Radio World - 12 June 1996
by Harold Hallikainen
SAN LUIS OBISPO, Calif. Last month, we used Thevenin equivalents to analyze a Wheatstone Bridge circuit (which, as you may recall, was first described by Samuel Hunter Christie in 1833). Let's convert the voltage sources in the Thevenin equivalent to current sources and move the Thevenin resistance from in series with the voltage source to across the current source. This is the Norton Equivalent, a technique described by E. L. Norton of Bell Telephone Laboratories in 1926. Next month we'll generalize this a little and get Millman's Theorem.
I recently received a fax from George Pfisterer, Jr of Huntingdon Valley, PA. Mr. Pfisterer recalls working in the Data Communications Department of Bell Telephone Laboratories in Murray Hill, NJ. A few doors down the hall in the building known as "Bitsburgh" was the office of Ed Norton, the originator of the theorem we all studied. Mr. Pfisterer recalls walking past the office and "seeing this elderly white-haired gentleman operating a giant 24 inch slide rule." I guess a large slide rule gives a digit or two more resolution? Anyway, Mr. Norton first described one of the fundamental theorems of circuit analysis and 40 years later was section head in data communications. The progress in electronics in one person's lifetime is truly phenomenal. I'm truly impressed by the bit of the history of our field I've lived through. How about you?
Let's go back to the simple three resistor circuit we've analyzed many times in this series. I hope that using a variety of techniques to analyze the same circuit we'll get the same answer each time (it would be nice!), and we'll develop a toolbox of analysis techniques.
Figure 1 shows the circuit to be analyzed. In figure 2, the circuit is broken into three parts. In figure 3, the left and right portions of the circuit are replaced with their Norton equivalents (while the center resistor remains unchanged).
Recall that in the Thevenin equivalent a complex circuit is replaced with a single voltage source and a series resistance. Further, the Thevenin voltage was the open circuit voltage of the original circuit. We determined this by placing an "imaginary voltmeter" across the two terminal we were modeling and measuring the voltage. Finally, we determined the Thevenin resistance by replacing the imaginary voltmeter with an imaginary ohmmeter, replacing all voltage sources with shorts, and replacing all current sources with opens. The imaginary ohmmeter then read the Thevenin resistance.
Similarly, in the Norton equivalent, a complex circuit is replaced by a single current source (a source that maintains a specified current independent of the load resistance) and a parallel Norton resistance. The Norton current is the short circuit current of the original circuit. Whereas we placed an imaginary voltmeter across the two terminals to find the Thevenin voltage, we place an immaginary ammeter across the two terminals for the Norton current. We determine the Norton resistance the same way we found the Thevenin resistance (which, incidentally, makes them the same!).
If we place an ammeter between the top of R1 and ground in figure 2 (disconnecting the top of R1 from the remainder of the circuit), we'll get the Norton current for the left side of the circuit. Using Ohm's law, we easily see that this is 12mA (12V/1K). The right side gives us a Norton current of 1 mA (3V/3K). Further, note that on the left side, conventional current would flow out of R1 (up), while on the right side, current flows into R3 (down) due to the polarities of the voltage sources. This difference of polarity can be represented either as shown in figure 3 (all current sources positive) or figure 4 (all current sources pointing the same direction). For now, let's stick with figure 4.
Current sources in parallel act exactly the same as voltage sources in series. Similarly, just as it is meaningless to put two voltage sources with different voltages in parallel (you'll get an infinite current around the loop as each source insists on setting the voltage), you cannot place different current sources in series (you'll get infinite voltage out of each as it tries to force the other to have the specified current). Just as voltage sources in series add algebraically, current sources in parallel add algebraically. Putting this all together, we get the circuit of figure 5, where the two current sources (12mA and -1mA) have added to yield 11mA.
In figure 5, we combine the three parallel resistors to find a single 11mA current source driving a single 545.5 ohm resistor. The voltage across that resistor is 11mA * 545.5 ohms, or 6.0 volts. Further, since the conventional current is going into the top end of the resistor, the top end of the resistor is positive with respect to the bottom (as shown). Finally, we can determine the voltage at VN by starting at ground and "winding our way around the circuit" to VN. In this case, we don't have far to go. Starting at ground, we go up the 545.5 ohm resistor, going up 6.0 volts (since we came out the positive end of the resistor). The voltage at VN is +6.0 volts. Luckily, this agrees with all our previous solutions to this problem!
The problem of several voltage sources driving a single point through
series resistors is a common one. Millman's theorem (which we'll discuss
next month) generalizes the approach we just took to cover any number of
voltage sources and series resistances. As we look at such a general
solution what do we do with a resistor that doesn't have a voltage
source in series with it (such as R2 in this circuit)? We say it has a
zero volt source in series with it! This a very useful concept! A short
(or a piece of wire) can be considered a zero volt voltage source. No matter
how much current goes throught the short, the voltage remains zero (ideally).
This is exactly the same as an ideal 12 volt battery. No matter how much
current goes through the battery, the voltage remains 12 volts. But, a
12 volt battery seems useful, while a zero volt one does not. The Spice
circuit analysis program (available on our web pages and our BBS) does
not have an ammeter element. If you want to measure current through some
particular point in a circuit, you insert a zero volt voltage source. The
current through that voltage source is calculated by Spice.
Next month we'll finish up DC circuit analysis with Millman's theorem, then move into AC circuit analysis. With AC, we'll use the same techniques we've developed so far, but use "funny numbers" to do the math. Eventually we'll move into active circuits, amplifiers, oscillators, modulation, and ideas about information theory (is FM the original spread spectrum system?). I look forward to your ideas on the route we should take on our trip.