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

Maybe, those students have a lot of material thrown at them in a short time. I've met both techs and engineers out of school who either "get" op-amp theory or not... it's really just common sense on the basics. For me, it was one practical app (and it was in a job interview that I flunked) to cause me to get my a$$ in gear and figure it out for good.

Haven't done this stuff in while (used to do elect design but no BSEE.. just an "almost" AS in EET), and was sick/bored today, so here's some bloated math... are you giving your potential candidates a couple of hours to give the answer in terms of the resistors? :-)

For the over-all load on VDC:

Req = (R1 + R2) / (1 + (R1 / R3))

For just the output network:

Req = ((R1 + R2) * R3) / R1

--------------------------------- Proof-

Since non-inverting input and VR3 will be at the same voltage:

VR3 = (R1 * VDC) / (R1 + R2)

Solving for VDC:

VDC = (VR3 * (R1 + R2)) / R1

.....

IR3 = VR3 / R3

Resistance just for the output side, since current through JFET and R3 are same:

Rout = VDC / IR3

Substituting from above:

Rout = ((VR3 * (R1 + R2)) / R1) / (VR3 / R3)

Rout = ((R1 + R2) * R3) / R1

Check:

Rout = ((1k + 3k) * 40) / 1k = 160 ohms

Considering the input network, Rin:

Rin = R1 + R2

For overall load, using conductance, easier to calculate parallel:

Geq = (1 / Rout) + (1 / Rin)

Geq = (R1 / ((R1 + R2) * R3)) + (1 / (R1 + R2))

Geq = 1 / ((1 / (R1 + R2)) * (1 + (R1 / R3)))

Req = 1 / Geq

Req = (R1 + R2) / (1 + (R1 / R3))

Check:

Req = (1k + 3k) / (1 + (1k / 40))

Req = 153.85

----------------------

Aww...i wanted to solve it without your hints. But a quick look at the schematic makes it obvious that Req (of the fet) is 120 ohms; took me a long 5 seconds (should have been less than one second, so my 70+ years seems to be slowing me down). Work? What work? ...

That statement is so significant, and so rarely understood...

Isaac

Maybe it's too easy. It is instantly obvious that Rfet has to be

3 times R3. So, an applicant may think it's a trick question, and be wracking his brains looking for the trick. OTOH, it could eliminate those who are not confident enough in their understanding of it to say "160 ohms in parallel with 4000 ohms" or "about 153.8 ohms".

Ed

Don't be so optimistic. Kids have a lot of trouble with controlled sources.

Tim

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

Indeed. National Semiconductor used to have an on-line course in op-amp circuit design, and this principle -- which should be the very first words out of the instructor's mouth -- is nowhere stated. Shame on you, Bob, shame on you.

In case the reason isn't obvious -- an ideal op-amp has infinite gain. If there were /any/ voltage difference between the inverting and non-inverting inputs, the op-amp's output would slam up against the positive or negative rail.

In practice, an op-amp has finite gain (usually between 100K and 1000K). This means the actual voltage difference has to be something other than zero. But it's is still so close to zero that it can be ignored for the purposes of analysis.

By the way, I cut my op-amp teeth nearly 40 years ago on the wonderful Philbrick brook. One of the greatest pieces of technical writing ever (I keep a copy for inspiration), and still a classic.

That "inputs the same thing" is good to think about when you run into integrators and differentiators too.... for the I the output is just having to drive against the cap none too quickly to keep the inputs from pulling away from each other while charging it. For D, the output gets slammed trying to kept up with the quick input change on the cap going from the input driving through the cap. My simple-minded way of thinking of it anyhow.

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.

NOW, what is less certain is the proper answer to the problem "Calculate the equivalent resistance of this programmable load." Given that R1, R2, and R3 are all part of the load, the proper answer to the original diagram is 153.846 ohms. Except that circuit does not show any evidence of being 'programmable'.

PlainBill

OOOhhhhh; you want a trick question based on real life? Ages ago i went to an interview for a job. They insisted that they had d'Arsonval meters that slowly decreased their sensitivity...toward zero (!!). The claim was that the Alnico magnets slowly lost their magnetism. I told them that was completely impossible; maybe a magnetic loss of one percent in a century, but not 90% in weeks. Did not get the job, because they "knew" they were right. I later found out those meters were in series with the plate of the RF final for their plywood dryer product line. Bypass on the low side of the meter was an electrolytic capacitor. Like in Groucho Mark's Bet your Life, when i heard that, the duck came down!

..like a few Curies of radium?

..

It is pretty easy if you know about how the op-amp will do whatever it can to make the voltage at pin 2 the same as pin 1.

Not if you have a title.

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.

You 'program' it by changing reistors.

--
Politicians should only get paid if the budget is balanced, and there is
enough left over to pay them.

"Slave" is a title.

;-)

On 10/24/2010 3:19 PM brent spake thus:

[snip quiz]

But is this really true? This sounds like it might be either a gross oversimplification or a possible falsehood.

DISCLAIMER: I'm just learning about this stuff. OK, I've been fooling around with electronics for, lessee, about 40 years now, but have had no formal training; I'm trying to rectify that by reading and studying.

So today I read up about op amps. Learned how a comparator works, generally speaking. How they differ from op amps (open loop).

My understanding of a comparator is that the output will be forced to one extreme or the other depending on the difference in voltage between noninverting and inverting inputs. Any significant voltage difference will drive the output to near the respective supply rail, positive or negative.

But I don't see how an op amp can, to quote brent, "do whatever it can to make the voltage[s the same]". After all, these are *inputs*, no? So at least in the case of a comparator, let's say that there exists a

1-volt difference between noninverting and inverting inputs (which I understand is a *huge* difference given the extremely high gain of the amp). Let's say the difference is positive: this will drive the output close to the + rail, correct? But the noninverting input will still be 1 volt positive w/respect to the inverting input, right? In other words, the change in output doesn't affect the inputs.

Now, this may be different in other configurations (operational or instrumentation amp), where there are connections between output and input instead of open loop. So is it true that in these cases the inputs will be forced to (near) equal? If so, how does that work?

This still sounds rather mysterious to me.

--
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)

It is a simplification, sort of. This simplification is assuming a few things:

1. It assumes a stable feedback arrangement. Knowing if something is stable is pretty deep into EE knowledge, but fortunately, using op- amps allows stability to be achieved in most cases without going through the stability analysis. This circuit feels stable by looking at it, but there are many simple ways that a feedback circuit can go unstable. But in a nutshell, anytime you have lots of gain (like in an op-amp) and you try to implement a feedback circuit with the gain (for real nice control and stability) you risk , or need to worry about instability. Instability is when the circuit reacts so fast and the delay is wrong and it can't make up its mind what to do so it oscillates back and forth. You can also just screw up and put the output on a rail too.

1. It is assuming an infinite gain opamp. This problem is a dc (static) problem, so in reality there might be a dc gain of 100000. This means that to achieve a voltage to control to FET there might be like 1 uVolt of difference in voltage between the two pins. Close enough to just say they are equal.

Correct.

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
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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

Sounds like you predate me -- but not by much. I cut my "op-amp teeth" on the earliest National Tech Notes (when they had that wonderful "NS" logo where both glyphs were identical, with one rotated and flipped).

Isaac