Download TL circuits with half- and quarter

33 TL circuits with half- and quarter-wave transformers
GENERATOR,
Circuit
• Last lecture we established that phasor solutions of telegrapher’s equations for TL’s in sinusoidal steady-state can be expressed as
V +ejβd − V −e−jβd
V (d) = V e + V e
and I(d) =
Zo
in a new coordinate system shown in the margin.
+ jβd
Input
port
We have replaced the short termination of the previous lecture with an
arbitrary load impedance
ZL = RL + jXL.
In this lecture we will discuss sinusoidal steady-state TL circuit problems having arbitrary reactive loads but with line
lengths l constrained to be integer multiples of λ4 (at the operation frequency).
The constraint will be lifted next lecture when we will develop the general analysis tools for sinusoidal steady-state TL
circuits.
1
−l
d
l
LOAD
+
Zo
V (d)
-
− −jβd
By convention the load is located on the right at z = 0 = d, and the
TL input connected to a generator or some source circuit is shown on
the left at d = l.
I(d)
l
ZL = RL + jXL
0 z
0
• In the TL circuit shown in the margin an arbitrary load ZL is connected
to a TL of length l = λ2 at the source frequency.
Given that
2π λ
2
λ
e±jβ 2 = e±j λ
= e±jπ = −1,
the general phasor relations
V +ejβd − V −e−jβd
and I(d) =
Zo
λ
Vin ≡ V ( ) = −V + − V − = −V (0) = −VL,
2
λ
−V + + V −
Iin ≡ I( ) =
= −I(0) = −IL.
2
Zo
We conclude that a λ2 -transformer
– inverts the algebraic sign of its voltage and current inputs at the
load end (and vice versa), and
– has an input impedance identical with the load impedance since
Zin ≡
IL = −Iin
Iin
V (d) = V +ejβd + V −e−jβd
imply
Half-wave transformer:
Vin −VL
=
= ZL.
Iin
−IL
These very simple results are easy to remember and use.
2
+
Vin
+
Zo
VL = −Vin
ZL
-
λ
2
-
0
• In the TL circuit shown in the margin an arbitrary load ZL is connected
to a TL of length l = λ4 at the source frequency.
Given that
e
±jβ λ4
=e
λ
±j 2π
λ 4
=e
±j π2
= ±j,
general phasor relations
Iin
IL = −j
+
+
ZL
Vin
V (d) = V +ejβd + V −e−jβd and I(d) =
imply
Quarter-wave transformer:
+ jβd
V e
− −jβd
−V e
Zo
-
VL = −jIin Zo
Zo
λ
4
Vin
Zo
-
0
ZinZL = Zo2
λ
Vin ≡ V ( ) = jV + − jV − = jI(0)Zo = jIL Zo,
4
λ
jV + + jV −
V (0)
VL
Iin ≡ I( ) =
=j
=j .
4
Zo
Zo
Zo
We conclude that a λ4 -transformer
– has an input impedance
Vin
jIL Zo
Zo2
Zo2
Zin ≡
=
=
=
,
Iin
jVL/Zo VL/IL ZL
Quarter-wave
current-forcing
equation:
IL = −j
Vin
.
Zo
– and provides a load current
Vin
IL = −j ,
Zo
Load voltage
VL = ZLIL
proportional to input voltage Vin but independent of load impedance
once IL is available
ZL.
from above equation.
3
Iin
+
+
Example 1: Given ZL = 50 + j50 Ω, what is Zin for a
λ
4
transformer with Zo = 50 Ω?
Solution: It is
Zo2
502
50
50 1 − j1
Zin =
=
=
=
= 25 − j25 Ω.
ZL
50 + j50 1 + j1 1 + j1 1 − j1
Notice that an inductive ZL has been turned into a capacitive Zin by
λ
4
ZL = 50 + j50 Ω
Vin
Z
=
50
Ω
- o
λ
0
4
Zo2
Zin =
ZL
trans-
former.
+
Vg = 100" 0 V
Example 2: The load and the transformer of Example 1 are connected to a source
with voltage phasor Vg = 100∠0o V at the input port. What is the load current
IL and what is the average power absorbed by the load?
Solution: Since Vin = Vg = 100∠0o V, the current-forcing formula for the quarter-wave
transformer implies
Vin
100
IL = −j
= −j
= −j2 A.
Zo
50
To find the average power absorbed, we first note that load voltage
VL = ZLIL = (50 + j50)(−j2) = 100 − j100 V.
Thus,
1
1
PL = Re{VL IL∗ } = Re{(100 − j100)(j2)} = 100 W.
2
2
4
+
-
λ
4
50 + j50 Ω
Zo = 50 Ω
-
0
I(d)
Zg = 25 Ω
+
Vg =
j10 V +-
Zo = 50 Ω
V (d)
-
Example 3: Load ZL = 100 Ω is connected to a T.L. with length l = 0.75λ. At the
generator end, d = 0.75λ, a source with open circuit voltage Vg = j10 V and
Thevenin impedance Zg = 25 Ω is connected. Determine VL and IL if Zo = 50 Ω.
Solution: First we determine input impedance Zin by noting that ZL = 100 Ω transforms to itself, namely 100 Ω at d = 0.5λ, but then it transforms from d = 0.5λ
to 0.75λ as
Zo2
502
Zin =
=
= 25 Ω.
Z(0.5λ) 100
Hence, using voltage division, we find,
Vin = Vg
Zin
25
= j10
= j5 V.
Zg + Zin
25 + 25
Next, using half-wave transformer rule, we notice that
V (0.25λ) = −Vin = −j5 V,
and finally applying the quarter-wave current forcing equation with V (0.25λ) we
get
V (0.25λ)
−j5
IL = −j
= −j
= −0.1 A.
Zo
50
Clearly, then, the load voltage is
VL = ZL IL = (100 Ω)(−0.1 A) = −10 V.
5
d
0.75λ
0.5λ
0.25λ
ZL =
100 Ω
0
IL2 = −j
Example 4: In the circuit shown in the margin, ZL1 = 50 Ω, ZL2 = 100 Ω, and Zo1 =
Zo2 = 50 Ω. Determine IL1 and IL2 if Vin = 5 V. Both T.L. sections are quarterwave transformers.
Solution: Using the current-forcing equation, we have
IL1 = IL2
Vin
5
= −j
= −j = −j0.1 A.
Zo
50
Consequently,
VL1 = IL1ZL1 = −j0.1 A × 50 Ω = −j5 V
and
VL2 = IL2ZL2 = −j0.1 A × 100 Ω = −j10 V.
Thus, total avg power absorbed is
1
1
∗
Re{VL1IL1
} + Re{VL2I ∗L2}
2
2
1
1
= = Re{−j5 × j0.1} + Re{−j10 × j0.1} = 0.75 W.
2
2
P =
6
Vin
Zo
ZL2
Zo2
IL1 = −j
+
Vin
Zo1
ZL1
-
λ
4
0
Vin
Zo