (c)
injection of the test current, (d) injection of the test voltage.
5
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
[10]
[11]
[12]
[13]
[14]
[15]
[16]
[17]
[18]
Amplitude
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50
SEL>>
0
((-V(VY,VXY)/V(VX,VXY))*(-I(iy)/I(ix))-1)/(-V(VY,VXY)/V(VX,VXY)-I
(iy)/I(ix)+2)
-I(RR)
0d
Phase
[1]
100
-50d
-100d
P(((-V(VY,VXY)/V(VX,VXY))*(-I(iy)/I(ix))-1)/(-V(VY,VXY)/V(VX,VXY)
-I(iy)/I(ix)+2))
P(-I(RR))
200m
ppm
REFERENCES
0
-200m
1.0mHz
1.0Hz
1.0KHz
1.0MHz
1.0GHz
1000000*(((-V(VY,VXY)/V(VX,VXY))*(-I(iy)/I(ix))-1)/(-V(VY,VXY)/V(
VX,VXY)-I(iy)/I(ix)+2)+I(RR))/I(RR)
1000000*(P(((-V(VY,VXY)/V(VX,VXY))*(-I(iy)/I(ix))-1)/(-V(VY,VXY)/
V(VX,VXY)-I(iy)/I(ix)+2))-P(-I(RR)))/P(-I(RR))
Frequency
Fig. 5. Measuring and comparing return ratios: (top and middle) amplitude
and phase of the return ratios measured by replacing the dependent source
with an equivalent independent one (the green curves) and by Middlebrook's
method (the red curves), (bottom) the relative difference (in ppm) between the
return ratios amplitude (the green curve) and phase measurements (the red
curve).
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for Feedback Systems," IEEE Microwave Magazine, vol. 7, pp. 50-63,
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Design of Analog Integrated Circuits, John Wiley & Sons, 2009.
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