Impedance Matching
05/21/2006
For a more complete understanding in how Impedance Matching works, you should also consider the associated topics "Thevenin's Equivalent Circuit" and "Norton's Equivalent Circuit". The reason for this comment is that every device has a "Source Resistance" or "Source Impedance". Now, having said that, let's continue with what we hope will a good approach to this very important subject:
- Consider if we have a 100V source that has a low internal resistance (like 40 ohms), and we want to use this source to deliver voltage, current, and power to a load. Now, if our load is something on the order of 1K ohm, it would be no trouble at all to deliver a pretty good 100V to that 1K ohm load. We would not however see a lot of current flowing through the load (like only 100V/1K ohm = 0.1A, or 100ma). Now if we calculate the power by any one of the three power formulas (P=EI, P=E2/R, P=I2R) we would arrive at 10W delivered into that particular load.
- Now if we were to take that 100V source and feed it into a much lower resistance, we would see a lot more current, but we would also find a voltage loss inside our source (that 40 ohms). For example, let's have a load of 10 ohms.
- The problem here, is that with that kind of a load, the total circuit resistance is the source of 40 ohms in series with the load resistance of the 10 ohms, to make a total of 50 ohms. Now our 100V is across the total series 50 ohms, and will deliver 2A of current to the load. Note that the voltage across the load will only be 10V, and that the remaining 80V is dropped across the internal resistance.
- If we do our power calculations, we will find that the power delivered to the load is 20W, and that there is a power loss inside the source of 160W. Note that there is more power loss in the source than there is gained in the load, and this often surprises folks to discover just how much power is lost inside of a power source.
- Now, lets use a load where the value matches that of the source. With 40 ohms across the source, we will find a total resistance of 80 ohms for our 100V. We can calculate that this will give us 1.25A, with half of the 100V across the equal 40 ohm load, and the other half lost in the source. We can also calculate the power delivered to the load as 62.5W, and we will also find that the power loss inside the source is the same 62.5W.
- We now find that with a light load we will get a good supply of voltage across the load, but not the maximum power delivered to that load. i.e. Lots of voltage, and very little current.
- We also find that with a very heavy load we will get a fair amount of current delivered to that load, but again not the maximum power delivered to that load. i.e. Lots of current, with a lot of voltage loss.
- Calculations will show us that only when we have a load that matches the internal source resistance will we be able to achieve maximum power transfer. We should still remember that there is an equal power loss inside the power source even while we are achieving this maximum power transfer to the load.
Where "Thevenin's Equivalent Circuit" and "Norton's Equivalent Circuit" play into these concepts is where we need to understand just what is a good voltage source, vs a good current source, vs what is a good power source (and remember that in these concepts, we are NOT evaluating these in terms of how much voltage, how much current, or even how much power is being considered).
