Op Amp Feedback and Design

[EN0]

Hey Everyone,

I’m having trouble understanding the basic analogy of op amp feedback networks. Due to my diminutive understanding of op amps, I can’t apply the math to analyze and apply the necessary output equations, and equations in general.

To edify myself in that regard, I thought it would be best if someone defined op amp feedback to begin with. I assume it is different than the feedback from a microphone and speaker via an audio amp?

Consider the following link and go to page 78-79 (Figure 5-6):

https://www.electro-tech-online.com/custompdfs/2010/04/slod006b.pdf

It might help to depict box diagrams that explain the feedback methods like in the link, but those equations I don’t entirely understand.

Next, I’m hoping that several of you can provide several visionary examples (in other words, schematics) that will elucidate the primary meaning of feedback, and how I can apply the equations based on resistor networks (start out with just resistors, no caps, inductors, etc.)?

Finally, I wanted to ask what most op amp designers need to know while designing various projects; what methods do you use the most? What do you need to know in order to perform those methods and equations?

I am exceedingly anxious to start designing multifarious projects with op amps, and I would appreciate any help that people have to offer!

Thanks,

Austin

 

[audioguru]

Acoustical feedback howling is positive feedback where the same sound goes around and around as an oscillator.
But an opamp has such a high amount of gain (100,000 to one million) that it uses negative feedback to reduce the gain to a useable amount, reduce the distortion and noise and increase the bandwidth.

An opamp also has inputs that use an extremely low input current (some fet-input opamps have no input current).

Because the gain of an opamp is so high then the voltage at each of its inputs is almost exactly the same. The negative feedback makes the output do what is neccessary for both inputs to have the same voltage. Then you use simple ohm's Law to calculate the current in each feedback resistor which also shows the output voltage.

 

[crutschow]

A good place to start understanding feedback is with the simple inverting op amp, shown on Section 3.3 (p44) of the TI document you referenced. For an ideal op amp the gain is so high and the input current is so low, that the negative feedback through the resistor from the output to the (-) input it will always try to keep (-) input the same voltage as the (+) input. In this case the (+) input is ground so the (-) input will also stay essentially at ground potential (thus the negative input in this configuration is often referred to as a virtual ground). That being the case the current from the input resistor, due to a signal, must equal the negative of the current through the feedback resistor. The gain of this circuit is thus just -Rf/Rg (the minus sign indicating that the output is the negative (inverted) voltage of the input). If you add more complex circuits such as inductors or capacitors to the input and/or feedback, the equation becomes -Zf/Zg where Z is the complex impedance of the network.

If you understand the basic principle of negative feedback in op amp circuits, then you will be able to understand the other op amp circuit configurations.

 

[MrAl]

Hello there,


If you already understand basic circuit theory then you can use this surprisingly very simple equation to derive the equation(s) for just about any op amp circuit:

Vout=(Vnoninv-Vinv)*Ao

where

Vout is the output voltage,
Vnoninv is the voltage at the non inverting terminal, and
Vinv is the voltage at the inverting terminal, and
Ao is the open loop gain.

What you do is develop the equations for both inputs, multiply by the open loop gain Ao, then solve for Vout. What you end up with is an equation that contains the open loop gain Ao, but to get the idealized equations you can simply take the limit of Vout as Ao approaches infinity.
As mentioned above, that approach will allow you to write the equations for any op amp circuit because that is one way of describing op amp behavior in general so it applies to any circuit.
If you want to include frequency roll off or other things you have to add them too of course, but the basic equation starts out just like that above.
If you want to include input offset add a small battery in series with one of the inputs.

In spice terms, the basic op amp block would look like a simple subtractor followed by a gain block with gain Ao.
That's how simple the op amp really is!
Of course in real life just like any other circuit there are limits that must be imposed upon the resulting equations, and one of those is of course the output voltage plus and minus limit which usually can not be allowed to exceed the source supply voltages.

 

[EN0]

Thanks Audioguru, now I understand the difference between acoustical and op amp feedback. When comparing negative and positive feedback for op amps, however, isn't it true that if I feed my output to the non-inverting input (+), it's positive feedback? Likewise, if I feed my output to the inverting input (-), it's negative feedback?

I believe some op amps have positive feedback for certain applications, what's the advantage of doing that instead of negative feedback? Isn't negative feedback more common?

Based on your points, I can now conclude mathematically two equations:

[latex]V_+ = V_-[/latex]

[latex]I_+ = I_- = 0[/latex] (You said op amp inputs have extremely low input current)

I appreciate the help Audioguru, you've helped me understand more about op amps.

Thanks,

Austin

 

[EN0]

Just so you all know, I consider myself a self-taught EE student in which I've had the help of several college students. So far I've learned basic electronic theory that may include: Ohm's Law, KVL, KCL, Superposition, Source Transformations, Nodal Analysis, Mesh Analysis, and Thevenin's Theorem. Op amps are next on my course, so that's what I'm learning about now.

Just wanted to let you all know where I am.

Thanks,

Austin

 

[audioguru]

When comparing negative and positive feedback for op amps, however, isn't it true that if I feed my output to the non-inverting input (+), it's positive feedback? Likewise, if I feed my output to the inverting input (-), it's negative feedback?

Yes.

I believe some op amps have positive feedback for certain applications, what's the advantage of doing that instead of negative feedback? Isn't negative feedback more common?

Positive feedback is used to make an oscillator and to make a Schmitt-trigger that switches its output quickly with a snap-action even if its input moves very slowly.
Negative feedback reduces the extremely high gain of an opamp to a useable amount, reduces distortion dramatically (some opamps have only 0.00008% distortion) and increases the bandwidth. I forgot, negative feedback also reduces the output impedance dramatically so that a load on an audio amplifier or on a voltage regulator does not cause the voltage to drop.

 

[EN0]

Yes.


Positive feedback is used to make an oscillator and to make a Schmitt-trigger that switches its output quickly with a snap-action even if its input moves very slowly.
Negative feedback reduces the extremely high gain of an opamp to a useable amount, reduces distortion dramatically (some opamps have only 0.00008% distortion) and increases the bandwidth. I forgot, negative feedback also reduces the output impedance dramatically so that a load on an audio amplifier or on a voltage regulator does not cause the voltage to drop.

Wow, op amps have really advanced!

Thanks Audioguru!

Austin

 

[EN0]

Example from TI Document:

Now that I understand feedback (or so I hope), let's try running through several examples:

To begin with, let's use conventional block diagrams as depicted in the TI link I showed. If that's too confusing for me or anyone else, we can by all means just form the equations via schematic.

On page 80, and from the second diagram, how do they derive their equations?

[latex]\frac{V_O_U_T}{V_I_N} = \frac{A}{1 + A\beta}[/latex]

[latex]E = \frac{V_I_N}{1 + A\beta}[/latex]

I'm particularly confused about the "1 + Aβ," where does the "1 + Aβ" come from? In the ∑ circle, you can have either ±, and they have -. Why would it be "+" instead?

Thanks,

Austin

 

[EN0]

Hi Everyone,

Good news! I've been able to understand and apply equations to simple circuits such as the non-inverting and inverting amplifiers and also the very simple buffer. The ease was due to the fact that I1 = I2, but now if I incorporate another resistor the difficulty arrises. Can someone please show me how to apply equations to an op amp with 3 or more resistors?

I'd appreciate any help tremendously!

Thanks,

Austin

 

[audioguru]

Since you do not post a schematic then we don't know what you are talking about.
What is I1 and I2?
What do the extra resistors do and where are they connected?

 

[indulis]

Opamp’s (ideal) are simple…

Before going to the "standard equations", consider the following.

1) They have zero output impedance
2) They have infinite input impedance
3) Both inputs want to be at the same potential
4) Whichever input is “higher” wins and the output will swing in that direction.
5) The output will do whatever it can to make both inputs equal

Apply the above to a simple inverting amplifier. As an example, take a circuit where the input resistor to the “-“ input is 1K from a 1V source, the “+” input is tied to gnd and a 10K resistor is connected between the “-“ input and the output.

According to #3, both inputs want to be the same… since the “+” input is tied to gnd, the “-“ input will want to be at gnd potential. So you have a 1V source connected to a 1K resistor which is connected to the “-“ input which wants to be at gnd potential (virtual ground), so that’s 1V/1K=1mA. Per #2, since the input has infinite input impedance, no current can flow into or out of an input, so the 1mA of current has to go somewhere… it goes thru the 10K “feedback” resistor to the output. Well, if 1mA flows thru 10K, that would be a 10V drop. If the “-“ input is at gnd potential and you have 10V across the 10K and current flows from a higher potential to a lower potential, that means that the output has to be at -10V for all the “rules” above to be true. You will find that the “rules” will apply to any opamp circuit… try applying them to a non-inverting circuit to convince yourself that they work. Now you know where the equations came from and have a better understanding of how opamps work and how to apply them.

 

[EN0]

Since you do not post a schematic then we don't know what you are talking about.
What is I1 and I2?
What do the extra resistors do and where are they connected?

Yea, I meant to say I+ = I-. Although with the non-inverting and inverting op amp configurations, somehow the current through R1 and R2 equal each other. Now I don't know how to apply equations to more than 3 resistors.

Austin

 

[EN0]

Opamp’s (ideal) are simple…

Before going to the "standard equations", consider the following.

1) They have zero output impedance
2) They have infinite input impedance
3) Both inputs want to be at the same potential
4) Whichever input is “higher” wins and the output will swing in that direction.
5) The output will do whatever it can to make both inputs equal

Apply the above to a simple inverting amplifier. As an example, take a circuit where the input resistor to the “-“ input is 1K from a 1V source, the “+” input is tied to gnd and a 10K resistor is connected between the “-“ input and the output.

According to #3, both inputs want to be the same… since the “+” input is tied to gnd, the “-“ input will want to be at gnd potential. So you have a 1V source connected to a 1K resistor which is connected to the “-“ input which wants to be at gnd potential (virtual ground), so that’s 1V/1K=1mA. Per #2, since the input has infinite input impedance, no current can flow into or out of an input, so the 1mA of current has to go somewhere… it goes thru the 10K “feedback” resistor to the output. Well, if 1mA flows thru 10K, that would be a 10V drop. If the “-“ input is at gnd potential and you have 10V across the 10K and current flows from a higher potential to a lower potential, that means that the output has to be at -10V for all the “rules” above to be true. You will find that the “rules” will apply to any opamp circuit… try applying them to a non-inverting circuit to convince yourself that they work. Now you know where the equations came from and have a better understanding of how opamps work and how to apply them.

Thanks indulis, I applied a schematic to your example in the attachement. I solved the inverting amplifier using nodal analysis:

Vin = V+ = V- = GND.

I1 = 1V - 0V/1k, I2 = 0V - Vout/10k. Since I1 = I1 (I'm not quite sure how?) The equation becomes ---> -10k/1k = Vout/1V ---> Vout = -10V.

Could you please provide an example with 3 resistors, and show how to solve for the Vout = G * Vin equation?

Thanks,

Austin

 

[audioguru]

Just use Ohm's Law to calculate the currents and voltages.
The input current is nearly the same as the current in the feedback resistor since the opamp has an extremely low input current (some have no input current) and because the voltage gain of an opamp is extremely high.

 

[EN0]

Just use Ohm's Law to calculate the currents and voltages.
The input current is nearly the same as the current in the feedback resistor since the opamp has an extremely low input current (some have no input current) and because the voltage gain of an opamp is extremely high.

What circuit analysis theorem would be best? If someone actually showed a step-by-step process of analyzing a op amp with 3 resistor values, I would be able to understand multiple resistors after that (I think). I will make a schematic of a op amp circuit with 3 resistors. Perhaps you could help me.

In the attachement you have 10V as the output, while it's actually an inverting amp. Thus, -10V should be correct?

Thanks,

Austin

 

[EN0]

3 Resistor Example

In the attachement is a op amp circuit with 3 resistors. Can someone show me how to calculate the gain and the Vout of the circuit?

I'd appreciate the help tremendously!

Thanks,

Austin

 

[audioguru]

Your opamp has positive feedback instead of negative feedback so it will simply latch its output as high as it can go or as low as it can go. it is not an amplifier.

 

[EN0]

Your opamp has positive feedback instead of negative feedback so it will simply latch its output as high as it can go or as low as it can go. it is not an amplifier.

Thanks for pointing that out! It's now fixed.

Austin

 

[audioguru]

R3 is the opamp's load that has nothing to do with its output voltage since most opamps can drive a load as low as 2k ohms perfectly. The 5k load is nothing.
The voltage gain is 1+ (R1/R2).