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1、UNIT 2 A: The Operational Amplifier One problem with electronic devices corresponding to the generalized amplifiers is that the gains, Au or A, depend upon internal properties of the two-port system (p, fl, R, Ro, etc.)? This makes design difficult since these parameters usually vary from device to
2、device, as well as with temperature. The operational amplifier, or Op-Amp, is designed to minimize this dependence and to maximize the ease of design. An Op-Amp is an integrated circuit that has many component part such as resistors and transistors built into the device. At this point we will make n
3、o attempt to describe these inner workings. A totally general analysis of the Op-Amp is beyond the scope of some texts. We will instead study one example in detail, then present the two Op-Amp laws and show how they can be used for analysis in many practical circuit applications. These two principle
4、s allow one to design many circuits without a detailed understanding of the device physics. Hence, Op-Amps are quite useful for researchers in a variety of technical fields who need to build simple amplifiers but do not want to design at the transistor level. In the texts of electrical circuits and
5、electronics they will also show how to build simple filter circuits using Op-Amps. The transistor amplifiers, which are the building blocks from which Op-Amp integrated circuits are constructed, will be discussed. The symbol used for an ideal Op-Amp is shown in Fig. 1-2A-1. Only three connections ar
6、e shown: the positive and negative inputs, and the output. Not shown are other connections necessary to run the Op-Amp such as its attachments to power supplies and to ground potential. The latter connections are necessary to use the Op-Amp in a practical circuit but are not necessary when consideri
7、ng the ideal 0p-Amp applications we study in this chapter. The voltages at the two inputs and the output will be represented by the symbols U+, U-, and Uo. Each is measured with respect t ground potential. Operational amplifiers are differential devices. By this we mean that the output voltage with
8、respect to ground is given by the expression Uo =A(U+ -U-) (1-2A-l) where A is the gain of the Op-Amp and U+ and U - the voltages at inputs. In other words, the output voltage is A times the difference in potential between the two inputs. Integrated circuit technology allows construction of many amp
9、lifier circuits on a single composite chip of semiconductor material. One key to the success of an operational amplifier is the cascading of a number of transistor amplifiers to create a very large total gain. That is, the number A in Eq. (1-2A-1) can be on the order of 100,000 or more. (For example
10、, cascading of five transistor amplifiers, each with a gain of 10, would yield this value for A.) A second important factor is that these circuits can be built in such a way that the current flow into each of the inputs is very small. A third important design feature is that the output resistance of
11、 the operational amplifier (Ro) is very small. This in turn means that the output of the device acts like an ideal voltage source. We now can analyze the particular amplifier circuit given in Fig. 1-2A-2 using these characteristics. First, we note that the voltage at the positive input, U +, is equa
12、l to the source voltage, U + = Us. Various currents are defined in part b of the figure. Applying KVL around the outer loop in Fig. 1-2A-2b and remembering that the output voltage, Uo, is measured with respect to ground, we have -I1R1-I2R2+U0=0 (1-2A-2) Since the Op-Amp is constructed in such a way
13、that no current flows into either the positive or negative input, I- =0. KCL at the negative input terminal then yields I1 = I2 Using Eq. (1-2A-2) and setting I1 =I2 =I, U0=(R1+R2)I (1-2A-3) We may use Ohms law to find the voltage at the negative input, U-, noting the assumed current direction and t
14、he fact that ground potential is zero volts: (U-0)/ R1=ISo, U-=IR1and from Eq. (1-2A-3), U- =R1/(R1+R2) U0Since we now have expressions for U+ and U-, Eq. (1-2A-l) may be used to calculate the output voltage,U0 = A（U+-U-）=AUS-R1U0/(R1+R2)Gathering terms, U0 =1+AR1/(R1+R2)= AUS (1-2A-4)and finally, AU = U0/US= A(R1+R2)/( R1+R2+AR1) (1-2A-5a)This is the gain factor for the circuit. If A is a very large number, larg
