Fundamentals-of-Microelectronics-Behzad-Razavi.pdf

Solution from fig 827 we recognize that the first

Info icon This preview shows pages 419–422. Sign up to view the full content.

View Full Document Right Arrow Icon
Solution From Fig. 8.27, we recognize that the first stage amplifies the offset by a factor of 100, generating a dc level of 200 mV at node (if the microphone produces a zero dc output). The second stage The reader can show that placing in series with the inverting input of the op amp yields the same result.
Image of page 419

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 395 (1) Sec. 8.4 Op Amp Nonidealities 395 10 k 100 X 10 k 100 out V A 1 A 2 Figure 8.28 Two-stage amplifier. now amplifies by another factor of 100, thereby attempting to generate V. If operates with a supply voltage of, say, 3 V, the output cannot exceed this value, the second op amp drives its transistors into saturation (for bipolar devices) or triode region (for MOSFETs), and its gain falls to a small value. We say the second stage is saturated. (The problem of offset amplification in cascaded stages can be resolved through ac coupling.) Exercise Repeat the above example if the second stage has a voltage gain of 10. DC offsets impact the inverting amplifier of Fig. 8.7(a) in a similar manner. This is studied in Problem 49. We now examine the effect of offset on the integrator of Fig. 8.10. Suppose the input is set to zero and is referred to the noninverting input [Fig. 8.29(a)]. What happens at the output? Re- call from Fig. 8.20 and Eq. (8.61) that the response to this “input” consists of the input itself [the unity term in (8.61)] and the integral of the input [the second term in (8.61)]. We can therefore express in the time domain as (8.70) (8.71) where the initial condition across is assumed zero. In other words, the circuit integrates the op amp offset, generating an output that tends to or depending on the sign of . Of course, as approaches the positive or negative supply voltages, the transistors in the op amp fail to provide gain and the output saturates [Fig. 8.29(b)]. The problem of offsets proves quite serious in integrators. Even in the presence of an input sig- nal, the circuit of Fig. 8.29(a) integrates the offset and reaches saturation. Figure 8.29(c) depicts a modification, where resistor is placed in parallel with . Now the effect of at the output is given by (8.9) because the circuits of Figs. 8.5 and 8.29(c) are similar at low frequencies: (8.72) For example, if mV and , then contains a dc error of 202 mV, but at least remains away from saturation..
Image of page 420
BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 396 (1) 396 Chap. 8 Operational Amplifier As A Black Box V V EE CC out V C 1 R 1 V os t V CC out V out V C 1 R 1 V os R 2 out V C 1 R 1 R 2 in V (d) (c) (a) (b) Figure 8.29 (a) Offset in integrator, (b) output waveform, (c) addition of to reduce effect of offset, (d) determination of transfer function. How does affect the integration function? Disregarding , viewing the circuit as shown in Fig. 8.29(d), and using (8.27), we have (8.73) Thus, the circuit now contains a pole at rather than at the origin. If the input signal frequencies of interest lie well above this value, then and (8.74) That is, the integration function holds for input frequencies much higher than . Thus, must be sufficiently small so as to minimize the amplified offset given by (8.72) whereas
Image of page 421

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 422
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern