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

R

RosemontCrest

Guest
I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&current=programmableload.jpg

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.

Thanks,
Brian
 
RosemontCrest wrote:
I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&current=programmableload.jpg

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.
Click to expand...

Then make your challenge on the proper newsgroup:
news:sci.electronics.design which is where you'll find the EEs.


--
Politicians should only get paid if the budget is balanced, and there is
enough left over to pay them.
 
"RosemontCrest" <rosemontcrest@yahoo.com> wrote in message
news:14528147-a327-4a43-a18b-de8eb69b88f7@l14g2000yqb.googlegroups.com...
I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&current=programmableload.jpg

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.

Thanks,
Brian
Click to expand...
4 * 40 ohms = 160 !! 4 kohms
 
In article
<14528147-a327-4a43-a18b-de8eb69b88f7@l14g2000yqb.googlegroups.com>,
RosemontCrest <rosemontcrest@yahoo.com> wrote:

I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&current=progr
ammableload.jpg

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.
Click to expand...
It's easier than either of the methods you propose; it's obvious by
inspection. Just notice that the two resistors in the op-amp's positive
input divider are in a 3:1 ratio, and so the same must be true for the
ones in the negative divider. So the FET must be acting as a 120 ohm
resistor.

Isaac
 
On Fri, 22 Oct 2010 23:04:26 -0400, Michael A. Terrell wrote:
RosemontCrest wrote:

I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&current=programmableload.jpg

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.

Then make your challenge on the proper newsgroup:
news:sci.electronics.design which is where you'll find the EEs.
Click to expand...
Sheesh! Graduate EEs can't get this? I'm "just a tech," but it took me
about 13 seconds to get the answer, assuming I remember correctly that
I=E/R:

((Vdc/4)/40) A.

Cheers!
Rich
 
On Oct 22, 8:04 pm, "Michael A. Terrell" <mike.terr...@earthlink.net>
wrote:
RosemontCrest wrote:

I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&curr....

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.

   Then make your challenge on the proper newsgroup:
news:sci.electronics.design which is where you'll find the EEs.

--
Politicians should only get paid if the budget is balanced, and there is
enough left over to pay them.
Click to expand...
Thank you Michael. I was not sure to which sci.electronics.* group I
should post. Thanks for adding sci.electronics.design to this thread.
 
On Oct 22, 8:31 pm, "tm" <the_obamun...@whitehouse.gov> wrote:
"RosemontCrest" <rosemontcr...@yahoo.com> wrote in message

news:14528147-a327-4a43-a18b-de8eb69b88f7@l14g2000yqb.googlegroups.com...



I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&curr....

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.

Thanks,
Brian

4 * 40 ohms = 160 !! 4 kohms
Click to expand...
It really is that simple, isn't it? My frustration that prompted my
post is that recent BSEE graduates don't seem to be able to answer
this question.
 
On Oct 22, 8:53 pm, RosemontCrest <rosemontcr...@yahoo.com> wrote:
On Oct 22, 8:31 pm, "tm" <the_obamun...@whitehouse.gov> wrote:



"RosemontCrest" <rosemontcr...@yahoo.com> wrote in message

news:14528147-a327-4a43-a18b-de8eb69b88f7@l14g2000yqb.googlegroups.com....

I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&curr...

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.

Thanks,
Brian

4 * 40 ohms = 160 !! 4 kohms

It really is that simple, isn't it? My frustration that prompted my
post is that recent BSEE graduates don't seem to be able to answer
this question.
Click to expand...

Graduates from where?

Nice carbine photos by the way.

(another) Michael
 
On Oct 22, 9:53 pm, Rich Grise <richgr...@example.net> wrote:
On Fri, 22 Oct 2010 23:04:26 -0400, Michael A. Terrell wrote:
RosemontCrest wrote:

I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&curr....

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.

   Then make your challenge on the proper newsgroup:
news:sci.electronics.design which is where you'll find the EEs.

Sheesh! Graduate EEs can't get this? I'm "just a tech," but it took me
about 13 seconds to get the answer, assuming I remember correctly that
I=E/R:

((Vdc/4)/40) A.

Cheers!
Rich
Click to expand...
It's not asking for I, but Req.

I agree with last poster as (160 * 4000) / (160 + 4000) ohms

But I would not get the answer under pressure, in an interview. I
guess I am one of the bad EEs.
 
On Fri, 22 Oct 2010 19:59:39 -0700 (PDT), RosemontCrest
<rosemontcrest@yahoo.com> wrote:

I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&current=programmableload.jpg

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.
Click to expand...

Since R2R1 is a voltage divider, the voltage on U1+ will be:

+VDC * R1
U1+ = -----------
R1 + R3

Now, since the voltage on U1- must be equal to the voltage on U1+, and
since R(Q1)R3 is another voltage divider,

(+VDC * R3) - ((U1+) * R3)
R(Q1) = ----------------------------
U1+

For the value, since the ratio of R2:R1 = 3 and the voltage across R1
and R3 are equal, the ratio of R(Q1) to R3 must also be 3, making the
FET's resistance 3 * R3 = 120 ohms.

---
JF
 
I'm posting this before looking at any of the other responses.

I'm always saying I'm smart because I look for the "fundamental principle"
underlying things. So... Let's see if I am...

This is a clever trick question. It assumes you understand /the/ basic rule
of op-amp circuit design -- if the circuit is stable, then the voltage
difference between the inverting and non-inverting inputs must be zero (or
in practice, vanishingly small).

Assuming Vdc is "stiff", then the voltage at the non-inverting input /must/
be Vdc/4. Right? The voltage at the inverting input /must/ be the same.
Ergo, the resistance of the JFET must be three times R3, or 120 ohms. Right,
too?

As the op amp draws no input current, the current through the JFET and R3
must be the same. Therefore, the load impedance must be 120 + 40.

QED?

Please note that only the most-trivial arithmetic is needed to solve the
problem. No fancy-shmancy algebra.

Bob Pease would be proud. I hope.



I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?


http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&current=programmableload.jpg

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.

Thanks,
Brian
Click to expand...
 
On Fri, 22 Oct 2010 23:31:04 -0400, "tm"
<the_obamunist@whitehouse.gov> wrote:

"RosemontCrest" <rosemontcrest@yahoo.com> wrote in message
news:14528147-a327-4a43-a18b-de8eb69b88f7@l14g2000yqb.googlegroups.com...
I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&current=programmableload.jpg

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.

Thanks,
Brian

4 * 40 ohms = 160 !! 4 kohms
Click to expand...

R2 Rx
---- = ---- = 120 ohms
R1 R3

---
JF
 
On Fri, 22 Oct 2010 23:04:26 -0400, Michael A. Terrell wrote:

RosemontCrest wrote:

I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&current=programmableload.jpg

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.


Then make your challenge on the proper newsgroup:
news:sci.electronics.design which is where you'll find the EEs.
Click to expand...
*Real* EEs only use nodal analysis where there isn't an easier alternative.

30 seconds inspection gives R effective = 160, in parallel with 4k =
153.85 ohms.

There's another issue, though. Depending on choice of JFET (most,
actually), the gate can go positive with respect to the source.

Nobody with any sense would use a JFET here. Rather a naďve example.

--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."
(Richard Feynman)
 
On Sat, 23 Oct 2010 05:02:38 -0500, John Fields wrote:

On Fri, 22 Oct 2010 19:59:39 -0700 (PDT), RosemontCrest
rosemontcrest@yahoo.com> wrote:

I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&current=programmableload.jpg

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.


Since R2R1 is a voltage divider, the voltage on U1+ will be:

+VDC * R1
U1+ = -----------
R1 + R3

Now, since the voltage on U1- must be equal to the voltage on U1+, and
since R(Q1)R3 is another voltage divider,

(+VDC * R3) - ((U1+) * R3)
R(Q1) = ----------------------------
U1+

For the value, since the ratio of R2:R1 = 3 and the voltage across R1
and R3 are equal, the ratio of R(Q1) to R3 must also be 3, making the
FET's resistance 3 * R3 = 120 ohms.
Click to expand...
It was the equivalent resistance of the *whole programmable load* that was
asked for.

I did it this way:

The voltage at u1 + input is 1/(3+1)*VDC =VDC/4

Feedback will make the voltage across R3 also VDC/4, hence the current in
Q1 and R3 will be VDC/(4*40*) = VDC/160

The resistance of that branch is therefore 160 ohms.

The whole load looks like 160ohms in parallel with 4000 ohms =
4000*160/(4000+160) = 153.85 ohms.

See my remarks in a previous post about Q1 gate going positive of its
source.

--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."
(Richard Feynman)
 
RosemontCrest <rosemontcrest@yahoo.com> wrote in
news:5e68a1c4-aaf0-45d4-b186-53e1734a953f@26g2000yqv.googlegroups.com:

On Oct 22, 8:04 pm, "Michael A. Terrell" <mike.terr...@earthlink.net
wrote:
RosemontCrest wrote:

I routinely use the following question to test candidates for EE or
TE positions. For many years, it continues to stump all but one of
many. Is it really that difficult to solve?
Click to expand...
No, it's trivial.

Since the voltage at the inputs of the opamp is 0.25 * Vdc, the
JFET drain/source current should be Vdc/160, or an equivalent
resistance of 160 ohms. The divider is in parallel.

So..

Req = 160 || 4000 ~= 153.8 ohms.


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
 
"Michael A. Terrell" <mike.terrell@earthlink.net> wrote in message
news:HaKdnQ3UkO1WzV_RnZ2dnUVZ_vednZ2d@earthlink.com...
RosemontCrest wrote:

I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&current=programmableload.jpg

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.
Click to expand...
To be honest I felt it wasn't obvious what the programmable load was. It's
also unusual to call a ground referenced load - a load, although I agree it
is, it does add to the confusion.

Also +VDC looks like a fixed voltage.

I would make it easier to interpret by saying:


VDC VDC
| |
circuit equiv to resistor
| |
GND GND


Apologies for non-fixed font if this gets messed up
 
"RosemontCrest" wrote in message
news:14528147-a327-4a43-a18b-de8eb69b88f7@l14g2000yqb.googlegroups.com...

I routinely use the following question to test candidates for EE
or TE
positions. For many years, it continues to stump all but one of
many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&current=programmableload.jpg

or

http://preview.tinyurl.com/2c8udf9

Thanks,
Brian
Click to expand...
To me question seems ambiguous. From a DC viewpoint, the JFET
needs to conduct enough current to make the voltage across 40
ohms equal 1/4 of VDC so the JFET have to be 120 ohms. From an AC
viewpoint, the JFET looks like zero ohms.

David
 
As I pointed out, this is a "trick" question intended to see whether the
applicant understands "the basic principle of op-amp circuit design".

When I briefly attended CalTech, one of our physics tests had a question
about the Doppler shift of a satellite transmitter passing directly
overhead. You'd be amazed at how many students wasted time calculating it.
 
Spehro Pefhany <speffSNIP@interlogDOTyou.knowwhat> wrote in
news:Xns9E1A6190B2594speffinterlogcom@69.16.186.50:

RosemontCrest <rosemontcrest@yahoo.com> wrote in
news:5e68a1c4-aaf0-45d4-b186-53e1734a953f@26g2000yqv.googlegroups.com:

On Oct 22, 8:04 pm, "Michael A. Terrell" <mike.terr...@earthlink.net
wrote:
RosemontCrest wrote:

I routinely use the following question to test candidates for EE
or TE positions. For many years, it continues to stump all but one
of many. Is it really that difficult to solve?

No, it's trivial.

Since the voltage at the inputs of the opamp is 0.25 * Vdc, the
JFET drain/source current should be Vdc/160, or an equivalent
resistance of 160 ohms. The divider is in parallel.

So..

Req = 160 || 4000 ~= 153.8 ohms.
Click to expand...
One of the "issues" with this circuit lies in the interpretation of
"ideal" for the op-amp. Here, I (and others) have ASS-U-MEd that
it has infinite gain and zero offset voltage in zero input bias
current, but also that it will swing negative using the single
+12/0 supply (for example, it might have a built-in charge-pump
voltage converter). Most real op-amps won't do that, they'll swing down
to somewhere near the lower rail. In which case, with a JFET,
the op-amp will not be able to balance until the current
exceeds Idss for the JFET. It also won't work much above Idss
(regardless of the op-amp functionality) because the gate will begin to
conduct, so it would have only a narrow range of operation over which it
"looks" like a fixed resistor.

It will also behave differently if "+VDC" happens to be a negative
voltage.

I don't think this is a very good "quiz" question, it leaves too
many questions open and uses non-standard nomenclature. The proper
answer to this one is probably "what are you trying to do?", the
subtext being "whatever it is, this probably isn't going to do it".


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
 
One of the "issues" with this circuit lies in the interpretation of
"ideal" for the op-amp. Here, I (and others) have ASS-U-MEd that
it has infinite gain and zero offset voltage in zero input bias
current, but also that it will swing negative using the single
+12/0 supply (for example, it might have a built-in charge-pump
voltage converter). Most real op-amps won't do that, they'll swing down
to somewhere near the lower rail. In which case, with a JFET,
the op-amp will not be able to balance until the current
exceeds Idss for the JFET. It also won't work much above Idss
(regardless of the op-amp functionality) because the gate will begin to
conduct, so it would have only a narrow range of operation over which it
"looks" like a fixed resistor.

It will also behave differently if "+VDC" happens to be a negative
voltage.

I don't think this is a very good "quiz" question, it leaves too
many questions open and uses non-standard nomenclature. The proper
answer to this one is probably "what are you trying to do?", the
subtext being "whatever it is, this probably isn't going to do it".
Click to expand...

I might be dead wrong, but it is a SUPERB question. It directly addresses
the question... "Does the applicant UNDERSTAND what goes on in a stable
op-amp circuit?"

The bottom line is that the "circuit" doesn't need the least bit of
"analysis" at all. If you understand that, in a stable circuit, the
inverting and non-inverting inputs have the same voltage on them, the
solution is utterly trivial.
 
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