An Op Amp Gain Bandwidth Product – Mastering Electronics Design
Chapters 7 to 10 reflect the dual role of the operational-amplifier circuit. zero diagrams and elementary relationships between the time and the frequency domain is .. back system to zero and breaking the signal path at any point inside the. So far we have explored the use of op amps to multiply a signal by a constant. For the .. In general the relationship between the input voltage and the output current is out in. I. SV. = .. diode is reverse biased and the feedback loop is broken. response of circuits that utilize our friendly IC, the operational amplifier. Before we get (indicated by the dB/decade rolloff), the bandwidth is the upper break Using the gain relationship developed last semester and substituting in the.
Manufacturers insert a dominant pole in the op amp frequency response, so that the output voltage versus frequency is predictable. Why do they do that? Because the operational amplifier, which is grown on a silicon die, has many active components, each one with its own cutoff frequency and frequency response. Because of that, the operational amplifier frequency response would be random, with poles and zeros which would differ from op amp to op amp even in the same family. As a consequence, manufacturers thought of introducing a dominant pole in the schematic, so that the op amp response becomes more predictable.
At the same time, it makes the op amp more user friendly, because its stability in a schematic becomes more predictable.
Op-Amp Bandwidth, Gain Bandwidth Product & Frequency Response
The dominant pole will make the op amp behave like a single-pole system, which has a drop of 20 dB for every decade of frequency, starting with the cutoff frequency.
Such a pole is made with a reactive element, usually a capacitor. The choice of a capacitor on the op amp die is because of the manufacturing process. A capacitor is easier to be grown on silicon, as opposed to an inductor. However, the capacitor value depends on its area, and since the die area is at a premium, capacitors can only be very small, in the picofarads range.
Since the pole is made with an RC time constant, we need a large resistor to bring the cutoff frequency at low values, hertz, or tens of hertz.
An Op Amp Gain Bandwidth Product
Without going into details, op amp manufacturers achieve these high resistors with active components, like the input resistance seen in a transistor base.
Having said that, the gain bandwidth product shows that the product between the op amp gain and frequency, in any point of the frequency response, is a constant. We can always calculate the bandwidth with the following formula. This is equivalent to saying that v-minus equals minus v-out over A. So what else can we write for this circuit? OK, let's look at these resistors. Let's call this plus and minus vR1; and we'll call this one plus minus vR2.
So there's a current flowing here, and that we'll call I. I equals vR1 over R1. Another way I can write that. What's this voltage here? So I can write this in terms of v-minus, and that equals v-in minus v-minus over R1. That's the current going through this guy here.
Now I'm gonna use something special. I'm gonna use something special that I know about this amplifier. What I know about an op-amp is that this current here is equal to zero. There's no current that flows into an ideal op-amp. So I could take advantage of that. What that means is that I flows in R2.
So let me write and expression for I based on what I find over here, based on R2. I can write I equals, let's do it, it's vR2 over R2. And I can write vR2 as: All right, so I took advantage of the zero current flowing in here to write an expression for current going all the way through. So now we're gonna set these two equal to each other. Now we're gonna make these two equal to each other. Let me go over here and do that. V-in minus v-minus over R1. That equals this term here, which is v-minus minus v-not, v-out rather, over R2.
How many variables do we have here? We have v-out, we have v-in, and we have v-minus. And what I want is just v-out and v-in, so I'm gonna try to eliminate v-minus; and the way I'm gonna do that is this expression over here. We're gonna take advantage of this statement right here to replace minus v-out over A. So I'll do that right here. So let me rewrite this.
- Inverting op-amp
It's gonna be v-in minus minus v-out over A, so I get to make this a plus, and this becomes v-out over A all divided by R1. V-minus is minus v-out over A minus v-out over R2. All right, let's roll down a little bit, get some room, and we'll keep going.
What am I gonna do next? Next, I'm gonna multiply both sides by A, just to get A out of the bottom there. There's a lot of algebra here, but trust me, it's gonna simplify down here in just a minute. All right, so now I'm gonna break this up into separate terms so I can handle them separately. Let's change colors so we don't get bored.
Next, what I'm gonna do is start to gather the v-out terms on one side and the v-in terms on the other side. So that means that this v-out term here is gonna go to the other side. Av-in over R1 equals, let's do minus v-out over R2 minus Av-out over R2.
op amp - Input Output Relationship of Inverting amplifier - Electrical Engineering Stack Exchange
And this term comes over as minus v-out over R1. Haven't made any sign errors yet. Now let me clear the R1.
We'll multiply both sides by R1. Yeah, the R1s cancel on that last term. Out of here I can factor this term here.
Minus R1 over R2 times v-not, I can factor that out of here and here. So I can do minus R1 over R2 v-not times one plus A minus v-not. So let's take a look at this expression and use our judgement to decide what to do next.
Now, because A is so huge, that means that this first term is gonna be gigantic compared to this v-not term here. V-not is some value like five volts or minus five volts or something like that.