# _interview question

Discussion in 'Electronic Basics' started by [email protected], Oct 15, 2003.

1. ### Guest

Interview question*

---/\/\/\/\--
| b |
\/-----|\ |
| \________|___
---|+/
| |/
|
-----
---
-

Q:What is the input "impedance" at b?
A:Very low because it is a virtual earth.
Q:So how much current would you expect to flow into it?
A:Very little because op-amp inputs take "no current".
Q:So if it takes very little current, what is the "impedance"?
A:High?!
Q:But you just said it was low, if the gain of the op-amp is 100,000
and the feedback resistor is 10,000 ohms, what is the "impedance" at
b?

Robin

*not original: from a magazine, possibly a reader's letter, 1979-1980?

2. ### Keith R. WilliamsGuest

No, the *input* impedance is very high. Looking out of the op-amp into
the network it's very low.

What's the input impedance of the '+' input. It must be zero since it
is tied to the real earth, no? No.
Current flow in the *input*? very little. The current flows in the
resistors.
The impedance at b (.1ohms) is different than the "input impedance" of
the op-amp (a parameter of the op-amp alone). The current flows
through the resistors (equally, more or less), not into the op-amp.

3. ### John LarkinGuest

That's the fallacy. "Takes very little current" is meaningless.

John

4. ### Jim LargeGuest

Node (b) is connected to the input of the op-amp, but it
is also connected to the output of the op-amp through the
feedback resistor. Practically no current can flow into
or out of the op-amp's input pin, but quite a lot can flow
into or out of the output pin.

The question, "What is the impedance at node x" usually
means, "If you attach an external circuit to node x and
force a small change in voltage, what is the ratio of the
voltage difference to the amount of current that the
external circuit has to source or sink?"

Without actually doing the math, I'd be inclined to guess
that the impedance at node (b) is:

Zin || ( R + Zout )

Where; Zin is the input impedance of the op-amp, R is the
value of the feedback resistor, and Zout is the output
impedance of the op-amp. Zin is practically infinite,
Zout is practically zero, so the answer is just R --
the value of the feedback resistor.

-- Jim L.

5. ### Guest

To find the "impedance" at b, we need, from ohm's law, V / I = R i.e.:

delta V / delta I

at b. If 'A' is the op-amp gain then a small change in V (delta V)
gives a big change (A * delta V) at the op-amp's ouput.

This changes the current in the feedback resistor by (very nearly)

delta I == A * delta V / R

So now we have both delta I and delta V. Using V / I == R:-

delta V / (A * delta V / R) == Z at b == R / A

i.e.

Z == R / A == 10,000 / 100,000 == 0.1 ohms (approx)

Robin

*not original: from a magazine, possibly a reader's letter, 1979-1980?

6. ### Keith R. WilliamsGuest

You have to divide by the open loop gain of the op-amp. If the output
of the op-amp is at 1V, the '-' input of the op-amp is at 1V/gain.
Solve for the resistor divider:

Rf Z-
Vout 0-\/\/-----vin-----\/\/---gnd
1V 10K 1V/1E5 ?? 0V

And you get ~.1ohm impedance at the "virtual ground" point.

7. ### Guest

<snip>

In the original article, the interviewer writes the interviewee's
he don't think at all? Why is this?

Robin

8. ### John LarkinGuest

Only below the opamp's rolloff corner, typically 10 Hz or thereabouts.
Above that, the node looks like an inductor.

John

9. ### Keith R. WilliamsGuest

Oh, *now* you want to bring in the "unreal" part of the equation.
Oh, sure, make life complicated! ;-)

I do believe we have to have a *lot* more information here. Just
what is the "inductance" at this node, given the specifications?

10. ### Keith R. WilliamsGuest

I don't see it this way at all. I was taught that this was a
"virtual ground" (italics/quotes even). It's not ground, but the
op-amp is going to try its best to make it as close to ground as
it can. It's not real and it's not imaginary, it's "virtual".

11. ### John LarkinGuest

Your concerns are entirely imaginary. Try not to be so reactive.
It's a tough job, but I do my best.
Depends on the opamp's open-loop rolloff corner. Assuming 10 Hz and
100K gain to start, L is about 1.6 millihenries. So the summing node
looks roughly like 1.6 mH in series with 0.1 ohms.

Until the second opamp pole kicks in, of course; then it becomes a
negative resistor.

John

12. ### Guest

My bet is that the interviewer (predicted and) got the pat answer
above without the underlying knowledge so it annoyed him.

Whereas someone saying "low impedance...at the low frequencies" would,
at the very least, be in with a better chance of a job.

Another problem with "virtual earth" is consistency e.g.:-

-----
|
|
|
\ ---/\/\/\--
R / | |
\ | |\ |
/ --|-\ |
| | >-------
|---------|+/
| |/
\
R /
\
/
|
|
|
-----
---
-

What is this? A half-rail virtual-earth ?

Robin

13. ### Keith R. WilliamsGuest

Well, yeah! ...as long as you put it in quotes!

....or would you prefer "virtual short"?

14. ### Jim LargeGuest

O.K., Stupid question time. I've built that
circuit, but in my version, I used zero ohms
for the feedback (i.e., no resistor). Was I
doing something wrong?

-- Jim L.

15. ### Costas VlachosGuest

No, but it is often good practice to limit the current in the feedback path.
See the following quote from the CA3240 op-amp data sheet:

"Moreover, some current-limiting resistance should be provided between the
inverting input and the output when the CA3240 is used as a unity-gain
voltage follower. This resistance prevents the possibility of extremely
large input signal transients from forcing a signal through the input
protection network and directly driving the internal constant current source
which could result in positive feedback via the output terminal. A 3.9 kOhm
resistor is sufficient."

cheers,
Costas

16. ### Guest

Ah, yes, this is too simplified (I should have left a little bit of
input resistor there too). It was only mean't as a partial drawing of
an *unbuffered* half_rail "feed" but as you say, it looks like a
buffered half_rail supply...

many apologies
Robin

17. ### Rich GriseGuest

OK, I fucking give up. The Apostrophe Police have officially been
crushed into ignominious defeat.

"Against Stupidity, The Gods Themselves Contend in Vain."
- Dr. Isaac Asimov

18. ### Guest

Ah yes, this is too bad. I meant mean't to be meant. Poor A'police but
I expect they will soon recover.

many apologies
Robin pain

"The Gods only survive in the minds of the stupid."
Milos The Orea Polythrona.

19. ### Sir Charles W. Shults IIIGuest

Gee, Rich... I feel your pain. But it will never end, as you know.

Cheers!

Chip Shults
My robotics, space and CGI web page - http://home.cfl.rr.com/aichip

20. ### Guest

Milos the Orea Polithrona: 378 to 440 BC He invented the reverse
living method where you are born old and grow young. This allows an
old head to exist on a young body and also for the old to have their
cake and eat it.

In his youth he uttered the now famous declaration "The Gods only
exist in the minds of the stupid." He was promptly stoned to death by
some stupid people who later on claimed that he was struck by a
thunder bolt thus avoiding justice and proving that they were not so
stupid afterall.