Analysis of the Tuggle Front End – Part II
We shall now consider the Tuggle tuner
delivering power to a load. First, we must account for the parallel RF losses
of the unloaded tuned circuits. Let R_{P} represent the losses of the tank
circuit comprising L_{1} and C, and R_{TNK} those of L_{2}
and C_{2} (please see Fig. 1).
The load consists of a diode detector D in
series with an audio load R_{L} (usually an audio transformer matching
a pair of 2k ohms DC resistance magnetic headphones or low-impedance sound
powered phones to the detector) and is coupled to the tuner via the magnetic
coupling existing between L_{1} and L_{2}, being M the mutual
inductance of the coils. The secondary is tuned to the same radian frequency as
the primary. An schematic for the load can be seen in Fig. 2. Usually, it is assumed that optimum RF power
transfer occurs when the antenna-ground system resonance resistance r is
matched to the unloaded-secondary resonance parallel resistance R_{TNK},
with the diode detector´s input resistance matched to this combination. Thus, _{} or _{}. This is the overall parallel RF resistance of
the secondary tank under matched conditions, and suggests that the unloaded Q
of this tank circuit has been reduced to ¼ of its value. In this case, the net parallel resistance to be
coupled to the primary is:
_{}
_{} …(1) Some circuit equivalents
In Fig.1, let´s replace L_{1} and the
coupled secondary circuit by the equivalent shown in Fig.3.a, which in turn can
be replaced by the transformer circuit shown in Fig.3.b. In the transformer circuit, the impedance
coupled to the primary side consists of a capacitance C_{2 }/ N^{2}
in parallel with a resistance N^{2}R_{2}. Both are in parallel
with the magnetizing inductance K^{2}L_{1} (please see
Fig.4.a). K^{2}L_{1} and C_{2} /
N^{2} resonate at a frequency _{ } _{ } _{} as _{}. The equivalent circuit of Fig.4.a reduces to
that of Fig.4.b, taking into account that for crystal set use, normally
K<<1. Up to this point, in the tuner side we have the
equivalent circuit depicted in Fig.5. The series-coupled resistance N^{2}R_{2}
can be transformed into a resistance R_{P1} in parallel with R_{P}.
Using the known series-to-parallel “loss resistance” transformation we get:
_{} …(2) Let _{}. Then:
_{} …(3) Next, we compute the equivalent resistance R_{T}
of the parallel combination of R_{P} and R_{P1}. It is given
by:
_{} Substituting R_{P1} by its equivalent
given by eq.(3):
_{}
_{} Letting _{} (unloaded Q of L_{1}-C
tank) we obtain:
_{} …(4) If _{} (please refer to part
I of this study) we can redraw the equivalent circuit of the tuner at resonance
as indicated by Fig.6.a, and, by virtue of the above inequality, the resonant
frequency _{} will still be given
by:
_{} …(5) Applying a parallel-to-series transformation,
the equivalent circuit of Fig.6.b is obtained. The series transformed resistance R_{se}
is given by:
_{}
_{}
_{} Let _{} and _{}. R_{s} is the series term due to R_{P}
and R_{s1}, that coming from the coupled resistance N^{2}R_{2}.
Now, maximum RF power transfer to N^{2}R_{2} occurs when:
_{} or when: _{} or _{} …(6) From part I of this study we know that:
_{}
_{} …(7) being
_{} Then: _{} _{} Eq.(6) is now written as: _{} …(8) Letting _{}, eq.(8) takes the more compact form:
_{} Solving for K we obtain:
_{} …(9) which gives the value of the coupling
coefficient for maximum RF power transfer to the secondary. Power calculationsThe RF power delivered to the secondary load of
Fig.1 will be at a maximum at resonance when eq.(6) is satisfied, this is, when
r + R_{s} = R_{s1}. The maximum available power is then: _{}
_{} …(10) where E_{a} is the peak value of the
voltage induced in the antenna. The power delivered to the secondary load R_{2}
is the same as that dissipated by the coupled resistance N^{2}R_{2}.
To compute this power we need the voltage across L_{1} at resonance.
This is the same as the voltage across L_{eq} in Fig.6.a. Then: _{} …(11) From Fig.5 we obtain for the voltage across N^{2}R_{2}:
_{} In crystal sets, _{}, due to the loose coupling between L_{1} and L_{2}.
Then:
_{}
_{} _{} Bearing in mind eq.(3):
_{} Substituting the value of E_{Leq} given
by eq.(11) into the above expression we obtain: _{} We can recall that:
_{} …(12) Then:
_{}
_{}
_{} Then, we obtain: _{} The power dissipated by N^{2}R_{2}
is:
_{}
_{} …(13) Eq.(12) can be written as follows: _{}
_{}
_{}
_{} or:
_{} Substituting this equivalence into eq.(13): _{} according to eq.(10). P_{MAX} is then dissipated by N^{2}R_{2}
and by consequence, this power is delivered to the secondary load. Some experimental resultsTwo coils, L_{1} and L_{2},
were wound on 4.5” diameter styrene forms using 660/46 Litz wire. L_{1}
measured 152 uH and L_{2}, 222 uH. A two-gang 475 pF variable capacitor
with bakelite insulation was used to tune L_{1}. L_{2} was
tuned with a 480 pF variable capacitor with ceramic insulators. Unloaded Qs for each of the tuned circuits were
measured at three frequencies. Accordingly, the corresponding RF losses were
calculated. Data is tabulated below.
C
: two-gang 475 pF variable capacitor with bakelite insulation C_{2 }: 480 pF variable
capacitor with ceramic insulation Q_{2UL }: unloaded Q of L_{2}-C_{2}
combination Q_{1}, Q_{2 }: defined in the text R_{P}, R_{TNK}: defined in the
text Using the tabulated data, values for the
optimum coupling coefficient K will be calculated for a working crystal set. f = 530 kHz L_{1 }= 152 uH Ca = 200 pF (assumed) r = 30 ohms (assumed) A = 1.555 x 10^{5} ohms Q_{1}Q_{2}A = 2.938 x
10^{10} ohms K = 4.165 x
10^{-3} Check: _{} f = 1
MHz L_{1} = 152 uH Ca = 200 pF (assumed) r = 30 ohms (assumed) A = 1.9 x 10^{5} ohms Q_{1}Q_{2}A =
2.334 x 10^{10} ohms K =
5.663 x 10^{-3} Check: _{} f = 1.7
MHz L_{1} = 152 uH Ca = 200 pF
(assumed) r = 30 ohms (assumed) A = 4.068 x 10^{5} ohms Q_{1}Q_{2}A = 1.29 x
10^{10} ohms K = 8.324 x
10^{-3} Check: _{} _{} CommentsThe values obtained for the coupling
coefficient K hold for _{}, as discussed previously. The transformed antenna-ground
system resonance resistance, as seen from the secondary, will be equal to R_{2},
or_{}, as we are dealing with maximum power transfer to R_{2.}
Under these conditions, the loaded Q of the secondary circuit will be 1/6 of
the unloaded value. For other load conditions, the respective data should be
entered into eq.(9).
Ramon
Vargas Patron Lima-Peru,
South America
April 1^{st} 2004 _{} _{} _{} |