自動(dòng)化專業(yè)外文翻譯--運(yùn)算放大器-其他專業(yè)-文庫(kù)吧在線文庫(kù)
【正文】 的外回路應(yīng)用基爾霍夫定律，注意輸出電壓 U0 指的是它與接零管 腳之間的電位，我們就可得到 I1R1I2R2+U0=0 (12A2) 因?yàn)檫\(yùn)算放大器是按照沒有電流流入正輸入端和負(fù)輸入端的原則制作的，即I =0。運(yùn)算放大器成功的一個(gè)關(guān)鍵就是許多晶體管放大器 “ 串聯(lián) ” 以產(chǎn)生非常大的整體增益。讓運(yùn)算放大器正常運(yùn)行所必需的其它一些管腳，諸如電源管腳、接零管腳等并未畫出。 運(yùn)算放大器的全面綜合分析超越了某些教科書的范圍。can39。P39。 or positive charges (P type). A single crystal of germanium or silicon treated with both Ntype dope and Ptype dope forms a semiconductor diode, with the working characteristics described. Transistors are formed in a similar way but like two diodes backtoback with a mon middle layer doped in the opposite way to the two end layers, thus the middle layer is much thinner than the two end layers or zones. Two configurations are obviously possible, PNP or NPN (Fig. 12Bl). These descriptions are used to describe the two basic types of transistors. Because a transistor contains elements with two different polarities (., 39。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 the gain is given by the expression AU =(R1+R2)/ R1 (12A5b) This shows that if A is very large, then the gain of the circuit is independent of the exact value of A and can be controlled by the choice of R1and R2. This is one of the key features of OpAmp design the action of the circuit on signals depends only upon the external elements which can be easily varied by the designer and which do not depend upon the detailed character of the OpAmp itself. Note that if A=100 000 and (R1 +R2)/R1=10, the price we have paid for this advantage is that we have used a device with a voltage gain of 100 000 to produce an amplifier with a gain of 10. In some sense, by using an OpAmp we trade off power for control. A similar mathematical analysis can be made on any OpAmp circuit, but this is cumbersome and there are some very useful shortcuts that involve application of the two laws of OpAmps which we now present. 1) The first law states that in normal OpAmp circuits we may assume that the voltage difference between the input terminals is zero, that is, U+ =U 2) The second law states that in normal OpAmp circuits both of the input currents may be assumed to be zero: I+ =I =0 The first law is due to the large value of the intrinsic gain A. For example, if the output of an Op Amp is IV and A= 100 000, then ( U+ U )= 10SV. This is such a small number that it can often be ignored, and we set U+ = U. The second law es from the construction of the circuitry inside the OpAmp which is such that almost no current flows into either of the two inputs. B: Transistors Put very simply a semiconductor material is one which can be 39。N39。 for negative in the case of an NPN transistor). This is also inferred by the reverse direction of the arrow on the emitter in the symbol for an NPN transistor, ., current flow away from the base. While transistors are made in thousands of different types, the nu
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