1 z v dt vout rc in and for a single fourier component


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Unformatted text preview: f an active lter. In practice, one may need to supply a resistor in parallel with the capacitor to give a DC path for the feedback. !RC C R - IN OUT + Figure 33: Op-amp integrator or low-pass lter. 6.5.2 Di erentiator The circuit of Fig. 34 can be analyzed in analogy to the integrator. We nd the following: vout = ,RC dvin dt G! = ,!RC 40 41 So this ideally represents a perfect di erentiator and an active high-pass lter. In practice, one may need to provide a capacitor in parallel with the feedback resistor. The gain cannot really increase with frequency inde nitely! 6.6 Negative Feedback As we mentioned above, the rst of our Golden Rules for op-amps required the use of negative feedback. We illustrated this with the two basic negative feedback con gurations: the inverting and the non-inverting con gurations. In this section we will discuss negative feedback in a very general way, followed by some examples illustrating how negative feedback can be used to improve performance. 39 R C - IN OUT + Figure 34: Op-amp di erentiator or high-pass lter. 6.6.1 Gain Consider the rather abstract schematic of a negative feedback ampli er system shown in Fig. 35. The symbol is meant to indicate that negative feedback is being added to the input. The op-amp device itself has intrinsic gain A. This is called the op-amp's open-loop gain since this is the gain the op-amp would have in the absence of the feedback loop. The quantity B is the fraction of the output which is fed back to the input. For example, for the non-inverting ampli er this is simply given by the feedback voltage divider: B = R1=R1 + R2. The gain of the device is, as usual, G = vout=vin. G is often called the closed-loop gain. To complete the terminology, the product AB is called the loop gain. v in a v out A + - B Figure 35: General negative feedback con guration. As a result of the negative feedback, the voltage at the point labelled a&quot; in the gure is va = vin , Bvout The ampli er then applies its open-loop gain to this voltage to produce vout: vout = Ava = Avin , ABvout Now we can solve for the closed-loop gain: vout=vin G = 1 +A AB 42 Note that there is nothing in our derivatio...
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