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Please provide an intuitive explanation of maximum power transfer.

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?

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  • \$\begingroup\$ "In my example it was not a good idea." Depends. If you're trying to amplify the voltage of a signal you might want a high input impedance. If you're trying to transfer maximum power, you want to match the load impedance to the source. Halving the voltage in that case IS the point where you're transferring the most power to the load. Write the equations and you can see. Lower load impedance is less V^2/R. Higher load impedance is less I2R. \$\endgroup\$ – John D Jun 7 at 23:09
  • \$\begingroup\$ Please give me an intuitive explanation between maximum power transfer and maximum load transfer. \$\endgroup\$ – rdeep Jun 7 at 23:14
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    \$\begingroup\$ The last sentence of my previous comment is kind of it. Lower load impedance is less V^2/R. Less voltage across the load and the power goes as square of voltage. Higher load impedance is less total current through the load or less I^2R, so again less power delivered to the load. The max power point is a balance of the two at source = load impedance. \$\endgroup\$ – John D Jun 7 at 23:30
  • \$\begingroup\$ I got a doubt. Power is the rate at which energy is converted from one form to another. For example, in voltage transfer, we have high input impedance (loud speaker as load) and low output impedance (amplifier as source). So, in this case we don't care about maximum power and we want to transfer maximum voltage to loud speaker. As power delivered is less (which means rate at which energy is transferred is also low), how this can turn on the loud speaker to work? Please explain. \$\endgroup\$ – rdeep Jun 7 at 23:59
  • \$\begingroup\$ In a voltage amplifier (say for example like an audio amp) you may not care about maximum power transfer. The power amplifier has its own power supply that provides the power to drive the speaker so the amount of power transferred from the source is irrelevant. For an RF amplifier transferring power from say an antenna to the first RF amplifier stage you want to transfer all the power possible, so matching source and load impedance is important. \$\endgroup\$ – John D Jun 8 at 0:07
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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.

You might do that if you are only interested in getting the most accurate measurement of open-circuit sensor voltage, but not if you want the best signal to noise ratio.

Suppose your amplifier has an equivalent input noise voltage of 1mV, and the signal is also 1mV. With a signal to noise ratio of 1:1 you will have a hard time separating them. But what if you use a transformer to match the source to the load? You want an impedance step-up ratio of 10M / 1k = 10000. The turns ratio required is the square root of the impedance ratio, ie. 100. Since the voltage step-up ratio is the same as the turns ratio, your signal voltage is now 100 times larger! Even after taking into account the 50% drop due to impedance matching, you still have 50 times more voltage than before (50mV vs 1mV). So by matching the impedances you have turned an abysmal 1:1 S/N ratio into a healthy 50:1 ratio (a 34dB improvement).

But what if you don't want to use a transformer? For a given temperature the thermal noise power in a resistor is constant, but the voltage is proportional to the square root of its resistance. So instead of increasing the impedance to match the amplifier, you could simply use an amplifier with lower input impedance - which (if properly designed) should have a proportionally lower input noise voltage.

The more signal power you can get into the amplifier's input, the more it has to play with for overcoming internal noise. Matching impedances produces maximum power transfer because although voltage reduces (compared to the open-circuit voltage), current increases. Though you lose 50% of the power in the source, you still get more out than with mismatched impedances.

Depending on how an amplifier is designed and the components it uses, the best S/N ratio may not occur when the impedances are precisely matched. As a general rule it will not be far off, but when comparing specific devices you need to examine the specs (or measure them). And there may be other criteria which are more important, such as sensor loading, linearity and dynamic range. Therefore we cannot say that matching impedances will produce the best results in all cases. Here are a couple of examples where matching impedance is not a good idea:-

  1. Audio power amplifier: You want high efficiency to reduce power consumption and cooling requirements. Therefore the amplifier should have much lower impedance than the speakers, so that most of the power is delivered to the load. The low output impedance also helps to damp mechanical resonances in the speakers, producing more accurate sound.

  2. Guitar amplifier. The inductive pickup in the guitar has a complex impedance that rises as it approaches its resonant frequency, but so does the output voltage. To get flattest frequency response the amplifier should have a fairly high input impedance (typically around 100kΩ) but an even higher impedance is often desirable to increase the peaking effect at higher frequencies. The guitar may also have tone controls that modify the response. Matching its impedance at all frequencies would dramatically change the way the instrument sounded.

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Decades ago I trained a team of people to perform RFIC design. But almost ALL the circuits were broadband, covering 1 or 2 octaves of input range or output range, in the 100MHz to 1,000MHz range. (OK twas one LO buffer with 2+ GHz requirement, and prescalers with similar requirement).

We faced this question: to match, or not to match. I explained we could view the onchip circuits, being about 50micron to 500 micron between RF source and RF load, as just broadband opamps. Given the broadband nature, and no termination nor reflections being part of our silicon layout challenges, we agreed we should not insert any 50 ohm resistors.

We had to face the interfaces as being voltage dividers, as already discussed. With very low Rout of the RF sources (bipolars running at 10mA => 2.6 ohms Rout), and high Rin (2.6 ohms * beta = 260 ohms), our interface losses were about 1% or about 0.1dB.

Had we terminated each interface, facing the 6dB loss, we'd have needed an additional stage or two in the receiver chip.

Thus you can view "no matching" as getting 6dBgain for free, in each stage.

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