Measuring Input and Output Resistance

Transcription

1 Measurg Input and Output esistance Introduction The objective of these notes is to allow you to quickly and accurately measure the put and output resistance of circuit elements so that you can concentrate on more advanced topics. Prelimary Note about Circuit Diagrams You're probably used to seeg circuit diagrams like this: 1 k Figure 1. A circuit diagram the style of an troductory physics textbook. However, electronics textbooks follow a different convention: 1k Figure 2. The same circuit the style of an electronics textbook. Figures 1 and 2 both represent exactly the same circuit! It's a 1k resistor with across it. Figure 2 takes a little less time to draw, so we prefer Figure 2. Input esistance Suppose you construct the followg circuit:

2 oltmeter Figure 3. A voltage source connected to a voltmeter through a resistor. QUESTION: Does the voltage recorded by the voltmeter depend on? WONG answer: No current ever flows to a voltmeter, so the current through is 0, so the voltage across is 0, so the voltmeter always sees across its termals. IGHT answer: ery little current flows to the voltmeter because its put resistance () is very high, but is not fite a real voltmeter. appears the circuit as follows: oltmeter Figure 4. The put resistance of the voltmeter is revealed. When <<, domates the series combation of and, which means that nearly the full appears across. However, when grows nearly as large as, we can no longer neglect. The current through the series combation is ()/( + ), so the voltage across (which is what the voltmeter measures) is measured. (1) You can see from Equation (1) that measured approaches as gets small, but measured drops as approaches. It seems that the higher is, the better; higher values allow to be higher before measured is affected. High put resistance is good.

3 Output esistance Now consider the followg circuit: oltmeter Figure 5. A voltage source connected to the parallel combation of a resistor and a voltmeter. QUESTION: Does the voltage recorded by the voltmeter depend on? WONG answer: Evidently, the voltage at one end of the voltmeter is always, and the voltage at the other end is always ground. IGHT answer: We must consider the output resistance of the voltage source: out oltmeter Figure 6. The output resistance of the voltage source is revealed. There are three resistors Figure 6. Let's first look at and. Let's assume that <<. In this case, domates the parallel combation, and we can neglect. Now we just have out series with. The current is then ()/(out + ), and the voltage across (which is measured by the voltmeter) is measured. (2) out You can see that when is much larger than out, measured approaches ; negligible voltage appears across out. However, when becomes as small as out, measured drops because some of the voltage appears across out. We don't want the voltage at the termals of our voltage source to depend on, so out should be as small possible;

4 smaller out values mean than can be smaller before measured is affected. Low output resistance is good. Measurg Input esistance Suppose I have the followg circuit: 1 out 2 Figure 7. A circuit element known as a voltage divider. Unlike the voltmeter Figure 4 (which only receives put) or the voltage source Figure 6 (which only supplies output), the circuit element Figure 7 has both an put and an output termal. Thus, this circuit element has both an put resistance and an output resistance. You might want to know the put resistance of this circuit element (if only because that's your assignment lab). This is the way to measure the put resistance of an arbitrary circuit element: 1. Model the entire circuit as a sgle resistor between the put and ground: Figure 8. The model used to determe. So whatever your circuit it made of, whether it contas a hundred resistors or a thousand transistors, you model it as a sgle resistor (just for this measurement). 2. Place a known resistor series with your circuit and apply a voltage, as shown Figure 9. (In this course, many circuit properties will depend on frequency, so we do NOT apply a dc voltage. Instead, we apply an ac voltage with a function generator. I refer to the amplitude of the ac signal as.)

5 ' Figure 9. The circuit used to measure. 3. Use an oscilloscope to measure the amplitudes and '. These two measurements can be done simultaneously or one at a time. Knowg, ', and, you can determe. The current through is I = ( ')/. The same current flows through : I = '/. We can combe these two equations to solve for : '. (3) ' Measurg Output esistance We just saw how to measure the put resistance of the circuit element Figure 7. We now want to know how to measure its output resistance. We follow the followg steps: 1. Apply a voltage signal at the put of the circuit element Figure 7. This voltage will not explicitly appear the model we use to determe out. 2. Model the entire circuit as a voltage series with out. (This is called a Theven model, and sometimes out is called Th.) The voltage this model is the OUTPUT voltage for an "unloaded" circuit (a circuit with nothg connected between its output and ground). out out Figure 10. The model used to determe out. 3. Measure the amplitude = out by connectg an oscilloscope to the output shown Figure 10. ( = out because negligible current flows to the oscilloscope, so the current through out is zero, so the voltage across out is zero.)

6 4. Now connect a known resistor between out and ground. Current now can flow through out, and out no longer equals. Measure the new out (which I'll call ') with an oscilloscope. The measurements of and ' cannot be done simultaneously! out ' Figure 11. The "loaded" circuit (the load is the resistance between the output and ground). It is extremely important to recognize that while the circuit is loaded (as shown Figure 11), it is IMPOSSIBLE to measure. You can only measure voltages that appear at the termals of the dotted box. In Figure 11, is "buried" side the model, and only ' can be measured. Conversely, Figure 10, only (= out this case) can be measured. Knowg, ', and, you can determe out. eferrg to Figure 11, the same current I flows through out and. The current through out is I = ( ')/out, and the current through is I = '/. We combe these equations to solve for out: out '. (4) '