362CHAPTER 8. OPERATIONAL AMPLIFIERSis to amplify the difference between the process variable and the setpoint with a differentialampliﬁer. In simple controller designs, the output of this differential ampliﬁer can be directlyutilized to drive the ﬁnal control element (such as a valve) and keep the process reasonablyclose to setpoint.• REVIEW:• A ”shorthand” symbol for an electronic ampliﬁer is a triangle, the wide end signifyingthe input side and the narrow end signifying the output. Power supply lines are oftenomitted in the drawing for simplicity.• To facilitate true AC output from an ampliﬁer, we can use what is called a splitor dualpower supply, with two DC voltage sources connected in series with the middle pointgrounded, giving a positive voltage to ground (+V) and a negative voltage to ground (-V).Split power supplies like this are frequently used in differential ampliﬁer circuits.• Most ampliﬁers have one input and one output. Differential ampliﬁershave two inputsand one output, the output signal being proportional to the difference in signals betweenthe two inputs.• The voltage output of a differential ampliﬁer is determined by the following equation:Vout = AV (Vnoninv - Vinv)8.3The ”operational” ampliﬁerLong before the advent of digital electronic technology, computers were built to electronicallyperform calculations by employing voltages and currents to represent numerical quantities.This was especially useful for the simulation of physical processes. A variable voltage, for in-stance, might represent velocity or force in a physical system. Through the use of resistivevoltage dividers and voltage ampliﬁers, the mathematical operations of division and multipli-cation could be easily performed on these signals.The reactive properties of capacitors and inductors lend themselves well to the simulationof variables related by calculus functions. Remember how the current through a capacitorwas a function of the voltage’s rate of change, and how that rate of change was designatedin calculus as the derivative? Well, if voltage across a capacitor were made to represent thevelocity of an object, the current through the capacitor would represent the force required toaccelerate or decelerate that object, the capacitor’s capacitance representing the object’s mass:iC = C dvdtF = m dvdtWhere,Where,iC = C =dvdt=Instantaneous current through capacitorCapacitance in faradsRate of change of voltage over timeF =m =dvdt=Force applied to objectMass of objectRate of change of velocity over time