自動化專業(yè)外文翻譯--運(yùn)算放大器-其他專業(yè)-展示頁
2025-01-31 02:44本頁面
【正文】 mple filter circuits using OpAmps. The transistor amplifiers, which are the building blocks from which OpAmp integrated circuits are constructed, will be discussed. The symbol used for an ideal OpAmp is shown in Fig. 12A1. Only three connections are shown: the positive and negative inputs, and the output. Not shown are other connections necessary to run the OpAmp such as its attachments to power supplies and to ground potential. The latter connections are necessary to use the OpAmp in a practical circuit but are not necessary when considering the ideal 0pAmp 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 respect to ground is given by the expression Uo =A(U+ U) (12Al) where A is the gain of the OpAmp 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 amplifier circuits on a single posite 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. (12A1) can be on the order of 100,000 or more. (For example, 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 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. 12A2 using these characteristics. First, we note that the voltage at the positive input, U +, is equal to the source voltage, U + = Us. Various currents are defined in part b of the figure. Applying KVL around the outer loop in Fig. 12A2b and remembering that the output voltage, Uo, is measured with respect to ground, we have I1R1I2R2+U0=0 (12A2) Since the OpAmp is constructed in such a way 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. (12A2) and setting I1 =I2 =I, U0=(R1+R2)I (12A3) We may use Ohm39。s law to find the voltage at the negative input, U, noting the assumed current direction and the fact that ground potential is zero volts: (U0)/ R1=I So, U=IR1 and from Eq. (12A3), U =[R1/(R1+R2)] U0 Since we now have expressions for U+ and U, Eq. (12Al) may be used to calculate the output voltage, U0 = A（ U+U） =A[USR1U0/(R1+R2)] Gathering terms, U0 =[1+AR1/(R1+R2)]= AUS (12A4) and finally, AU = U0/US= A(R1+R2)/( R1+R2+AR1) (12A5a) This is the gain factor for the circuit. If A is a very large number, large enough that AR~ (R1+R2),the denominator of this fraction is dominated by the AR~ term. The factor A, which is in both the numerator and denominator, then cancels out and t
教學(xué)課件相關(guān)推薦
鄂ICP備17016276號-1