November 5, 2008 Educational Radio Net, PSRG 24th session, Lee Bond N7KC
The impedance series is now history. During the course of that 13 week series we looked at several of the most fundamental ideas in the physics of electrical phenomenon and, hopefully, gained some practical knowledge of how these ideas link together to form a basis for our understanding of all things electrical. Let's exercise some of this earlier impedance series material and see how it can be applied to solve practical problems which are routinely encountered on the bench. The first study examined the potentiometer or "pot" and its behavior when used as a voltage divider. This second study will firstly examine how energy is moved from a source to a load and secondly, consider the effect of a transmission line in this process.
First, we need to understand a very elementary concept in describing mathematical plots. Imagine that we are walking along a straight path which starts to curve uphill. We notice that the walking is getting tougher as the path curves upward. We might make the observation that this is a steep upward slope. As we continue our walk along the path, it levels out and immediately starts to slope downward and we must hold back to avoid running. We might make the observation that this is a steep downward slope. Looking back on our route we see that the high point on the walk was at the highest point on the hill and, further, that the slope was actually zero at that point. So it is with graphical plots. A maximum point (or minimum for that matter) on a graph always occurs at exactly zero slope. If you are skilled with your math then it is an easy matter to set the slope to zero and determine the conditions which will then cause the maximum or minimum on the plot.
It seems to be common knowledge that one must match the antenna impedance to the transmission line to transfer maximum energy per unit time (power) across the connection. What is not so widely known is that we can use a resistive voltage divider to demonstrate the idea directly and with ease.
Let’s set up a demonstration circuit to test the idea. We will set a powerful oscillator to a frequency to 10 Mhz and adjust the output voltage to 100 volts rms. Consider this to be a "perfect" voltage source with zero internal impedance. This means that our oscillator will stubbornly maintain the 100 vrms at its output without regard to load. Now, let’s convert this oscillator to a real world device by adding 50 ohms to the output. This is the equivalent to your radio transmitter which has a 50 ohm output.
Next, we have a large carbon resistor and we can change the resistance value from zero ohms to 200 ohms by merely turning a calibrated knob. Since we suspect that heating of the resistor might be an interesting thing to watch let’s attach a thermometer to the resistor to see how its temperature changes during the demonstration.
Finally, attach the carbon load resistor to the 50 ohm output of our demonstration oscillator to complete the circuit. Let’s also attach an RF voltmeter to the load resistor so that we can log some numbers during the demo process. (see spreadsheet data and plot at end of this article)
So, we are ready to start the test and take some data. To make our point and to keep this short we will just do 3 measurements so set the load resistor to 30 ohms and we notice that the voltage across the load resistor is 37.5 volts. We know from Joule’s Law that power is just voltage squared divided by the resistance so the power (energy per time) dissipated in the load is 46.88 watts or 46.88 joules per second. Checking the thermometer we see that it has moved upscale from room temperature to level 1.
Next, set the load resistor to 50 ohms and we notice that the voltage across the load resistor is 50 volts. Applying Joule’s Law once again we see the dissipated power to be 50 watts or 50 joules per second. The thermometer is now reading higher than level 1 from the first measurement.
Finally, set the load resistor to 80 ohms and we notice that the voltage across the load resistor is 61.5 volts. Applying Joule’s Law once again we see the dissipated power to be 47.34 watts or 47.34 joules per second. The thermometer is now reading very close to level 1 from the first measurement.
Taking a look at our data we see that the load resistor temperature was highest at 50 ohms and dropped off either side of 50. If we were to take multiple data points and plot them on graph paper, such that the vertical axis, the ordinate, represented power and the horizontal axis, the abscissa, represented values of load resistance then we would show a "hill" much like the hiking hill we traversed earlier. The maximum value would occur at zero slope (top of the hill), when the "source" resistance of the oscillator equaled the load resistance. By extension, resistance can be replaced with impedance and you will obtain exactly the same results.
We have taken the graphical approach here but mathematically this is a very clean problem. One simply writes the equation for power dissipated in the load in terms of the simple voltage division associated with the source and load impedance. Compute the slope, set it to zero, and notice that both source and load impedance must be equal to achieve zero slope.
Fine you say but what happens when I separate the source impedance and load impedance with a transmission line? Now things start to become very interesting. If, instead of RF energy, we had used DC then the transmission line could be a simple wire to complete the circuit. I deliberately used RF in the demonstration example to make a point about transmission lines. Although transmission lines are tagged with an impedance value they are not resistors. If you were to connect an ohmmeter across an open 50 ohm transmission line the meter would indicate an open circuit. If you connected the same meter across a 50 ohm resistor then the meter would read exactly 50 ohms. The solitary mission of a resistor is to convert electrical energy to heat. The mission of a transmission line is to transfer energy from point A to point B with minimum loss of energy. For example, the 50 ohm transmitter provides the energy, the 50 ohm transmission line directs the energy with intended minimum loss, and the 50 ohm load consumes the energy. So, what is going on here when the line is in play?
To answer this question we need to understand that transmission lines, coaxial cables for example, have distributed inductance and capacitance throughout. The, so called, characteristic impedance of a transmission line is defined by the square root of the ratio of the distributed inductance over the distributed capacitance of a tiny cross section of the line and is resistive. Our 50 ohm line simply scales the line current such that the ratio of line voltage to line current is 50 and no energy is lost in the process. However, there are two primary energy losses in a transmission line which we need to address. Skin effect currents flowing in the copper conductors and dielectric absorption losses in the insulating material between conductors produces heat loss. Both losses vary as a function of frequency.
There are three cases which we need to consider for the transmission line between the source and load in the demonstration example.
First, consider the infinitely long 50 ohm transmission line. Energy entering and moving down the line is constantly reduced by the skin effect losses and dielectric losses and, eventually, this energy is reduced to zero. Input energy is totally dispersed nearest the input end of the infinite length line. Our demo system behaves as if the line were a 50 ohm resistor and maximum energy is transferred from the source. No useful work has been done.
Secondly, consider a very much shorter 50 ohm line which is terminated with a 50 ohm resistor. This line is so short that skin effect losses, and others, are small so that almost the entire input energy is converted to heat in the load resistor. The 50 ohm termination matches the line so there is no impedance discontinuity and there is no reflected energy. The line and load are matched and maximum power is transferred by the line to the load. Maximum useful work has been done.
Thirdly, using the same short line as above we arrange for the load to be something other than a 50 ohm resistor. Perhaps a 60 ohm resistor is now the load. When the incident energy first encounters the 50 ohm line the characteristic impedance of the line scales the current appropriately for the 50 ohms. Then, at some later time, the traveling energy encounters the 60 ohm termination. Obviously there is now an impedance mismatch and some small fraction of the incident energy is reflected back toward the generating end. Given these circumstances, with the reflected energy in play, it is clear that maximum energy transfer can never be achieved. Less than maximum useful work has been done.
In summary, one can show either graphically or mathematically that maximum energy is transferred when the source impedance equals, or matches, the load impedance. Transmission lines do not dissipate energy as do resistors. The entire system must be matched to realize maximum energy transfer. Transmission line to load mismatches cause energy reflections which always reduce system throughput and degrade performance.
This concludes the set up for the discussion of impedance matching. Are there any questions or comments?
This image is a scan of spreadsheet data and associated plot for the demonstration circuit described above.
Double click the image to see a larger version.
This is N7KC for the Wednesday night Educational Radio Net.
Wednesday, November 5, 2008
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