Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Understanding basic Opamps

Status
Not open for further replies.

Psytrox

New Member
Hello,

I'm having problems understanding basic opamps, and what "decides" the gain.

I'm able to look at examples, such the models for inverting and non inverting amplifiers and etc. and calculate a gain. Basicly, I can look at previous examples and change out the nessecery variables to get the write answere. But in reality, I don't really know what I'm doing.

I don't understand exactly what decides what the gain is...? Is it the difference between the + and - terminals of the opamp? Or is it the voltage at the point where to resistors are in parallell (sorry, bad description but cant think of any better way of explaining it). Or is it different for all the different models?


When solving problems with opamps at higher levels, say for example you want to find Vout in this **broken link removed**. How would one go around doing it? Do we use those basic models, and find one that "fits" and looks simuler?

Hope someone can help me :)


- Psytrox
 
The important point is that, in a normal negative feedback type op amp circuit, the op amp (ideal op amp with very high gain) tries to keep the plus and minus inputs so there is zero volts between them, thus you base your calculations on that.

Thus for a simple inverter with the + terminal to ground and the - terminal connected to an input resistor (Rin) and a feed-back resistor (Rfb) to the output, the gain (input resistor to op amp output) will be Rfb divided by Rin since that is the condition that will keep the - terminal voltage equal to the + terminal voltage. In other words the current through the input resistor exactly cancels the current flowing through the feedback resistor. (With the + terminal at ground, the - terminal is often called a "virtual ground" since its voltage will stay very near 0V as long as the op is normally operating).

Edit: To reiterate, feedback will cause the amp to have an output voltage such that all the currents into the minus junction sum to zero (which keeps zero voltage at that point).

In the circuit you posted, there is an inverting circuit and a non-inverting circuit. You just calculate the gain of each separately and then add them together to get the total gain at Vout.

When I first studied op amps I had the same problem as you, I didn't really understand what was happening. The light bulb for me was realizing that the high open-loop gain of the op amp with negative feedback will always try to keep the voltage across the plus and minus terminals very near zero. You just start from there to determine the operation of any op amp circuit with feedback. It's how the circuit is connected to the op amp that "decides" the gain. Hope that works for you also.
 
Last edited:
There are multiple forms of gain, so it can be confusing sometimes. All op-amps have an "open-loop gain" of some order of magnitude, sometimes ranging up to one million. This open-loop gain is without feedback, so it's very unstable. You can find the open-loop gain in an op-amp datasheet.

"Closed-loop gain" is where you apply feedback to your amplifier. When using negative feedback, for instance, it reduces your open-loop gain by feeding a portion of the output to the negative input terminal. I’ll give you an analogy of how negative feedback invokes stability of the op amp. Consider an inverting amplifier, as shown in the attachment. If the voltage output of your amplifier increases spontaneously, then the V- input terminal becomes more positive. Thus, it is more positive than the V+ input terminal, which yanks the voltage output back down.

In the attachment I show you how to derive the equations for a inverting amplifier.
 

Attachments

  • Inverting Amplifier Equations.PNG
    Inverting Amplifier Equations.PNG
    43.2 KB · Views: 398
Hello,

I'm having problems understanding basic opamps, and what "decides" the gain.

I'm able to look at examples, such the models for inverting and non inverting amplifiers and etc. and calculate a gain. Basicly, I can look at previous examples and change out the nessecery variables to get the write answere. But in reality, I don't really know what I'm doing.

I don't understand exactly what decides what the gain is...? Is it the difference between the + and - terminals of the opamp? Or is it the voltage at the point where to resistors are in parallell (sorry, bad description but cant think of any better way of explaining it). Or is it different for all the different models?


When solving problems with opamps at higher levels, say for example you want to find Vout in this **broken link removed**. How would one go around doing it? Do we use those basic models, and find one that "fits" and looks simuler?

Hope someone can help me :)


- Psytrox


Hi there,


I dont know how much circuit analysis work you have done in the past so it's hard to say what is best for you, but if you want to learn how to analyze ANY op amp circuit then one of the simplest methods is to treat the op amp as a simple voltage controlled voltage source. It's that simple. You can add imperfections later once you understand this basic idea.
Usually the output of the op amp is referenced to ground too, so you ground one terminal of the voltage controlled voltage source's output. This makes the op amp look like this:
(note vn is the inverting terminal voltage and vp is the non inverting terminal of the op amp)
Vout=Ao*(vp-vn)

and that's how simple it can be. Ao is the open loop gain of the op amp, which can typically be 1000 to maybe 10 million, but for theoretical purposes we can let Ao approach infinity and so the output becomes:
Vout=limit[Ao-->+inf](Ao*(vp-vn))

Once you put the op amp into a circuit it's basic circuit analysis after that, and you can later calculate the gain.
If you need help applying this, just yell and i'll add more info.
 
Last edited:
Hello,
Thanks for the replies... Allthough i still have some questioins.
Vout=limit[Ao-->+inf](Ao*(vp-vn))
That doesnt really make sense. Since (vp - vn) * INF = INF? Or is it something I didn't quite understand...? Also, isn’t vp = vn? In that case isnt vp – vn = 0;
Which would make Vout = inf * 0 = 0?
Is vn and vp the "equation" for voltage at those terminals Looking at the figure posted by En0, is vn the voltage over Rg?
in which case, is vn = (Rg/(Rg+Rf))Vin and vp = 0 ? Is it possible to get a current/voltage out of ground, so that vp = vn? I don’t think i quite understand what ground is...

In the circuit you posted, there is an inverting circuit and a non-inverting circuit. You just calculate the gain of each separately and then add them together to get the total gain at Vout.
How do I calculate them seperatly? Do i set the different input signals to ground? Or do i put the whole pluss or minus terminal to ground?

Again, looking at the figure EN0 posted. What would happen if say the ground connected to the positive terminal was switched out with V2?
Would Vout = (-(Rf/Rg) * Vin) - V2)?

op amp with negative feedback will always try to keep the voltage across the plus and minus terminals very near zero

If the input resistance is INF, wont that make the voltage over the plus and minus terminals very high, or INF? I can agree that the current going through the resistor is almost 0.


I’m going to try and calculate the basic modells again (such as the inverting and non-inverting etc.), and maybe I can get a hold of it. It looks so easy watching the teacher do it on the board. But when I actually have to do it myself, I guess I just end up confusing myself, or forgetting something important.
 
Hello,
Thanks for the replies... Allthough i still have some questions.

"Vout=limit[Ao-->+inf](Ao*(vp-vn))"
That doesnt really make sense. Since (vp - vn) * INF = INF? Or is it something I didn't quite understand...? Also, isn’t vp = vn? In that case isnt vp – vn = 0;
Which would make Vout = inf * 0 = 0?


In that statement we are taking the limit of a function so if you try to look at it at face value we would have to think of it as an indeterminate form, at least for now. For example we can not calculate the limit of f(x) without knowing what the function f is first. We find out later after we use the model in a real circuit. As soon as it gets into a circuit it all comes together as there are other things that enter into the equation that allow us to finally calculate the limit. Perhaps you would like an example of this.
Keep in mind that as Ao *tends* to infinity, (vp-vn) *tends* to zero. We cant always calculate limits by simply replacing the variable with the limit value. It might help to look up indeterminate forms and L'Hopital's rule, but we probably dont need that just yet.
When we do a circuit after we take the limit we end up with a function for Vout like Vout=Vin*A/(B+C) or something like that where the constants are obtained from the circuit component values. We can then calculate the gain by dividing Vout by Vin:
Gain=Vout/Vin
and this method works for every circuit out there with some limitations.
 
Last edited:
How do I calculate them seperatly? Do i set the different input signals to ground? Or do i put the whole pluss or minus terminal to ground?
If it helps you can connect one of the inputs to ground to perform the calculations, but that's not necessary. Just treat each op amp as a separate circuit to calculate each one's gain. Then simply add them together.

If you are referring to the plus and minus op amp terminals, then no, you would not connect them to ground. The circuit would not operate.

If the input resistance is INF, wont that make the voltage over the plus and minus terminals very high, or INF? I can agree that the current going through the resistor is almost 0.
The ideal op amp input impedance is infinity. That's not what determines the voltage there. That is determined by the feedback and input resistors.

The currents through the input and feedback resistors are not almost 0. It is determined by the resistor value and the voltage across them.

For a simple inverter circuit with the op amp plus terminal to ground, the input resistor current is equal to the input voltage divided by the input resistance. To balance this, the op amp gain will then drive the output voltage so that the feedback resistor current is equal but opposite to the input current. Thus the net current (and voltage) at the summing (minus) junction is zero.

The important point is that, with negative feedback, the voltage across the plus and minus op terminals will be very near zero, and the net current into the summing junction is also zero. Once you understand that, you will be able to determine the operation of most negative feedback op amp circuits (the exception being circuits that operate open loop or use positive feedback such as comparator circuits).
 
Hello,
Thanks for all the help :) I'm starting to understand how to use opamps. Which is good because my exams is actually this comming wensday.
So I'm sure there are more questions to follow :p
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top