Why in RF receiver we need impedance matching for maximum power transfer but not maximum voltage? And why there is no Impedance matching or Power matching in audio engineering?
Is it depends on the signal frequency to chose impedance matching or impedance bridging?
Electronic – impedance matching and impedance briging
audioimpedance-matchinginput-impedancepowerRF
Related Solutions
What causes the reflection? Why doesn't it occur with other frequencies? (if it doesn't)
Reflections occur at all frequencies when there is a mismatch in impedances. At low frequencies, such as audio, these reflections are difficult to see but they are there all the same. Reflections are generally said to be significant when the frequency is high enough AND the interconnection between sender and receiver is long enough. Somewhere in the order of about a tenth of a wavelength or bigger is a general rule of thumb.
At 20 kHz, the wavelength (in 100% speed of light cable) is about 15 kilometres and if you had a cable of about 1.5 km length you might start to see the effect of reflections.
However, if you had a 100 MHz transmitter, you might see the effect of reflections at 300 mm.
Consider a battery and a lightbulb. The lightbulb is connected to the battery with a switch. The battery and switch are at one end of a lossless 10 km cable and the bulb is at the other end. When the switch closes, how much current is drawn from the battery? - how can the battery know how much current to supply in that instant? The answer is it can't - it supplies what the cable demands and, for a 50 ohm cable an appropriate current is supplied. If the voltage were 10V then the current would be 200mA.
This travels down the cable (at a power of 2W) until it hits the lightbulb. The bulb may have an impedance of (say) 100ohms - it only wants 100mA at 10V but it gets 200mA - there is a mismatch and the excess power gets reflected back up the cable to the battery and switch. This power can't be dissipated in the battery so it gets relfected back and forth. Of course, cable has real losses and these eat away at this reflection and the system stabilizes with 100mA flowing down the cable. This is a simplifed explanation.
Does this help you understand?
Well, I don't know if the schematic you uploaded is what you have in your case, or it is only an example. Checking it, it has a mistake, and maybe it is the origin of the issue.
Let me explain it from the begining:
For maximum output power, you need to match the external components of the Source to be its conjugate. That means that the real part seen from the Source must be 10 ohm, and the imaginary part must be the oposite than the produced by the 2nF internal capacitance.
As you have 100ohm Rload, you need to convert it to an equivalent 10ohm. The only way to reduce the equivalent value of a resistor is adding another in parallel. That is why you add an inductor in parallel to R2. The L value is calculated to obtain 10 ohm as real part of this parallel equivalent reactance.
For obtaining a Requivalent of 10ohm, you use a inductor that introduce a complex inductance, so you need to add a capacitor in serie (that works as a negative imaginary value) to remove this complex part.
When you calculate the C needed to remove the imaginary part, you must take into account the existing C on the source.
To solve that, there is a few formulas used, that includes the Q value concept as explained in previous answers.
Calculations:
The 1,06nF required is not what you need to place outside the Source, this is the equivalent C needed in all the LC network. As you have an internal 2nF capacitance, you need to calculate the outside C to get an equivalent value of 1,06nF.
The equivalent capacitance of 2 capacitors in serie es calculated as:
So what you need is an 2,255 nF capacitor instead of the 1nF that you have placed outside the source.
Related Topic
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Best Answer
Audio in fact did used to be done as constant impedance (~600 ohms was typical) as a holdover from Bell Telecomm who developed most of the early tech for long distance phone lines.
Consider that something like an audio amp has an output impedance that is basically a current limited voltage source (near zero output impedance), given an electrically short cable with sufficient copper this condition will exist at the other end of the cable as well, and all is good. Now lets extend that cable into being a transcontinental long distance phone line... Now there is some audio frequency at which that cable is a 1/4 wavelength long, and that low Z source is transformed by the cable to high Z at that frequency (with a corresponding increase in voltage), thus our mismatched line has become a filter, probably not what was wanted. If instead the load matched the cable impedance then you get no filtering effect.
At audio frequencies the lines are (almost) always electrically short (as in less then maybe 1/10 wavelength long) so a low Z source will be seen as such at the other end of the cable irrespective of the cable characteristic impedance.
This is not the case at RF where the cable can easily be a significant fraction of a wavelength long and are often well over a wavelength long.
Note also that the audio band covers approximately 3 Decades of bandwidth, good luck designing a conjugate match for that, bridging is just so much easier for the 99% of cases where it will work!
Regards, Dan.