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 RP represent the losses of the tank circuit comprising L1 and C, and RTNK those of L2 and C2 (please see Fig. 1).

The load consists of a diode detector D in series with an audio load RL (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 L1 and L2, 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 RTNK, 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:






We shall work on this later.

Some circuit equivalents

In Fig.1, let´s replace L1 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 C2 / N2 in parallel with a resistance N2R2. Both are in parallel with the magnetizing inductance K2L1 (please see Fig.4.a).












K2L1 and C2 / N2 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 N2R2 can be transformed into a resistance RP1 in parallel with RP. Using the known series-to-parallel “loss resistance” transformation we get:





Let . Then:



Next, we compute the equivalent resistance RT of the parallel combination of RP and RP1. It is given by:




Substituting RP1 by its equivalent given by eq.(3):






Letting  (unloaded Q of L1-C tank) we obtain:




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:




Applying a parallel-to-series transformation, the equivalent circuit of Fig.6.b is obtained.







The series transformed resistance Rse is given by:








Let  and .

Rs is the series term due to RP and Rs1, that coming from the coupled resistance N2R2. Now, maximum RF power transfer to N2R2 occurs when:




or when:









From part I of this study we know that:















Eq.(6) is now written as:




Letting , eq.(8) takes the more compact form:




Solving for K we obtain:




which gives the value of the coupling coefficient for maximum RF power transfer to the secondary.


Power calculations      

The 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 + Rs = Rs1. The maximum available power is then:





where Ea is the peak value of the voltage induced in the antenna.

The power delivered to the secondary load R2 is the same as that dissipated by the coupled resistance N2R2. To compute this power we need the voltage across L1 at resonance. This is the same as the voltage across Leq in Fig.6.a. Then:




From Fig.5 we obtain for the voltage across N2R2:




In crystal sets, , due to the loose coupling between L1 and L2. Then:








Bearing in mind eq.(3):




Substituting the value of ELeq given by eq.(11) into the above expression we obtain:







We can recall that:












Then, we obtain:




The power dissipated by N2R2 is:






Eq.(12) can be written as follows:













Substituting this equivalence into eq.(13):



according to eq.(10).

PMAX is then dissipated by N2R2 and by consequence, this power is delivered to the secondary load.

Some experimental results

Two coils, L1 and L2, were wound on 4.5” diameter styrene forms using 660/46 Litz wire. L1 measured 152 uH and L2, 222 uH. A two-gang 475 pF variable capacitor with bakelite insulation was used to tune L1. L2 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.














354.32 kohms

598.816 kohms

450 pF

406 pF

1 MHz




558.70 kohms

878.766 kohms

100 pF

114 pF

1.7 MHz




487.072 kohms

754.065 kohms

31 pF

39.5 pF


C            : two-gang 475 pF variable capacitor with bakelite insulation

C2               : 480 pF variable capacitor with ceramic insulation

Q2UL         : unloaded Q of L2-C2 combination

Q1, Q2     : defined in the text

RP, RTNK: 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                         

L1 = 152 uH                        Ca = 200 pF (assumed)               r = 30 ohms (assumed)                            

A = 1.555 x 105 ohms        Q1Q2A = 2.938 x 1010 ohms        K = 4.165 x 10-3



f  = 1 MHz

L1 = 152 uH                        Ca = 200 pF (assumed)               r = 30 ohms (assumed)

A = 1.9 x 105 ohms             Q1Q2A = 2.334 x 1010 ohms       K = 5.663 x 10-3



f  = 1.7 MHz

L1 = 152 uH                        Ca = 200 pF (assumed)               r = 30 ohms (assumed)                        

A = 4.068 x 105 ohms         Q1Q2A = 1.29 x 1010 ohms          K = 8.324 x 10-3




The 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 R2, or, as we are dealing with maximum power transfer to R2. 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 1st 2004