Operational Amplifier
Application of Non-Inverting Amplifier
We can build on our model of the non-inverting op-amp to produce another very useful circuit. Recall that the gain of the circuit is given by: Av=vo/vin=1+R2/R1. Suppose we want the gain to approach unity. How could we achieve that?
One way is to let the value of the resistor R2 go to zero. Another way we could achieve the same result is to let the value of the resistor R1 approach infinity. Actually, since both of these approaches achieve the result we desire, why not use both? If we let R1 approach infinity and R2 approach zero, we get the following circuit.
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This circuit is called a unity gain buffer (because the voltage gain is exactly 1). It’s also often called a voltage follower as the output voltage exactly follows the input voltage.
OK, that's cool, but what’s the point? Why go to all the work of creating a circuit that does nothing?
Well, the answer to that question is that the circuit actually does do something and it’s something that’s very useful. Let’s take a closer look.
Consider the network in the following figure:
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Here we have a voltage source VS in series with a resistor RS. These components are driving some unknown resistive network at the right of the circuit (the gray blob). It should be clear that the voltage Vo measured across the resistive network is not equal to VS but rather is given by Vo=VS-IRS. Because the current I flowing through Rs is non-zero there's a voltage drop across RS which leads to Vo being different from VS. This is because the resistive network 'loads' the voltage source Vs.
Now suppose that we use a unity gain buffer or voltage follower to isolate the source circuitry from the resistive network as shown in the following figure.
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We know from what we’ve studied previously that the current I flowing into the op-amp will be zero (or very nearly so for real op-amps). We also know that the output voltage of the op-amp will be equal to the source voltage VS as there will be no voltage drop across the resistor RS.
The question you may be asking yourself is: “Where does the energy that is dissipated in the resistive network come from?” The answer is that the energy supplied to the resistive network comes from the power supplies that are driving the op-amp.
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