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

Probably written by Bob Widlar. Unfortunately, NS is still using his documents, which are now rather dated. Nothing wrong with continued respect for his genius, but /someone/ needs to update them.

--- 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)

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:

and solve for E2:

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:

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:

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:

Best regards, Spehro Pefhany

" 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

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.

op-amp tries to force both inputs to the same voltage. Since it was stipulated the op-amp is a 'classic, ideal' op amp, we can assume it has none of the defects found in the real world. As a result the voltage across R3 will also be 1/4 of VDC. The only way that can happen is if the effective resistance of Q1 is 3 times the resistance of R3, or 120 ohms.

No kidding? You replace a simple variable power resistor, which only requires a screw driver to change the resistance with three resistors, an op-amp (which requires a separate power supply), and a mosfet. To change the value of the virtual resistor you have to change a resistor?

It would seem to me a potentiometer would improve usability greatly.

PlainBill

Geese. It was just a quiz to see if an applicant understood how an opamp works.

BTW, you didn't get the job.

tm

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

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 ...

On 10/26/2010 9:57 AM tm spake thus:

Yabbut, it says right there on the diagram "programmable load". So is it or isn't it? To me, "programmable" means (or at least implies) changeable by changing voltages or some other electronic parameter, not by physically substituting components. Yes, a potentiometer would seem to be a better choice--even if it is "just a quiz".

I didn't want it anyway.

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.

I was just pulling your chain. It was a s***ty job anyway. Imagine working for someone that asked you that question on a job interview. I would more like to be asked what have I done that made someone some money. It's just business anyway.

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.