..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 is needed to preserve POWER (watts, milliwatts etc)
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.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?
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?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.
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.There is an AES Preprint from 1980 which discusses the transition in broadcast facilities between older power matched systems and newer voltage transmission systems.
Almost is the key word here.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?
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.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?
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?
we are really more interested in transferring the maximum signal level (voltage)
I am matching impedances all the way through my signal chain
should I transform the 68 ohm output to match the 200 ohm or 600 ohm input
I know tube preamps from the 40s and 50s were designed be impedance matched (both input and output)
some vintage equipment requires to be loaded with the proper termination for linearization of frequency response
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.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:
I'm thinking that Maximum Power is kind of the wrong term if thinking purely in terms of electrical efficiency.
If we stick purely to electrical efficiency then Maximum Power Transfer as a term is a little confusing.
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?
Maximum power transfer sounds like it is the most efficient case. No power transfer would be the least efficient case.Probably why no one has brought up the term "efficiency" to this point in the thread.
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. \
There is no power transfer unless there is an R in the equation. That much I know.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.
But the days of transferring power are long gone. Impedance matching source and load wastes half the signal voltage
I’m not understanding something more fundamental. Why are equal source and load impedances said to transfer “maximum power”?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).
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.I’m not understanding something more fundamental. Why are equal source and load impedances said to transfer “maximum power”?
Maximum power transfer theorem - Wikipedia.P=VA. Losses in either voltage or current diminish power. Why do equal source and load impedance’s minimize losses?
Enter your email address to join: