PopularFX
Home Help Search Login Register
Welcome,Guest. Please login or register.
OverUnity Research > Electrical / Electronic Theory and Learning Center > EM Theory > Impedance matching
2024-11-27, 10:38:58
News: Registration with the OUR forum is by admin approval.

Pages: [1]
Author Topic: Impedance matching  (Read 6891 times)
Topic Options img
img  GibbsHelmholtz
Impedance matching « on: 2012-08-03, 20:13:33 »
Group: Guest

I've been looking up information about impedance matching.  It seems like a method to get most power a source can give (no reflection).  A well article here:

http://www.bcae1.com/trnimpmt.htm

I'm looking for information about impedance matching and efficiency but could not find any.  It is probably assumed that it does not matter if the impedance is matched or not matched, output/input=1 ideally.  I'd like to consult the experts here if any knowledge or information are given.   The reason I put this in question because our feed back circuit reduced overall input.  It should not be this way in theory.  What if we are to depart from matching impedance. 
   
img  forcefield
Re: Impedance matching « Reply #1 on: 2012-08-03, 23:27:33 »
Group: Guest
You are correct - impedance matching the load to the source maximizes the power to the load for a given source.  "Efficiency" will be 50%.  That is to say that half the power will be absorbed by the load, and half by the source.

In some cases, impedance matching is an important part of protecting the source.  If the load resistance is too low, the source might burn out due to over-current.  If the load resistance is too high, it might burn up due to excessively high voltages.

When dealing with "long" distances between the source and load, the source might "see" loads that are very different from what the load actually is.  For example, an open circuit might "look like" a short circuit.  In a system with a resistive source and load that are matched to the characteristic impedance of the transmission line, the length of the transmission line won't affect what resistance the source sees.

Depending on exactly what you mean by "efficiency" in a particular situation, the best efficiency is when the source has low resistance, and the load has high resistance.
   
img  ion
Re: Impedance matching « Reply #2 on: 2012-08-04, 14:24:32 »
Group: Elite
Hero Member
******

Posts: 3537
It's turtles all the way down
 I look at impedance matching in the simple form as optimized coupling between source and load taking into account the requirements of each.

e.g a valve amplifier swings hundreds of volts at low current, but a speaker may require low voltage swings at high current.

The output transformer provides the necessary voltage / current transformation to match the requirements of each device.

Like the transmission in an auto, the engine likes high RPM's but cannot deliver high torque at lower RPM's so a transmission provides the RPM / Torque transformation to match the engine (source) to the wheels (load).  As the auto gains speed, the impedance changes so we shift gears to change the transformation ratio.

Ideally the impedance matching device should not dissipate up any power. In practice transformers have ohmic and eddy losses, and transmissions have some friction, but this is usually small compared to the amount of power transferred / transformed.


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
img  GibbsHelmholtz
Re: Impedance matching « Reply #3 on: 2012-08-06, 05:14:25 »
Group: Guest
Thank you for both input.  I'm delighted.

To me a transformer is like torque converter trading distance for force (current and voltage).  A source is like a pendulum and can connect to either end of the length.  Think of this as a Milkovic pendulum.  If we do not load the transformer, energy would retains as swinging just as we tie one end of the mechanical oscillator.  Loading a transformer is like loading a Milkovic pendulum.  From here we can use EM's math to visualize OU.  A change in magnetic field creates a voltage (centrifugal force). 

It is important to put our differences aside and try to visualize this... for a greater good. 
   
img  exnihiloest
Re: Impedance matching « Reply #4 on: 2012-08-08, 10:44:18 »
Group: Guest
Quote from: GibbsHelmholtz on 2012-08-06, 05:14:25
...
Loading a transformer is like loading a Milkovic pendulum.  From here we can use EM's math to visualize OU.
...

Force, voltage, current... are not energy. A transformer doubles the voltage but divides the current by two, or vice versa: no energy gain. A lever having a arm twice the length of the other doubles the force but divides the displacement by two: no gain.
Using math, we get OU=0.

Quote
A change in magnetic field creates a voltage (centrifugal force).

A centrifugal force doesn't exist in a referential frame at rest, that one where we hope to recover energy. A centrifugal force exists only in the non-inertial referential of an accelerating body. The energy is referential dependent. You can't mix a force in a referential (and worse, non-inertial) with the displacement in another one, while expecting for a correct result.
The Milkovic pendulum is a conventional device. I had modelized it with Workingmodel 2005, it works according to mechanics laws. It is just a useful idea, especially when used by Milkovic as a manual pump, well fitting the human force (= "impedance matching"). The J. Marjanovic's math about the Milkovic pendulum is full of basic errors. I can hardly believe that he is an engineer (or he falsified the matter).

« Last Edit: 2012-08-08, 11:49:53 by exnihiloest »
   
img  GibbsHelmholtz
Re: Impedance matching « Reply #5 on: 2012-08-08, 14:52:56 »
Group: Guest
Quote from: exnihiloest on 2012-08-08, 10:44:18
Force, voltage, current... are not energy. A transformer doubles the voltage but divides the current by two, or vice versa: no energy gain. A lever having a arm twice the length of the other doubles the force but divides the displacement by two: no gain.
Using math, we get OU=0.

A centrifugal force doesn't exist in a referential frame at rest, that one where we hope to recover energy. A centrifugal force exists only in the non-inertial referential of an accelerating body. The energy is referential dependent. You can't mix a force in a referential (and worse, non-inertial) with the displacement in another one, while expecting for a correct result.
The Milkovic pendulum is a conventional device. I had modelized it with Workingmodel 2005, it works according to mechanics laws. It is just a useful idea, especially when used by Milkovic as a manual pump, well fitting the human force (= "impedance matching"). The J. Marjanovic's math about the Milkovic pendulum is full of basic errors. I can hardly believe that he is an engineer (or he falsified the matter).



I think the mistake is to make judgement on a half full glass (now I'm being bias lol). 

Just take it easy Exn.  It is whatever it is. 

   
Pages: [1]
« previous next »


 

Home Help Search Login Register
Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u 2024-11-27, 10:38:58