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[Components]
Match Loop Antennas Via Mutual Inductance
This approach simplifies matching printed-circuit loop antennas to transmitter and receiver circuits in small electronic devices for one- and two-way UHF applications.

Alan Bensky  |  ED Online ID #22325 |  January 2010

Loop antennas are widely used in small wireless products, particularly for UHF bands between 300 and 1000 MHz. They are small in size relative to wavelength, independent from a ground plane, and relatively immune to the influence of nearby objects. They are also easily implemented in printedcircuit form with corresponding low cost. But their low radiation resistance makes them difficult to match and subject to low efficiency. They exhibit high quality factor (Q) and associated narrow bandwidth that can make performance nonuniform over a normal range of matching component tolerances. Fortunately, mutual inductance techniques can help reduce loop-antenna losses and achieve optimal impedance matching.

In many applications, a simple circuit matches the low-resistance component of a loop antenna to the generally higher output impedance of a transmitter or the input impedance of a receiver [Fig. 1(a)]. Figure 1(b) is an equivalent-circuit representation with the loop impedance shown as loop inductance LL and resistance R1. R1 is the sum of the radiation resistance, antenna losses, and the series resistance of capacitance C1. The antenna efficiency is the ratio of the radiation resistance to R1. In Fig. 1(a), C1 is shown in series with the loop; L1 in Fig. 1(b) is the equivalent of the loop inductance LL, reduced by the negative reactance of C1. The equations for L1 and C2 to match R1 to R2 are

The values of R1 and LL must be estimated from the dimensions of the loop, and the impedance to which the loop is matched, and R2 (typically 50 Ohms) must be known.1 The value of C1 can then found from

where

f = the frequency of operation.

Loss resistance R1 is the sum of the radiation resistance, loop conductor resistance, component losses, and losses due to surrounding objects. Equation 4 provides the radiation resistance (Rrad) in Ohms:

where

A = the area enclosed by the loop

and

 = the wavelength with dimension units corresponding to A.

The loop conductor loss (in Ohms) can be approximated by

where

len = the loop perimeter,

w = the conductor width in the same units as len, and

freq = the frequency in MHz.

Equation 6 gives a good approximation for the inductance of a printed rectangular loop (in nH) with sides s1 and s2 and conductor width w, (with units in mm):2

The matching configuration of Fig. 1 (has the disadvantage that, for a given set of entry parameters, only one value each is found for C1 and C2, and these results may not even be close to standard capacitor values. Matching configurations in Figs. 2(a) and 2(b) provide an additional degree of freedom so that at least one of the matching component values can be chosen as a standard value, preferably C1 since it is most critical, and the others calculated accordingly (Eqs. 7-13). As before, the value of C1 is found using Eq. 3. The resulting component values are varied by adjusting the chosen Q according to Eq. 7:

From Fig. 2 (a),

with C2 and C3 in Eqs. 9 and 10, and from Fig. 2 (b),

and C2 and L2 in Eqs. 12 and 13.

Although it is necessary to know the loop inductance and total resistance to calculate matching-network component values, this knowledge may not come routinely. While there are fairly accurate formulas for coil inductance, the actual inductance is often higher than the calculated value due to spurious capacitance across the coil terminals. This capacitance can be due to nearby conducting objects or printed-circuit-board (PCB) traces referenced to ground. An example helps to demonstrate the effect of the spurious capacitance on the effective impedance and resistance of the loop.

Assume a printed square loop measuring 30 x 30 mm with 2-mm trace width and frequency of 434 MHz. From Eqs. 4, 5, and 6, R = 0.274 Ohm and L = 82 nH. Now, assume a spurious capacitance of C = 0.8 pF, and the circuit then has the form of Fig. 3. The impedance of this circuit at 434 MHz can be found to be Z = 1 + 436j Ohms. The inductance of the circuit, found from the reactance of 436 Ohms, is 160 nH. The resistance to be matched to the transmitter or receiver output or input impedance is 1 Ohm, plus the resistive loss of the series capacitor of the matching circuit, which will be approximately 0.3 Ohms.3

The spurious capacitance is difficult to measure directly, but it can be determined by calculation from the resonant frequency of the loop when it is connected in parallel to two different capacitors. Measurements are performed as follows. In the circuit of Fig. 1(a), a short circuit is used in place of C2. A standard value should be chosen for C1 that is close to the value that resonates with the value of LL that is calculated from the loop dimensions; this capacitance is called Ca. The true resonant frequency is measured using a vector network analyzer (VNA) or scalar network analyzer (SNA) configured for one-port measurements. A small round wire loop of approximately 10 mm diameter is attached to the end of a coaxial cable connected to the analyzer test port. When the coil is brought near to the loop, a narrow dip will be observed on the analyzer’s display (Fig. 4). The trough of the dip is the first resonant frequency, fa. The next step is to replace Ca with a capacitance value that is about 30-percent lower, called Cb, and measuring the resulting second resonant frequency, fb. The spurious capacitance and loop inductance are calculated as follows:

Then, the apparent inductance at the desired frequency f is

For example, assume a 2.2-pF parallel capacitor with observed resonance at 363.8 MHz. The capacitor is changed to 1.5 pF and the resonance is then observed at 415 MHz. From Eqs. 14 and 15, the spurious capacitance is 0.8 pF and the true inductance is 63.8 nH. At an operating frequency of 434 MHz, Eq. 16 shows that the apparent inductance is 102.8 nH. The effective loop resistance, derived from Fig. 3, is

In this example, R = 0.274 Ohms and, from Eq. 17, R’ = 0.71 Ohms.

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The article is very instructive, BUT the downloding of 3 article sections is sooo poor its the worst scheme I have ever met.(20y +) It's not printer friendly. How can each graphic appear on a seperate page within the magazine? (I was a subscriber at one time and that, just does not happen). Additionally in downloading 15 or so items: I know that may not have been inserted the download of every element into the file I opened up for the article. Life is short and I won't check for stray files here now.

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