Op Amp Design
Dieter Knollman
the procedures are a mess. For negative gains, the inverting amplifier
formula Rf/Ri applies. For positive gains greater than one, the
non-inverting gain formula of 1+Rf/Ri applies. Multiples gains are ok only
if they are all negative. Other cases, such as mixed gains are considered
difficult. Is this complexity really needed?
In the mid 80s, when teaching a class at a local community college, I
discovered that the Rf/Ri formula could be used for ALL cases. You just
need to add a ground input if the gains do not add up to one. Rf is the
feedback resistor which is connected from the Op-Amp output to the "-"
input. Ri is the input resistor. Positive gain inputs connect to the "+"
input, negative gain inputs connect to the "-" input. It's really Dog Gone
Simple.
I have tried to get text authors to include this procedure in their next
edition. Typical response is "The students will get a better understanding
with the individual procedures."
The EET students that I had were not interested in theory. They just wanted
a simple procedure that worked.
The Design Procedure was published in EDN. Due to the poor editing, I'm
embarrassed to have my name appear on the article. Please ignore this
article.
The procedure was available on the web at Xoom and later at Nbci. It is now
available at http://members.save-net.com/dj...@save-net.com/k9analysis/.
Just click on Design.
A couple of disclaimers:
The Rf/Ri formula applies to all gains only if the circuit has been designed
to balance the inputs (minimize bias current errors). The procedure always
produces such a design. If you have an existing design, you may need to
apply a fudge factor to positive gains. Click on Plato at the k9analysis
site to get the complete gain formula.
The procedure does not always create designs that have the minimum number of
components. For example, an inverting amplifier will typically have the "+"
input grounded. Some Analog experts will argue "+" input should be
connected to ground via a resistor. The resistor is selected to balance the
inputs. The procedure automatically includes this resistor. If you don't
need balance, use Ozzie's rule to identify components that can be replaced
with shorts.
The main disadvantage of the procedure is that it is too simple. To guard
against users who occasionally hit the wrong key on their calculator, a
simple check is included.
The site is intended to be non-commercial. The procedure is free for
individual use. For other applications, I ask that a donation be made to a
Canine charity. If you like the procedure, rescue a four legged creature
from you local shelter.
Jim Thompson
"Dieter Knollman" <dj...@netzero.com>,
In Newsgroup: sci.electronics.design,
Article: <a6oh4h$kuj$1...@news.chatlink.com>,
Entitled: "Op Amp Design",
Wrote the following:
|If you use the Legacy Op-Amp design procedures, you must have noticed that
|the procedures are a mess. For negative gains, the inverting amplifier
|formula Rf/Ri applies. For positive gains greater than one, the
|non-inverting gain formula of 1+Rf/Ri applies. Multiples gains are ok only
|if they are all negative. Other cases, such as mixed gains are considered
|difficult. Is this complexity really needed?
|
Well, Duh! Dog-gone, that's for sure ;-)
Let's hear it for anything but understanding!
...Jim Thompson
--
(If replying by E-mail please observe obscure method of anti-spam.)
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| Jim-T@analog_innovations.com Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |
For proper E-mail replies SWAP "-" and "_", and remove the obvious.
"Things turn out best for those who make the best of how things turn out."
Helmut Sennewald
"Dieter Knollman" <dj...@netzero.com> schrieb im Newsbeitrag
news:a6oh4h$kuj$1...@news.chatlink.com...
> If you use the Legacy Op-Amp design procedures, you must have noticed that
> the procedures are a mess. For negative gains, the inverting amplifier
> formula Rf/Ri applies. For positive gains greater than one, the
> non-inverting gain formula of 1+Rf/Ri applies. Multiples gains are ok only
> if they are all negative. Other cases, such as mixed gains are considered
> difficult. Is this complexity really needed?
>
Hello Dieter,
I looked over your webpage and I have never seen so much complexity for
basically two easy formulas.
I cannot see any advantage by your approach and understand the
resistance of the EE-students. Shure, they don't have the knowledge
to judge your idea, but I think you have made that thing more complex
for shure.
Best Regards
Helmut
John Larkin
>If you use the Legacy Op-Amp design procedures, you must have noticed that
>the procedures are a mess. For negative gains, the inverting amplifier
>formula Rf/Ri applies. For positive gains greater than one, the
>non-inverting gain formula of 1+Rf/Ri applies. Multiples gains are ok only
>if they are all negative. Other cases, such as mixed gains are considered
>difficult. Is this complexity really needed?
Complexity? Where? If you understand how an opamp works (and *that* is
very simple) none of this is complex. If you don't understand it, you
do indeed have to memorize a million equations and hope you've chosen
the right one.
John
Jim Weir
verbal description of the opamp process, I've never before seen one that so
hopelessly complexes up the obvious.
Good Lord...
Jim
Jim Thompson
"Jim Weir" <j...@rst-engr.com>,
In Newsgroup: sci.electronics.design,
Article: <u8vlm8j...@corp.supernews.com>,
Entitled: "Re: Op Amp Design",
Wrote the following:
|I must admit, in all the years I've spent grading papers that ask for a
Amen, Brother ;-)
Kevin Aylward
"Dieter Knollman" <dj...@netzero.com> wrote in message
news:a6ohgb$l57$1...@news.chatlink.com...
> If you use the Legacy Op-Amp design procedures, you must have noticed that
> the procedures are a mess. For negative gains, the inverting amplifier
> formula Rf/Ri applies. For positive gains greater than one, the
> non-inverting gain formula of 1+Rf/Ri applies. Multiples gains are ok only
> if they are all negative. Other cases, such as mixed gains are considered
> difficult. Is this complexity really needed?
>
> In the mid 80s, when teaching a class at a local community college, I
> discovered that the Rf/Ri formula could be used for ALL cases. You just
> need to add a ground input if the gains do not add up to one. Rf is the
> feedback resistor which is connected from the Op-Amp output to the "-"
> input. Ri is the input resistor. Positive gain inputs connect to the "+"
> input, negative gain inputs connect to the "-" input. It's really Dog Gone
> Simple.
>
> I have tried to get text authors to include this procedure in their next
> edition. Typical response is "The students will get a better understanding
> The EET students that I had were not interested in theory. They just wanted
> a simple procedure that worked.
>
> The Design Procedure was published in EDN. Due to the poor editing, I'm
> embarrassed to have my name appear on the article. Please ignore this
> article.
> The procedure was available on the web at Xoom and later at Nbci. It is now
> available at http://members.save-net.com/dj...@save-net.com/k9analysis/.
> Just click on Design.
>
> A couple of disclaimers:
> The Rf/Ri formula applies to all gains only if the circuit has been designed
> to balance the inputs (minimize bias current errors). The procedure always
> produces such a design. If you have an existing design, you may need to
> gain formula.
> The procedure does not always create designs that have the minimum number of
> components. For example, an inverting amplifier will typically have the "+"
> input grounded. Some Analog experts will argue "+" input should be
> connected to ground via a resistor. The resistor is selected to balance the
> inputs. The procedure automatically includes this resistor. If you don't
> need balance, use Ozzie's rule to identify components that can be replaced
> with shorts.
>
> The main disadvantage of the procedure is that it is too simple. To guard
> against users who occasionally hit the wrong key on their calculator, a
> simple check is included.
>
> The site is intended to be non-commercial. The procedure is free for
> individual use. For other applications, I ask that a donation be made to a
> Canine charity. If you like the procedure, rescue a four legged creature
> from you local shelter.
>
What I will say here is that you are probably unaware just how real, complex
analogue design gets done in practise, and why explicit expressions for gains of
said complicated circuits, are to all an intents and purposes, useless.
Kevin Aylward , Warden of the Kings Ale
ke...@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
Ken Smith
Dieter Knollman <dj...@netzero.com> wrote:
>If you use the Legacy Op-Amp design procedures, you must have noticed that
>the procedures are a mess. For negative gains, the inverting amplifier
>formula Rf/Ri applies. For positive gains greater than one, the
>non-inverting gain formula of 1+Rf/Ri applies. Multiples gains are ok only
>if they are all negative. Other cases, such as mixed gains are considered
>difficult. Is this complexity really needed?
There is no complexity that needs elimination. All you need to know, for
90% of op-amp design, is ohms law and how an op-amp works. If you are
teaching your students anything else you are doing them harm. Turning a
simple situation of understanding into a complex memorization project is
one of the worst things you can do to a student. Once the students
understand how to deal with op-amps and resistors you can introduce
capacitors into the circuit and teach them to do filters etc.
Your students should have no problem with this circuit:
6.8K 4.7K
1V -----/\/\/\/----------\/\/\/\/------- Vout
! ! !
! ! !\ !
! --!-\ !
! ! --------------
!----------------!+/
!/
If they can't tell you what Vout is in less than 2 minutes you have
failed them.
--
--
kens...@rahul.net forging knowledge
Jim Thompson
Ken Smith <kens...@rahul.net>,
In Newsgroup: sci.electronics.design,
Article: <a6ql1s$skp$1...@samba.rahul.net>,
Entitled: "Re: Op Amp Design",
Wrote the following:
|In article <a6oh4h$kuj$1...@news.chatlink.com>,
Now, now, Ken. You've done gone and confused the poor bastard... he
doesn't have a ready-made equation ;-)
And I want the students to answer the question for arbitrary values:
Replace 6.8K with R1 and 4.7K with R2, 1V with Vin; and write me an
equation for Vout versus Vin <*very* big grin>
Goran Tomas
<Jim-T@analog_innovations.com.invalid> wrote:
>| 6.8K 4.7K
>| 1V -----/\/\/\/----------\/\/\/\/------- Vout
>| ! ! !
>| ! ! !\ !
>| ! --!-\ !
>| ! ! --------------
>| !----------------!+/
>| !/
>|
>|
>|If they can't tell you what Vout is in less than 2 minutes you have
>|failed them.
>
>doesn't have a ready-made equation ;-)
>
>And I want the students to answer the question for arbitrary values:
>Replace 6.8K with R1 and 4.7K with R2, 1V with Vin; and write me an
>equation for Vout versus Vin <*very* big grin>
OK, I'll take the bait (I hope I'll learn something :)
My answer to Ken's question would be: Vout=0
And to Jim's generalization: Vout=R2/R1*[V(+)-V(-)]
So?
Kind regards,
Goran Tomas
Jim Thompson
goran...@radio101.hr (Goran Tomas),
In Newsgroup: sci.electronics.design,
Article: <3c91117b...@news.hinet.hr>,
Entitled: "Re: Op Amp Design",
Wrote the following:
|On Thu, 14 Mar 2002 10:45:22 -0700, Jim Thompson
Sorry! You don't get to be this summer's intern ;-) Your answers
were incorrect for both questions.
Klaus Vestergaard Kragelund
news:3c91117b...@news.hinet.hr...
hmm - my guess would be something like Vout = Vin! (or Vout = 1Volt in the
first case)
Thanks
Klaus
>
>
> Kind regards,
> Goran Tomas
Helmut Sennewald
"Jim Thompson" <Jim-T@analog_innovations.com.invalid> schrieb im Newsbeitrag
news:1n529ucgea4vob6g7...@4ax.com...
Hello,
thanks for the circuit. I will keep this circuit in mind for the
next job applicant.
By the way, there is really an application based on this circuit.
This would then be the second question to the applicant:
What happens if the switch is open or closed?
10k 10k
1V -----/\/\/\/----------\/\/\/\/------- Vout
! ! !
! ! !\ !
! --!-\ !
! 10k ! --------------
!--/\/\/\/-------!+/
! !/
!
/ switch
|
---
Best Regards
Helmut
Jim Thompson
"Helmut Sennewald" <HelmutS...@t-online.de>,
In Newsgroup: sci.electronics.design,
Article: <a6r5qm$3rr$04$1...@news.t-online.com>,
I first used that trick around 1973. I'll not give away the answer.
Let's see how many intern-quality people there are out there ;-)
Frank Bemelman
"Jim Thompson" <Jim-T@analog_innovations.com.invalid> schreef in bericht
news:k4o19ukg7keb8snm3...@4ax.com...
> On 14 Mar 2002 17:01:48 GMT,
> Ken Smith <kens...@rahul.net>,
> In Newsgroup: sci.electronics.design,
> Article: <a6ql1s$skp$1...@samba.rahul.net>,
> Entitled: "Re: Op Amp Design",
> Wrote the following:
>
> |
> |
> | 6.8K 4.7K
> | 1V -----/\/\/\/----------\/\/\/\/------- Vout
> | ! ! !
> | ! ! !\ !
> | ! --!-\ !
> | ! ! --------------
> | !----------------!+/
> | !/
> |
> |
> |If they can't tell you what Vout is in less than 2 minutes you have
> |failed them.
> |
> |
> |--
>
> Now, now, Ken. You've done gone and confused the poor bastard... he
> doesn't have a ready-made equation ;-)
>
> And I want the students to answer the question for arbitrary values:
> Replace 6.8K with R1 and 4.7K with R2, 1V with Vin; and write me an
> equation for Vout versus Vin <*very* big grin>
I'm not a candidate for your summer's home entertainer, since these
kind of questions are at the limits of my capabilities, or even
beyond <G> , if the answer is wrong:
Vout = 1,69V
Vout = Vin * (1 + (R2/R1))
--
Thanks,
Frank Bemelman
(remove 'x' & .invalid when sending email)
Jim Thompson
"Frank Bemelman" <beme...@euronet.nl.invalid>,
In Newsgroup: sci.electronics.design,
Article: <a6r6gr$1epd$1...@scavenger.euro.net>,
Entitled: "Re: Op Amp Design",
Wrote the following:
[snip]
|I'm not a candidate for your summer's home entertainer, since these
|kind of questions are at the limits of my capabilities, or even
|beyond <G> , if the answer is wrong:
|
|Vout = 1,69V
|Vout = Vin * (1 + (R2/R1))
Frank! Frank! Frank! Back to school with you (or at least to the
lab :-)
Frank Bemelman
"Jim Thompson" <Jim-T@analog_innovations.com.invalid> schreef in bericht
> On Thu, 14 Mar 2002 22:48:04 +0100,
> "Helmut Sennewald" <HelmutS...@t-online.de>,
> In Newsgroup: sci.electronics.design,
> Article: <a6r5qm$3rr$04$1...@news.t-online.com>,
> Entitled: "Re: Op Amp Design",
> Wrote the following:
>
> |thanks for the circuit. I will keep this circuit in mind for the
> |next job applicant.
> |
> |By the way, there is really an application based on this circuit.
> |This would then be the second question to the applicant:
> |What happens if the switch is open or closed?
> |
> | 10k 10k
> | 1V -----/\/\/\/----------\/\/\/\/------- Vout
> | ! ! !
> | ! ! !\ !
> | ! --!-\ !
> | ! 10k ! --------------
> | !--/\/\/\/-------!+/
> | ! !/
> | !
> | / switch
> | |
> | ---
> |
> |Best Regards
> |Helmut
> |
> |
>
> I first used that trick around 1973. I'll not give away the answer.
> Let's see how many intern-quality people there are out there ;-)
>
Increase the gain by 1, if the switch is open?
Is that all it takes, to become an intern?
Jim Thompson
"Frank Bemelman" <beme...@euronet.nl.invalid>,
In Newsgroup: sci.electronics.design,
Article: <a6r6pj$1i05$1...@scavenger.euro.net>,
Frank, You need to stop and think for a minute... you're shooting from
the hip with your eyes crossed ;-)
Spehro Pefhany
> Increase the gain by 1, if the switch is open?
>
> Is that all it takes, to become an intern?
<Bzzzzzzzzt>
Best regards,
--
Spehro Pefhany --"it's the network..." "The Journey is the reward"
sp...@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
9-11 United we Stand
Frank Bemelman
"Jim Thompson" <Jim-T@analog_innovations.com.invalid> schreef in bericht
> On Thu, 14 Mar 2002 22:59:47 +0100,
> "Frank Bemelman" <beme...@euronet.nl.invalid>,
> In Newsgroup: sci.electronics.design,
> Article: <a6r6gr$1epd$1...@scavenger.euro.net>,
> Entitled: "Re: Op Amp Design",
> Wrote the following:
>
> [snip]
> |I'm not a candidate for your summer's home entertainer, since these
> |kind of questions are at the limits of my capabilities, or even
> |beyond <G> , if the answer is wrong:
> |
> |Vout = 1,69V
> |Vout = Vin * (1 + (R2/R1))
>
> Frank! Frank! Frank! Back to school with you (or at least to the
> lab :-)
I didn't want to make it look too easy for the others ;)
Eh, Vout is -0.31V and Vout = Vin * ((R2/R1)-1)
BTW, this is all brainwork <G>, no lab at home, no simulator
too.
Spehro Pefhany
> BTW, this is all brainwork <G>, no lab at home, no simulator
> too.
*simulator* ???
Jim Thompson
"Frank Bemelman" <beme...@euronet.nl.invalid>,
In Newsgroup: sci.electronics.design,
Entitled: "Re: Op Amp Design",
Wrote the following:
[snip]
|> [snip]
|> |I'm not a candidate for your summer's home entertainer, since these
|> |kind of questions are at the limits of my capabilities, or even
|> |beyond <G> , if the answer is wrong:
|> |
|> |Vout = 1,69V
|> |Vout = Vin * (1 + (R2/R1))
|>
|> Frank! Frank! Frank! Back to school with you (or at least to the
|> lab :-)
|
|I didn't want to make it look too easy for the others ;)
|Eh, Vout is -0.31V and Vout = Vin * ((R2/R1)-1)
|BTW, this is all brainwork <G>, no lab at home, no simulator
|too.
No Algebra, or intuition ????
Think.
Think fundamentals.
Like inductors can't have an instantaneous change in current.
Like capacitors can't have an instantaneous change in voltage.
And OpAmps with feedback that are in their linear operating region ??
Cone Killer
"Frank Bemelman" <beme...@euronet.nl.invalid> wrote in message
news:a6r762$1kue$1...@scavenger.euro.net...
Regards, Doug
Rich Grise
Going by the "seat of the pants" explanation of opamp behavior
that I heard in some class or another (or read somewhere),
an opamp does whatever it needs to do at its output to
make the inputs equal. So you've got +1V at +in, it wants
+1v at -in. Lo and behold, without the 4K7, there's already
1V at -in. So there's no current through the 4K7, so Vout
is 1V.
--
Cheers!
Rich
"We have met the enemy, and he is us!"
- Pogo Possum, ca. 1950's
Jim Thompson
Rich Grise <rich...@earthlink.net>,
In Newsgroup: sci.electronics.design,
Article: <3C9124...@earthlink.net>,
Entitled: "Re: Op Amp Design",
Wrote the following:
|Ken Smith wrote:
That's the FUNDAMENTAL OpAmp behavior that I was referring to in a
earlier post.
Frank Bemelman
"Jim Thompson" <Jim-T@analog_innovations.com.invalid> schreef in bericht
...have 0 volt between + and - inputs. So it's a gain of -1!
Hmm, was that within 2 minutes, I didn't look at the clock, you see ;)
Jim Thompson
In Newsgroup: sci.electronics.design,
Article: <k4o19ukg7keb8snm3...@4ax.com>,
Entitled: "Re: Op Amp Design",
Wrote the following:
[snip]
|
|And I want the students to answer the question for arbitrary values:
|Replace 6.8K with R1 and 4.7K with R2, 1V with Vin; and write me an
|equation for Vout versus Vin <*very* big grin>
|
| ...Jim Thompson
And the winners are (in order of posting):
Klaus Vestergaard Kragelund
Helmut Sennewald
Rich Grise
Frank Bemelman
"Cone Killer" <NOS...@hotmail.com> schreef in bericht
news:Qr9k8.13258$4d.18...@news1.west.cox.net...
No! that was the first answer I gave ;)
Frank Bemelman
"Jim Thompson" <Jim-T@analog_innovations.com.invalid> schreef in bericht
> And the winners are (in order of posting):
There are 8 winners, 4 of them are under a small roof,
and the other 4 only have 1 umbrella. They don't get
wet, how is that possible?
Andreas Hadler
>And the winners are (in order of posting):
>
>Klaus Vestergaard Kragelund
>Helmut Sennewald
>Rich Grise
And Frank, if you'll come to the CEBIT this year, I'll offer you the
consolation prize for the lousiest hip-shooting ever seen ;-)
cu, Andreas
Andreas...@t-online.de---------------------------------------+
|The trouble with computers is: |
|they do what you tell them to do, not what you want them to do. |
+----------------------------------------------------------------+
Andreas Hadler
>
>"Jim Thompson" <Jim-T@analog_innovations.com.invalid> schreef in bericht
>news:ml929u8can80h7epc...@4ax.com...
>
>> And the winners are (in order of posting):
>
>There are 8 winners, 4 of them are under a small roof,
>and the other 4 only have 1 umbrella. They don't get
>wet, how is that possible?
What do they do with the umbrella, when there's no rain at all?...
cu, Andreas
--
Jim Thompson
Andreas Hadler <Andreas...@t-online.de>,
In Newsgroup: sci.electronics.design,
Article: <4u929ukhdnc43kdnh...@4ax.com>,
Entitled: "Re: Op Amp Design",
Wrote the following:
|Jim Thompson <Jim-T@analog_innovations.com.invalid> wrote:
ROTFLMAO!
Cone Killer
"Frank Bemelman" <beme...@euronet.nl.invalid> wrote in message
I did notice MY error though. The equation
should be Vi*((Rf/Ri) + 1). I said R1. Only
R2 enters in the gain equation so I should
have said V0 = Vi * ((R2/Ri) + 1) where
Ri = sideways 8.
Sideways 8 is not on my keypad (Infinity)
That is my reasoning, right or wrong. The
hard answer is still +1V.
Regards, Doug
Jim Thompson
Jim Thompson <Jim-T@analog_innovations.com.invalid>,
In Newsgroup: sci.electronics.design,
Entitled: "Re: Op Amp Design",
Wrote the following:
|On Thu, 14 Mar 2002 10:45:22 -0700,
|Jim Thompson <Jim-T@analog_innovations.com.invalid>,
|In Newsgroup: sci.electronics.design,
|Article: <k4o19ukg7keb8snm3...@4ax.com>,
|Entitled: "Re: Op Amp Design",
|Wrote the following:
|
|[snip]
||
||And I want the students to answer the question for arbitrary values:
||Replace 6.8K with R1 and 4.7K with R2, 1V with Vin; and write me an
||equation for Vout versus Vin <*very* big grin>
||
|| ...Jim Thompson
|
|And the winners are (in order of posting):
|
|Klaus Vestergaard Kragelund
|Helmut Sennewald
|Rich Grise
|
Actually one more: "Cone Killer" Doug, His
"Vo = Vi ((R2/R1) + 1) (R1 = sideways 8)" is also correct in an
oblique sort of way ;-)
Frank Bemelman
"Andreas Hadler" <Andreas...@t-online.de> schreef in bericht
news:eja29u4flm7b4umlj...@4ax.com...
> "Frank Bemelman" <beme...@euronet.nl.invalid> wrote:
>
> >
> >"Jim Thompson" <Jim-T@analog_innovations.com.invalid> schreef in bericht
> >news:ml929u8can80h7epc...@4ax.com...
> >
> >> And the winners are (in order of posting):
> >
> >There are 8 winners, 4 of them are under a small roof,
> >and the other 4 only have 1 umbrella. They don't get
> >wet, how is that possible?
>
> What do they do with the umbrella, when there's no rain at all?...
Easier than opamps eh? I heard this one yesterday, solved it
in less than a second. But I need to practice a bit more
on opamps, after all, I *do* use them occasionally. And rather
succesfully, really ;) Fun stuff!
Frank Bemelman
"Jim Thompson" <Jim-T@analog_innovations.com.invalid> schreef in bericht
> On Thu, 14 Mar 2002 23:46:23 +0100,
> Andreas Hadler <Andreas...@t-online.de>,
> In Newsgroup: sci.electronics.design,
> Article: <4u929ukhdnc43kdnh...@4ax.com>,
> Entitled: "Re: Op Amp Design",
> Wrote the following:
>
> |Jim Thompson <Jim-T@analog_innovations.com.invalid> wrote:
> |
> |>And the winners are (in order of posting):
> |>
> |>Klaus Vestergaard Kragelund
> |>Helmut Sennewald
> |>Rich Grise
> |
> |And Frank, if you'll come to the CEBIT this year, I'll offer you the
> |consolation prize for the lousiest hip-shooting ever seen ;-)
> |
> |cu, Andreas
> |
> |Andreas...@t-online.de---------------------------------------+
> ||The trouble with computers is: |
> ||they do what you tell them to do, not what you want them to do. |
> |+----------------------------------------------------------------+
>
> ROTFLMAO!
I don't know what the prize is, otherwise I may go and collect it ;)
Andreas Hadler
Not really. I had to think. And not about a precarious hidden
intellectual pitfall.
BTW: Though there's no rain in Hannover right now, an umbrella will be
ok in this lousy wheather here.
Cone Killer
"Cone Killer" <NOS...@hotmail.com> wrote in message
news:dZ9k8.13337$4d.18...@news1.west.cox.net...
Geez, I need to pay attention. You did indeed give
+1 as your first answer. The rest of my explanation
still stands.
Incidentally, in the 'real' world one can not expect
Ri to equal infinity. Without proper board cleaning
I have seen this value < 1 Mohm, which does affect
the performance of the circuit in many applications,
obliquely or otherwise.
From 'sunny' southern CA.
Best Regards, Doug
(ref. AoE 2nd Edition, page 178, par.. 4.05)
Andreas Hadler
Ever seen such a fast thread?
Tony Williams
Ken Smith <kens...@rahul.net> wrote:
> Your students should have no problem with this circuit:
> 6.8K 4.7K
> 1V -----/\/\/\/----------\/\/\/\/------- Vout
> ! ! !
> ! ! !\ !
> ! --!-\ !
> ! ! --------------
> !----------------!+/
> !/
And when they've done that.......
R R
Vin ---+-/\/\/\/-----+----\/\/\/\/-------+ Vout
| ! !
| ! !\ !
\ --!-\ !
/ ! --------------
RV1= R \<---------------!+/
/ /|\ !/
\ k*R
/ |
| \|/
--+----------------------------------0v
RV1= a pot, overall value R.
k= pot position, varying from 0 to 1.
--
Tony Williams.
Jim Thompson
Tony Williams <to...@ledelec.demon.co.uk>,
In Newsgroup: sci.electronics.design,
Article: <4b174dc...@ledelec.demon.co.uk>,
Entitled: "Re: Op Amp Design",
Wrote the following:
[snip]
| And when they've done that.......
|
| R R
| Vin ---+-/\/\/\/-----+----\/\/\/\/-------+ Vout
| | ! !
| | ! !\ !
| \ --!-\ !
| / ! --------------
| RV1= R \<---------------!+/
| / /|\ !/
| \ k*R
| / |
| | \|/
| --+----------------------------------0v
|
| RV1= a pot, overall value R.
|
| k= pot position, varying from 0 to 1.
I won't give an answer... it's *much* more amusing to watch what
others post ;-)
John Larkin
>
>"Jim Thompson" <Jim-T@analog_innovations.com.invalid> schrieb im Newsbeitrag
>news:1n529ucgea4vob6g7...@4ax.com...
>> On Thu, 14 Mar 2002 21:14:33 GMT,
>> goran...@radio101.hr (Goran Tomas),
>> In Newsgroup: sci.electronics.design,
>> Article: <3c91117b...@news.hinet.hr>,
>> Entitled: "Re: Op Amp Design",
>> Wrote the following:
>>
>> |On Thu, 14 Mar 2002 10:45:22 -0700, Jim Thompson
>> |<Jim-T@analog_innovations.com.invalid> wrote:
>> |>| 6.8K 4.7K
>> |>| 1V -----/\/\/\/----------\/\/\/\/------- Vout
>> |>| ! ! !
>> |>| ! ! !\ !
>> |>| ! --!-\ !
>> |>| ! ! --------------
>> |>| !----------------!+/
>> |>| !/
>> |>|
>> |>|
>> |>|If they can't tell you what Vout is in less than 2 minutes you have
>> |>|failed them.
>> |>
>> |>Now, now, Ken. You've done gone and confused the poor bastard... he
>> |>doesn't have a ready-made equation ;-)
>> |>
>> |>And I want the students to answer the question for arbitrary values:
>> |>Replace 6.8K with R1 and 4.7K with R2, 1V with Vin; and write me an
>> |>equation for Vout versus Vin <*very* big grin>
>> |
>> |OK, I'll take the bait (I hope I'll learn something :)
>> |
>> |My answer to Ken's question would be: Vout=0
>> |
>> |And to Jim's generalization: Vout=R2/R1*[V(+)-V(-)]
>> |
>> |So?
>> |
>> |
>> |Kind regards,
>> |Goran Tomas
>>
>> Sorry! You don't get to be this summer's intern ;-) Your answers
>> were incorrect for both questions.
>>
>
>Hello,
>thanks for the circuit. I will keep this circuit in mind for the
>next job applicant.
>
>By the way, there is really an application based on this circuit.
>This would then be the second question to the applicant:
>What happens if the switch is open or closed?
>
> 10k 10k
> 1V -----/\/\/\/----------\/\/\/\/------- Vout
> ! ! !
> ! ! !\ !
> ! --!-\ !
> ! 10k ! --------------
> !--/\/\/\/-------!+/
> ! !/
> !
> / switch
> |
> ---
>
>Best Regards
>Helmut
>
>
Helmut,
synchronous demodulator!
John
John Larkin
Jim,
the intern situation looks pretty grim for this coming summer.
John
John Larkin
<Jim-T@analog_innovations.com.invalid> wrote:
>On Thu, 14 Mar 2002 10:45:22 -0700,
>Jim Thompson <Jim-T@analog_innovations.com.invalid>,
>In Newsgroup: sci.electronics.design,
>Article: <k4o19ukg7keb8snm3...@4ax.com>,
>Entitled: "Re: Op Amp Design",
>Wrote the following:
>
>[snip]
>|
>|And I want the students to answer the question for arbitrary values:
>|Replace 6.8K with R1 and 4.7K with R2, 1V with Vin; and write me an
>|equation for Vout versus Vin <*very* big grin>
>|
>| ...Jim Thompson
>
>And the winners are (in order of posting):
>
>Klaus Vestergaard Kragelund
>Helmut Sennewald
>Rich Grise
>
> ...Jim Thompson
Jim,
my favorite quickie quiz is this:
+10v
|
|
C
+5V-------------B <== NPN transistor
E
|
|
R = 1K
|
|
GND
OK, just roughly, without serious calculating,...
what's the base voltage?
what's the collector voltage?
what's the emitter voltage?
what's the emitter current?
what's the base current?
any other comments?
I get the most amazing statements, even from EEs with loaded resumes.
Even to the first two questions! This one has made some very smug
people get very shook up, fast.
John
Jim Thompson
John Larkin <jjlarkin@highlandSNIP_THIStechnology.com>,
In Newsgroup: sci.electronics.design,
Article: <zEeRPF3zxLaQ56...@4ax.com>,
Entitled: "Re: Op Amp Design",
Wrote the following:
[snip]
|Jim,
|
|my favorite quickie quiz is this:
|
|
|
| +10v
| |
| |
| C
| +5V-------------B <== NPN transistor
| E
| |
| |
| R = 1K
| |
| |
| GND
|
|
|
|OK, just roughly, without serious calculating,...
|
|what's the base voltage?
|
|what's the collector voltage?
|
|what's the emitter voltage?
|
|what's the emitter current?
|
|what's the base current?
|
|any other comments?
|
|
|I get the most amazing statements, even from EEs with loaded resumes.
|Even to the first two questions! This one has made some very smug
|people get very shook up, fast.
|
|John
|
My favorite....
+----+-----> +VCC
| |
R |
| |
| C
+---B <== NPN transistors
| E
C |
B---+-------> VOUT
E |
| |
| |
R R
| |
| |
GND GND
Assuming BETA is large enough to ignore base currents what is the
expression for VOUT and what is its TC?
Ian Field
"John Larkin" <jjlarkin@highlandSNIP_THIStechnology.com> wrote in message
news:zEeRPF3zxLaQ56...@4ax.com...
---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
Version: 6.0.338 / Virus Database: 189 - Release Date: 3/15/02
Christopher R. Carlen
> | And when they've done that.......
> |
> | R R
> | Vin ---+-/\/\/\/-----+----\/\/\/\/-------+ Vout
> | | ! !
> | | ! !\ !
> | \ --!-\ !
> | / ! --------------
> | RV1= R \<---------------!+/
> | / /|\ !/
> | \ k*R
> | / |
> | | \|/
> | --+----------------------------------0v
> |
> | RV1= a pot, overall value R.
> |
> | k= pot position, varying from 0 to 1.
>
> I won't give an answer... it's *much* more amusing to watch what
> others post ;-)
>
> ...Jim Thompson
Yeah, I can't wait to see what Frank B. will post.
Here's my take:
Let:
Vn be the voltage on the inverting amplifier input node.
Vp is the voltage on the non-inverting amplifier input node.
We have Vp = k * Vin from the voltage divider.
Then:
Vn = Vp = k * Vin for the ideal amp operating in the linear region.
The sum of the currents leaving the node Vn must be zero, so
(Vn - Vout)/R + (Vn - Vin)/R = 0 , or
Vout = 2 * Vn - Vin
Substituting the expression Vn = k * Vin gives:
Vout = 2 * k * Vin - Vin , simplified:
Vout = Vin * (2 * k - 1)
Now does it make sense?
If k = 0 then the circuit looks like an inverting amplifier. In that
case my equation gives Vout = -Vin, which agrees with what we know about
an inverting amp with equal input and feedback resistors. Why does this
happen qualitatively? Since the amp is operating linearly, the Vout
must be such that the node Vn = 0. This will occur when the output
voltage is equal in magnitude but opposite in sign to the input voltage,
because the input and feedback resistors are equal, forming a voltage
divider between Vin and Vout.
If k = 1 then the circuit looks like the thing that started this whole
ridiculous thread. In that case my equation gives Vout = Vin. That
makes sense because the linear amp has Vn = Vp as always, so if Vp = 1
then Vn = 1 as well. If this is so then no current flows in the input
resistor. That means that no current flows in the feedback resistor as
well, so Vout = Vn = Vin.
The interesting case occurs when k = 1/2. Then the equation predicts
Vout = 0 no matter what the input. This is logical as well considering
the voltage divider action of the input and feedback resistors.
Good day!
____________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA
crc...@sandia.gov
Terran Melconian
Frank Bemelman <beme...@euronet.nl.invalid> wrote:
>Increase the gain by 1, if the switch is open?
For very large values of "1"?
qrk
You failed Rich. Your explanation made sense. Try to obscure your
answer in lots of steps and arithmetic and then report back with a 10
page document to explain your procedure(s).
-
Mark Chun
Santa Barbara, CA
Bob Penoyer
<HelmutS...@t-online.de> wrote:
>By the way, there is really an application based on this circuit.
>This would then be the second question to the applicant:
>What happens if the switch is open or closed?
>
> 10k 10k
> 1V -----/\/\/\/----------\/\/\/\/------- Vout
> ! ! !
> ! ! !\ !
> ! --!-\ !
> ! 10k ! --------------
> !--/\/\/\/-------!+/
> ! !/
> !
> / switch
> |
> ---
At a time when I was regularly interviewing people for entry-level
engineering positions, I would always present them with this circuit
and ask what the output was for the two switch positions. None had
ever seen the circuit before. Some answered correctly, though most
didn't. The only non-graduate (i.e., still in school) found the answer
with the greatest ease and confidence. The others who got the right
answers sort of stumbled into the answers.
To those who are attempting to answer the questions using equations, I
will tell you that if you understand op-amps, you don't need equations
to figure out how this circuit operates.
Spehro Pefhany
> My favorite....
>
> +----+-----> +VCC
> | |
> R |
> | |
> | C
> +---B Q1 <== NPN transistors
> | E
> C |
> Q2 B---+-------> VOUT
> E |
> i2 | | i1
> | |
> R R
> | |
> | |
> GND GND
>
> Assuming BETA is large enough to ignore base currents what is the
> expression for VOUT and what is its TC?
Vout = i1 *R = Vcc - i2 *R - Vbe1
i1 - i2 = Vbe2/R => i2 = i1 - Vbe2/R
substituting, i1 * R = Vcc - (i1 - Vbe2/R) *R - Vbe1
i1 * 2 *R = Vcc + Vbe2 - Vbe1
So, Vout = Vcc/2, TC ~= 0 if Q1, Q2 track
Neat!!!
Best regards,
--
Spehro Pefhany --"it's the network..." "The Journey is the reward"
sp...@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
9-11 United we Stand
Cone Killer
"Bob Penoyer" <rpen...@NOSPAMieee.org> wrote in message
news:01t29u0sevihjf74o...@4ax.com...
And I quote:
---------------------------------------------------------------------
>"And I want the students to answer the question for arbitrary values:
>Replace 6.8K with R1 and 4.7K with R2, 1V with Vin; and write me an
> ^^^^^^^^^^^^^^^
>equation for Vout versus Vin <*very* big grin>"
>^^^^^^^^
> ...Jim Thompson
---------------------------------------------------------------------
Best Regards, Doug
Charles Johnson
"Jim Thompson" <Jim-T@analog_innovations.com.invalid> wrote in message
news:oe729ukk0tdnbnrs1...@4ax.com...
> On Thu, 14 Mar 2002 22:59:47 +0100,
> In Newsgroup: sci.electronics.design,
> Article: <a6r6gr$1epd$1...@scavenger.euro.net>,
> Entitled: "Re: Op Amp Design",
> Wrote the following:
>
> [snip]
> |kind of questions are at the limits of my capabilities, or even
> |beyond <G> , if the answer is wrong:
> |
> |Vout = 1,69V
> |Vout = Vin * (1 + (R2/R1))
>
> Frank! Frank! Frank! Back to school with you (or at least to the
> lab :-)
>
Kevin Aylward
news:01t29u0sevihjf74o...@4ax.com...
> <HelmutS...@t-online.de> wrote:
>
> >By the way, there is really an application based on this circuit.
> >This would then be the second question to the applicant:
> >What happens if the switch is open or closed?
> >
> > 10k 10k
> > 1V -----/\/\/\/----------\/\/\/\/------- Vout
> > ! ! !
> > ! ! !\ !
> > ! --!-\ !
> > ! 10k ! --------------
> > !--/\/\/\/-------!+/
> > ! !/
> > !
> > / switch
> > |
> > ---
>
> At a time when I was regularly interviewing people for entry-level
> engineering positions, I would always present them with this circuit
> and ask what the output was for the two switch positions.
And in my view, this would be a right daft question to ask at an interview. This
sort of stuff is your typical, "I know more then you question, and I want to
impress myself". Interviews are very artificial situations and asking people do
do new things has no relevance as to how they can do new things in a real
situation. Its very easy to to simple hit tempoary blind spots and fail simple
questions due to the pressure of the situation.
For, me I ask questions on things that I belive that the person should _already_
know, i.e. things that he should have come accross in the past. If I was to
present a newish problem, I would not expect a solution, only an outline of how
he would go about solving the problem.
Kevin Aylward , Warden of the Kings Ale
ke...@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
Charles Johnson
"Charles Johnson" <qru...@oneimage.com> wrote in message
news:a6s84e$fa7$1...@suaar1aa.prod.compuserve.com...
Tony Williams
Christopher R. Carlen <crc...@sandia.gov> wrote:
> > | R R
> > | Vin ---+-/\/\/\/-----+----\/\/\/\/-------+ Vout
> > | | ! !
> > | | ! !\ !
> > | \ --!-\ !
> > | / ! --------------
> > | RV1= R \<---------------!+/
> > | / /|\ !/
> > | \ k*R
> > | / |
> > | | \|/
> > | --+----------------------------------0v
> Vout = Vin * (2 * k - 1)
Give the man a chocolate fish!
I used that circuit years ago, for the f.panel
zeroing adjustments of the Sin/Cos signals
in an airborne magnetometer. The customer could
not accept that something so simple would do the
required linear +/- 1 adjustments. He refused to
commit to layout until a working breadboard of it
had been demonstrated..... cheeky blighter.
> The interesting case occurs when k = 1/2. Then the equation predicts
> Vout = 0 no matter what the input. This is logical as well considering
> the voltage divider action of the input and feedback resistors.
We could patent that. Need a snappy name for it
though. How about "2-resistor_plus_centred-pot
differential amplifier"?
--
Tony Williams.
Spehro Pefhany
> We could patent that. Need a snappy name for it
> though. How about "2-resistor_plus_centred-pot
> differential amplifier"?
"Williams bridge" ;-)
Brad Albing
original poster (Goran, I think -- it's been a while since he posted). Guess he
ran off with his tail 'twixt his legs....
Tony Williams
Spehro Pefhany <sp...@interlog.com> wrote:
> "Williams bridge" ;-)
Now, now Speff..... That's going a "Bridge Too Far".
--
Tony Williams.
Frank Bemelman
"John Larkin" <jjlarkin@highlandSNIP_THIStechnology.com> schreef in bericht
news:zEeRPF3zxLaQ56...@4ax.com...
> Jim,
>
> my favorite quickie quiz is this:
>
>
>
> +10v
> |
> |
> C
> +5V-------------B <== NPN transistor
> E
> |
> |
> R = 1K
> |
> |
> GND
>
>
>
> OK, just roughly, without serious calculating,...
>
> what's the base voltage?
+5V
> what's the collector voltage?
+10V
> what's the emitter voltage?
+4.3V (roughly)
> what's the emitter current?
4.3mA (roughly)
> what's the base current?
4.3mA/Beta or... collector current minus emitter current.
> any other comments?
> I get the most amazing statements, even from EEs with loaded resumes.
> Even to the first two questions! This one has made some very smug
> people get very shook up, fast.
>
> John
--
Frank Bemelman
"Christopher R. Carlen" <crc...@sandia.gov> schreef in bericht
news:3C9162EF...@sandia.gov...
> Jim Thompson wrote:
>
> > | And when they've done that.......
> > |
> > | R R
> > | Vin ---+-/\/\/\/-----+----\/\/\/\/-------+ Vout
> > | | ! !
> > | | ! !\ !
> > | \ --!-\ !
> > | / ! --------------
> > | RV1= R \<---------------!+/
> > | / /|\ !/
> > | \ k*R
> > | / |
> > | | \|/
> > | --+----------------------------------0v
> > |
> > | RV1= a pot, overall value R.
> > |
> > | k= pot position, varying from 0 to 1.
> >
> > I won't give an answer... it's *much* more amusing to watch what
> > others post ;-)
> >
> > ...Jim Thompson
>
>
> Yeah, I can't wait to see what Frank B. will post.
>
> Here's my take:
I *quickly* deleted the rest of your post, and I won't
read it before I have my indepth analysis finished <G>.
Frank Bemelman
"Frank Bemelman" <beme...@euronet.nl.invalid> schreef in bericht
news:a6tajj$3cq$1...@scavenger.euro.net...
It amplifies between -1 and +1, depending on the pot's
position. Very nice! If someone would ask me to design
something like that, I would start with a potentiometer
with it's wiper connected to ground, an inverting opamp
on one side, a non-inverting on the other side, a summing
amplifier and what have you, resulting in quite a number
of opamps.
Well, at least I make my component suppliers happy ;)
John Woodgate
<to...@ledelec.demon.co.uk> wrote (in <4b17838...@ledelec.demon.co.
uk>) about 'Op Amp Design', on Fri, 15 Mar 2002:
>We could patent that. Need a snappy name for it
> though. How about "2-resistor_plus_centred-pot
> differential amplifier"?
"Isn't it a Williams Bridge?" he wrote, just **before** opening Spehro's
post! ......(;-)
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!
Spehro Pefhany
> I read in sci.electronics.design that Tony Williams
> <to...@ledelec.demon.co.uk> wrote (in <4b17838...@ledelec.demon.co.
> uk>) about 'Op Amp Design', on Fri, 15 Mar 2002:
>>We could patent that. Need a snappy name for it
>> though. How about "2-resistor_plus_centred-pot
>> differential amplifier"?
>
> "Isn't it a Williams Bridge?" he wrote, just **before** opening Spehro's
> post! ......(;-)
Done, then. Maybe we could generalize the usage to refer to any
non-trivial circuit that provides no output regardless of the input.
Win, could this term be included in the 3rd Edition AofE?
Goran Tomas
No, I'm still here... I was just waiting for everybody to stop
grinning and amusing themselves with the wrong answers and somebody to
give that common sense explanation, so that I can actually learn
something.
Regards,
Goran Tomas
Ken Smith
John Larkin <jjlarkin@highlandSNIP_THIStechnology.com> wrote:
>>
>> 10k 10k
>> 1V -----/\/\/\/----------\/\/\/\/------- Vout
>> ! ! !
>> ! ! !\ !
>> ! --!-\ !
>> ! 10k ! --------------
>> !--/\/\/\/-------!+/
>> ! !/
>> !
>> / switch
>> |
>> ---
>>
.... and/or modulator
For extra points, replace the switch with a capacitor. What very useful
function does the circuit now perform.
Jim Thompson
Entitled: "Re: Op Amp Design",
Wrote the following:
|On Fri, 15 Mar 2002 13:27:55 GMT, Brad Albing
Rich Grise did in:
Message-ID: <3C9124...@earthlink.net>
As I alluded to in:
Message-ID: <v9829u8eviu07g0p7...@4ax.com>
The trick is OpAmps (except when railed) *always* have Vin+ = Vin-
Ken Smith
Frank Bemelman <beme...@euronet.nl.invalid> wrote:
[....]
>> | 6.8K 4.7K
>> | 1V -----/\/\/\/----------\/\/\/\/------- Vout
>> | ! ! !
>> | ! ! !\ !
>> | ! --!-\ !
>> | !----------------!+/
>> | !/
>> |
>> |
[...]
>Vout = 1,69V
>Vout = Vin * (1 + (R2/R1))
Ask your self the following questions:
(1) What voltage is on the (+) input of the op-amp?
(2) Assuming all is well what is on the (-) input of the op-amp?
(3) Given the answer in (3) what is the voltage drop on the 6.8K?
(4) Knowing the voltage drop, what is the current in the 6.8K?
(5) How much current flow in/out of the input of an op-amp?
(6) Knowing (4) and (5) how much current flows in the 4.7K?
(7) Knowing (6) what is the voltage drop on the 4.7K?
(8) Knowing (2) and (7) what is the voltage at Vout?
Ken Smith
Tony Williams <to...@ledelec.demon.co.uk> wrote:
[....]
> And when they've done that.......
[... a different op-amp circuit ....]
>
No, I figure they now how a op-amp works if they answer that one and I go
on to seeing if they know logic gates with:
12KHz ------------------------------------
! XOR ----- Out
!--- !---
! XOR-------- !
---- ! XOR----- !
--- ! XOR----
----
The question asks what is at the output.
Then I ask about two simple RC circuits with a step going in to
see if they can sketch what's on the output.
I then ask "why is a car battery so heavy?" etc.
These BTW are questions I use on the technician test. If its an
engineering position I expect them to be able to come up with circuits and
ideas as well as know what an existing circuit would do.
Kevin Aylward
"Ken Smith" <kens...@rahul.net> wrote in message
news:a6thqq$hct$1...@samba.rahul.net...
I'll think I'll pass on that one.
Ken Smith
Tony Williams <to...@ledelec.demon.co.uk> wrote:
[....]
> I used that circuit years ago, for the f.panel
> zeroing adjustments of the Sin/Cos signals
> in an airborne magnetometer.
I assume you mean the compensator and not the magnetometer its self. That
is exactly the circuit that would be used with a 10 turn pot and a counter
dial. You paint the front panel with paint that "100 mile an hour" tape
sticks to so the user can skick some on there and write down the correct
values, because you know he's going to do something like that anyway.
> The customer could
> not accept that something so simple would do the
> required linear +/- 1 adjustments. He refused to
> commit to layout until a working breadboard of it
> had been demonstrated..... cheeky blighter.
>
I once had a problem like that with a manager. He just would not believe
that if you followed an op-amp with a common emitter stage, it was
negitive feed back to run a capacitor from the collector to the (+) input
of the op-amp.
Kevin Aylward
"Frank Bemelman" <beme...@euronet.nl.invalid> wrote in message
news:a6r6gr$1epd$1...@scavenger.euro.net...
>
> "Jim Thompson" <Jim-T@analog_innovations.com.invalid> schreef in bericht
> news:k4o19ukg7keb8snm3...@4ax.com...
> > On 14 Mar 2002 17:01:48 GMT,
> > Ken Smith <kens...@rahul.net>,
> > In Newsgroup: sci.electronics.design,
> > Article: <a6ql1s$skp$1...@samba.rahul.net>,
> > Entitled: "Re: Op Amp Design",
> > Wrote the following:
> >
> > |Your students should have no problem with this circuit:
> > |
> > |
> > | 6.8K 4.7K
> > | 1V -----/\/\/\/----------\/\/\/\/------- Vout
> > | ! ! !
> > | ! ! !\ !
> > | ! --!-\ !
> > | ! ! --------------
> > | !----------------!+/
> > | !/
> > |
> > |
> > |If they can't tell you what Vout is in less than 2 minutes you have
> > |failed them.
> > |
> > |
> > |--
> >
> > Now, now, Ken. You've done gone and confused the poor bastard... he
> > doesn't have a ready-made equation ;-)
> >
> > And I want the students to answer the question for arbitrary values:
> > Replace 6.8K with R1 and 4.7K with R2, 1V with Vin; and write me an
> > equation for Vout versus Vin <*very* big grin>
>
> I'm not a candidate for your summer's home entertainer, since these
> kind of questions are at the limits of my capabilities, or even
> beyond <G> , if the answer is wrong:
>
> Vout = 1,69V
> Vout = Vin * (1 + (R2/R1))
>
> --
> Thanks,
> Frank Bemelman
> (remove 'x' & .invalid when sending email)
>
>
>
The + input is at Vin therefore so is the negative input, therefore the voltage
across the input resistor is 0V, therefore the current in the input resistor is
zero, therefore the current through the feedback resistor is zero, therefore the
voltage drop across the feedback resistor is zero, therefore the voltage at the
negative input appears at the output via the feedback resistor, therefore the
output voltage is the input voltage, therfore the gain is 1.
John Larkin
Ken,
I know what it does; I just can't think how it would be useful!
John
Aubrey McIntosh
but
translated into math.
d := R1 / (R1 + R2) ; just a handy notation to get voltage divider drop.
Vout := A ( V+ - V-) ;"fundamental op amp behavior"
V+ := Vin
V- := Vin - d * ( Vin - Vout ); Voltage at divider network
Vout := A ( Vin - [Vin - d*(Vin-Vout)]) ; substitute all
Vout := A [ d*(Vin-Vout)] ; work, work, work ...
Vout := (1+Ad) = AdVin
Vout := A*d / ( 1+A*d ) * Vin
limit A-> big Vout -> Vin
so A*d has to be lots bigger than 1.
"Rich Grise" <rich...@earthlink.net> wrote in message
news:3C9124...@earthlink.net...
Spehro Pefhany
> For extra points, replace the switch with a capacitor. What very useful
> function does the circuit now perform.
P+D controller.
Jim Thompson
Entitled: "Re: Op Amp Design",
Wrote the following:
|In article <ukaRPC4lI35cC9...@4ax.com>,
AllPass
Rich Grise
>
> On Thu, 14 Mar 2002 10:45:22 -0700,
> Jim Thompson <Jim-T@analog_innovations.com.invalid>,
> In Newsgroup: sci.electronics.design,
> Article: <k4o19ukg7keb8snm3...@4ax.com>,
> Entitled: "Re: Op Amp Design",
> Wrote the following:
>
> |
> |And I want the students to answer the question for arbitrary values:
> |Replace 6.8K with R1 and 4.7K with R2, 1V with Vin; and write me an
> |equation for Vout versus Vin <*very* big grin>
> |
>
> And the winners are (in order of posting):
>
> Klaus Vestergaard Kragelund
> Helmut Sennewald
> Rich Grise
>
> ...Jim Thompson
Well, I din't answer the whole question - I missed the "equation"
part, so here goes: Vout = Vin. ;-)
--
Cheers!
Rich
"We have met the enemy, and he is us!"
- Pogo Possum, ca. 1950's
Rich Grise
>
> "Jim Thompson" <Jim-T@analog_innovations.com.invalid> schreef in bericht
>
> > And the winners are (in order of posting):
>
> and the other 4 only have 1 umbrella. They don't get
> wet, how is that possible?
>
> --
> Thanks,
> Frank Bemelman
> (remove 'x' & .invalid when sending email)
It's not raining. ;-)
Ken Smith
John Larkin <jjlarkin@highlandSNIP_THIStechnology.com> wrote:
>On 15 Mar 2002 19:25:14 GMT, Ken Smith <kens...@rahul.net> wrote:
>
>>In article <ukaRPC4lI35cC9...@4ax.com>,
>>John Larkin <jjlarkin@highlandSNIP_THIStechnology.com> wrote:
>>>>
>>>> 10k 10k
>>>> 1V -----/\/\/\/----------\/\/\/\/------- Vout
>>>> ! ! !
>>>> ! ! !\ !
>>>> ! --!-\ !
>>>> ! 10k ! --------------
>>>> !--/\/\/\/-------!+/
>>>> ! !/
>>>> !
>>>> / switch
>>>> |
>>>> ---
>>>>
>>>synchronous demodulator!
>>
>>.... and/or modulator
>>
>>For extra points, replace the switch with a capacitor. What very useful
>>function does the circuit now perform.
>>
>>--
>
>Ken,
>
>I know what it does; I just can't think how it would be useful!
(1)
Back in the days before the 24 bit delta modulation converter we used to
make things called Instantanious Floating Point converters. These
converters had to measure the signal twice. The first conversion was to
see where the "binary point" was going to be and the second was to do the
mantissa part of the value.
In most designs of IFP converters, the two converters follow a sample and
hold. The exponent or "binary point" is determined and then the correct
gain is inserted between the sample and hold and the ADC. This design has
a problem in that it tends to increase the effects of distortion in the
sample and hold.
Other designs insert the gain before a second sample and hold. This
normally leads to another problem because the exponent conversion is not
done on the same signal as the mantissa conversion and could therefor
begiving the wrong value of exponent. If the exponent calls for too
little gain not much harm is done. If it asks for too much the mantissa
ADC will clip and some serious distortion will happen.
Combining the circuit described above with some other clever electronics
it is posible to make a IFP converter with the gain before the sample and
hold that can never overdrive its mantissa ADC.
(2)
If you want to do single side band communication, one way to construct a
single sideband signal or to demodulate one is to use the properties of
multiplying sines and cosines.
In such a system more than one of the above circuit is used. The circuits
are matched to make two outputs that are at 90 degrees to each other and
at the same amplitude over some desired band.
(3)
If you combine some of the more extreme versions of the circuit with some
variable gain amplifiers, you can get some cool audio effects.
Ken Smith
Spehro Pefhany <sp...@interlog.com> wrote:
>The renowned Ken Smith <kens...@rahul.net> wrote:
>
>> For extra points, replace the switch with a capacitor. What very useful
>> function does the circuit now perform.
>
>P+D controller.
Don't guess. You spotted the poles and zeros but you didn't look at where
they landed.
You could download switcher cad from www.linear.com and try it.
Ken Smith
Jim Thompson <Jim-T@analog_innovations.com.invalid> wrote:
[....]
>AllPass
Give that man a chocolate rabbit.
Spehro Pefhany
(some stuff that disappeared off my server)
Ah, yes. I just did simulate it. My bad, I did mix that one up. Quite
non-intuitive. Thanks!
Jim Thompson
Ken Smith <kens...@rahul.net>,
In Newsgroup: sci.electronics.design,
Article: <a6tjcq$hlc$1...@samba.rahul.net>,
Entitled: "Re: Op Amp Design",
Wrote the following:
|In article <4b174dc...@ledelec.demon.co.uk>,
|Tony Williams <to...@ledelec.demon.co.uk> wrote:
|
|[....]
|> And when they've done that.......
|[... a different op-amp circuit ....]
|>
|
|No, I figure they now how a op-amp works if they answer that one and I go
|on to seeing if they know logic gates with:
|
|
|
| 12KHz ------------------------------------
| ! XOR ----- Out
| !--- !---
| ! XOR-------- !
| ---- ! XOR----- !
| --- ! XOR----
| ----
[snip]
12KHz, one gate delay later.
Allan Herriman
>In article <4b174dc...@ledelec.demon.co.uk>,
>Tony Williams <to...@ledelec.demon.co.uk> wrote:
>
>[....]
>> And when they've done that.......
>[... a different op-amp circuit ....]
>>
>
>No, I figure they now how a op-amp works if they answer that one and I go
>on to seeing if they know logic gates with:
>
>
>
> 12KHz ------------------------------------
> ! XOR ----- Out
> !--- !---
> ! XOR-------- !
> ---- ! XOR----- !
> --- ! XOR----
> ----
>
>The question asks what is at the output.
How many of them gave the oversimplified answer "12kHz" (or even
"12KHz, one gate delay later" as another poster said) ?
(Think about what happens if the delay to each of the XOR inputs isn't
quite the same.)
Allan.
John Larkin
>(1)
>Back in the days before the 24 bit delta modulation converter we used to
>make things called Instantanious Floating Point converters. These
>converters had to measure the signal twice. The first conversion was to
>see where the "binary point" was going to be and the second was to do the
>mantissa part of the value.
This, don't know why, reminds me of the classic minimalist 24-bit
serial DAC:
serial digital data in -----R----*---------- analog out
|
|
C
|
|
gnd
Just apply the clocked serial data stream to the input, lsb first, and
extract the analog out after the 24th bit has been transfered.
R*C should be about 0.693*T where T is the clock period.
John
Frank Bemelman
"Allan Herriman" <allan_herrim...@agilent.com> schreef in bericht
news:3c92c036...@netnews.agilent.com...
> On 15 Mar 2002 19:51:54 GMT, Ken Smith <kens...@rahul.net> wrote:
>
> >In article <4b174dc...@ledelec.demon.co.uk>,
> >Tony Williams <to...@ledelec.demon.co.uk> wrote:
> >
> >[....]
> >> And when they've done that.......
> >[... a different op-amp circuit ....]
> >>
> >
> >No, I figure they now how a op-amp works if they answer that one and I go
> >on to seeing if they know logic gates with:
> >
> >
> >
> > 12KHz ------------------------------------
> > ! XOR ----- Out
> > !--- !---
> > ! XOR-------- !
> > ---- ! XOR----- !
> > --- ! XOR----
> > ----
> >
> >The question asks what is at the output.
>
> How many of them gave the oversimplified answer "12kHz" (or even
> "12KHz, one gate delay later" as another poster said) ?
>
> (Think about what happens if the delay to each of the XOR inputs isn't
> quite the same.)
I can think all day long, but I still come to the oversimplified
answer. The first XOR at the bottom produces only '0', and so
will the 2nd and 3rd. The last XOR get's a '0' and 12Khz.
What's the catch?
Rich Grise
Presumably, there's some finite time delay from the time
the edge reaches input 1 of the first Xor and input 2. Depending
on how long the lead is, you could get from 1-100 pS, if the
gate will switch that fast.
>
> --
> Thanks,
> Frank Bemelman
> (remove 'x' & .invalid when sending email)
--
Spehro Pefhany
>
> I can think all day long, but I still come to the oversimplified
> answer. The first XOR at the bottom produces only '0', and so
> will the 2nd and 3rd. The last XOR get's a '0' and 12Khz.
>
> What's the catch?
I don't think there is one, this seems like a "trick question" in
that the circuit doesn't really do anything useful.
Jim Thompson
Rich Grise <rich...@earthlink.net>,
In Newsgroup: sci.electronics.design,
Article: <3C934A...@earthlink.net>,
Entitled: "Re: Op Amp Design",
Wrote the following:
|Frank Bemelman wrote:
OK, I'll amend my answer to 12KHz, with *possibly* ratty edges.
However I do note the it takes at least 2ns of skew to get a sliver
thru a 74HC86.
And, since the 74HC86 is constructed of inverters and transmission
gates, that's not likely.
John Fields
--
---
I haven't been following this thread, but at first glance I'd say 24kHz.
---
John Fields
Professional circuit designer
http://www.austininstruments.com
John Woodgate
<sp...@interlog.com> wrote (in <qbJk8.9325$5Id....@news2.bloor.is>)
about 'Op Amp Design', on Sat, 16 Mar 2002:
>I don't think there is one, this seems like a "trick question" in
>that the circuit doesn't really do anything useful.
Just like the Williams Bridge or the 'all-stop filter'.
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!
Jim Thompson
presently 83-deep, and *no* name-calling!
Win Hill
>
> This thread is amazing... we must have set a new record, the
> thread is presently 83-deep, and *no* name-calling!
>
> ...Jim Thompson
Why you #^@$^%^%$*&^%(&#q^$%_(q^#$! There I feel better already.
___
Thanks, /.-.\
((( ))
- Win \\\//
\\\
Winfield Hill //\\\
Rowland Institute for Science /// \\\
Cambridge, MA \/ \/
John Larkin
<jfi...@austininstruments.com> wrote:
> I haven't been following this thread, but at first glance I'd say 24kHz.
16.8 KHz, and that's my final offer.
John
Bob Penoyer
<ke...@anasoft.co.uk> wrote:
>> At a time when I was regularly interviewing people for entry-level
>> engineering positions, I would always present them with this circuit
>> and ask what the output was for the two switch positions.
>
>And in my view, this would be a right daft question to ask at an interview.
I disagree. Read on.
>This
>sort of stuff is your typical, "I know more then you question, and I want to
>impress myself".
That was neither my intent nor my motivation.
>Interviews are very artificial situations and asking people do
>do new things has no relevance as to how they can do new things in a real
>situation.
Really?! Then why interview anyone? Just read his resume and decide
his suitability for the job based on that. Remember, I was talking
about interviews for entry-level jobs where people had little or no
experience.
In addition to learning about the guy's background, don't you want to
get some sense of what potential he might have to do the sorts of
things you will need him to do if you hire him?
It's amazing to me how often it happens that, given a simple
statement, people can draw all sorts of unconnected conclusions as you
have done here. I didn't ask the question in a vacuum. Also, I didn't
ask the question as a stand-alone question. We would discuss simple
op-amp circuits along with some of the non-ideal conditions that would
affect their operation. We would also talk about things you shouldn't
do with an op-amp, etc. This was generally a very painless exercise
for the interviewee. If he stumbled a bit, I would offer a little
coaching (since I understood that he was nervous and "on the spot").
Somewhere in this process, after some buildup, I would pose the
circuit you and I are discussing. Now, you might ask, what purpose
could this question serve? It gave me a chance to watch/hear the guy
reason his way through a problem he hadn't faced in school.
Nevertheless, it was a simple problem with a non-obvious answer that
would make him think. If he appeared stuck, I would offer hints such
as, "Think about the currents. What can you say about them?"
You need to understand that this question was not a show stopper. His
candidacy was not directly affected by his answer. However, if some
people couldn't answer it, if others stumbled their way to the answer
with some coaching, or if someone could answer the question via some
innate, cogent, knowledgeable reasoning, I would view these people
accordingly. BUT this aspect of the interview was only part of an
overall opinion based on our entire conversation.
>Its very easy to to simple hit tempoary blind spots and fail simple
>questions due to the pressure of the situation.
No question about it. I've done that myself when I've been
interviewed, so I'm sensitive to that problem.
>For, me I ask questions on things that I belive that the person should _already_
>know, i.e. things that he should have come accross in the past.
Me, too. But I also want to get a sense of how bright the person is.
For example, even if he doesn't have an answer to a new problem, does
he attack the problem in an astute way?
>If I was to
>present a newish problem, I would not expect a solution, only an outline of how
>he would go about solving the problem.
That's what I was looking for: given a new problem, how does the
interviewee deal with it? If he finds the right answer, so much the
better.
John Larkin
>>For, me I ask questions on things that I belive that the person should _already_
>>know, i.e. things that he should have come accross in the past.
>
>Me, too. But I also want to get a sense of how bright the person is.
>For example, even if he doesn't have an answer to a new problem, does
>he attack the problem in an astute way?
>
>>If I was to
>>present a newish problem, I would not expect a solution, only an outline of how
>>he would go about solving the problem.
>
>That's what I was looking for: given a new problem, how does the
>interviewee deal with it? If he finds the right answer, so much the
>better.
I don't think it's fair, or useful, to ask 'trick' qiestions in an
interview, but it's certainly prudent to ask some simple,
straightforeward circuit questions and expect intelligent discussion
and simple answers. I don't want an outline of how to solve an R:3R
voltage divider problem, I expect the answer, reasonably fast. If they
can handle that, move on to harder stuff (hey, maybe an RC lowpass)
until you zero in on their skill level. And once you know that, you
can figure if there is a useful place for them in your company.
Lots of impressive-resume interviewees can't even solve the voltage
divider problem, and I sure don't want them as an engineer. And it
does neither of us any good if I have to fire them after six months.
This reminds me of the new qualification test they give to California
public school teachers. It checks to see if that can do something like
5th grade level math and reading. The teachers' union was in revolt
because they said it was an insult and didn't measure teaching skills.
Something like 20% of the teachers flunked it.
John
Win Hill
>
> "Helmut Sennewald" <HelmutS...@t-online.de> wrote:
>
>> By the way, there is really an application based on this circuit.
>> This would then be the second question to the applicant:
>> What happens if the switch is open or closed?
>>
>> 10k 10k
>> 1V -----/\/\/\/----------\/\/\/\/------- Vout
>> ! ! !
>> ! ! !\ !
>> ! --!-\ !
>> ! 10k ! --------------
>> !--/\/\/\/-------!+/
>> ! !/
>> !
>> / switch
>> |
>> ---
>
> engineering positions, I would always present them with this circuit
> ever seen the circuit before. ...
They would have if they'd read the Art of Electronics opamp chapter.
It's near the beginning on page 183. And the G = +1 through -1
continuously-variable gain circuit is on page 253.
Frank Bemelman
"Jim Thompson" <Jim-T@analog_innovations.com.invalid> schreef in bericht
> This thread is amazing... we must have set a new record, the thread is
> presently 83-deep, and *no* name-calling!
Ehhh, someone called me a lousy hip-shooter! <G>
It was good he offered me a prize, though ;)
Frank Bemelman
"Rich Grise" <rich...@earthlink.net> schreef in bericht
news:3C934A...@earthlink.net...
On the bottom row of XOR's not much interesting is going on.
After the first XOR it is all zero's you will find. You could
tie 100 XOR's in a row, but what would it matter? No switching
occurs in that string of XOR's.
Bob Penoyer
<jjla...@highlandSNIPTHIStechnology.com> wrote:
>I don't think it's fair, or useful, to ask 'trick' qiestions in an
>interview, but it's certainly prudent to ask some simple,
>straightforeward circuit questions and expect intelligent discussion
>and simple answers. I don't want an outline of how to solve an R:3R
>voltage divider problem, I expect the answer, reasonably fast. If they
>can handle that, move on to harder stuff (hey, maybe an RC lowpass)
>until you zero in on their skill level. And once you know that, you
>can figure if there is a useful place for them in your company.
It's just a matter of method, John. You favor one. I favor another.
>Lots of impressive-resume interviewees can't even solve the voltage
>divider problem, and I sure don't want them as an engineer. And it
>does neither of us any good if I have to fire them after six months.
Fair enough. But, given that someone successfully answers fundamental
questions, if you find he can deal successfully with so-called trick
questions, maybe he's someone you REALLY want to hire.
>This reminds me of the new qualification test they give to California
>public school teachers. It checks to see if that can do something like
>5th grade level math and reading. The teachers' union was in revolt
>because they said it was an insult and didn't measure teaching skills.
>Something like 20% of the teachers flunked it.
On average, among all entering freshmen, prospective teachers get the
lowest SAT scores. On entering graduate school, teachers, on average,
achieve the lowest GRE scores. Too often, teachers are drawn from the
worst of the people who enter college. Education students learn about
teaching tools, teaching methods, the psychology of teaching, etc. BUT
they typically don't learn anything to teach; that is, in American
grade schools, teachers of English, math, or history, for example,
have no specialized or advanced training in those or other subjects.
In American high schools, where teachers are supposed to have
single-subject credentials in the courses they teach, too often a
history major is teaching math and the baseball coach is teaching
biology. Generally speaking, American public schools are a disaster.
(You've hit a nerve here, John. The problems with teaching are a pet
peeve of mine.)