Lecture 12: Operational Amplifiers

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. v v - + G a in : A + v - o u t 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. 1 c D.Rowell 2008 copyright � 12–1 • 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: o ffs e t b a la n c e in v e r tin g in p u t 1 8 2 + n o n in v e r tin g in p u t 3 7 - -V s s 4 + V s s o u tp u t 6 5 o ffs e t b a la n c e 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: is as a unity gain “buffer” amplifier: The simplest configuration of the op-amp v in + vo u t in p u t + - -1 5 v 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 o u tp u t and 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 above circuit may be modified to feed back only a fraction of the output voltage by including a two resistor voltage divider. R 2 R 1 + v in R 1 R 2 vo u t + in p u t + 1 5 v - o u tp u t -1 5 v In this case R1 and R2 act as a voltage divider, and R2 vout R1 + R 2 v− = Then vout = A(v+ − vin ) � � R2 = A vin − vout R1 + R 2 A(R1 + R2 ) = vin R1 + R2 + AR2 and when A � 1 R1 + R2 vin R2 For example if R1 = 47 kΩ, and R2 = 10 kΩ, the amplifier voltage gain will be 5.7. vout = The Voltage-Controlled Current-Source: duce a current source. The above circuit can be modified to pro­ v + i - L o a d R 12–3 Assume 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 1 i= R � � 1 vin − vout A and if A � 1 1 vin R 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. i= Inverting Configurations The Inverting Amplifier: below: R s u m m in g ju n c tio n ( v ir tu a l g o u n d ) v in R i in in + - The basic inverting amplifier has a configuration as shown R f f if v o u t G n d R in + - + 1 5 v o u tp u t -1 5 v in p u t 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 Since the junction is at the virtual ground (v− ≈ 0) we can simply use Ohm’s law: vin − v− Vout − v− vin Vout + ≈ + =0 Rin Rf Rin Rf or Rf vin Rin In other words the voltage gain is defined by the ratio of the two resistors. The term invert­ ing amplifier comes about because of the sign change. vout = − The Inverting Summer: inputs: s u m m in g ju n c tio n ( v ir tu a l g o u n d ) v v 2 1 R 1 i1 R 2 + - R The inverting amplifier may be extended to include multiple R f f if v o u t i2 + G n d -1 5 v - + 1 5 v o u tp u t R 2 R 1 in p u ts As before we assume that the inverting input is at a virtual ground (v− ≈ 0) and apply Kirchoff’s current law at the summing junction i1 + i2 + if = 0 v1 v2 Vout + + = 0 R1 R2 Rf or Rf Rf v1 − v2 R1 R2 which is a weighted sum of the inputs. If R1 = R2 = Rin the output is proportional to the sum of the input voltages. Rf vout = − (v1 + v2 ) Rin The summer may be extended to many inputs by simply adding input resistors Ri vout = − vout = − n � Rf i=1 Ri vi If all of the input resistors are equal (Ri = Rin ) vout = − n Rf � vi Rin i=1 12–5 The Integrator: If the feedback resistor in the inverting amplifier is replaced by a capac­ itor C the amplifier becomes an integrator. v in R in i in C C s u m m in g ju n c tio n ( v ir tu a l g o u n d ) if - v + o u t G n d R in + + 1 5 v - o u tp u t -1 5 v in p u t At the summing junction we apply Kirchoff’s current law as before but the feedback current is now defined by the elemental relationship for the capacitor: vin Rin iin + if = 0 dvout +C = 0 dt Then dvout 1 =− vin Rin C dt or � t 1 vout = − vin dt + vout (0) Rin C 0 As above, the integrator may be extended to a summing configuration by simply adding extra input resistors: � � � n 1 t � vi vout = − dt + vout (0) C 0 i=1 Ri and if all input resistors have the same value R � � t �� n 1 vi dt + vout (0) vout = − RC 0 i=1 The Differentiator: If the input resistor Rin in the inverting amplifier is replaced by a capacitor, the circuit becomes a differentiator: R s u m m in g ju n c tio n ( v ir tu a l g o u n d ) v in i C in + - R f f if v o u t + G n d -1 5 v C in p u t 12–6 - + 1 5 v o u tp u t At the summing junction we apply Kirchoff’s current law as before but the input current is now defined by the elemental relationship for the capacitor: iin + if = 0 dvin vout C + + = 0 dt Rf Then 1 dvin Rin C dt Pure differentiators are rarely used because of their susceptibility to noise. In practice it is common to use a pseudo-differentiator by including a resistor in series with the capacitor. vout = − A General Linear Transfer Function: The inverting configuration may be extended to synthesize many general linear transfer functions by including appropriate networks in the feedback and input paths. Consider the general circuit shown below with elements described by impedances Zin (s) and Zf (s) in the input and output paths: Z f (s ) s u m m in g ju n c tio n ( v ir tu a l g o u n d ) Z in ( s ) v in i - i in f v + o u t At the summing junction iin + if = 0 Vin (s) Vout (s) + + = 0 Zin (s) Zf (s) so that the amplifier becomes a linear system with a transfer function H(s) = Vout (s) Zf (s) =− Vin (s) Zin (s) This configuration may be used to synthesize lead-lag compensators for control, active filters for signal processing, and many other functions. The Differential Amplifier A differential amplifier produces an output that is proportional to the difference between two inputs. 2 v 2 v1 R 1 if - i in v + R 1 R 2 12–7 o u t At the non-inverting input R2 v1 R1 + R2 If the op-amp gain is assumed to be infinite, v− = v+ . The input current at the inverting amplifier is: v+ = v2 − v− v2 − v+ = R1 R1 1 R2 = v2 − v1 R1 (R1 + R2 ) R1 iin = Similarly the feedback current is vout − v− vout − v+ = R2 R2 1 1 = vout − v1 R2 (R1 + R2 ) if = Applying Kirchoff’s current law at the summing junction iin + if = 0, then yields vout = R2 (v1 − v2 ) R1 so that the output is proportional to the voltage difference between the two inputs. Practical Considerations: The above simplified description does not include many practical considerations that must be taken into account when designing high performance op-amp circuits, including: • Most op-amps are low current devices. Most circuits are designed to operate with signal currents in the mA or µA range. This means that resistor values should be in the range 10 kΩ and above. • An op-amp cannot supply voltage levels greater than the supply voltage levels. Care must be taken to ensure that the output voltage remains within the limits defined by the supply voltage. • Our simplified analysis has not taken bandwidth limitations into account. Practical op­ amps exhibit a high-frequency roll-off of gain because of internal capacitances within the chip. • A similar phenomenon is know as ”slew-rate” limiting. In practice, op-amps have a limit on the slope (v/s) of the output voltage. This nonlinear effect has a limiting effect on the high frequency performance. 12–8