..again, it's not so much about preservation of power, but about being a transmission-line that'll work even when your cable is longer than your wavelength..
/Jakob E.
/Jakob E.
Impedance matching / bridging between tube gear and modern audio interfaces
- Thread starter mrtnasty
- Start date
Help Support GroupDIY Audio Forum:
Cqwet Dbdfte
Well-known member
- Joined
- Oct 2, 2023
- Messages
- 411
mmmkay....
There is an AES Preprint from 1980 which discusses the transition in broadcast facilities between older power matched systems and newer voltage transmission systems.
Author: Hess, Richard L.
Affiliation: American Broadcasting Company, New York, NY
AES Convention: 67 (October 1980)
Paper Number: 1708
Publication Date: October 1, 1980
Permalink: AES E-Library » Voltage Transmission for Audio Systems
Voltage Transmission for Audio Systems
The benefits of using a voltage audio transmission system in broadcast facilities is investigated. State of the art microphone preamplifier requirements and an ideal voltage system distribution amplifier are outlined. The application of the Peak Program Meter to the new systems and the modified installation at WABC-TV are covered.Author: Hess, Richard L.
Affiliation: American Broadcasting Company, New York, NY
AES Convention: 67 (October 1980)
Paper Number: 1708
Publication Date: October 1, 1980
Permalink: AES E-Library » Voltage Transmission for Audio Systems
That is only the case when you have no control over the source impedance, but you can control the transmission line and load.
If you have a fixed load impedance but you can control the source, then you transfer the most power by reducing the source impedance as much as possible so that there is no power dissipated in the source.
Isn't low source impedance and high load impedance by definition a voltage transfer system? Hardly any power is transferred between devices with a bridging load.
Cqwet Dbdfte
Well-known member
- Joined
- Oct 2, 2023
- Messages
- 411
This issue is partially semantics. A voltage driven system implies current will be no problem.
The specific phrase I used was "fixed load impedance," meaning that is not something you control. It does not matter if the load impedance is "high" or "low" using whatever arbitrary scale you have in your head for that, the most power gets transferred to that fixed load (for a given source voltage) when you do not dissipate any appreciable power in the source end, so by minimizing the source impedance.
If the source impedance is fixed and you cannot control that, then the most power is transferred when you make the load match the source. Lower than that and excessive power is dissipated in the source, higher than that and you decrease the current which decreases total delivered power.
I guess I'm still confused. If the load impedance is fixed at 10K and the source impedance is say 50 ohm there is almost no power transferred, right? How does this jive with the above?
Probably an interesting reading. However, I'm not ready to give $33 to the Plunkett heir for a mere confirmation of a practice that has long proved its worth and practicality.
Almost is the key word here.
It could be said that impedance matching is a very precise thing; the source and load impedance should be equal. Actually in practice there is some tolerance, which should be appreciated on an individual basis. A short connection between transformerless units is very tolerant, with level variation the only consequence of mismatch. On the contrary, a landline with transformers requires accurate impedance matching for proper frequency response. In RF applications, accurate matching is also required, at the receveing end because it impacts directly sensitivity and S/N ratio, at the transmitting end because any loss of power is dissipated either in the transmitter electronics or in the cable. Not unsignificant when you deal with a 50 kW+ transmitter.
OTOH bridging is a rather extensible notion that the load should be significantly larger than the source. How much is "significantly larger"?
It is commonly admitted that 10x is a typical value, but actually for microphones it may be closer to 5x, when for many line level connections, it's often about 100x.
Indeed, 100x results in better power preservation than 5x.
In many cases, "power preservation" is not the most significant concern.
You're correct when you infer that having the source impedance much lower than the load results in minimum losses=> optimum efficiency => power preservation.
ruffrecords
Well-known member
Just because the power is very small it does not mean it is not the maximum power. If you reduce the source from 50 ohms to 25 ohms, a little more power will be transferred. Reduce it to zero and that will be maximum power. Increase it to 1K and the power drops.
The maximum power transfer theorem can be very confusing. This confusion starts because the theorem begins with the assumption that you have a source of known impedance and then it works out what the load needs to be to transfer the maximum power. The answer is the load impedance must equal the source impedance. So in your example of a 50 ohm source the maximum power will be transferred into a 50 ohm load. But in the real world almost all sources are incapable of driving a load equal to their own source impedance. And in the world of audio we are really more interested in transferring the maximum signal level (voltage) which means low source impedances and relatively high load impedances.
Cheers
Ian
I don't want to take the discussion too far afield, but the point is that even in a tranformer coupled pair of devices, you get better results (in this context higher signal level so typically better SNR) by minimizing the source impedance, no need to build out to 600 Ohms.
And of course for a fixed load impedance you get the maximum power at the load when the voltage is maximized, tying the concepts back together.
A lot of people just do things without thinking through the fundamental electrical principles at play, so trying to tie back to the original posting:
Really there is just one relevant consideration as I can see it:
I've been doing that. Possibly not very well. I'm thinking that Maximum Power is kind of the wrong term if thinking purely in terms of electrical efficiency. I know there are other issues involved that complicate things. If we stick purely to electrical efficiency then Maximum Power Transfer as a term is a little confusing. After all 1000V at 1A has the same power as 1000A at 1V.
The most efficient way to implement 1000V at 1A is with a voltage transfer system with a low source impedance and high load impedance. In the second case we have a current transfer system. I am not familiar with this type of system but I think an electroplating bath fits in this category. I'm assuming a high source impedance and low load impedance.
When source and load impedances are the same both voltage and current are in equilibrium. This is why this is said to be most efficient. Yes?
Probably why no one has brought up the term "efficiency" to this point in the thread.
Why would we "stick to" a topic which has not been brought up before? Seems like the wrong term to use. Maybe "if we switch to the different topic of electrical efficiency then...."
I don't believe anyone said it was the most efficient. I'm not even sure what voltage and current being "in equilibrium" means, you will have to define what you mean by using that term in this context.
Don't try to over complicate things, you can work this out with a pencil and half a sheet of paper. Or Pick a voltage and a source impedance. Make them round numbers so the math is easy, 1V and 1 Ohm, 1V and 100 Ohms, whatever works for you. Pick a load impedance lower than, equal to, and greater than the source impedance. Calculate the power in the load based on the voltage divider formed by the source and load for each of those three conditions.
Which value is highest?
That is all that is claimed by the statement that maximum power is delivered when source and load impedance are matched. And most people leave out the qualifying factor "when source impedance is fixed and cannot be changed."
The follow on in an audio context is whether that is relevant in the context of transformers. Fixed source impedance is not the expected case on I/O connectors, but for things like transformer coupling from a moving ribbon or moving coil transducer it might be relevant.
Maximum power transfer sounds like it is the most efficient case. No power transfer would be the least efficient case.
I’m thinking about for a given amount of power. When source and load impedance are matched there is equal impedance to both voltage and current transfer between source and load. As the source impedance goes down and the load impedance goes up there is less current transfer and more voltage transfer. For a given amount of power. As source impedance goes up and load impedance goes down there is more current transfer and less voltage transfer.
There is no power transfer unless there is an R in the equation. That much I know.
Last edited:
Contemporary signal interfaces are designed to transfer maximum signal voltage, not power. The earpiece in an old passive telephone system responded to power, so those systems were impedance-matched for that reason. 600 ohms just happened to be the characteristic impedance of the wires strung across the country (USA) for telegraph systems, so it was adopted as a "standard" for subsequent telephone systems (telephone lines are physically long enough to become true electrical transmission lines). "Bridging" was a term that meant a load with impedance high enough not to cause a significant level drop on a 600-ohm line - generally taken to mean 10 k-ohms or more. That equipment and standard was borrowed by the earliest vacuum-tube systems for radio broadcast and recording studios. But the days of transferring power are long gone. Impedance matching source and load wastes half the signal voltage available at the source. Therefore, modern interfaces are designed to deliver maximum signal voltage - which requires the lowest practical source impedance and the highest practical load impedance (IEC standards call for sources 100 ohms or less and loads 10 k-ohms or more). In the case of microphones, impedances are often transformed to a higher value to become the optimum value for lowest total noise in a particular amplifier (the point where the amplifier self voltage noise is equal to the voltage noise caused by its self current noise flowing in the source impedance).
I’m not understanding something more fundamental. Why are equal source and load impedances said to transfer “maximum power”?
P=VA. Losses in either voltage or current diminish power. Why do equal source and load impedance’s minimize losses?
ruffrecords
Well-known member
That is not what is said. What is said is that for a given source impedance the maximum power will be transferred when the load impedance equals that source impedance.
Maximum power transfer theorem - Wikipedia.
Cheers
Ian
Imagine a black box with a 5 V battery in series with a 50 ohms resistor and that's the output terminal. Do the math about how much output voltage will be with various resistor loads. Remember that P = E squared divided by R. You will find that P reaches it's maximum when R is 50 ohms ( the same as the source resistance inside the black box!
Similar threads
