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## OPERATIONAL AMPLIFIERS## Operational AmplifiersAn operational amplifier, commonly called an Figure 1: Operational Amplifier Block DiagramFigure 2: Operational Amplifier SymbolThe input terminals have been labeled "Inverting" and "non-inverting" to describe what the op amp output polarity is with respect to the input polarity. The term "non-inverting" means that the polarity is not reversed, or non-inverted. The use of both a positive supply and a negative supply allows the op amp to produce either a positive or a negative output. A general understanding of the three main sections is beneficial in understanding the basic characteristics and limitations of an op amp. Figure 3: Simplified Op Amp Internal CircuitThe input signals to the inverting and non-inverting terminals are directly coupled to high impedance, differential amplifier composed of transistors Q` and Q. Transistor Q} senses the negative and positive supply voltages and functions as a constant current source to regulate against voltage supply variations. Transistors Q? and Qr form another differential amplifier to provide additional voltage gain. Transistor Qq controls the biasing of transistors Qo and Qp, which are connected as emitter-followers to provide a high current gain and low output impedance. Diodes CR` and CR provide temperature stability. The actual operating characteristics of an op amp come very close to the characteristics of an "ideal amplifier." A typical op amp compares with these ideal characteristics, as indicated in Table 1: Comparison of Op Amp CharacteristicsAs shown, one of the most limiting characteristics of a real op amp is the output current capability. The low value is predominately due to the size and, therefore, the heat-dissipating capability. Op amps are normally considered voltage devices. If significant amounts of current are needed, a power or current amplifier is usually added as the next stage after an op amp. Menu
## Open Loop ConfigurationAs shown in Figure 4: Open Loop Op Amp CircuitWhen a voltage potential is applied to the input terminals, a voltage difference exists across the internal differential amplifier and the op amp output increases. Because the positive voltage is applied to the inverting input terminal with respect to ground, the output is negative with respect to ground. As described earlier, the gain of the op amp is approximately 20,000. This means that if the input is equal to 1 millivolt, the op amp attempts to increase the output to negative 20 volts. This can be shown mathematically using the standard gain equation shown below: The negative polarity is due to the inverting action of the op amp. In reality, the output does not increase to negative 20 volts. The voltage cannot be larger than the supply voltage, which as shown is positive and negative 15 volts. What does happen is the op amp "saturates" at an output voltage very near the negative supply voltage. Figure 5: Open Loop Amp Circuit Waveforms## Closed Loop Configuration
Figure 6: Positive and Negative FeedbackPositive feedback exists when a portion of the output is sent back to a summing junction such that it is of the same polarity, and aids the input signal. As shown in the left side of ## ComparatorWithout negative feedback, an op amp goes into saturation with only a small input applied.Figure 7 shows an op amp comparator circuit that uses this characteristic. Assume initially that the signal applied to the inverting input is a positive 2 volts to exactly match the reference voltage. The op amp sees zero differentials across the input terminals and, therefore, the output is zero volts. At time T1, the input increases to a positive 2.1 volts. The op amp sees a positive 0.1 volt differential and, due to the high gain value, the op amp goes into negative saturation. At time T2, the input is positive 1.9 volts. The op amp sees a negative 0.1 volt differential, and the op amp goes into positive saturation.Ãƒâ€šÃ‚Â Figure 7: Comparator## BufferFigure 8 shows an op amp buffer circuit that makes use of negative feedback. A buffer provides signal isolation between two other circuits. Op amps with their high input impedance are well suited to provide this isolation. A good buffer also should not attenuate the signal. An op amp buffer circuit has a gain of 1 or unity, and so results in no signal loss or gain.Figure 8: BufferInitially, the positive 3 volts applied to the non-inverting input causes the op amp to see a 3-volt potential difference across the inputs. The output then rapidly increases in the positive direction. A negative feedback loop has been provided which sends back a voltage equal to the output voltage. When the output voltage reaches positive 3 volts, the potential difference across the inputs is zero volts, and the output stops increasing. At this point, the op amp is "satisfied"; it has increased its output sufficiently to reduce the input differential to zero. As can be seen, the output voltage exactly follows the input voltage. Another name then for a buffer is a ## Inverting Amplifier
Figure 9: Inverting AmplifierAt this point, it is important to keep in mind the difference between the operational circuit and the operational amplifier. The operational amplifier is represented by the triangle-like symbol, while the operational circuit includes the resistors and any other components as well as the operational amplifier. Ein is the input to the operational circuit. However, the signal at the inverting input of the operation amplifier is determined by the feedback signal as well as the circuit input signal. The term Because the op amp essentially draws no current, all the current flows through feedback resistor Rf and input resistor Rin to the input voltage. The output voltage rises until the voltage felt between the input pins of the op amp approaches zero. The amount of voltage required to drive the voltage between the input pins to zero is determined by the ratio of Rf/Rin. Kirchhoffs Voltage Law for closed loops can be used to determine the voltage potential at the summing junction. Voltages in a loop must algebraically equal zero when summed together. (Ein) + (Erf ) + (E summing junction to ground) = 0 In this case, the output continues to increase until the feedback current is large enough to cause the voltage drop across Rf to be 5 volts. At that point, the input voltage will be reduced to 0 volts at the summing junction because the opposing 5 volts across Rin totally cancels out the 5 volt input. This can be verified by using Kirchhoffs Voltage Law. Proceeding clockwise form the negative terminal of Ein, voltages are assigned to the equation as: (-5V) + (+5V) + (Es.j. - gnd.) = 0 (Es.j. - gnd.) = 0 When the summing junction is at 0 volts, the op amp is satisfied. The output then stops increasing and remains at an output level to maintain the summing junction at 0 volts. Kirchhoff can again be used to determine what the amplitude of voltage output is at this point. By knowing the voltage across R` must be 5 volts to cause the summing junction to be 0 volts, the current flow through R` is determined to be: ERU/RU = I thru RU {1k}/{1k} = IRU IRU = 5 mA This 5 mA also flows through R1 to give a voltage drop of: The voltage loop can be analyzed at this point using the equation of: (Ein) + (ER`) + (ER1) + (Eout) = 0 Proceeding clockwise from the negative terminal of Ein, the voltages are assigned to the equation as: (-5V) +(+5V) + (+5V) + (Eout) = 0 Eout = -5V
Figure 10: Inverting Amplifier at Final OutputThe gain of the inverting amplifier is represented by: The negative sign represents the inverting action of the op amp. Figure 11: Inverting AmplifierThe voltage across R` must be equal to the input voltage for the summing junction to be 0 volts, and the op amp to be satisfied. Initially, the positive 2 volts is sensed at the inverting terminal and the op amp output increases negative. As the output increases, the feedback current through R` increases. When the voltage across R` equals 2V to cancel the input voltage, the current through R` is: The 2 mA flows through R to cause a voltage of: I x R` = ER1 (2mA) x (5k) = ER1 ER1 = 10V Again, using Kirchhoff: (Ein) + (ER`) + (ER1) + (Eout = 0 (-2V) + (+2V) +(+10V) + (Eout) = 0 Eout = -10V The gain of this inverting amplifier is: Notice that by only changing the value of the resistor in the feedback loop, the gain of the inverting amplifier was changed, and a larger output was provided. The output voltage is dependent on the ratio of the resistance in the feedback loop versus the resistance in the input circuit. For inverting amplifiers, this is expressed as: Eout = -Ein R1/R` Transposing the output voltage equation to: ## Non-Inverting Amplifier
Figure 12: Non-Inverting AmplifierFor the op amp to be satisfied in this circuit, the voltage at the summing junction must be a positive 2 volts. The feedback current through Rf provided by the output when the op amp is satisfied is: This 2 mA flows through Rin to provide a voltage of: Using Kirchhoff, and proceeding from the grounded side of R`: The output voltage to ground is also the voltage seen across R1 plus R` with respect to ground, and the input voltage equals the voltage across R`. Therefore, the gain of the non-inverting amplifier is: ## Summing Amplifier
Figure 13: Summing AmplifierAny one of the input circuits can be used with the voltage across R? to determine the output voltage: (E1) + (ER1) + (ER?) + (Eout)= 0 (-2V) + (+2V) + (+6V) + (Eout)= 0 Eout = -6V This 3-input summer is really just an inverting amplifier that has two additional input circuits. Recall that the output voltage for an inverting amplifier is: The output voltage determined by the three inputs is: Each of the inputs contributes to the output according to the gain it sees: If it is desired to use only a portion of one or more input signals to determine a desired output voltage, the gain for the one or more inputs could be changed by changing the value of their respective input resistors. This is called "input scaling." For example, if it is desired that input E` has a larger effect on the output as compared to the other inputs, the resistance value of R` could be changed as shown in Figure 14: Scaling AmplifierInput E1 is divided by R1 and R/. To determine the voltage at the non-inverting terminal, the voltage across R/ must be determined. The current through R/ is: The 1 mA through R/ provides a positive 1-volt at the non-inverting terminal. For the op amp to be satisfied, a positive 1-volt at the inverting terminal summing junction must be provided. To obtain this, a 2-volt drop across R` must be provided by the feedback current: The difference amplifier is essentially an inverting amplifier and a non-inverting amplifier providing an output that is the difference between the two input signals. ## IntegratorOperational amplifiers can be arranged so as to perform the mathematical function of integration. Integration is the process of summing the input signal over time. The op amp integrator is shown in Figure 16: IntegratorAn op amp integrator converts a voltage input to a ramp output, whose slope is a function of the magnitude of input voltage and the RC time constant. As the amplitude of the input waveform is increased, the resultant output at any point in time is increased, or simply the rate of change of the output is increased. A square-wave can be viewed as a step change in voltage. The integrator takes this step change and converts it to a slower rate of change. The new rate of change is dependent on the amplitude of the original step change. A current entering the summing junction through R due to Ein, must be cancelled with a current from the output of the amplifier in order to keep the summing point at 0 volts. The only way to get a current through a capacitor is by having a changing voltage across it. A constantly increasing voltage on the output of the amplifier will produce such a constant current through the capacitor. Therefore, as long as current from Ein exists, Eout must continue to change. At T1 the input step changes to a positive 1 volt. A voltage potential exists across the op amp input terminals, and the output increases in the negative direction. The capacitor attempts to charge to the potential of the input voltage, and the RC time constant determines the rate of which the capacitor charges. Therefore, the voltage across the capacitor at T2 is expressed as: The 1-volt across the resistor opposes the input voltage and would appear to reduce the summing junction to 0 volts, which would stop any further output increase. However, once any capacitor is charged, charging current stops. Without charging current, there would be no voltage developed across the resistor to oppose the input voltage. This would result in the positive input again being applied to the op amp to cause a further negative increase in the output. In actuality, a constant current is needed through the resistor to oppose the input voltage. To obtain this, the voltage across the capacitor must be increased as seen by the capacitor voltage equation in terms of I: Time is not a constant term; it is continually increasing. To provide a constant value of current I, the voltage V must be increased because capacitance C is a constant value. The op amp Integrator circuit characteristically operates different than a simple series RC integrator circuit, in that the output does not stop at a value equal to the voltage across the capacitor. The output of the op amp continues to increase in the negative direction. The increasing voltage across the capacitor maintains a constant charging current through the resistor to develop a voltage that opposes the input. The output continues to increase until it reaches negative saturation. However, the output at the saturation amplitude is not the portion of the output that is useful. The portion of the output that is useful is the ramp voltage. Different amplitude input voltage causes different rate of change ramp output voltages. This can be shown by converting the integrator output equation: ## DifferentiatorA differentiator is basically an inverting amplifier with the input resistor replaced by a capacitor, as shown in Figure 17: DifferentiatorAny change in input voltage will cause a current to flow through C, and this change will be cancelled by a current through R. As long as Ein is changing, Eout will be non-zero. The value of Eout depends upon the rate of change of Ein. The output voltage is directly dependent on the amplitude of input voltage and the RC time constant. This determines how rapidly the capacitor charges and, therefore, the amplitude of the charging current through the resistor. When a ramp voltage is applied to a differentiator, the capacitor sees this as a constantly increasing input at a fixed or constant rate of change. Current flows through the resistor at a constant rate because the capacitor is being charged by the input in a linear manner. Initially, the capacitor is not charged. At time T1 the input ramps positive at a rate of change equal to 1 volt per second. This positive going input is coupled by the uncharged capacitor to the op amp, and the output voltage is driven negative. The capacitor charges at a rate determined by the RC time constant, and the charging current flows through the resistor. At time T2, the capacitor has charged to 1 volt. |