MIT 2 004 - Introduction to the Operational Amplifier

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MIT OpenCourseWare http://ocw.mit.edu 2.004 Dynamics and Control II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.+-vvvo u t+-G a i n : A� Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008 Lecture 121 Reading: • Class Handout - Introduction to the Operational Amplifier Nise: pages 55–58 (Operational Amplifiers) • This lecture is inserted in the course at this point to coincide with Laboratory Session 3, which requires the construction of a proportional controller using a differential amplifier. 1 Introduction to the Operational Amplifier The integrated-circuit operational-amplifier (op-amp) is a fundamental building block for many electronic circuits, including analog control systems. In this hand-out we examine some of the basic circuits that can be used to implement control systems. We take simplified approaches to the analysis, and this discussion is by no means complete or exhaustive. What is an operational amplifier? It is simply a very high gain electronic amplifier, with a pair of differential inputs. Its functionality comes about through the use of feedback around the amplifier, as we show below. The op-amp has the following characteristics: It is basically a “three terminal” amplifier, with two inputs and an output. It is a • differential amplifier, that is the output is proportional to the difference in the voltages applied to the two inputs, with very high gain A, vout = A(v+ − v−) where A is typically 104 – 105, and the two inputs are known as the non-inverting (v+) and inverting (v ) inputs respectively. In the ideal op-amp we assume that the gain A−is infinite. 1copyright cD.Rowell 2008 12–112345678- V+ Vo u t p u to f f s e t b a l a n c es ss sn o n i n v e r t i n g i n p u ti n v e r t i n g i n p u to f f s e t b a l a n c e+-+-+ 1 5 v- 1 5 vo u t p u ti n p u t+-vvo u ti n• In an ideal op-amp no current flows into either input, that is they are voltage-controlled and have infinite input resistance. In a practical op-amp the input current is in the order of pico-amps (10−12) amp, or less. • The output acts as a voltage source, that is it can be modeled as a Thevenin source with a very low source resistance. Op-amps come in many forms and with a bewildering array of specifications. They range in cost from a few cents to many dollars, depending on the specs. These specifications include input impedance, input bias current, output offset voltage, external power requirements, etc. Higher grade amplifiers are known as precision, or instrumentation amplifiers. Op-amps come in a variety of packages. A common inexpensive op-amp, the 741, has an 8 pin DIP (dual in-line package) form. Many amps use a common basic pin-out for this package as shown below: The pins are numbered counter-clockwise from the top left as shown above. (Note that pin 1 is identified by a notch at the top or a dot beside pin 1.) The basic amplifier is connected between pins 2, 3 and 6. The amplifier requires a pair of external supply voltages to operate, these are typically ±15 volts and are are connected to pins 7 (positive) and 4 (negative). Pins 1 and 5 are usually used for optional external offset nulling circuitry - the actual connection is dependent on the type. We will not use this feature if we can get away without it. Non-Inverting Configurations The Unity Gain Non-Inverting amplifier: The simplest configuration of the op-amp is as a unity gain “buffer” amplifier: The output is connected directly to the inverting input. Then vout = A(v+ − v−) = A(v+ − vout) A = v+1 + A 12–2+-+ 1 5 v- 1 5 vo u t p u ti n p u t+-vvo u ti nRR12RR21+-L o a dRivand if, as noted above, A is typically 104 or much higher we may simply say vout = v+ The unity gain amplifier is used as a buffer to minimize loading on a circuit, because it has a very high input impedance (draws no current) yet provides a low output resistance output. The Non-Inverting amplifier with gain: The ab ove circuit may be mo dified to feed back only a fraction of the output voltage by including a two resistor voltage divider. In this case R1 and R2 act as a voltage divider, and R2 v = vout− R1 + R2 Then vout = A(v+ − vin)� R2 � = A vin − R1 + R2 vout A(R1 + R2) = vinR1 + R2 + AR2 and when A � 1 R1 + R2 vout = vinR2 For example if R1 = 47 kΩ, and R2 = 10 kΩ, the amplifier voltage gain will b e 5.7. The Voltage-Controlled Current-Source: The above circuit can be modified to pro-duce a current source. 12–3+-+ 1 5 v- 1 5 vo u t p u t+-vvo u ti nRRi nfiii nfs u m m i n g j u n c t i o n( v i r t u a l g o u n d )G n di n p u tRRi nfAssume that the op-amp drives a load of unknown resistance. A small resistor R is placed in series with the load. Then the inverting input is v = iR. At the output of the amplifier − vout = = A (v+ − v−) A (vin − iR) Then i = 1 R � vin − 1 A vout � and if A � 1 i = 1 vinR which is independent of the load. For example, if R = 0.1Ω, then the current in the load will be i = 10vin amps. It should be realized, however, that typical op-amps are not capable of supplying more than a few milli-amps of current. A power op-amp or an additional power amplifier will generally be necessary to supply significant current. Inverting Configurations The Inverting Amplifier: The basic inverting amplifier has a configuration as shown below: The non-inverting input is connected to ground and a pair of resistors, Rin and Rf (the feedback resistor) is used to define the gain. To simplify the analysis we make the following assumptions: (a) The gain A is very high and we let it tend to infinity. Because vout = A(v+ − v−) and v+ = 0, this means that v = limA→∞ vout/A = 0. In other words, the inverting input −voltage is so small (µvolts) that we declare it to be a “virtual ground”, that is in our simplified analyses we assume v = 0. − (b) No current flows into either of the inputs. This allows us to apply Kirchoff’s current law at the junction of the resistors. With the above assumptions, when we apply Kirchoff’s current law at the junction of the resistors (the inverting input terminal): iin + if = 0 12–4+-+ 1 5 v- 1 5 vo u t p u t+-vvo u tRfifs


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