# Op Amp Applications - Integrator

#### Op Amps and Op Amp Circuits

Video Lectures created by Tim Fiegenbaum at North Seattle Community College.

Lecture Transcript:
11.3 and we're looking at Op Amp applications. The next form we're going to look at is called the integrator. An integrator is a circuit among other things. You can use an integrator if you have a, for example, positive or a square wave coming in, you can get a shape that is almost rectangular coming out of it, and that's the kind of circuit we are going to be looking at here. They can be built around an AC time constant with R1 and C1. Here is the value of R1 and here is C1 up here, so 33K and ten Nano-Farads are the components this is actually built around. Let's take a look at this … let's see … we've got R3 here. This is going to be Rb so we can basically ignore that. We have a signal source coming in and in this case, it is going to be a rectangular square wave. It's going to be fed through R1 and this going into an inverting amplifier. The output is as the input goes positive, the output will actually go negative and note that as it goes in here, then here is the output and the output is fed to a feedback loop through R2 and C1. The idea here is that we're going to have the signal come in, it's going to see the virtual ground right here, so there will be that voltage across this resistance, there will be a current and since it's a square wave it will be a constant current for as long as it is positive and as long as it goes negative. That current is going to be fed up to the output and it will charge this capacitor and so that line that you see, that rectangular shape you see, is actually the charging of the capacitor in the feedback loop. OK, a few things we need to consider here, let's see, we looked at R1 and C … OK … the input voltage is felt across R1 and … OK we've talked about that … the current is fed through the feedback loop and charges the can based on the time of the input. OK this is an important thing: based on the time of the input. We have to calculate what the time is. Now if this were a 1000 Hertz signal then the time of one would be, you know, one over one thousand and that would be one millisecond. Since we're looking at a square wave and this would be a whole cycle, and we're going to be looking at the charging time of the positive cycle and then the charging time of the negative cycle, but we need to have to separate them out. We're going to take this divided by two so we would have 0.5 milliseconds would be our time and so when we put this into a formula we need to be able to specify the right time that we're going to be using.
Now the next page here is this same circuit, it's always large so you can actually see it, and we have the signal source here and we have the O-scope redone and this is the simulation from Workbench where I had captured the output and the actual circuit. Now this is the formula we'll be using and this is going to be Vi times T so that would be the input voltage times the time and then that would be divided by this RC time constant and that will tell us the output voltage. All we're looking at is we'll going to be sending a specific voltage for a specific amount of time and we'll going to see what will that develop through in the charging of this capacitor, what kind of voltage will we see? You'll see that the results over here: the square wave is the input and this rectangular shape that you see here is the output. Let's do the formula and so let's pull up on our calculator and let's see what we're going to get here … Okay, first of all, we need to look at the time and so we said we had 1000 Hertz as our input, so let's just do one over and that gives us one millisecond that is the time of one cycle and then we'll divide that by two. That is the time … that is our time right here. If we look at our square wave here, it would go down here and observe that this is a one volt peak signal so we would take that times one, and gets on the … well the time's going to be the same value. That is our numerator, and then we would divide that by, and this is going to be the RC time constant, so that would be R1, which is 33 exponent three, times and this is ten Nano-Farads, so that would be ten exponent minus nine in parenthesis … and then equals. You'll notice that that value is 1.51 volts; let's write that down, 1.51 volts. What that's saying is that during this positive cycle there is going to be a current through this device. It's going to charge this capacitor and it's going to develop a voltage. Now during this time, you'll notice that the voltage appears to be going negative. Well the reason it is going negative is because it's fed into the inverting side so as this is positive the output's going negative, which you would expect. It has it that as the peak voltage goes negative, you see the charge in that cap going positive. Now the question here is, “what is the value of this triangular signal key?” Go down here you can see right here that the value's 1.447 volt. That's from the bottom of the triangle to the top of the triangle. Remember our calculated was actually quite close to that value and when you're doing this the input must be peak-to-peak value and V1 would be the peak value. This is a quick look at a circuit call and integrator.