Operational Amplifier basics

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Yup! :-)

JF

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

Right. The direction of the feedback is such that it forces the opposite of what the output does. If you kick the output positive, the feedback forces a negative to compensate for it.

You have it somewhat backwards. The amplifier is designed to have an "infinite" differential gain. That property is useful, using negative feedback, to make an amplifier with a gain controlled solely by external resistors. The three properties I've repeated many times (because they're all important) is what allows the OpAmp to be useful. It's not because someone said "hey, wouldn't it be useful to have an amplifier that tries to balance its inputs".

Again, you have it backwards. The input voltage is zero *because* of the negative feedback. Negative feedback is what makes the OpAmp useful. By itself the (ideal) OpAmp has the three properties I've stated before.

IOW, we analyze the circuit assuming the input voltage is zero because the output, via negative feedback, forces it to be so.

No. There is often a DC offset in amplifiers. You'll often find DC blocking caps (or AC coupling caps, your choice of terms) between amplifiers that do not have a zero operating point. If you run an OpAmp on a single rail you'll see the same thing because a DC offset is added to the signal to keep the signal away from the rails.

No, it is *precisely* feedback that is allowing you to analyze the circuit assuming the input voltage is zero. Without the negative feedback it certainly won't be (zero).

The concept of an "ideal" op-amp with infinite gain strikes me as a far from ideal way to teach op-amp theory. It must be very confusing for many students.

Real op-amps have a large but finite gain 'A'

Vout = A*(VP-VN)

Feedback makes the differential input voltage small, but not zero.

If I build a unity-gain non-inverting voltage follower, I have :

Vout = A*(Vin-Vout)

(A+1) * Vout = A*Vin

Vout ~= Vin

In article , snipped-for-privacy@nospam.co.uk says...>

How is your "almost ideal" less confusing than going without the almost part? The three characteristics of an ideal Op-Amp work very well to analyze almost any OpAmp circuit, at least as a first pass. Indeed the idealized OpAmp works better and is a whole lot simpler than your "almost ideal" version.

Yes, equating VP=VN is often the best/easiest way to analyze op-amp circuits.

I just don't like infinity*zero = the correct answer.

In article , snipped-for-privacy@nospam.co.uk says...>

Almost infinity is hardly correct for anything. Infinity is easy because all your "almosts" drop out, greatly simplifying the math. Since no one will measure the input voltage [*], infinity is close enough.

[*] other non-idealized OpAmp issues come in *way* before this.

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http://www.philbrickarchive.org/k3_series_components.htm

there is no such thing as an ideal op-amp it's impossible for a number of reasons. this can initially be a real distraction for those who have more experience with real hardware than with mathematics,

Complete hogwash. The "normal" operation of a real OpAmp is closer to it's ideal than pretty much any other component. If you can't deal with the simplicities of "zero" and "infinity" get out of the business.

Okay I think I've got it. With feedack V- = V+ occurs. That happens with dc amplifiers. Get a very high gain dc amplifier, have very high input inpedance, low ouput impedance, great isolation between input and output (internally), and when you connect it in a certain manner, principally where the output is fed into the input to counteract changes in the input, you get a situation where V- and V+ will become equal. That's it. Correct?

My electronics knowledge comes from studying ham radio. I've never studied dc amplifiers. In my usual experience, when there is no input, there is no output. So, when I think of this, it looks like something other than feedback it at play in *addition to the things already mentioned*. I got to thinking about setpoints in fact.

It's the bit about having an output when there is no input voltage that throws me.

But of course if you do kind of have an output with "ac amplifiers" when input is zero.. In the sense that there is a dc voltage on the other side of the output capacitor. That dc voltage rises sinosoidally (say). But when there is no input it goes to a quiesent value. Of course, it it's sinusoildal voltages that you want mangnify, and dc voltage on the other side of the output capacitor (or transformer, whatever) is of no significance to the amplification of the input.

Is this getting there? Rich

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

Where the feedback is of the opposite polarity as the output, yes. That is, a change in the output is fed back as the opposite signal to the input, the inputs must balance or the output will "rail" ("infinite" voltage) trying.

Even your RF amplifiers have an output with no input. The (DC) output is generally blocked with capacitors. Here we have no capacitors, though for an AC amplifier one could and often does use them.

You are just ignoring the output of your RF amplifier, with no input. It's there.

Yep! If you're concerned about DC you have to consider it. If you aren't, block and ignore. ;-)

--- Yes. :-)

One thing you have to consider if you only have a single unipolar power supply and you want to amplify a bipolar ac signal is biasing the opamp so that the output voltage is quiescently between the rails.

For the inverting case what you'd do is connect a couple of resistors across the rails and then connect their junction to the opamp's non inverting input:

12V-----------+------------+ | | [R1] +---|--[R4]--+ | | | | Vin>--[C1]----|--[R3]--+--|-\\ | | | >------+-[C2]--->Vout +-----------|+/ | | [R2] | | | GND>----------+------------+

With a 12V supply you'd want the output to be at 6V with no AC signal in, (so you could get a symmetrical +/-6V swing at Vout) so by making R1 and R2 equal, the + input would be at 6V and the output would have to swing to 6V to make the to opamp input voltages equal.

Assuming Xc of C1

I made three points and I know you can't disprove any of them.

Since you snipped my response, I'll assume you'd rather lie to cover your inadequacy than discuss the issues. No, I don't approve of liars, or incompetents for that matter

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Well, if: 

"this can initially be a real distraction for those who 
have more experience with real hardware than with mathematics,"

was one of the points, then that one is not only complete, it\'s complete
and _utter_ hogwash.

JF

If you are interested in the history, you might see about finding a copy of "Op Amp Applications," edited by Walt Jung, published by Analog Devices, Inc. The history will help inform you a little more about the "why did someone want to do that?" question.

You can get the PDFs for it at: :

In particular, you might want to look at Chapter H and perhaps even more particularly on the section called "Black's Feedback Amplifier."

Jon

Brilliant. Thanks.

Again you go off on a tangent instead of dealing with what I wrote,

You do that too much.

Nope.

Wrong, as always.

Yep. You're a proven liar. So be it known.