Re: OT: Is this question too challenging for a BSEE graduate?

Correct, if the outputs and inputs are isolated from one another. In this (open-loop) configuration, an op amp will behave pretty much like a comparator... and not a terribly good one (it'll be slow to reverse its output state after it has been driven into saturation).

Correct. What you do (in the usual closed-loop op amp configuration) is to feed back a portion of the output, to the inverting input. This may be a direct connection (for a unity-gain noninverting circuit), or there may be other components between the output and inverting input, and other connections to the inverting input as well.

In any case, the effect of this "feedback" is to create the sort of "input forcing" you're referred to. Specifically, if the inverting input voltage is below that of the noninverting input, the output voltage will rise towards the positive rail... and a portion of this voltage increase, going back through the feedback network, will raise the voltage at the noninverting input, reducing the difference between the two inputs. This process continues until (to a first approximation) the two inputs are at equal voltage.

Don't feel back about that. I understand that when the concept of the feedback-looped high-forward-gain operation amplifier was first presented to the U.S. Patent Office, the examiner rejected the invention, claiming that it couldn't work and thus couldn't possibly be useful.

--
Dave Platt                                    AE6EO
Friends of Jade Warrior home page:  http://www.radagast.org/jade-warrior
  I do _not_ wish to receive unsolicited commercial email, and I will
     boycott any company which has the gall to send me such ads!

snippers

You are neglecting (negative) feedback. In its simplest form, the output connects back to the negative (inverting) input. Now it is the basic voltage follower. Used as a buffer in real life.

So, given your initial conditions, the positive (non-inverting) input is set to 1 volt. The op-amp will drive the output in the positive direction until both inputs are equal, or at 1 volt.

Regards, tm

No, it's fact. It is, as I said, /the/ fundamental principle of op-amp circuit design.

I can't think of a book that discusses this in a fairly simple way. Even the Philbrick book -- which is hard to find these days -- doesn't address the matter as directly as I'd like. But, trust me. Most op-amp circuits can be analyzed by assuming the voltage is the same, then applying simple circuit analysis. You might start with the basic op-amp inverting amplifier, and see what happens.

--- If you think about a comparator as being nothing more than an opamp with a huge amount of gain operating open-loop, then for its output to swing to either rail requires an almost _insignificant_ difference in voltage between its inputs.

Here: (View in Courier)

+10V | |U1 Vin+>-------|+\ | >--Vout Vin->-------|-/ | | -10V

Let's say U1 is an ideal opamp with a gain of one million and that if Vin+ goes more positive than Vin-, Vout will rise toward +10V, and if it goes more negative than Vin-, Vout will fall toward -10V.

Now, if Vin+ = Vin-, Vout will sit between the rails at 0V, but since U1 has a gain of one million and we have a +10V rail, all it'll take for the output to rise to the positive rail is for Vin+ to be 10 microvolts (or more) more positive than Vin-.

Conversely, all it'll take for Vout to fall to the negative rail is for Vin+ to be >=10µV more negative than Vin-.

Also, it's important to realize that it doesn't matter what voltage Vin+ and Vin- are sitting at, what matters is the difference between them.

--- Correct.

---

--- Yes

---

--- Consider, first, the magical voltage divider: ;)

E1 | [R1] | +---E2 | [R2] | GND

Where, if GND is at 0V,

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

Just for grins, let's set this up:

10V E1V | [9k] R1 | +---E2 | [1k] R2 | GND

and solve for E2:

10V * 1kR E2 = ----------- = 1V 9kR + 1kR

Next, let's hook up the the voltage divider to an opamp in the non-inverting configuration and apply 1V to the + input to see what'll happen:

+V | 1V>-------|+\ | >--+-- E1 +--|-/ | | | [9k] R1 | -V | +--------+-->E2 | [1k] R2 | gnd

Now, since the + input is sitting at 1V, the output will have to move with sufficient magnitude and in the proper direction to make the voltage on the - input equal to the voltage on the + input.

Since the + input is at +1V, the output will have to swing positive, and for E2 to to get to +1V, we'll rearrange the voltage divider equation to solve for E2, and we'll get:

E2R1 + E2R2 (1V * 9kR) + (1V * 1kR) E1 = ------------- = ------------------------- = 10V R2 1kR

So there you have it: a non-inverting opamp with its gain of one million throttled down to 10. :-)

To adjust the gain, then, all that's necessary is to adjust the ratio R1:R2 and the output will go wherever it has to in order to make the inputs of the opamp equal to each other.

The gain of the circuit can be described as, simply,:

R1 + R2 Av = --------- R2

---

--- Even now? ;)

--- JF

One of three principles:

1: Gain is infinite Vout/((V+) - (V-)) = infinity 2: Input impedance is infinite 3: Output impedance is zero

Of course none of these is really true, but close enough for most purposes.

The difference between an opamp and comparitor is that the comparitor is designed to be operated with its output burried in the rails (some don't have a current sourcing capability, either). Comparitors can be made out of most opamps but they're generally pretty slow because they don't like their outputs saturated.

" snipped-for-privacy@att.bizzzzzzzzzzzz" wrote in news: snipped-for-privacy@4ax.com:

4: Phase shift is zero at all frequencies

Best regards, Spehro Pefhany

--
"it's the network..."                      "The Journey is the reward"
speff@interlog.com       Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info :  http://www.speff.com

" snipped-for-privacy@att.bizzzzzzzzzzzz" wrote in news: snipped-for-privacy@4ax.com:

A comparator also designed to operate with a significant voltage between the inputs. Many modern op-amps will allow significant current to flow if there is more than a diode drop or so between the inputs. At least one older op-amp would exhibit Vos drift over time in the presence of continous voltage between the inputs.

Best regards, Spehro Pefhany

--
"it's the network..."                      "The Journey is the reward"
speff@interlog.com       Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info :  http://www.speff.com

Actually, you can do it with bipolar transistor circuits too, to a first approximation:

Base => inverting input Emitter => non-inverting input Collector => output

Of course, the two inputs have rather different impedances, but for figuring out gains and so on, it works pretty well.

Isaac

I think you might be missing the fact that the circuit has feedback, from the output to the inverting input. Take a look at the schematic again.

The op amp "sees" a voltage on the + input and does whatever it can to make the - input the same voltage. The output of the op amp is connected back to the - input, so when the op amp raises or lowers the voltage on the output pin, that voltage appears on the - input. Thus, if you put X volts on the non-inverting (+) input, you'll get X volts on the inverting (-) input.

Ed

On 10/25/2010 10:37 PM ehsjr spake thus:

Now that I understand things a little better, yes, I do get the feedback here, and in other op amp circuits.

But just a small quibble with the way you and others have described what's going on here. You say "the op amp ... does whatever it can to make the - input the same voltage" (as the + input). In fact, it does no such thing: the input is, after all, just an input.

What you might ought have said is that the *circuit*, including the feedback loop, forces the inverting input to (virtually) the same voltage as the noninverting input, right? The op amp, in and of itself, doesn't "do" anything to (that is, out of) either input. It's only by virtue of the feedback that this action occurs.

Maybe just a semantic quibble. Other than that I'm with you here. Thanks to all for explaining.

--
The fashion in killing has an insouciant, flirty style this spring,
with the flaunting of well-defined muscle, wrapped in flags.

- Comment from an article on Antiwar.com (http://antiwar.com)

Yes... and of course you have to get the feedback "right" as well (negative for the simple sorts of applications we're discussing here) -- the astute EE

101 student will point out that using the rules about infinite input impedances and the inverting/non-inverting voltages being the same, you could swap the inverting and non-inverting inputs and everything should still work, yes?

At least when I took the appropriate course, it was only about a week or so between "here's the absolutely ideal op-amp model and use these rules to figure out the gain" and "here's a real-world op-amp with finite gain" and then a few more days to "...and finite frequency response, and offset voltages, etc." -- so you didn't have to feel uneasy about the initial hand-waving for too long. :-)

On 10/26/2010 10:53 AM Joel Koltner spake thus:

Yep. That stuff about the ideal op-amp--infinite gain, infinite input impedance, zero output impedance, bandwidth to the far edges of the electromagnetic spectrum--makes it sound like a perpetual-motion machine ...

--
The fashion in killing has an insouciant, flirty style this spring,
with the flaunting of well-defined muscle, wrapped in flags.

- Comment from an article on Antiwar.com (http://antiwar.com)

--
David,

Did you get the article I posted for you?

I'd say the fundamental principle is high gain DC-and-up amplification; the balance of input potentials is derived from this, and is, rather, a productive approximation. Useful, yes, but not fundamental.

It's this kind of 'principle' you have to drop when something more important calls for your attention.

For many uses the idealized model works extremely well. It's kinda like the ideal resistor; it doesn't really exist but for most problems reality is close enough to practice that the difference doesn't matter.

Positive infinity is equal to negative infinity on your planet? ;-)

I don't think I ever took a formal class using opamps. I learned about them in a "special problems" classes, first with (tube/servo-multiplier) analog computers then the IC versions.

Something like that...

Someone here once posted a link to a commercial op-amp data sheet (Maxim?) where the example circuit had +/- swapped!

At the U. of Wisc where I was, it was covered in the first real EE class you took. It went something like... DC analysis (Ohm, Norton, Thevenin), All About Phasors, Op-Amps, general AC circuit analysis, and finally RC/RL/RLC circuits (although the letter "Q" never came up once -- that was for later).

The teacher was a very good teacher, although unfortunately his knowledge didn't extend much outside of the textbook. Strangely, he was a full tenured guy, whereas the much more "practical" fellow teaching the "Circuits You Might Actually Find Yourself Building In The Real World" classes missed obtaining tenure several times and eventually retired early.

I occasionally wonder whatever became of some of my old professors and teachers; one looks at them so differently when you're in school than once you've been out in the working world for awhile.

---Joel

you

ll

C

r).

Q is for Qan't Quite Qalculate the Quiirky parts of the Quantity Qorrectly