assignment08232016s

With at least six, 20 ohm, 200 watt, non-inductive resistors in hand, the way I approached it was to begin putting them in series to see how close to 50 ohms I could get. Two in series adds to 40 ohms, and three in series adds to 60 ohms. So we have to figure how to either add 10 more ohms to two in series, or subtract 10 ohms from three in series.

We know that resistors in parallel have a total resistance smaller than the smallest resistor. We also know that resistors of equal value in parallel, give a total resistance of one of them, divided by the number of them. Hence, two, 20 ohm resistors in parallel yields a total resistance of 10 ohms, while three in parallel would yield a total resistance of 20/3 = 6.67 ohms. You can use the equation:

So it jumps out at us that if we add (put in series) the combination of two, 20 ohm resistors in parallel with two 20 ohm resistors in series, we get a total equivalent resistance of 50 ohms as desired:

I could also approach the problem from three resistors in series, by noting that if I put the last resistor in a series string of three, in parallel with another 20 ohm resistor, I would get the same combination as above.

Another way, if I had a total of 10, 20 ohm resistors, would be to make two series strings of five resistors each. So each string is a total of 100 ohms. Then, if I put the two strings of five in parallel, I would divide the that resistance in half and get 50 ohms. But this would require 10 resistors instead of just four. However, there might be an advantage to the 10 resistor array in terms of the total transmitter power I might apply without overstressing the power rating (200 watts) of any one resistor. We'll see below.

First, for the first "extra credit question", let's calculate the maximum power I can apply to the four resistor network. We know that the current entering at A and exiting at B must be the same. By inspection, we also know that the total current flows through the two resistors in series, and divides equally between the two resistors in parallel because the latter are of equal value. As a result, we know that the two resistors in series will each dissipate the same power; moreover, that power is greater than either of the resistors in parallel have to dissipate.

First let's calculate the maximum power either of the two resistors in series can tolerate at their maximum power rating of 200 watts. Power (P) = I squared x R,

But 3.162 A is flowing through a total equivalent circuit resistance of 50 ohms, so calculating how much power 3.162 A dissipates in 50 ohms:

Even then, the resistors would get warm if not hot, and it would be wise to blow some air across them to prevent deterioration. And if you really blew some air flow across them, you could get away with applying more than 500 w, at least for some period of time.

Now what about the 10 resistor network (two strings of five in series, connected in parallel) totaling 50 ohms? From the four resistor array solution, we know that a series string of 20 ohm resistors can tolerate 3.162 A without exceeding the 200 watt rating of any one resistor. If I have two such strings in parallel, each one having the same series resistance, the total current divides in half for each string. So if I apply a total of 3.162 x 2 = 6.324 A to the input, I know that each series string of five will only see 3.162 A, which is allowed. That means that the maximum power I can apply to the entire resistor network would be:

Note that power goes up in proportion to the square of current, and because we could double the current, the power allowed goes up by a factor of four.

Let's assume the transmitter has a source resistance of 50 ohms and is driving the 50 ohm dummy load. We know that E = I x R, and we previously calculated that 3.162 A is the max current we should apply. Calculating the applied voltage, E = I x R = 3.162 x 100 ohms ( the sum of source and load resistances) yields E = 316.2 volts. This is the maximum value the voltage source can be without exceeding the power rating of any resistor. In class, Alex showed that maximum power is transferred to a load when the load resistance matches ( is equal to ) the source resistance. So, with 312.6 volts applied, from the first question, we know the load is dissipating the maximum of 500 watts.

Draw the simple series circuit with a voltage source driving both the source and load resistances in a series circuit to get the picture.

Normally, the source we are dealing with has a fixed internal (source) resistance and we can vary to the load resistance to match it, and hence maximize the power transfer. But what if the load resistance is fixed and we can't vary it to achieve a match,( and don't have a device such as an antenna system tuner to achieve a match to the source)? What if instead, we can vary the source resistance, which is effectively in series with the fixed load, and we still have a voltage source or 316.2 volts?

Let's make the source resistance 25 ohms in instead of 50 ohms. That makes the total series resistance of the circuit = 25 + 50 = 75 ohms. Then, calculating the total power dissipated in the 50 ohm load:

The point to be made is that our ability to maximize power to a load is determined by what is fixed and what is variable in the specific practical circuit situation. We normally have a fixed source resistance and so have to vary the load resistance to match the source and thus maximize power transfer. But if we have a fixed load and can reduce the source resistance we can actually increase power transfer. Also note, however, that if we could also still decrease the load resistance to match the reduced source impedance, we could further increase, indeed maximize, the power transferred.

An example of this is having a stereo system whose speaker load resistance is essentially fixed. We can actually increase power from the amplifier to the speakers, by reducing the amp's source resistance. This can be done by using alternative amp design approaches. Tube amplifiers with transformer coupled outputs had less flexibility to reduce source resistance; but, solid state designs made it easier to do so. Lower amp source resistance also had the benefit of increasing the "damping factor", which reduced the natural resonances of the electro-magnetic speakers and improving fidelity, especially by reducing boominess in the bass range.

Later in the course, when we get into AC and RF circuit theory, we'll see that there is an analogy between the speaker system case and the case of having an antenna with relatively fixed load characteristics, and using an antenna system "tuner" to achieve maximum power transfer.