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TWO-PORT NETWORKS
The Impedance Parameters Z1 and Z0
For the two-port system of Fig. 1, Zi is the input impedance between terminals 1 and 1', and
Z0 is the output impedance between terminals 2 and 2'. For networks an impedance level can
be defined between any two (adjacent or not) terminals of the network.
Io
Ii
1
Two-Port
system
Ei
Eo
1
Zi
Zi
The input impedance is defined by Ohm’s law in the following form:
Ei
ohms                         (1)
Ii
with Ii the current resulting from the application of a voltage Ei. The output impedance Zo is
defined by
E
Z o  o ohms                         (2)
Io
with Io the current resulting from the application of a voltage E0 to the output terminals, with
E set to zero.
Zi 
Note that both Ii and Io are defined as entering the package. This is common practice for a
number of system analysis methods to avoid concern about the actual direction for each
current and also to define Zi and Zo as positive quantities in Eqs. (1) and (2), respectively. If
Io were chosen to be leaving the system, Zo as defined in Eq.(2) would have to have a
negative sign.
An experimental set-up for determining Zi for any two input terminals is provided in Fig. 3.
The sensing resistor Rs is chosen small enough not to disturb the basic operation of the
system or require too large a voltage Eg to establish the desired level of Ei. Under operating
conditions, the voltage across Rs is Eg – Ei, and the current through the sensing resistor is
I Rs 
Eg - Ei
VRs

Rs
Rs
but I s  I R s and Z, 
Ei Ei

Ii
I Rs
1
IRs
VRs
1
Ii
Rs
Eg
~
Ei
Two-Port
system
1
Zi
FIG. 2 Determining Z1.
The sole purpose of the sensing resistor, therefore, was to determine Ii sing purely voltage
measurements.
As we progress through this chapter, keep in mind that we cannot use an ohmmeter to
measure Zi or Zo since we are dealing with ac systems whose impedance may be sensitive to
the applied frequency.
Ohmmeters can be used to measure resistance in a dc or ac network, but recall that
ohmmeters are employed only on a de-energised network, and their internal source is a dc
battery.
The output impedance Z0 can be determined experimentally using the set-up of Fig. 4. Note
that a sensing resistor is introduced again, with Eg being an applied voltage to establish
typical operating conditions. In addition, note that the input signal must be set to zero, as
defined by Eq 2. The voltage across the sensing resistor is Eg - E0, and the current through
the sensing resistor is
I Rs 
VR S
RS

Eg  Eo
RS
but I o  I R S and Z o 
Eo Eo

Io
I RS
Fig 3 Determining Zo
For the majority of situations Zi and Zo will be purely resistive resulting in an angle of zero
degrees for each impedance. The result is that either a DMM or a scope can he used to find
the required magnitude of the desired quantity. For instance, for both Zi and Zo, VRs can he
measured directly with the DMM, as can the required levels of Eg, E1, or Eo. The current for
2
each case can then be determined using Ohm's law, and the impedance level can be
determined using either Eq. (1) or Eq. (2).
If we use an oscilloscope, we must be more sensitive to the common ground requirement. For
instance, in Fig. 4, Eg and Ei can be measured with the oscilloscope since they have a
common ground. Trying to measure VRs directly with the ground of the oscilloscope at the
top input terminal of E would result in a shorting effect across the input terminals of the
system due to the common ground between the supply and oscilloscope. If the input
impedance of the system is "shorted out," the current I can rise to dangerous levels because
the only resistance in the input circuit is the relatively small sensing resistor Rs. If we use the
DMM to avoid concern about the grounding situation, we must be sure the meter is designed
to operate properly at the frequency of interest. Many commercial units are limited to a few
kilohertz.
If the input impedance has an angle other than zero degrees (purely resistive), a DMM cannot
be used to find the reactive component at the input terminals. The magnitude of the total
impedance will be correct if measured ag described above, but the angle from which the
resistive and reactive components can be determined will not be provided. If an oscilloscope
is used, the network must be hooked up as shown in Fig. 5. Both the voltage E g and VRs can
be displayed on the oscilloscope at the same time, and the phase angle between Eg and VRs
can be deter-mined. Since VRs and I are in phase, the angle determined will also be the angle
between Eg and Ii. The angle we are looking for is between Ei and Ii not between Eg and Ii
but since R~ is usually chosen small -enough, we can assume that the voltage drop across Rs
is so small compared to Eg that Ei  Eg. Substituting the peak, peak-to-peak, or rms values
from the oscilloscope measurements, along with the angle just -determined, will permit a
determination of the magnitude and angle for ZI from which the resistive and reactive
components can be determined using a few basic geometric relationships. The reactive nature
(inductive or capacitive) of the input impedance can be determined when the angle between E
and I is computed. For a dual-trace oscilloscope, if Eg leads VRs (Ei leads Ii), the network is
inductive; if the reverse is true, the network is capacitive.
Fig 4 Determining ZI, using an oscilloscope
To determine the angle associated with Zo, the sensing resistor must again be moved to the
bottom to form a common ground with the supply Eg. Then, using the approximation Eg  Eo,
the magnitude and angle of Zo can be determined.
3
EXAMPLE 1 Given the DMM measurements appearing in Fig. 5, determine the input
impedance Z for the system if the input impedance is known to be purely resistive.
Solution : VR S  Eg - E i  l00 - 96  4 mV
I1  I R S -
VR S
Zi  R i 
RS

4mV
 40A
100 
Ei
96

 2.5k
Ii
40 x 10 -6
Fig 5
EXAMPLE 2 Using the provided DMM measurements of Fig.6, determine the output
impedance Z0 for the system if the output impedance is known to be purely resistive.
Solution :
VR S  E g - E 0  2 V - 1.92 V  0.08 V  80 mV VRA 80mV
I o  I RS 
Zo 
VR S
RS
80 x 10 3

 40A
2000
E0
1.92

 48k
Io
40 x 10 6
Fig 6
EXAMPLE 3 The input characteristics for the system of Fig. 7(a) are unknown. Using the
oscilloscope measurements of Fig. 7(b), determine the input impedance for the system. If a
reactive component exists, determine its magnitude and whether it is inductive -or capacitive.
4
Solution : The magnitude of Z i :
I i ( p  p )  I R S (p-p) 
Zi 
VR S (p.p)
RS

2mV
 200 A
10
E 5OmV

 250
I i 250A
The angle of Z i : The phase angle between E g and VR S (or I R S  I i ) is 180 0 - 150 0  30 0
with E g leading I i so the system is inductive. Therefore,
 250   - 30 0
 216.51   j125   R  jX L
5
If we expand Eq.(5) as follows:
A VT 
Eo Eo
E E  E E

(1)  o  i   o . i
Eg Eg
Eg  Ei  Ei Eg
A VT  A v
then
A VT  A v NL
or
Ei
(if loaded )
Eg
Ei
(if unloaded )
Eg
The relationship between Ei and Eg can be determined from Fig 8 if we recognise that Ei is
across the input impedance ZI-+ and thus apply the voltage divider rule as follows:
Ei 
Z i E g 
Zi  R g
Ei
Zi

E g Zi  R g
Substituting into the above relationships will result in
A VT  A v
or
Zi
(if loaded )                          (6)
Zi  R g
A VT  A v NL
Zi
(if unloaded )                          (7)
Zi  R g
A two-port equivalent model for an unloaded system based on the definitions of ZI, Z0, and
AvNL is provided in Fig.9. Both Zi and Zo appear as resistive values since this is typically the
case for most electronic amplifiers. However, both Zi and Zo can have reactive components
and not invalidate the equivalency of the model.
FIG. 9 Equivalent model for two-port amplifier.
6
The input impedance is defined by Zi = Ei/Ii and the voltage E o  A v NL E i in the absence of a
load, resulting in A v NL  E o /E i as defined. The output impedance is defined with Ei set to
zero volts, resulting in A v NL E i  0 , which permits the use of a short-circuit equivalent for the
controlled source. The result is Z0 = E0/I0, as defined, and the parameters and structure of the
equivalent model are validated.
If a load is applied as in Fig. 10, an application of the voltage divider rule will result in
Eo 
Av 
R L A v NL E i

RL  RO
Eo
RL
 A v NL
             (8)
Ei
RL  RO
For any value of RL or R0, the ratio RL (RL + Ro) must be
less than mandating that Av is always less than AvNL as
stated earlier. Further,
Figure 10 Applying load to the output of fig
for a fixed output impedance (R0), the larger the load resistance (RL), closer the loaded gain
to the no-load level.
An experimental procedure for determining R0 can be developed if we solve Eq. (8) for the
output impedance R0:
A v  A v NL
RL
RL  RO
or A v R L  R O   A v NL R L
A v R L  A v R O  A v NL R L
A v R O  A v NL R L  A v R L
 Av  Av
R o  R L  NL
Av

or



 Av

R o  R L  NL 1                        (9)
 Av

Equation (9) reveals that the output impedance R0 of an amplifier be determined by first
measuring the voltage gain E0/E1 without a load in place to find AvNL and then measuring the
gain with a load RL to fin Av. Substitution of AvNL, Av and RL into Eq. (9) will then provide
the value for Ro.
7
9
Example 4 For the system of Fig 11(a) employed in the loaded amplifier of Fig.11(b):
Figure 11
a.
b.
c.
d.
Determine the no-load voltage gain AvNL.
Find the loaded voltage gain Av.
Calculate the loaded voltage gain AvT.
Determine R0 from Eq. (26.9) and compare it to the specified value Fig 11.
Solutions :
Eo
 20

 5000
E i 4mV
(a)
A VNL 
b 
A V  A VNL
RL
2.2k
  5000
 210.73
RL Ro
2.2k  50k
(c)
A VT  A VNL
Zi
1k
  210.73
 105.36
Zi  R g
1k  1k
 Av

  5000

(d ) R o  R L  NL 1  2.2k
1  50k as specified .
  210.73 
 Av

8
THE CURRENT GAINS Ai AND AiT. AND THE POWER GAIN AG
The current gain of two-port systems is typically calculated from volt age levels. A no-load
gain is not defined for current gain since the absence of RL requires that Io = Eo/RL = 0 A and
Ai = Io/Ii = 0
For the system of Fig.112, however, a load has been applied and
Io 
 Eo
RL
with I i 
Ei
Zi
Fig 12 Defining AI and AiT
Note the need for a minus sign when Io is defined, because the defined polarity of Eo would
establish the opposite direction for Io through RL.
The loaded current gain is
 Eo
Io
R L  E o  Zi

Ai  

Ei
Ii
E i  R L
Zi
and A i 



 E o Zi
                       (10)
Ei R L
In general, therefore, the loaded current gain can be obtained directly from the loaded voltage
gain and the ratio of impedance levels, Zi over RL.
If the ratio AiT = Io/II were required, we would proceed as follows:
9
The result obtained with Eq. (10) or (11) will be the same since Ig = Ii, but the option of
which gain is available or which you choose to use is now available.
Returning to Fig 9. (repeated in Fig.13), an equation for the current gain can be determined in
terms of the no-load voltage gain.
Fig 13 Developing an equation for Ai in terms of AvNL
Through Ohm's law:
Io 
 Av NL E i
RL  Ro
but E i 
Ii
Ri
and I o 
 Av NL I i R i 
RL  Ro
so that A i 
Io
 Av NL R i

                       (12)
Ii
RL  Ro
The result is an equation for the loaded current gain of an amplifier in terms of the nameplate
no-load voltage gain and the resistive elements of the network.
Recall an earlier conclusion that the larger the value of RL, the larger the loaded voltage gain.
For current levels, Eq. (12) reveals that
the larger the level of RL, the less the current gain of a loaded amplifier.
In the design of an amplifier, therefore, one must balance the desifed voltage gain with the
current gain and resulting ac output power level.
For the system of Fig.13, the power delivered to the load is determined by E o2 /R L , whereas
the power delivered at the input terminals is E i2 /R L .
10
The power gain is therefore defined by
Po E o2 /R L E o2  R i

AG 
 2

Pi
E i /R i E i2  R L
and A G  A 2v



Ri
                           (13)
RL
Expanding the conclusion,

R 
A G  A v  A v i   A v  A i 
RL 

A G   A v A i                      (14)
Don't be concerned about the minus sign. AV or Ai will be negative to ensure that the power
gain is positive, as obtained from Eq. (13).
If we substitute A v  - A i R L /R I [from Eq. (10)] into Eq. (14) we will find

R 
A G   A v A i   - A i L A i
RI 

A G A i
2
RL
                         (15)
RI
which has a format similar to Eq. (13), but now AG is given in terms of the current gain of
the system.
The last power gain to be defined is the following:
E2 / R L
E2 / R L
E2  Rg  Ri
P
A GT  L  o
 2 o
 o2 
Pg
Eg Ig
Eg / R g  R i Eg  R L
 Rg  Ri
or A G T  A 2VT 
 RL
expanding




                    (16)

Rg  Ri

A G T  A VT  A VT
RL




and A G T  A VT A i T                        (17)
11
EXAMPLE 5 Given the system of Fig 14 with its nameplate data:
a.
b.
c.
d.
e.
f.
Determine AV.
Calculate Ai.
Increase RL to double its current value and note the effect on AV and Ai.
Find A I T .
Calculate AG.
Determine Ai from Eq. (1) and compare it to the value obtained in part (b).
Solutions:
a.
A v  A v NL
b. A i  A v NL


RL
4.7 k
  - 100.94
 (-960)
RL  RO
4.7
k


40
k





RL
2.7 k
  - 57.99
 (-960)
RL  RO
 4.7 k  40k 
c. R L  2(4.7 k)  9.4k
A v  A v NL


RL
9.4 k
  - 182.67 versus - 100.94, which is a significan t increase.
 (-960)
RL  RO
 9.4 k  40k 
A i  A v NL


RL
2.7 k
  52.47 versus 57.99
 (-960)
RL  RO
 40 k  9.4k 
which is a drop in level but not as significan t as the change in Av
 Rg  Ri
d A i T  A v T 
 RL
e A g  A 2v
 
R i  R g  R i  
Ri 
 2.7k 
  A v
  57.99

  A v
  (100.94)
 4.7k 
  R i  R g   R L   R L 
Ri
2.7k
 (100.94) 2
 5853.19
RL
4.7k
12