Transformers in Single-Tuned and Double-Tuned Filters
14.17
and 0.2 MHz, respectively. However, the signal frequency components in 0.9MHz to 1.1MHz range
get a variable gain – varying from 100% to 70.7%. This will lead to
waveform distortion in the
demodulated waveform. The signal after demodulation will be a distorted version of signal that was
transmitted. The gain offered to components in the frequency range 0.9MHz to 1.1MHz must be a
constant if this kind of signal distortion is to be avoided. We can achieve this by placing two resonant
humps at two frequencies that are nearby in frequency response of the tuned amplifier. This is achieved
in a double-tuned amplifier.
A double-tuned amplifier is a voltage-driven amplifier and uses a
weakly coupled transformer and
two capacitors – one in series with the primary and one in series with the secondary and load. We
consider a 1:1 transformer with equal capacitors in the primary and secondary for illustration. The
secondary is loaded resistively with a resistor equal in value to the source resistance in the primary
side. We ignore the winding resistances for simplifying the analysis. The double-tuned stage is shown
in Fig. 14.5-4. Transistor circuitry provides the input voltage signal to this filter.
R
R
L
L
C
v
s
(
t
)
v
o
(
t
)
C
+
+
−
−
1:1
k
< 1
Fig. 14.5-4
Doubled-tuned band-pass filter
We first study the AC steady-state input impedance of the transformer to understand how two
tuning frequencies appear in the circuit. We assume that the resistor
R is small compared to
L
C
and
ignore the resistor in this study. See Fig. 14.5-5.
L
L
C
Z
(
j
w
)
j
w
(
L
1
−
M
)
j
w
(
L
2
−
M
)
j
w
M
1
j
w
C
1:1
(a)
(b)
k
< 1
Fig. 14.5-5
Simplified circuit for study of input impedance
The circuit in Fig. 14.5-5 (a) is translated to the phasor equivalent circuit in Fig. 14.5-5 (b) using
conductive equivalent circuit for the transformer. We require the input impedance of the circuit in Fig.
14.5-5 (b) for sinusoidal steady-state. This may be obtained by applying impedance series parallel
combination rules. The result is given in the following:
Z j
j L
LC
k
LC
(
)
[
(
)]
[
]
w
w
w
w
=
−
−
−
1
1
1
2
2
2
We have used the relation
M
k L L
kL
=
=
1 2
in arriving at this impedance function. The impedance
is always reactive. It is an inductive reactance for 0
1
≤ <
w
LC
. It is
a capacitive reactance for
14.18
Magnetically Coupled Circuits
1
1
1
2
LC
LC
k
< <
−
w
(
)
. It is an inductive reactance again for
w
<
−
1
1
2
LC
k
(
)
. It becomes zero at
w
=
−
1
1
2
LC
k
(
)
. It changes from an infinite valued inductive reactance to infinite valued capacitive
reactance as
w
crosses
1
LC
from left to right.
The plot of this reactance function is shown in Fig. 14.5-6. Also shown in the same figure in dotted
curve is the magnitude of capacitive reactance of the capacitor connected in series with the primary.
L
=
115
m
H,
C
=
200pF and
k
=
0.5 for the circuit for which these reactance curves were prepared. The
resonant frequency of
L and
C is 1 MHz.
2000
1500
1000
500
−
500
0.5
A
B
1
1.5
2
f
in MHz
−
1000
−
1500
−
2000
Z
(
j
w
)/
j
Fig. 14.5-6
Input reactance in a double-tuned circuit with no load resistance
The input impedance of the circuit in Fig. 14.5-5 (a) is the sum of solid curve and negative of the dotted
curve in Fig. 14.5-6. This sum will go zero at frequencies corresponding to A and B in Fig. 14.5-6. The
corresponding frequencies are 0.835 MHz and 1.45 MHz. The circuit is resonant at these two frequencies.
When
the load resistance R is connected in the secondary, the resonant frequencies will shift.
However,
the shift will be small if R <<
L
C
.
The circuit in Fig. 14.5-4 can be analysed to obtain the steady-state frequency response function
H(
j
w
) relating the output voltage to input voltage. It is possible to show that the frequency response
function between load voltage and source voltage is
H j
k
j
LC RC
LC R C
LC
k
j
RC
(
)
(
)(
)
[
)
(
) (
)]
(
)(
w
w
w
w
w
=
−
−
+
+
−
−
3
2
2
2
4
2
2
1 2
1
2
1
−−
w
2
LC
)
The term within the square brackets in the denominator goes to zero at
w
w
1
2
1
1
≈
−
≈
+
k
LC
k
LC
and
if
R
L
C
<<
. The
gain magnitude of H(
j
w
) will exhibit a peak value of 0.5 at these two frequencies.
The plot of magnitude of frequency response function between source voltage and output voltage for
single-tuned filter and double-tuned filter is shown in Fig. 14.5-7. The source resistance was 700
W
and
load resistance was 700
W
in both cases. Desired centre frequency is 1.05 MHz and desired bandwidth
Analysis of Coupled Coils Using Laplace Transforms
14.19
is 165 kHz. The single-tuned design makes use of a 1:1 transformer with
k
≈
1, primary and secondary
inductance of 8
m
H each and a 2.8nF tuning capacitor connected across primary. Secondary is loaded
with a 700
W
resistor. Gain magnitude for this filter is shown in dotted curve in Fig. 14.5-7. The double-
tuned filter uses a 1:1 transformer with
k
=
0.11, primary and secondary inductance of 1.045mH and
two capacitors of value 22pF each in series with the windings. A 700
W
in series with the capacitor and
secondary winding load the filter. Gain magnitude for this filter is shown in solid curve in Fig. 14.5-7.
Two resonant peaks due to double tuning can be seen clearly in the frequency response plot. Note
that while keeping the centre frequency and bandwidth the same as in single-tuned filter, pass-band
gain variation in double-tuned filter is lesser than in the case of single-tuned filter. The skirts on either
side of pass-band are steeper. The stop-band performance of double-tuned filter too is superior to that
of single-tuned filter.
0.55
0.5
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
Frequency in MHz
Gain
Magnitude
Double-tuned
Single-tuned
Fig. 14.5-7
Magnitude of frequency response function for single-tuned and double-tuned
band-pass filters
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