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

[isw]

In article <ia141b$kad$1@news.eternal-september.org>,
"William Sommerwerck" <grizzledgeezer@comcast.net> wrote:

The principle is this... In a stable op-amp circuit, the
feedback forces the inverting and non-inverting inputs
/to have the same voltage/. The rest is trivial arithmetic.

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

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.

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

 

[William Sommerwerck]

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.

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

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.

 

[William Sommerwerck]

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

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).

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.

 

[John Fields]

On Sun, 24 Oct 2010 18:48:08 -0700, David Nebenzahl
<nobody@but.us.chickens> wrote:

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

[snip quiz]

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.

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.

---
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 >=10ľV 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.

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.

---
Correct.
---

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?

---
Yes
---

If so, how does that work?

---
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
---

This still sounds rather mysterious to me.

---
Even now?

---
JF

 

[krw@att.bizzzzzzzzzzzz]

On Mon, 25 Oct 2010 00:43:24 -0700, "William Sommerwerck"
<grizzledgeezer@comcast.net> wrote:

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.

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

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

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.

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.

 

[krw@att.bizzzzzzzzzzzz]

On Sun, 24 Oct 2010 18:48:08 -0700, David Nebenzahl <nobody@but.us.chickens>
wrote:

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

[snip quiz]

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.

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 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.

 

[Spehro Pefhany]

"krw@att.bizzzzzzzzzzzz" <krw@att.bizzzzzzzzzzzz> wrote in
news:mv9cc6pn3fclvj5k6certg2e1mj443p3d8@4ax.com:

On Mon, 25 Oct 2010 00:43:24 -0700, "William Sommerwerck"
grizzledgeezer@comcast.net> wrote:

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.

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

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

One of three principles:

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

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

 

[Spehro Pefhany]

"krw@att.bizzzzzzzzzzzz" <krw@att.bizzzzzzzzzzzz> wrote in
news:25acc616cisme9t9pgb37aojp1vv4r88fa@4ax.com:

On Sun, 24 Oct 2010 18:48:08 -0700, David Nebenzahl
nobody@but.us.chickens> wrote:

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

[snip quiz]

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.

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 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.

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

 

[isw]

In article <ia3ced$no0$1@news.eternal-september.org>,
"William Sommerwerck" <grizzledgeezer@comcast.net> wrote:

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.

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

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.

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

 

[ehsjr]

David Nebenzahl wrote:

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

[snip quiz]

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.


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.

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

 

[David Nebenzahl]

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

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.

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)

 

On Sun, 24 Oct 2010 21:14:00 -0400, "Michael A. Terrell"
<mike.terrell@earthlink.net> wrote:

PlainBill47@yawho.com wrote:

On Sat, 23 Oct 2010 13:05:23 -0700, "William Sommerwerck"
grizzledgeezer@comcast.net> wrote:

Any electronics technician should be able to solve this
by inspection; no calculator necessary. 3K/1k = x/40,
therefore x= 120 ohms. Come up with more difficult ones
next time.

What do you mean by "inspection"? Are you applying a formula you memorized?
Or do you /understand/ what's involved?
Probably better than you do. The voltage across R1 is 1/4 of +VDC. An 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.

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'.


You 'program' it by changing reistors.
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

 

[tm]

<PlainBill47@yawho.com> wrote in message
news:nvudc65qpifjekd8bv7o0hr4grtc3slntk@4ax.com...

On Sun, 24 Oct 2010 21:14:00 -0400, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


PlainBill47@yawho.com wrote:

On Sat, 23 Oct 2010 13:05:23 -0700, "William Sommerwerck"
grizzledgeezer@comcast.net> wrote:

Any electronics technician should be able to solve this
by inspection; no calculator necessary. 3K/1k = x/40,
therefore x= 120 ohms. Come up with more difficult ones
next time.

What do you mean by "inspection"? Are you applying a formula you
memorized?
Or do you /understand/ what's involved?
Probably better than you do. The voltage across R1 is 1/4 of +VDC. An
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.

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'.


You 'program' it by changing reistors.
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.

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

BTW, you didn't get the job.


tm

 

[Joel Koltner]

"David Nebenzahl" <nobody@but.us.chickens> wrote in message
news:4cc67064$0$2444$822641b3@news.adtechcomputers.com...

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.

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.

 

[David Nebenzahl]

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

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.

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 Nebenzahl]

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

PlainBill47@yawho.com> wrote in message
news:nvudc65qpifjekd8bv7o0hr4grtc3slntk@4ax.com...

On Sun, 24 Oct 2010 21:14:00 -0400, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:

You 'program' it by changing reistors.

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.

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

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".

BTW, you didn't get the job.

I didn't want it anyway.


--
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)

 

[John Fields]

On Tue, 26 Oct 2010 11:18:58 -0700, David Nebenzahl
<nobody@but.us.chickens> wrote:

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

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.

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 ...

---
David,

Did you get the article I posted for you?

---
JF

 

[tm]

"David Nebenzahl" <nobody@but.us.chickens> wrote in message
news:4cc71c48$0$2445$822641b3@news.adtechcomputers.com...

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

PlainBill47@yawho.com> wrote in message
news:nvudc65qpifjekd8bv7o0hr4grtc3slntk@4ax.com...

On Sun, 24 Oct 2010 21:14:00 -0400, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:

You 'program' it by changing reistors.

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.

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

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".

BTW, you didn't get the job.

I didn't want it anyway.

I was just pulling your chain. It was a shitty 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.

 

[whit3rd]

On Oct 25, 12:43 am, "William Sommerwerck"
<grizzledgee...@comcast.net> wrote:

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

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

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.

 

[krw@att.bizzzzzzzzzzzz]

On Tue, 26 Oct 2010 11:18:58 -0700, David Nebenzahl <nobody@but.us.chickens>
wrote:

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

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.

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 ...

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.