this attenuation is undesirable and that we want the entire input signal V s to appear across the load. By inserting a voltage follower in between R s and R L , we can accomplish just that! Since the current i p = 0, the voltage v p = v s . Therefore, we have v s = v p = v n = v o (8.53)
218 CHAPTER 8. OPERATIONAL AMPLIFIERS 8.4.2 Instrumentation Amplifier Figure 8.22: Instrumentation amplifier An instrumentation amplifier as shown in Figure 8.22 is used to amplify a small difference between two signals. This type of circuit is often used in sensors to detect deviations from a nominal value. If one can relate a physical quantity such as temperature, pressure, humidity, etc. to a voltage, then by using an instrumentation amplifier a sensor can be designed to detect when, for example, room temperature/humidity/pressure/etc. deviates from an acceptable value. In Engr40M, we will be using an instrumentation amplifier as part of a larger circuit that can monitor your heart rate! To express v o in terms of v 1 and v 2 , we first note that the the third amplifier is a difference amplifier with inputs v o 1 and v o 2 . Therefore, using the result we derived earlier, we have the following equation: v o = - R 4 R 5 v o 1 + R 4 + R 5 R 5 R 4 R 4 + R 5 v o 2 (8.54) This can be simplified to: v o = R 4 R 5 ( v o 2 - v o 1 ) (8.55) Next, we determine the intermediate output voltages v o 1 and v o 2 . We first write a KCL equation at v n 1 , the negative input of the first amplifier and apply the op-amp “golden rules:” v 1 - v 2 R 2 + v 1 - v o 1 R 1 = 0 (8.56) This can be simplified to: v o 1 = v 1 + R 1 R 2 ( v 1 - v 2 ) (8.57) Next, we write a KCL equation at v n 2 , the negative input of the second amplifier and apply the op-amp “golden rules:” v 2 - v 1 R 2 + v 2 - v o 2 R 3 = 0 (8.58) This can be simplified to: v o 2 = v 2 + R 3 R 2 ( v 2 - v 1 ) (8.59)
8.5. SUMMARY 219 Next, we subtract equation ( 8.57 ) from equation ( 8.59 ) to obtain: v o 2 - v o 1 = v 2 - v 1 + R 3 R 2 ( v 2 - v 1 ) + R 1 R 2 ( v 2 - v 1 ) (8.60) This can be simplified to: v o 2 - v o 1 = R 1 + R 2 + R 3 R 2 ( v 2 - v 1 ) (8.61) Finally, we can obtain the overall expression for v o in terms of v 1 and v 2 by substituting equation ( 8.61 ) into equation ( 8.55 ), and we obtain the following: v o = R 4 R 5 R 1 + R 2 + R 3 R 2 ( v 2 - v 1 ) (8.62) In the case where all the resistors are equal except for R 2 , the expression for the output voltage becomes: v o = 1 + 2 R R 2 ( v 2 - v 1 ) (8.63) In this case, the gain of the instrumentation amplifier is set by R 2 . This configuration can be very useful if one desires to have a variable gain since R 2 can be be implemented as a potentiometer. 8.5 Summary The Golden Rules for Ideal Op-Amps • A = ∞ • R i = ∞ • R o = 0 • i p = i n = 0 • v n = v p (for negative feedback configurations) The output voltage of an op-amp is limited by the supply voltage for the op-amp: | v o | ≤ V dd (8.64) 8.6 Analysis of non-ideal Op Amp (bonus material) The previous sections analyzed an ideal op-amp behavior. Analyzing the behavior of a non-ideal op-amp is only a little more difficult and is done in the next sections.
220 CHAPTER 8. OPERATIONAL AMPLIFIERS 8.6.1 Noninverting Amplifier Figure 8.23: A noninverting amplifier using the equivalent circuit model of an op-amp A noninverting amplifier can be used to amplify an input signal without inverting its sign. In order
- Spring '19
- Hassan Kasfy