# How to Perform Complex Processing with Op Amps

If you understand the basic building blocks of op amp circuits, you’re ready to tackle complex processing actions with op amps. Using op amp circuits, you can analyze an instrumentation amplifier, solve mathematical equations, or create systems for signal processing, instrumentation, filtering, process control, or digital-to-analog/analog-to-digital conversion.

## Analyze an instrumentation amplifier

The instrumentation amplifier is a differential amplifier suited for measurement and test equipment. Here is the input stage of an instrumentation amplifier. Your goal is to find the voltage output *v** _{O}* proportional to the difference of the two inputs,

*v*

_{1}and

*v*

_{2}. Getting the desired output requires some algebraic gymnastics, but you can handle it.

At Node C2, you apply KCL (*i*_{1} + *i*_{2} = 0) and Ohm’s law (*i* = *v/R*) and wind up with

At Node C1, the KCL equation (–*i*_{2} + *i*_{3} = 0) with Ohm’s law leads you to

The sample circuit shows the noninverting input connected to independent voltages *v*_{1} and *v*_{2}. Use the op amp voltage constraint *v** _{P}* =

*v*

*to get the following:*

_{N}Substitute *v*_{1} and *v*_{2} into KCL equations, which gives you

Now solve for *v*_{B}_{2} and *v*_{B}_{1}, because the output voltage *v** _{O}* depends on these two values:

The output voltage *v** _{O}* is the difference between the

*v*

_{B}_{1}and

*v*

*:*

_{B2}Cool! Resistor *R*_{2} can be used to amplify the difference *v*_{2} – *v*_{1}. After all, it’s easier to change the value of one resistor *R*_{2} than of two resistors *R*_{1}.

## Implement mathematical equations electronically

As an example of how op amps can solve equations, consider a single output and three voltage input signals:

You can rewrite the equation in many ways to determine which op amp circuits you need to perform the math. Here’s one way:

The equation suggests that you have an inverting summer with three inputs: –*v*_{1}, –*v*_{2}, and *v*_{3}. You need an inverting amplifier with a gain of –1 for *v*_{1} and *v*_{2}. Input *v*_{1} has a summing gain of –10, input *v*_{2} has a summing gain of –5, and input *v*_{3} has a summing gain of –4.

You can see one of many possible op amp circuits in the top diagram of this sample circuit. The dashed boxes indicate the two inverting amplifiers and the inverting summer.

The outputs of the two inverting amplifiers are –_{ }*v*_{1} and –* **v*_{2}, and they’re inputs to the inverting summer. The third input to the summer is *v*_{3}. Adding up the three inputs with required gains entails an inverting summer, which you see in the sample circuit.

For input *v*_{1}, the ratio of the inverting summer’s feedback resistor of 200 kΩ to its input resistor of 20 kΩ provides a gain of –10. Similarly, for input *v*_{2}, the ratio of the feedback resistor of 200 kΩ to its input resistor of 40 kΩ gives you a gain of –5.

Finally, for input *v*_{3}, the ratio of the feedback resistor of 200 kΩ to its input resistor of 50 kΩ provides a gain of –4. You can use other possible resistor values as long as the ratio of resistors provides the correct gains for each input.

Reducing the number of op amps during the design process helps lower costs. And with some creativity, you can reduce the number of op amps in the circuit by rewriting the math equation of the input-output relationship:

This suggests you need two op amps. One input is a combination of inputs *v*_{1} and *v*_{2} formed by an inverting summer. When you take the output of the first summer and feed it and another input to a second inverting summer, the result is proportional to *v*_{3} with gain –4. The bottom diagram of the sample circuit shows one way to implement this equation.

For *v*_{1}, the ratio of the feedback resistor of 100 kΩ to the input resistor of 20 kΩ produces a gain of –5. For *v*_{2}, the input resistor of 40 kΩ gives you a gain of –2.5. The output of the first summer is then multiplied by –2 because of the ratio of the second inverting summer’s feedback resistor of 100 kΩ to the input resistor of 50 kΩ.

The input *v*_{3} to the second summer is multiplied by –4 because of the ratio between the 100-kΩ feedback resistor to the 25-kΩ resistor.

## Create systems with op amps

Op amp circuits are basic building blocks for many applications in signal processing, instrumentation, process control, filtering, digital-to-analog conversion, and analog-to-digital conversion.

For example, you can do a digital-to-analog conversion (DAC) using the inverting summer. The primary purpose of this common device is to convert a digital signal consisting of binary 1s and 0s (perhaps coming from your personal computer) to an analog and continuous signal (to run your DC motor in your remote-control toy). The device has extensive applications in robotics, high-definition televisions, and cellphones.

This example is simplified by focusing on 3-bit devices (even though most applications use 8- to 24-bit DACs). DACs have one output voltage *v*_{o} with a number of digital inputs (*b*_{0}, *b*_{1}, *b*_{2}), along with a reference voltage *V** _{REF}*. Here you see a block diagram of a 3-bit input.

The following equation gives you the relationship between the digital input and analog output:

Bit *b*_{2} is the most significant bit (MSB) because it’s weighted with the largest weight in the sum; bit *b*_{0} is the least significant bit (LSB) because it has the smallest weight.

To implement a DAC, you can use an inverting summer, as shown in the figure. Also shown are the digital inputs that can have only one of two voltage values: A digital 1 is equal to *V** _{REF}*, and a digital 0 is equal to 0 volts.

The inputs *v*_{1}, *v*_{2}, and *v*_{3} to the summer are weighted appropriately to give you the voltage output *v** _{O}* based on the three inputs. Input

*v*

_{1}has the most weight, and input

*v*

_{3}has the least.