Impedance matching

Impedance matching to minimize reflections is achieved by making the load impedance equal to the source impedance. If the source impedance, load impedance and transmission line characteristic impedance are purely resistive, then reflection-less matching is the same as maximum power transfer matching.[1]

Complex conjugate matching is used when maximum power transfer is required, namely

${\displaystyle Z_{\mathsf {load}}=Z_{\mathsf {source}}^{*}\,}$

where a superscript * indicates the complex conjugate. A conjugate match is different from a reflection-less match when either the source or load has a reactive component.

If the source has a reactive component, but the load is purely resistive, then matching can be achieved by adding a reactance of the same magnitude but opposite sign to the load. This simple matching network, consisting of a single element, will usually achieve a perfect match at only a single frequency. This is because the added element will either be a capacitor or an inductor, whose impedance in both cases is frequency dependent, and will not, in general, follow the frequency dependence of the source impedance. For wide bandwidth applications, a more complex network must be designed.

Transformers are sometimes used to match the impedances of circuits. A transformer converts alternating current at one voltage to the same waveform at another voltage. The power input to the transformer and output from the transformer is the same (except for conversion losses). The side with the lower voltage is at low impedance (because this has the lower number of turns), and the side with the higher voltage is at a higher impedance (as it has more turns in its coil).

One example of this method involves a television balun transformer. This transformer converts a balanced signal from the antenna (via 300-ohm twin-lead) into an unbalanced signal (75-ohm coaxial cable such as RG-6). To match the impedances of both devices, both cables must be connected to a matching transformer with a turns ratio of 2 (such as a 2:1 transformer). In this example, the 75-ohm cable is connected to the transformer side with fewer turns; the 300-ohm line is connected to the transformer side with more turns. The formula for calculating the transformer turns ratio for this example is:

${\displaystyle {\text{turns ratio}}={\sqrt {\frac {\text{source resistance}}{\text{load resistance}}}}}$

Resistive impedance matches are easiest to design and can be achieved with a simple L pad consisting of two resistors. Power loss is an unavoidable consequence of using resistive networks, and they are only (usually) used to transfer line level signals.

Most lumped-element devices can match a specific range of load impedances. For example, in order to match an inductive load into a real impedance, a capacitor needs to be used. If the load impedance becomes capacitive, the matching element must be replaced by an inductor. In many cases, there is a need to use the same circuit to match a broad range of load impedance and thus simplify the circuit design. This issue was addressed by the stepped transmission line,[2] where multiple, serially placed, quarter-wave dielectric slugs are used to vary a transmission line's characteristic impedance. By controlling the position of each element, a broad range of load impedances can be matched without having to reconnect the circuit.

Filters are frequently used to achieve impedance matching in telecommunications and radio engineering. In general, it is not theoretically possible to achieve perfect impedance matching at all frequencies with a network of discrete components. Impedance matching networks are designed with a definite bandwidth, take the form of a filter, and use filter theory in their design.

Applications requiring only a narrow bandwidth, such as radio tuners and transmitters, might use a simple tuned filter such as a stub. This would provide a perfect match at one specific frequency only. Wide bandwidth matching requires filters with multiple sections.

Basic schematic for matching R₁ to R₂ with an L pad. R₁ > R₂, however, either R₁ or R₂ may be the source and the other the load. One of X₁ or X₂ must be an inductor and the other must be a capacitor.

L networks for narrowband matching a source or load impedance Z to a transmission line with characteristic impedance Z₀. X and B may each be either positive (inductor) or negative (capacitor). If Z/Z₀ is inside the 1+jx circle on the Smith chart (i.e. if Re(Z/Z₀)>1), network (a) can be used; otherwise network (b) can be used.[3]

A simple electrical impedance-matching network requires one capacitor and one inductor. In the figure to the right, R₁ > R₂, however, either R₁ or R₂ may be the source and the other the load. One of X₁ or X₂ must be an inductor and the other must be a capacitor. One reactance is in parallel with the source (or load), and the other is in series with the load (or source). If a reactance is in parallel with the source, the effective network matches from high to low impedance.

The analysis is as follows.[4] Consider a real source impedance of ${\displaystyle R_{1}}$ and real load impedance of ${\displaystyle R_{2}}$. If a reactance ${\displaystyle X_{1}}$ is in parallel with the source impedance, the combined impedance can be written as:

${\displaystyle {\frac {jR_{1}X_{1}}{R_{1}+jX_{1}}}}$

If the imaginary part of the above impedance is canceled by the series reactance, the real part is

${\displaystyle R_{2}={\frac {R_{1}X_{1}^{2}}{R_{1}^{2}+X_{1}^{2}}}}$

Solving for ${\displaystyle X_{1}}$

${\displaystyle \left\vert X_{1}\right\vert ={\frac {R_{1}}{Q}}}$.

${\displaystyle \left\vert X_{2}\right\vert =QR_{2}}$.

where ${\displaystyle Q={\sqrt {\frac {R_{1}-R_{2}}{R_{2}}}}}$.

Note, ${\displaystyle X_{1}}$, the reactance in parallel, has a negative reactance because it is typically a capacitor. This gives the L-network the additional feature of harmonic suppression since it is a low pass filter too.

The inverse connection (impedance step-up) is simply the reverse—for example, reactance in series with the source. The magnitude of the impedance ratio is limited by reactance losses such as the Q of the inductor. Multiple L-sections can be wired in cascade to achieve higher impedance ratios or greater bandwidth. Transmission line matching networks can be modeled as infinitely many L-sections wired in cascade. Optimal matching circuits can be designed for a particular system using Smith charts.

In a transmission line, a wave travels from the source along the line. Suppose the wave hits a boundary (an abrupt change in impedance). Some of the wave is reflected back, while some keeps moving onwards. (Assume there is only one boundary, at the load.)

Let

${\displaystyle V_{i}\,}$ and ${\displaystyle I_{i}\,}$ be the voltage and current that is incident on the boundary from the source side.

${\displaystyle V_{t}\,}$ and ${\displaystyle I_{t}\,}$ be the voltage and current that is transmitted to the load.

${\displaystyle V_{r}\,}$ and ${\displaystyle I_{r}\,}$ be the voltage and current that is reflected back toward the source.

On the line side of the boundary ${\displaystyle V_{i}=Z_{c}I_{i}\,}$ and ${\displaystyle V_{r}=-Z_{c}I_{r}\,}$ and on the load side ${\displaystyle V_{t}=Z_{L}I_{t}\,}$ where ${\displaystyle V_{i}\,}$, ${\displaystyle V_{r}\,}$, ${\displaystyle V_{t}\,}$, ${\displaystyle I_{i}\,}$, ${\displaystyle I_{r}\,}$, ${\displaystyle I_{t}\,}$, and ${\displaystyle Z_{c}\,}$ are phasors.

At a boundary, voltage and current must be continuous, therefore

${\displaystyle V_{t}=V_{i}+V_{r}\,}$

${\displaystyle I_{t}=I_{i}+I_{r}\,}$

All these conditions are satisfied by

${\displaystyle V_{r}=\Gamma _{TL}V_{i}\,}$

${\displaystyle I_{r}=-\Gamma _{TL}I_{i}\,}$

${\displaystyle V_{t}=(1+\Gamma _{TL})V_{i}\,}$

${\displaystyle I_{t}=(1-\Gamma _{TL})I_{i}\,}$

where ${\displaystyle \Gamma _{TL}\,}$ the reflection coefficient going from the transmission line to the load.

${\displaystyle \Gamma _{TL}={Z_{L}-Z_{c} \over Z_{L}+Z_{c}}=\Gamma _{L}\,}$[5][6][7]

The purpose of a transmission line is to get the maximum amount of energy to the other end of the line (or to transmit information with minimal error), so the reflection is as small as possible. This is achieved by matching the impedances ${\displaystyle Z_{L}}$ and ${\displaystyle Z_{c}}$ so that they are equal (${\displaystyle \Gamma =0}$).

At the source end of the transmission line, there may be waves incident both from the source and from the line; a reflection coefficient for each direction may be computed with

${\displaystyle -\Gamma _{ST}=\Gamma _{TS}={Z_{s}-Z_{c} \over Z_{s}+Z_{c}}=\Gamma _{S}\,}$,

where Zs is the source impedance. The source of waves incident from the line are the reflections from the load end. If the source impedance matches the line, reflections from the load end will be absorbed at the source end. If the transmission line is not matched at both ends reflections from the load will be re-reflected at the source and re-re-reflected at the load end ad infinitum, losing energy on each transit of the transmission line. This can cause a resonance condition and strongly frequency-dependent behavior. In a narrow-band system this can be desirable for matching, but is generally undesirable in a wide-band system.

${\displaystyle Z_{in}=Z_{C}{\frac {(1+T^{2}\Gamma _{L})}{(1-T^{2}\Gamma _{L})}}\,}$ [8]

where ${\displaystyle T\ ,}$ is the one-way transfer function (from either end to the other) when the transmission line is exactly matched at source and load. ${\displaystyle T\,}$ accounts for everything that happens to the signal in transit (including delay, attenuation and dispersion). If there is a perfect match at the load, ${\displaystyle \Gamma _{L}=0\,}$ and ${\displaystyle Z_{in}=Z_{C}\,}$

${\displaystyle V_{L}=V_{S}{\frac {T(1-\Gamma _{S})(1+\Gamma _{L})}{2(1-T^{2}\Gamma _{S}\Gamma _{L})}}\,}$

where ${\displaystyle V_{S}\,}$ is the open circuit (or unloaded) output voltage from the source.

Note that if there is a perfect match at both ends

${\displaystyle \Gamma _{L}=0\,}$ and ${\displaystyle \Gamma _{S}=0\,}$

and then

${\displaystyle V_{L}=V_{S}{\frac {T}{2}}\,}$.

Telephone systems also use matched impedances to minimise echo on long-distance lines. This is related to transmission-line theory. Matching also enables the telephone hybrid coil (2- to 4-wire conversion) to operate correctly. As the signals are sent and received on the same two-wire circuit to the central office (or exchange), cancellation is necessary at the telephone earpiece so excessive sidetone is not heard. All devices used in telephone signal paths are generally dependent on matched cable, source and load impedances. In the local loop, the impedance chosen is 600 ohms (nominal). Terminating networks are installed at the exchange to offer the best match to their subscriber lines. Each country has its own standard for these networks, but they are all designed to approximate about 600 ohms over the voice frequency band.

Typical push–pull audio tube power amplifier, matched to loudspeaker with an impedance-matching transformer

Audio amplifiers typically do not match impedances, but provide an output impedance that is lower than the load impedance (such as < 0.1 ohm in typical semiconductor amplifiers), for improved speaker damping. For vacuum tube amplifiers, impedance-changing transformers are often used to get a low output impedance, and to better match the amplifier's performance to the load impedance. Some tube amplifiers have output transformer taps to adapt the amplifier output to typical loudspeaker impedances.

The output transformer in vacuum-tube-based amplifiers has two basic functions:

• Separation of the AC component (which contains the audio signals) from the DC component (supplied by the power supply) in the anode circuit of a vacuum-tube-based power stage. A loudspeaker should not be subjected to DC current.
• Reducing the output impedance of power pentodes (such as the EL34) in a common-cathode configuration.

The impedance of the loudspeaker on the secondary coil of the transformer will be transformed to a higher impedance on the primary coil in the circuit of the power pentodes by the square of the turns ratio, which forms the impedance scaling factor.

The output stage in common-drain or common-collector semiconductor-based end stages with MOSFETs or power transistors has a very low output impedance. If they are properly balanced, there is no need for a transformer or a large electrolytic capacitor to separate AC from DC current.

Similar to electrical transmission lines, an impedance matching problem exists when transferring sound energy from one medium to another. If the acoustic impedance of the two media are very different most sound energy will be reflected (or absorbed), rather than transferred across the border. The gel used in medical ultrasonography helps transfer acoustic energy from the transducer to the body and back again. Without the gel, the impedance mismatch in the transducer-to-air and the air-to-body discontinuity reflects almost all the energy, leaving very little to go into the body.

The bones in the middle ear provide impedance matching between the eardrum (which is acted upon by vibrations in air) and the fluid-filled inner ear.

Horns in loudspeaker systems are used like transformers in electrical circuits to match the impedance of the transducer to the impedance of the air. This principle is used in both horn loudspeakers and musical instruments. Because most driver impedances are poorly matched to the impedance of free air at low frequencies, loudspeaker enclosures are designed to both match impedance and minimize destructive phase cancellations between output from the front and rear of a speaker cone. The loudness of sound produced in air from a loudspeaker is directly related to the ratio of the diameter of the speaker to the wavelength of the sound being produced: larger speakers can produce lower frequencies at a higher level than smaller speakers. Elliptical speakers are a complex case, acting like large speakers lengthwise and small speakers crosswise. Acoustic impedance matching (or the lack of it) affects the operation of a megaphone, an echo and soundproofing.

A similar effect occurs when light (or any electromagnetic wave) hits the interface between two media with different refractive indices. For non-magnetic materials, the refractive index is inversely proportional to the material's characteristic impedance. An optical or wave impedance (that depends on the propagation direction) can be calculated for each medium, and may be used in the transmission-line reflection equation

${\displaystyle r={Z_{2}-Z_{1} \over Z_{1}+Z_{2}}}$

to calculate reflection and transmission coefficients for the interface. For non-magnetic dielectrics, this equation is equivalent to the Fresnel equations. Unwanted reflections can be reduced by the use of an anti-reflection optical coating.

If a body of mass m collides elastically with a second body, maximum energy transfer to the second body will occur when the second body has the same mass m. In a head-on collision of equal masses, the energy of the first body will be completely transferred to the second body (as in Newton's cradle for example). In this case, the masses act as "mechanical impedances", which must be matched. If ${\displaystyle m_{1}}$ and ${\displaystyle m_{2}}$ are the masses of the moving and stationary bodies, and P is the momentum of the system (which remains constant throughout the collision), the energy of the second body after the collision will be E₂:

${\displaystyle E_{2}={\frac {2P^{2}m_{2}}{(m_{1}+m_{2})^{2}}}}$

which is analogous to the power-transfer equation.

These principles are useful in the application of highly energetic materials (explosives). If an explosive charge is placed on a target, the sudden release of energy causes compression waves to propagate through the target radially from the point-charge contact. When the compression waves reach areas of high acoustic impedance mismatch (such as the opposite side of the target), tension waves reflect back and create spalling. The greater the mismatch, the greater the effect of creasing and spalling will be. A charge initiated against a wall with air behind it will do more damage to the wall than a charge initiated against a wall with soil behind it.