Operational Amplifier
Example Problem
Now lets work through an example.
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Notice in this case we have two voltage sources vA and vB both of which are connected to the inverting input through the resistors RA and RB respectively. This can be handled just as we did before. The difference in this case is that there will be two currents coming into node A - one for each voltage source.
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We begin by doing KVL around the loop indicated by the red arrow. This results in the following equation -vA+iARA=0 which can rearranged to give us the current iA=vA/RA. The other voltage source can be handled in the same way as shown.
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Resulting in a similar equation iB=vB/RB. As was done previously KCL is done at node A resulting in the following equation iA+iB-if=0 which can be solved for the current through the feedback resistor giving us if=iA+iB.
KVL can now be used to relate the output voltage vo to the current through the feedback resistor if.
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The equation that results is Rfif+vo=0 which leads to vo=-Rfif. Plugging in the previous reults for if results in vo=-Rf(iA+iB)=-Rf((vA/RA)+(vB/RB)) or vo=-(RfvA/RA)-(RfvB/RB) [Equation 1].
This is an interesting and powerful result. Note that the output voltage vo is a function of both of the input voltages vA and vB, but each of those input voltages is subject to a potentially different gain because the resistors RA and RB need not be the same.
Let’s take this example a little bit further. Suppose we needed to create the following functional relationship between the input voltages and the output voltage: vo-10vA-5vB. How might we do that? Assume at this point that you can pick pretty much any value you want for each of the resistors. Since the feedback resistor Rf is common to both terms in Equation 1 let's start by picking a value for that. Let's go with 10 K ohms. Having picked that value for the feedback resistor it should be obvious that the values for the resistors RA and RB need to be 1 K ohms and 2 k ohms respectively. We can see that this generates the desired result:
vo=-(10kΩ/1kΩ)vA-(10kΩ/1kΩ)vB→ vo=-10vA-5vB
That’s pretty cool! We can actually use op-amps to perform algebraic operations on voltages. Actually, that’s where the word ‘operational’ in the phrase operational amplifier comes from. As we continue, we’ll see that there are many other mathematical operations that op-amps can perform.
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