Here in Part 3, we take off in another direction. We are going to analyze a method of bandspreading where the bandspread capacitor is connected to a tap on the tank inductor, and a bandset capacitor is connected across the entire coil. This configuration is shown in Figure 1.
This is a system that has been used on both commercial and homebrew shortwave receivers dating back to the 1930’s. The circuit is as shown in figure 1. In this circuit, the main “Bandset” variable capacitor is CA, and may be either fixed or variable, while the “Bandspread” capacitor CB is variable.
In essence, the goal is to find a relationship between circuit impedance across terminals T1 & T2 and frequency for given values of capacitance and inductance.
Before we go any further, we may ask ourselves why we want to do something that looks quite a bit more complicated than the bandspread arrangements we have already looked at. The answer, as we will discover, is that this gives a remarkably linear relationship between frequency and tuning capacitance, as the following graph illustrates:
In this analysis we will treat the coil as an autotransformer keeping in mind that traditional transformer theory diverges somewhat from loosely coupled coil theory. With transformer theory, the coupling coefficient between windings is nearly unity, which allows for a few simplifications. Therefore, in this analysis, we will need to adjust the model to account for the fact that the sections of the coil are loosely coupled. (This work is based largely on the information provided in the classic texts: ‘Radio Engineering’ and ‘Radio Engineer’s Handbook,’ both by Frederick E. Terman.)
When two coils are inductively coupled, then a current in one coil will induce a magnetic flux in the other coil and vice versa. When the current in one coil changes, the flux changes, and the changing flux will induce a voltage in the other coil according to this relationship:
where the subscripts 1 and 2 designate the parameters of the first and second coil respectively. M is the mutual inductance between the two coils, and the voltages and currents are sinusoidal at an angular frequency ω = 2πf. (ω is the lowercase Greek letter omega.) If the two coils are perfectly coupled, then
If they are not perfectly coupled then we introduce a coupling coefficient into the relationship thus:
where k ranges between zero (no coupling at all) and one (perfect coupling).
From Figure 1:
L3 is the inductance of the upper section of the coil, and has N3 turns
L2 is the inductance of the bottom section of the coil, and has N2 turns
L1 is the total inductance of the whole coil and has N1 = N2+N3 turns
If the coupling coefficient is one, then there is a direct relationship between number of turns and inductance, and the following relationships apply:
However, except as specifically noted below we will not make any assumptions about the value of the coupling coefficient.
Note that L1 is not equal to L2+L3 (except when k = 0). The inductance of the entire coil is, in fact, given by:
We will leave the circuit of Figure 1 for the time being, and deal with an equivalent coupled circuit consisting of completely separate windings as shown in Figure 2.
In this circuit, we will still assume that even though the coils are separate, they are still inductively coupled. Notice also, we keep the same terminology between figures 1 and 2 such that CA is still connected across L1 and CB is still connected across L2. We will also assume that L1 has the same number of turns in both figures and L2 has the same number of turns in both figures. Consequently, for the purpose of this analysis, both circuits will behave essentially the same. The reason for separating the bandspread section of the circuit into an electrically isolated winding is to make the circuit analysis a bit simpler to follow.
Now, let us assume that a sinusoidal alternating voltage VT is applied across terminals T1 and T2. This causes an alternating current i1 to flow in the primary circuit. This is illustrated in figure 3, which shows the circuit components as impedances.
From the previous definition of mutual inductance, then this alternating current induces a voltage VZ2 in the secondary circuit across Z2, which in turn causes a current i2 to flow in the secondary circuit.
Now, current i2 in the secondary also induces a voltage VZ1 back in the primary across Z1.
The voltage and current relationships are given by the following formulae:
The second term in equation (12) shows that the effect of the secondary circuit corresponds to a voltage drop. We can substitute in equations (13) and (14) to get this voltage drop in terms of i1.
From this we can define Z', the coupled impedance as:
Using equation (16) we can simplify the circuit of figure 3 into a single loop as shown in figure 4.
In summary then, the effect that the current flowing in the secondary circuit has on the primary circuit can be treated as another voltage drop VZ' due to an impedance Z' applied to the primary circuit as shown in figure 4. Here, Z' is placed in series with the other primary circuit components, and completely accounts for the effect due to the secondary circuit. We have now reduced the original two loop circuit to a single loop circuit.
The net impedance ZN of the primary circuit is:
where || indicates components combined in parallel.
Substituting (11) into (17) we put M in terms of L1 L2 and k.
If we neglect any resistance in the circuit, then the impedances are purely reactive. The impedance of an inductor is given as:
Taking everything on the right to a common denominator:
Remembering that j2 = −1:
Notice that in the numerator, the first and last terms will cancel each other out when k = 1. Let's assume for the moment that this is the case. Then we are left with:
Also when k = 1, the n2 relationship exists between inductance values as previously mentioned:
where we define n as the turns ratio. Substituting L2/n2 for L1
Notice that we now have a fraction in the form of AB/(A+B) which is the formula used when combining parallel impedances. Therefore, this is equivalent to having A and B in parallel. Hence:
So, for the case of k = 1 (which would generally apply in the situation of a coil on a ferromagnetic core), we can redraw our circuit as shown in figure 5.
Note that when we convert from impedance to capacitance, 1/n2 becomes n2. Also, since n = N2/N1, and we assume that because N2 represents the tapped part of the coil in the original circuit, it is smaller than N1. Hence, n will be less than 1. Therefore, when CB is transferred to the primary side, its effective value becomes smaller.
We have demonstrated that if the coupling coefficient is 1, then the equivalent circuit is reduced to the secondary impedance multiplied by the turns ratio and transferred to the primary in parallel with the primary inductance and capacitance, which agrees with traditional transformer theory. The capacitor CB on the secondary side can be scaled by multiplying by n2 and placed in parallel with CA on the primary side. However, it's worth noting one important fact. In traditional transformer theory, the self inductance of the primary is generally considered much higher than the transferred impedance from the secondary side, and is often neglected. However, we can't neglect it in our case, since it forms the inductive part of the resonant circuit.
Since we have carried the effect of the coupling coefficient k through to formula (21), we can go back a few steps and see the effect of a coupling coefficient less than 1.
Starting with (21) again:
First we combine the common ω2L1L2 terms
Next, we invert the formula so that we can combine the parallel impedances:
Also note that the reciprocal of the impedance is the admittance, and at parallel resonance the admittance 1/ZN (neglecting any resistive components) will be zero. Hence:
Capacitive reactance is given by:
Substituting this for the ZCx terms in the formula:
Multiplying everything by the denominator of the fraction:
and then multiplying by jωCB we get a simpler formula:
Since there are no ω or ω3 terms we can solve this as yet another quadratic equation:
Note, that a quadratic equation will generally have two solutions, because the square root portion can be either positive or negative. For normal situations, the form given above will provide the correct frequency. The other result is equally valid mathematically, and will give a much higher resonant frequency which we may interpret as a parasitic. Therefore, it may be prudent to ensure that related circuit components are selected to prevent operation at undesired frequencies.
Now, we can plug the component values into the above formula and solve for ω2, and consequently solve for ω and frequency.
The only value that will not be readily known is the coupling coefficient. However it can be calculated from the other coil parameters.
If we are dealing with the isolated windings of the circuit in figure 2, then we can measure the inductance of the two windings separately, and then connect them in series, and measure them again. Using the following formula, the mutual inductance can be calculated:
The contribution due to M12 will be positive or negative depending on whether the polarity of the series connections are aiding or opposing, and the sign and value of M12 will be self-evident.
If we are dealing with the single tapped winding of the circuit in figure 1, then we can directly measure the values of L1, L2 and L3.
Then the mutual inductance between L2 and L3, is
And it can be shown (see Appendix below) that the mutual inductance between L1 and L2 is
Now that we have determined M12 by one of the above methods, the coupling coefficient is then given by:
or combining these two formula, we get:
This now gives us all the information required to calculate resonant frequency of the bandspread tank circuit.
The relationship between component values, and resonant frequency of a tapped coil bandspread circuit has been given by formula (31):
where the coupling coefficient k is given by:
and then (31) can be solved for frequency using the usual quadratic formula:/p>
Having calculated x from the above, the resonant frequency F will be
where F is in Hz, and the other component values are given in Farads and Henries.
For convenience, the calculation may be done using picofarads and microhenries, but the resulting frequency will then be in Gigahertz, and therefore will need to be multiplied by 1,000,000 to get kHz. Hence:
In contrast to Parts 1 and 2, where we selected an arbitrary frequency range and then developed formulae to find the necessary components values, here we have developed a formula to find the resonant frequency from a given set of components. To develop formulae to find the required component values is beyond the scope of this discussion. In this case we are better off availing ourselves of a spreadsheet using these formulae, and then plugging in various values until we have the desired frequency range. If it turns out that the practical values of inductor and bandspread capacitor we have to work with don’t give a narrow enough frequency range, we can of course, reduce the range of the bandspread capacitor as we did in Parts 1 and 2 by adding a padder and/or trimmer.
For this example we will specify a tuning range of 5500 to 6500 kHz, using a variable capacitor CB with a range of 10 to 140 pF. We will then determine the coil inductance tap position and bandset capacitance CA which will give us the specified tuning range. As mentioned in the above derivation, the coupling coefficient plays an important part in determining frequency. We can't arbitrarily pick a value for this without it affecting the inductance relationships between the various sections of the coil. Our approach will be to model a real coil. Therefore, we will define the coil by its length, diameter, and number of turns, and from this, we can calculate L1, L2, L3 and k12.
The circuit is modeled in this spreadsheet. It is an OpenOffice/LibreOffice .ods open document spreadsheet file, which is compatible with current versions of most spreadsheet applications including Excel.
The input parameters are shown with a yellow background. Below the input parameters are the inductance calculations.
The inductance formula is an optimized version of Wheeler's 1982 continuous inductance formula. It is accurate to 0.002%. In addition, round wire corrections are included to account for wire spacing. The overall inductance calculation accuracy will be at least as good as our homebrew coil making ability.
Below the coil calculations are the tapped coil bandspread frequency calculations. There is a calculation for both the lower and upper frequencies.
Below that is a calculation showing the difference between the actual and target values, plus a net error calculation (shown with aqua background) which will be equal to zero when we are on target.
Finally, there is a graph showing the tuning characteristic.
There are five different coil input parameters, all of which will affect all of the coil's electrical properties. We don't really want to have to tweak all of the values. So, the approach will be to pick some starting dimensions which put us in the correct approximate range, and then adjust only one coil parameter. Therefore we will pick an initial value for the bandset capacitor of 85 pF. To resonate with this at the upper end of band, 6500 kHz, we will need an inductance of about 7 µH. (We estimate the components for the top of the band, because the effect of the bandspread capacitor will be to lower the frequency.)
With the aid of an inductance calculator, we can design a suitable coil with a 2.5 cm diameter, length of 3 cm, and 20 turns of 1 mm diameter wire. These are entered into the spreadsheet. A rule of thumb for the tap position is to pick a position slightly below the halfway point. We can select 7, 8 or 9, and if these give too narrow or too wide a tuning range, we can try other tap positions. Now, once a coil has been constructed, an easy way to fine tune the inductance is to change the winding length by squeezing or stretching it. Coil length appears to be a good parameter to adjust in the calculation as well. The benefit of this is that in the circuit which is finally constructed, we will have the same physical ability to adjust the coil.
To proceed, we enter these parameters and then adjust the value of CA, and the coil length until we reach the optimum. As we near the optimum, the graph will show the tuning characteristic move into the correct position, and the Net Error value will decrease and approach zero. It's simply a matter of trial and error. With a tap position of 9, we eventually find the optimum bandset capacitance to be 79.29 pF, and the coil length to be 2.82 cm. We can set the coil tap to 8 and re-optimize. This time the required bandset capacitance is 60.64 pF, and required coil length is 1.65 cm. So, either solution will work. Following is the resulting tuning characteristic for the coil tap at position 9:
And for coil tap at position 8:
As can be seen, the two are indistinguishable. With coil tap positions of 7 or 10, it appears that the linearity begins to suffer, although I must admit my method of measuring linearity is to hold a straight edge to the computer screen and compare.
Because the required bandset capacitance is not a standard value, we would use a trimmer, or combination of fixed capacitor plus trimmer in the final circuit, in order to set the tuning range accurately.
A useful discovery, from working through the numbers: changing the bandset capacitance has the effect of shifting both the lower and upper frequencies, while having only a small effect on the tuning range (i.e., the difference between lower and upper frequency). Changing the coil length also shifts the both frequencies but has a greater effect on the tuning range. The tuning range varies inversely with the coil length. Hence, compressing the coil expands the tuning range, while expanding the coil compresses the tuning range. Perhaps this mnemonic may be of some use to us when we find ourselves stuck in the iterative process of aligning a homebrew receiver.
One final note: In Section 3a of the discussion of Numerical Methods, I present an easy automated way to do the calculation discussed here, using a Solver macro in the spreadsheet.
Now, let's verify that the numbers from the spreadsheet are correct, by substituting them back into the quadratic solution derived from formula (31).
The results from the spreadsheet are:
CA = 79.29 pF
L1 = 7.3748 µH
L2 = 2.6006 µH
L3 = 3.4085 µH
Then the coupling coefficient is:
At the lowest frequency, the setting of the bandspread capacitor will be:
Substituting these values into the formulae for the quadratic coefficients:
And then substituting these into the quadratic formula:
Since we used units of µH and pF, we will have to use a factor of 1,000,000 in the final frequency formula:
which is the required lower frequency in kHz.
For the upper end of the band, the setting of the bandspread capacitor will be:
Substituting the values into the formulae for the quadratic coefficients:
And then substituting these into the quadratic formula:
And then the upper frequency, in kHz, is:
which is the required upper frequency. Therefore, we have verified that these component values will give the desired tuning range.
It was mentioned above that the quadratic formula has two possible solutions, and we have chosen the one which gives the lower frequency. Let's change the sign of the square root term to see what the other solution gives.
For CB=140 pF:
which gives the low end frequency in kHz:
And then, for CB = 10 pF:
which gives the high end frequency in kHz:
As previously mentioned these two alternative frequencies are mathematically valid, and there appears to be no reason, electronically why the circuit would not resonate at these frequencies. Therefore, it would be wise to design the rest of the circuit with this in mind, using whatever means necessary, to snub out parasitic resonances.
I would like to express my thanks to Mike Tuggle who provided valuable feedback and also performed SPICE simulations verifying the formulae.
Using the same notation as used previously, L1 is the inductance of the entire coil, L2 is the inductance of the bottom section, and L3 is the inductance of the top section.
If an AC voltage source v2 is impressed across L2, the current i2 will be:
The voltage v3 induced in L3 will be:
The voltage v1 induced in L1 will be:
Since voltages in series must add up, the voltage across the entire coil v1 will be:
Putting these voltages in terms of i2 from the previous relationships, we get
Taking out the common factor jωi2, this simplifies to
which is what we set out to demonstrate.
Continue to: Part 4