Operational Amplifier basics

In article , snipped-for-privacy@emailo.com says...>

There are only three "rules" to know about the ideal op amp

All else can be derived.

^ input

It only has one. From rule #2 above, assuming a non infinite output voltage, we conclude the difference between the inputs is 0V.

No. It can have *NO* difference in the input voltage. A non-idel amplifier will have some difference but it can be assumed to be zero (for most purposes).

No. They *must* be run as a differential amp, with 0V across the input. You may only USE them in a single-ended configuration, though the Op-Amp is still a differential amplifier.

"It" doesn't know the difference.

Whatever floats your boat. I usually label them "+15V" and "-15V" (even though they're really +/-12V ;-).

No, they're not (so designed). As stated above, they don't know the difference. They only see the two rails. They aren't connected to the "half-way" voltage.

Use ground (or really, anything you like) as your reference points. Abide by the three rules above, sum the currents in the inputs and feedback (output to -In), and you're done.

Yes, but I'm looking at things from a practical wiring point of view. Being able to point out where the two points are for an input or ouput voltages.

Maybe it's just me, but I've got to be able to, see the two points where the input and ouput voltages are meant to be measured from.

I'm looking at something like this:

And saying to myself for the differential amplifier there are two input points, V1 and V2, whose reference points are -VEE. Yet below in the equivalent circuit there is just V1 and V2. From a practical point of view I'm thinking both V1 and V2 will have a potential with respect to -VEE. But you would not see this in the equivalent circuit.

In article , snipped-for-privacy@emailo.com says...>

You can use V- as your reference. You can use V+, or anything else you choose. If you look at the difference between '+' and '-' it doesn't matter what '+' and '-' are referenced to. The

Does the circuit really work because of R4?

We need two points to measure a voltage.

This circuit will amplify the voltage difference between V- and V+.

So, how can you do that with two inputs?

The input voltage comes fron the input voltage sources.

So, where are the two points for the voltages on V1 and V2?

From the input source side, (i.e. looking left as it were into the sources) point V1 and Ground and V2 and Ground. There is a circuit in the input source that goes from V1 and V2 and ground. (Like a battery, there is in internal circuit in the battery).

But what about lookng into the OP Amp devic, where is that cicuit?

Looking to the right, into the Op Amp, the internal circuit starts at V1 and ends at V2.

So, from a sources POV, one source sees it's termination on as R1, through the internal resistor inside the Op amp (infinity) then through R4 to ground.

The other source sees it's termination as R3, then through R4.

I'm probably wrong, but that's the way I'm thinking.

I think it only amplifies the difference between V1 and V2 because both input sources are connected to a common reference point, Ground. True.

Yet the Op Amp itself just sees a voltage between V- and V+. Which is an amplification of the voltage difference from the sources.

Is this getting near?

An op-amp _does_ have two inputs, normally called "inverting" and "non-inverting".

An op-amp _always_ amplifies the voltage difference between its two inputs.

A circuit using an op-amp can be a "regular" (single input) amplifier, but the op-amp still has differential inputs.

The convention in electronics is that "ground" is always the "reference point" for voltage measurements.

No - "ground" is always the reference point. The type of transistors used in the op-amp do not change this.

wrong.

It is generally necessary to bias both input pins to about mid-supply, whether using a single supply or two supplies.

In article , snipped-for-privacy@emailo.com says...>

Without R4 it works but doesn't do the same thing as it does with R4.

Certainly. However, the "reference" is chosen to make the math easier. The circuit doesn't care what you name things.

No. The voltage between '+' and '-' is (almost) zero. The idea of this circuit is to amplify the voltage between V1 and V2.

You just said it takes two points to measure (define) a voltage. Choose V1 and V2. It may be easier for you to see what the output voltage is because of V1 (to ground) and V2 (to ground), independently, then add the two together.

That's why we call them the "input". ;-)

Pick one, ut be consistent. To make the math easier pick the ground ('0V'). REMEMBER, there is zero voltage from '+' to '-'.

Ok, I'm no really following you, but...

It's a black box, that has the characteristics of infinite input impedance, infinite gain ((V+ - V-)/Vout) = infinity, and zero output impedance.

It's a black box. Don't care what's in there. If you want to analyze the black box forget what's outside.

Forget this side for now.

Yep, the input impedance is infinite, so forget it. Thus: V+ = V2(R4/(R3+R4)) You now have half the answer.

Now, since the gain is infinite, the voltage on the input must be zero. Therefor V- = V+. Now you can solve for the current in R1. Since the input impedance of the Op-Amp is infinite, that same current must be in R2.

V1-V+(from above) V+ - Vout ----------------- = ----------- R1 R2

Your conclusion is correct, but your reasoning is flawed. The "reference" point doesn't matter. It's a philament of your imagination. ;-)

The Op-Amp (ouput) forces its input to be 0V. The rest falls out in the math.

Read the last line again (and again, and...). This is the *key* point. The input voltage *MUST* be zero.

Here are some synonyms for "ground": Common Common Return Path Common Reference Point Zero-Volts Point

Never mind all that stuff; here, in a nutshell, is how an opamp works:

The opamp output does whatever it has to to make the voltages on the inputs equal.

That's basically it, but what does it mean?

(View in Courier:)

Let's say that you've got a circuit set up that looks like this:

. E1 . / . +-------------+ . | | .+------+------+ [R1] .| VARIABLE | | E2 E3 .|POWER SUPPLY | |/ / +--------+ .+------+------+ +---[VOLTMETER]---|+ | . | | | FIXED | . | [R2] | POWER | . | | | SUPPLY | . +-------------+-----------------|- | . +--------+

Let's also say that R1 is equal to zero ohms, that R2 is equal to 1000 ohms, that E1 is equal to 1 volt and that E3 is equal to 2 volts.

So how do you make the voltmeter read zero volts?

You crank up the variable supply until E1 equals E3, thereby making the voltage _difference_ across the voltmeter zero.

Now lets change R1 to 1000 ohms while leaving everything else the same.

What will that do? It'll increase the amount of voltage the variable supply has to put out to get 0V into the voltmeter.

How much? R1R2 is what's called a voltage divider, and the relationship between the voltages and resistances is given by:

E1 * R2 E2 = --------- R1 + R2

We want to find E1, so we rearrange like this:

E2(R1 + R2) E1 = ------------- R2

and solve:

If we were to leave E3 alone and change E1 from 1000 ohms to 10000 ohms in 1000 ohm steps, and solve for E1 at each step, we'd have a table that looks like this:

R1 E1 OHMS VOLTS

-------+------- 0 2 1000 4 2000 6 3000 8 4000 10 5000 12 6000 14 7000 16 8000 18 9000 20

Let's go back to when we had 1000 ohms for R1 and see what would happen if we changed E3.

So now let's make a table of V1 VS V3 for V3 = 0V to 10V when E1 = 1000 ohms:

E3 E1 VOLTS VOLTS

-------+------- 0 0 2 4 3 6 4 8 5 10 6 12 7 14 8 16 9 18 10 20

If we say E3 is the input and E1 is the output, then the circuit has a voltage gain of 2

But how does this apply to opamps?

OK let's redraw things a little:

Now pretend that R1 and R2 are both 1000 ohms, that E3 = 1V, and that all that matters is that the opamp makes the voltage at E1 be whatever it needs to be to make E2 equal to E3, which will make the voltmeter read zero.

What would that voltage be?

JF

Midnight here. I'll figure it all out tomorrow. :c)

E3 is set to 1V, that means E2 needs to be at 1V.

E1 = 1V (1000R + 1000R) -------------------------- = 2V 10000

The Invering amplifier which was drawn seeks to get V- and V+ to be equal. When there is a tendency for E3 to go up, the voltage of the "Variable PSU" goes down, thus stabalizing the system. The system reaches stability when E2 = E3. And at that point there is a certain voltage at E1. The value of that particular voltage is a function of the make-up of the voltage divider network. It cannot exceed supply. Of course the amplifier is acting as if it's a "variable PSU", like a generator in fact.Correct?

Not sure how the op amp itself settles when V- = V+. I just know that the amp acts like a variable PSU and what voltage E1 needs to be to get V- to equal V+.

Actually, it's like the OP Amp has an internal setting. A- = A+ is like a setpoint which stays put all of the time. You get your desired results by messing about with the external networks.

Correct.

That's all you need to know (and that the input impedance is infinite and output zero).

Negative feedback. If the output wanted to be too high, the current in R1 would be too high forcing V- high, causing the output to go lower. ...until the voltage between + and - is zero.

If it's not (the negative feedback can't stabilize it) the output has to go to one rail or the other ("infinite gain").

In article , snipped-for-privacy@emailo.com says...>

No, the output, via the negative feedback (R1 feeds back from output to the '-' input, thus "negative") forces the inputs to be at the same voltage. If the inputs are forced to be different the output voltage will try to go to infinity (either positive or negative). In real life this means the output will limit at the positive or negative rail (known as the output "railing"). Feedback is lost at this point so the OpAmp no longer functions as an OpAmp.

The method to stabilize is feedack. Feedback through a resistor ensures that V- = V+, but is it not an internal fuction of the OP Amp that Vout stabilizes are the required voltage (Set by extermnal networks) when V- = V+? I suppose you could make an Op Amp where Vout stabilzes when there is 1V across V- and V+?

In article , snipped-for-privacy@emailo.com says...>

Right. If V- V+ the output is infinite, so they must be equal or the feedback isn't doing its job.

You wouldn't. To do so one would need a 1V reference, something that is completely unnecessary and unwanted.

What I'm thinking about is this: If I had an audio amplifier speaking into the microphone I might drive the ouput transistor into distortion.

But if I had some negative feedback I could ensure that the distortion was less. But, the amplifier does not seek to drive the input to zero and achieve stabilization there.

Actually the circuit is a non inverting amplifier circuit.

In article , snipped-for-privacy@emailo.com says...>

Ok, if your drive exceeds the OpAmp's rails, sure. But that doesn't have anything to do with the voltage across the OpAmp's input.

If the output voltage needed to drive the input voltage to zero exceeds the power supply rails (minus some bit for most real OpAmps), yes there will be distortion and, yes, the inputs will no longer have 0V across them. It will *seek* (i.e. try) to drive the input voltage to zero, but it can't.