This is a conceptual question therefore I want to ask with an example. I'm wondering in which cases impedance matching is used.

For example, if one has a sensor which can be modeled as voltage source call it Vs. Imagine it is outputting mV level voltage signals and it has 1k output resistance. And if we want to amplify it we would use an amplifier with a very big input impedance like 10Meg so that most of the Vs reach the amplifier input.

But for maximum power transfer the input impedance of the amplifier must be equal to 1k which would halve the Vs reaching to the amplifier input.

If we have a signal to be amplified why would we match the source output impedance to receiver's input impedance? In my example it was not a good idea. And I know that in radio transmission impedance matching is used. But again I'm stuck at the same point. There must be a difference in two cases where we need maximum voltage at the receiver or we need maximum power at the receiver.

Can this be explained by two examples for each case?

| improve this question | | | | |

Indeed there is a difference.

In your sensor example, the amplifier works on the voltage it receives from the sensor. So it makes sense to maximise the voltage by using a high input impedance amplifier.

Likewise sensors like photo diodes output a current that is proportional to their input (light for example) so then it makes sense to maximize the current by using a (current) amplifier with a low input impedance: the Trans Impedance Amplifier.

Both cases above achieve maximum power mismatch. Yes MISmatch! But that is on purpose to maximize the voltage or current.

Power matching is mostly done when we need to maximize the power transfer. An example of this is a device which delivers a certain power and will only deliver that if the load has the correct impedance.

A low frequency example is an audio power amplifier. If the audio amplifier can deliver an undistorted signal of up to 10 V, 2.5 A, what load impedance would result in maximum power? Simply apply ohm's law: 10 V / 2.5 A = 4 ohms. That results in 10 V * 2.5 A = 25 W.

If we loaded that amplifier with an 8 ohm speaker, the voltage is the limiting factor: 10 V / 8 ohms = 1.25 A which equals to 12.5 W (at 2.5 A through 8 ohms, we need 2.5 A * 8 ohm = 20 V => outside the 10 V limit)

If we loaded that amplifier with a 2 ohm speaker, the current is the limiting factor: 2.5 A * 2 Ohms = 5 V which equals to 12.5 W (at 10 V across 2 ohms, 10 V / 2 ohm = 5 A would be needed => outside the 2.5 A limit)

Another example is an RF transmitter feeding power into an Antenna. To avoid signal reflections (explained below), RF power amplifiers have a certain output impedance, they need to "match" the cable's characteristic impedance. Also antennas only work properly (minimal reflections) when they are matched to their cable. You might think an antenna is "just a piece of metal" but there's more, it is actually an electrical interface to the air. This air also has an impedance just like the cable feeding the signal to the antenna. All these impedances need to be such that no reflections occur.

There is another reason for power matching, when we use RF signals it is possible that the wavelength of the signal comes close to or is smaller than the dimensions of the conductors (cables etc.) we use. Then conductors and cables behave like a transmission line.

A transmission line must be terminated properly to avoid signal reflections. Signal reflections disturb the signal, they distort it. Proper termination (the right impedance at both ends of the transmission line) is needed to avoid reflections and allow for undistorted transportation of the signal.

This is why many coaxial cables have a "characteristic impedance", for example 50 ohms. The cable does not "have" this impedance (you cannot measure it with an ohm meter) but you do need to terminate the cable with 50 ohms at both ends to avoid reflections.

As mentioned, this only applies when we use signals which have a wavelength which comes close to or is smaller than the dimensions of the conductors. For most sensors this is not the case, the signal does not change rapidly, it is a low frequency signal. Therefore it has a very long wavelength (can be hundreds of meters long) and reflections are not an issue.

| improve this answer | | | | |
  • \$\begingroup\$ Very nice answer. I understand the part related to wavelength and transmission lines which we minimize/avoid reflections.One point is blur to me though. You mentioned "Power matching is mostly done when we need to maximize the power transfer. An example of this is a device which delivers a certain power and will only deliver that if the load has the correct impedance." Can you give a more explicit example on that? An example would help a lot. Especially the antenna example. \$\endgroup\$ – user1999 May 16 '19 at 12:02
  • \$\begingroup\$ I added some explanations. \$\endgroup\$ – Bimpelrekkie May 16 '19 at 12:24
  • \$\begingroup\$ Is it just me who's not considering f=c/"hundreds of meters" (> 300kHz!) a "low frequency" sensor signal? - Maybe change that to something like "ranging from hundreds of meters up to millions of kilometers or more". \$\endgroup\$ – JimmyB May 16 '19 at 12:30
  • \$\begingroup\$ @JimmyB A 300 kHz sinewave has a wavelength of 1 km. So for a sinewave up to 300 kHz "a few hundred meters" would still be OK. However, if your signal is 300 kHz but not a sinewave then it contains harmonics which have shorter wavelengths. A 300 kHz squarewave can have harmonic frequencies up to whatever frequency you want. So it is actually the total bandwidth of a signal that matters. \$\endgroup\$ – Bimpelrekkie May 16 '19 at 12:39
  • \$\begingroup\$ I don't doubt that, but, assuming an actual square wave, the wavelengths of the harmonics approach 0, not hundreds of meters. I'd say that most things considered as "sensors" output signals at sub-Hertz (e.g. temperature sensors) up to maybe kilo-Hertz (e.g. microphones) frequencies. \$\endgroup\$ – JimmyB May 16 '19 at 12:48

Power matching is said to be used to match an output impedance to a physical system. ex, a precision led with a 27pF capacitance or a moving coil with a physical atenuation factor to it's free movement (A subwoofer). Also when you have a very large power source and very small loads in parallel which could have very different resistances each causing one to source a bigger current draw when in parallel with larger resistance...
a precision led is a led which it's conductance and capacitance are in anti-symmetry in a way none of them work in the continuous conduction of current model (a resistance) such as any of those characteristics produce a deviation in the frequency the thermal junction sources energy (photons) thus deviating it's output characteristic (heat, photons) if not power matched using a physical model equation (the power matching itself). Something similar happens when a Lasing medium requires power matching for different gas flow conditions in a tube which its initial and continuous lasing characteristics need a single power matched current modulation.
Thus voltage is a parameter of the system not a characteristic like those determined physically when built.

| improve this answer | | | | |

In some systems, the parasitic capacitance is the dominant absorber of the signal energy (such as Miller Capacitance is a problem for amplifiers).

When capacitance is a problem, we use bootstrapping/cascoding to reduce Cin, or we include inductors to achieve some resonance and some matching.

The use of inductors (of stublines, which are reactances of narrowband use) provides a whole new world of design methods.

For narrow-band systems, we can resonate. And achieve low power consumption. And we match for best use of the signal energy, as from antennas.

For broadband systems (radar pulse amplifiers, video amplifiers, etc) we have no choice but to use expensive and high power circuits. And we may use termination resistors that waste signal energy but also preserve waveform integrity.

Now an example: ESD structures are a major problem for radios. Some ICs (bipolars) are shipped without ESD, with collaboration between the foundary and the PCB fab house, so major impairments to the NoiseFigure are avoided should the ESD diodes have been included and then shunted major amounts of the RF current into the (energy-absorbing) silicon substrate.

For MOS, the ESDs are unavoidable. Thus resonance using tuners is the circuit approach.

| improve this answer | | | | |
  • \$\begingroup\$ This answer bears little relation to that what is asked. \$\endgroup\$ – Bimpelrekkie May 16 '19 at 13:00
  • \$\begingroup\$ when the capacitance becomes the primary energy sink, then maximum voltage is not longer the design criterion. \$\endgroup\$ – analogsystemsrf May 16 '19 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.