Wow thanks Mr AI. That's also very enlightening to me. The fact that the nf is used to control the accuracy of Vo wrt Vin. I mean, I have read so much about opamps in the past 3 weeks but its all theoretical.

Hi,

Vout will go down with Vn (the inverting input) because that input always causes Vout to go down when it goes higher in voltage. But the reason for negative feedback (rather than just applying a separate voltage) is to make the amplifier characteristics better.

One of the main advantages to negative feedback is accuracy of the output voltage with respect to the input voltage at (in this case) the non inverting input Vp. The negative feedback makes up for the deficiencies of the open loop gain which is not well specified and is not stable over temperature. The closed loop gain is highly predicable and stable, and that comes from negative feedback. The drawback is we loose some of the original gain in the process. Even so, it was probably one of the most important inventions in electronic history.

A better overall view of the op amp is that it is an "error" amplifier, and different circuit configurations have different advantages.

 

Wow thanks Mr AI. That's also very enlightening to me. The fact that the nf is used to control the accuracy of Vo wrt Vin. I mean, I have read so much about opamps in the past 3 weeks but its all theoretical.

Did you even read my post

http://forum.allaboutcircuits.com/t...-negative-feedback-works.125564/#post-1017539

Where I showed how the feedback works and how the gain of them amplifier results in Vo being closer to Vin as the gain increases such that the error in most opamps is on the order fractions of a millivolt?

You seemed happier with an explanation that just said that things happen magically, so perhaps not.

 

I did read the post but got a little confused by the calculation because you wrote:

"What if Av = 100?
Then if Vp = 2.000 V and Vn = 2.100 V, Vo would be 10.0 V. While if Vp = 2.100 V and Vn =2.000 V, Vo would be -10.0 V."

Wouldn't the signs be reversed? -10V for the first case and +10V for the second case?

 

You are correct, the signs of the output are reversed in that example.

Once you understand the op amp, you will realize that it is one of the most elegant things ever invented in any field of engineering. And it is all due to use of negative feedback to control the behavior. Unlike a transistor, or even a diode, it's behavior (with negative) feedback is nearly perfectly described by a trivial equation. The only time this equation if violated is when you violate one of the limits of a real amplifier, as opposed to the ideal one, i.e. expecting an output greater than the max it can output, or setting an input outside the common mode limits, or trying to get too high a gain for the frequency. The op amp is the wheel of analog electronics.

Bob

 

..................... The op amp is the wheel of analog electronics.

I call it the Swiss-army-knife of analog electronics.

 

I call it the Swiss-army-knife of analog electronics.

Hi,

That's cool too

I call it the "electronic balance scale equivalency" because it 'weighs' the two voltages and provides an indication of the imbalance at the output. I'll get into this more in my next post.

 

\o/ YEAY! I was correct once in 3 weeks! Thanks you guys. Im slow ok... Im sold on many things in electronics being truly remarkable! I got into Arduino about 1 year ago and I was amazed by the BJTs and MOSFETs. So yes, op amps are impressing me and I don't even understand how they work yet

Also that post is confusing me (Im not saying the post is confusing, Im saying Im confused, so please don't get upset.) What confused me were 2 things and Im sorry WBahn, I just havent had time to reply to all posts, but your example was feeding signal into the Vp, when most of the examples I have been using to try and understand have had input on the Vn. But I got thrown by the fact that you applied a 2V at Vp, and using the formula for Vo = Vp(100/101), you get that Vo = 1.98V. But then you said that the 20mV diff between Vp(2V) and Vn(1.98V) was generating the 2V at Vo...but didnt we just conclude that Vo=Vn=1.98V?

I think my confusion comes from the time relation here. I realized this while reviewing Jony's post yesterday. So Im working on a step by step (image aided) sequence of what happens so I can post it and you can correct me where Im wrong.

You guys have been very helpful. Thanks so much.

 

Wow thanks Mr AI. That's also very enlightening to me. The fact that the nf is used to control the accuracy of Vo wrt Vin. I mean, I have read so much about opamps in the past 3 weeks but its all theoretical.

Hi,

You are welcome, and you might want to note that WBahn was providing a numerical example and although the signs were incorrect it's a good example. Once the signs are corrected then you can read it again.

One of the most basic view of the op amp shows it's actual operation, from the time the power is turned on to the time it becomes stable. This simple view explains to a large degree how the op amp actually works in a real circuit.

First, the output of the op amp is at zero, then the power is turned on. The op amp detects the inputs, then starts to ramp up to the voltage that forces the inverting terminal to become almost equal to the non inverting terminal.
So if after turn on we have 1v on the non inverting terminal, the output starts to ramp up. As it ramps up the inverting terminal (with direct feedback) ramps up too. Eventually the inverting terminal becomes equal to the non inverting terminal and so the output no longer ramps up but stays at that voltage. So in the end, we have both inputs being almost equal to each other while the output remains stable at some voltage. This is DC operation of course.

So the main point is that with feedback the op amp will push the output up or down until both inputs have the same voltage on them. It's not exactly the same, but very close, so for many theories of the op amp they are assumed to be exactly equal. If you start with this view you will get a feel for the op amp, then you can move on to the more detailed view where the inputs are not exactly the same. You can get pretty far by assuming they are the same, but remember the two views are used independently of one another.

We could look at more examples if you like. Once you look at several examples you start to get a good feel for how this works. It takes a little time but it's worth it.

 

Once you understand that, with negative feedback, an op amp will always adjust the output so that the voltage across its two inputs is essentially 0V (within the linear voltage constraints of the output), then you can analyze just about any op amp circuit.
Understanding that was the "eureka" moment for me when I started using op amps.

 

Yep. Equations do not get much simpler than x = y.

Bob

 

Im having trouble posting pictures. I uploaded it to flickr here:

https://www.flickr.com/photos/43666730@N04/28176206265/

Can you guys see that?

Because I get a broken image icon when I put the url into the image code like this:

 

Im having trouble posting pictures. I uploaded it to flickr here:
https://www.flickr.com/photos/43666730@N04/28176206265/

You are correct. You have hit on the next level of subtle detail about how opamps really work, something most people never get to on their own. Gold star.

Disclaimer: This is a stripped-down discussion is about a basic, standard inverting or non-inverting gain stage, not a filter, comparator, oscillator, gyrator, current source, or any of the hundred other opamp circuit configurations.

Assuming perfect input stage transistor matching, the voltage difference at the inverting and non-inverting inputs of a normal opamp gain stage *never* equals 0 V. This is because the internal gain of an opamp is not infinity, and *never changes.* We talk about the opamp gain being set by the feedback resistors, but it really is the overall circuit gain that is adjustable. The opamp device internal gain is whatever is on the datasheet, period. For the old LM741 it is around 300,000 to 500,000, and of course is much higher for newer devices. This is called the open loop gain, and is specified either in dB, or in V/uV or V/mV for older parts. V/uV tells the story - output voltage per input differential microvolt. So if an opamp has an open loop gain of 1,000,000, or 1 V/uV, or 1,000 V/mV, or 120 dB (always specified at DC), and a theoretically perfect input stage, then in order for it to have an output of +1.0V, the differential input voltage at the device pins must be 1 uV, not 0 V.

Here is how a real opamp gain stage works: If you want to amplify a small DC voltage by 1000, to turn a 1 mV input signal into a 1 V output signal, and your opamp device has an open loop forward gain of 120 dB, then...

The 1 mV input is *reduced* from 1 mV down to 1 uV at the input pin (by subtracting a portion of the output voltage), and then that remaining 1 uV is amplified by 120 dB to produce 1 V at the output pin. The two feedback resistors form an attenuator. They attenuate the amount of the output that is subtracted from the input before the input is amplified. It is a loop, and the explanation makes sense only in steady-state operation. There is a *very* brief period at power up before things stabilize where the output is incorrect for the input.

Most opamps have an internal frequency response shaping capacitor which is why the open loop gain decreases as the frequency being amplified increases, but that's another story.

ak

 

I did read the post but got a little confused by the calculation because you wrote:

"What if Av = 100?
Then if Vp = 2.000 V and Vn = 2.100 V, Vo would be 10.0 V. While if Vp = 2.100 V and Vn =2.000 V, Vo would be -10.0 V."

Wouldn't the signs be reversed? -10V for the first case and +10V for the second case?

Yep, the signs are reversed. Thanks for catching the typo. I'll go fix it.

 

Ok I changed the image because the left side is my interpretation of 2V at the inverting terminal that Jony gave in his post. It gave me a small confusion at the end. Then I compared it (on the right of the image), with the explanation WBahn gave using 2V at the non-inverting terminal. I ran into the same confusion at the end. Basically, according to the formula, Vo should be 1.98V (for WBahn's example) but he states its 2V. There must obviously be a real differential if the amp is putting anything out Vo.



And here is my setup from last night. Ill try to post the results soon but I remember my biggest doubt was whether to connect the grounds from the Vin and the split battery supply because it made a huge difference in results.

 

Hello,

The two images in your post do not show up.
Please use the "Upload a File" button to upload the images to the forum.

Bertus

 

Ok I changed the image because the left side is my interpretation of 2V at the inverting terminal that Jony gave in his post. It gave me a small confusion at the end. Then I compared it (on the right of the image), with the explanation WBahn gave using 2V at the non-inverting terminal. I ran into the same confusion at the end. Basically, according to the formula, Vo should be 1.98V (for WBahn's example) but he states its 2V. There must obviously be a real differential if the amp is putting anything out Vo.

Note that I said, "So if we apply 2.000 V to Vp, the Vo (and Vn) will be 1.980 V." But much of what followed was using round numbers because I thought the point had been made. I've gone back and made them more exact.

 

Ok here is my confusion and a question:

1. You wrote a 19.80V in there but I think it was meant to be a 1.98V right?

2. This is the bit that confuses me, first you write that Vo (and Vn) are 1.98V according to the formula. This produces a 19.80mV differential which gets amplified into 2V at the output. But wasn't the output, Vo, 1.98V?

 

Ok here is my confusion and a question:

1. You wrote a 19.80V in there but I think it was meant to be a 1.98V right?

Yes, that should be 1.980 V. It's been corrected.

2. This is the bit that confuses me, first you write that Vo (and Vn) are 1.98V according to the formula. This produces a 19.80mV differential which gets amplified into 2V at the output. But wasn't the output, Vo, 1.98V?

I'm not sure where you are referring to now. Here is what is in the text after the prior adjustment:

"So if we apply 2.000 V to Vp, the Vo (and Vn) will be 1.980 V. That 20 mV difference is what is producing the 2.000 V at the output. (It would actually be more like a 19.80 mV difference producing the 1.980 V at the output.)"

 

Right, there it is:
"So if we apply 2.000 V to Vp, the Vo (and Vn) will be 1.980 V"
This means Vo = 1.98V right?

Then:
"That 20 mV difference is what is producing the 2.000 V at the output."
Output means Vo, doesn't it? So Vo = 2.000V?

 

Right, there it is:
"So if we apply 2.000 V to Vp, the Vo (and Vn) will be 1.980 V"
This means Vo = 1.98V right?

Then:
"That 20 mV difference is what is producing the 2.000 V at the output."
Output means Vo, doesn't it? So Vo = 2.000V?

Uh... Hello???? Read the next part: "(It would actually be more like a 19.80 mV difference producing the 1.980 V at the output.)"