Op Amp Design

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Dieter Knollman

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Mar 13, 2002, 4:42:56 PM3/13/02
to
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
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

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Mar 13, 2002, 4:56:05 PM3/13/02
to
On Wed, 13 Mar 2002 14:42:56 -0700,
"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

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Mar 13, 2002, 5:29:33 PM3/13/02
to

"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

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Mar 13, 2002, 5:34:53 PM3/13/02
to
On Wed, 13 Mar 2002 14:42:56 -0700, "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?


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

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Mar 13, 2002, 5:46:44 PM3/13/02
to
I must admit, in all the years I've spent grading papers that ask for a
verbal description of the opamp process, I've never before seen one that so
hopelessly complexes up the obvious.

Good Lord...


Jim


Jim Thompson

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Mar 13, 2002, 7:15:43 PM3/13/02
to
On Wed, 13 Mar 2002 14:46:44 -0800,
"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

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Mar 14, 2002, 2:24:58 AM3/14/02
to

"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
> with the individual procedures." They need to learn the theory.

> 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 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.
>

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

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Mar 14, 2002, 12:01:48 PM3/14/02
to
In article <a6oh4h$kuj$1...@news.chatlink.com>,

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

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Mar 14, 2002, 12:45:22 PM3/14/02
to
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:

|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

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Mar 14, 2002, 4:14:33 PM3/14/02
to
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

Jim Thompson

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Mar 14, 2002, 4:34:11 PM3/14/02
to
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

Sorry! You don't get to be this summer's intern ;-) Your answers
were incorrect for both questions.

Klaus Vestergaard Kragelund

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Mar 14, 2002, 10:40:52 PM3/14/02
to
"Goran Tomas" <goran...@radio101.hr> wrote in message
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

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Mar 14, 2002, 4:48:04 PM3/14/02
to

"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

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Mar 14, 2002, 4:58:45 PM3/14/02
to
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>,

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

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Mar 14, 2002, 4:59:47 PM3/14/02
to

"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)

Jim Thompson

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Mar 14, 2002, 5:03:23 PM3/14/02
to
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 :-)

Frank Bemelman

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Mar 14, 2002, 5:04:28 PM3/14/02
to

"Jim Thompson" <Jim-T@analog_innovations.com.invalid> schreef in bericht
news:u5729u8iepmuqbs5e...@4ax.com...

> 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:
>
> |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
> |
> |
>
> 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

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Mar 14, 2002, 5:07:21 PM3/14/02
to
On Thu, 14 Mar 2002 23:04:28 +0100,
"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

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Mar 14, 2002, 5:10:40 PM3/14/02
to
The renowned Frank Bemelman <beme...@euronet.nl.invalid> wrote:

> 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

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Mar 14, 2002, 5:11:00 PM3/14/02
to

"Jim Thompson" <Jim-T@analog_innovations.com.invalid> schreef in bericht
news:oe729ukk0tdnbnrs1...@4ax.com...

> 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

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Mar 14, 2002, 5:16:12 PM3/14/02
to
The renowned Frank Bemelman <beme...@euronet.nl.invalid> wrote:

> BTW, this is all brainwork <G>, no lab at home, no simulator
> too.

*simulator* ???

Jim Thompson

unread,
Mar 14, 2002, 5:21:00 PM3/14/02
to
On Thu, 14 Mar 2002 23:11:00 +0100,

"Frank Bemelman" <beme...@euronet.nl.invalid>,
In Newsgroup: sci.electronics.design,
Article: <a6r762$1kue$1...@scavenger.euro.net>,

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

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Mar 14, 2002, 5:25:52 PM3/14/02
to

"Frank Bemelman" <beme...@euronet.nl.invalid> wrote in message
news:a6r762$1kue$1...@scavenger.euro.net...
Vo = Vi ((R2/R1) + 1) (R1 = sideways 8)

Regards, Doug


Rich Grise

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Mar 14, 2002, 5:32:07 PM3/14/02
to

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

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Mar 14, 2002, 5:38:24 PM3/14/02
to
On Thu, 14 Mar 2002 22:32:07 GMT,
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

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Mar 14, 2002, 5:41:21 PM3/14/02
to

"Jim Thompson" <Jim-T@analog_innovations.com.invalid> schreef in bericht
news:v9829u8eviu07g0p7...@4ax.com...

...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

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Mar 14, 2002, 5:42:04 PM3/14/02
to
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

Frank Bemelman

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Mar 14, 2002, 5:43:02 PM3/14/02
to

"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

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Mar 14, 2002, 5:48:48 PM3/14/02
to

"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?

Andreas Hadler

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Mar 14, 2002, 5:46:23 PM3/14/02
to
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. |
+----------------------------------------------------------------+

Andreas Hadler

unread,
Mar 14, 2002, 5:56:45 PM3/14/02
to
"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?...

cu, Andreas
--

Jim Thompson

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Mar 14, 2002, 6:00:10 PM3/14/02
to
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:

ROTFLMAO!

Cone Killer

unread,
Mar 14, 2002, 6:01:29 PM3/14/02
to

"Frank Bemelman" <beme...@euronet.nl.invalid> wrote in message
news:a6r91u$20g2$1...@scavenger.euro.net...
The first answer you gave had a (-1).

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

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Mar 14, 2002, 6:04:22 PM3/14/02
to
On Thu, 14 Mar 2002 15:42:04 -0700,

Jim Thompson <Jim-T@analog_innovations.com.invalid>,
In Newsgroup: sci.electronics.design,
Article: <ml929u8can80h7epc...@4ax.com>,

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

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Mar 14, 2002, 6:06:56 PM3/14/02
to

"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

unread,
Mar 14, 2002, 6:08:32 PM3/14/02
to

"Jim Thompson" <Jim-T@analog_innovations.com.invalid> schreef in bericht
news:5ra29ug8tcp21vjju...@4ax.com...

> 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

unread,
Mar 14, 2002, 6:21:07 PM3/14/02
to
"Frank Bemelman" <beme...@euronet.nl.invalid> wrote:

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

unread,
Mar 14, 2002, 6:31:23 PM3/14/02
to

"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

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Mar 14, 2002, 6:38:03 PM3/14/02
to
Just about 30 posts in one and a half hour?

Ever seen such a fast thread?

Tony Williams

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Mar 14, 2002, 5:53:34 PM3/14/02
to
In article <a6ql1s$skp$1...@samba.rahul.net>,
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

unread,
Mar 14, 2002, 7:35:42 PM3/14/02
to
On Thu, 14 Mar 2002 22:53:34 +0000 (GMT),
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

unread,
Mar 14, 2002, 7:57:10 PM3/14/02
to
On Thu, 14 Mar 2002 22:48:04 +0100, "Helmut Sennewald"
<HelmutS...@t-online.de> wrote:

>
>"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

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Mar 14, 2002, 7:58:42 PM3/14/02
to


Jim,

the intern situation looks pretty grim for this coming summer.

John

John Larkin

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Mar 14, 2002, 8:12:07 PM3/14/02
to
On Thu, 14 Mar 2002 15:42:04 -0700, Jim Thompson
<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

unread,
Mar 14, 2002, 8:28:24 PM3/14/02
to
On Thu, 14 Mar 2002 17:12:07 -0800,
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

unread,
Mar 14, 2002, 8:33:05 PM3/14/02
to
I got all the answers right! - what did I win?


"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

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Mar 14, 2002, 9:56:47 PM3/14/02
to
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:

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

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Mar 14, 2002, 9:46:20 PM3/14/02
to
In article <a6r6pj$1i05$1...@scavenger.euro.net>,

Frank Bemelman <beme...@euronet.nl.invalid> wrote:
>Increase the gain by 1, if the switch is open?

For very large values of "1"?

qrk

unread,
Mar 14, 2002, 10:33:25 PM3/14/02
to
On Thu, 14 Mar 2002 22:32:07 GMT, Rich Grise <rich...@earthlink.net>
wrote:

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

unread,
Mar 14, 2002, 11:31:13 PM3/14/02
to
On Thu, 14 Mar 2002 22:48:04 +0100, "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
> |
> ---

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

unread,
Mar 15, 2002, 1:12:06 AM3/15/02
to
The renowned Jim Thompson <Jim-T@analog_innovations.com.invalid> wrote:

> 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

unread,
Mar 15, 2002, 1:44:22 AM3/15/02
to

"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

unread,
Mar 15, 2002, 2:32:30 AM3/15/02
to
Vout=Vin-Vin(1-r1/r2)
"Jim Thompson" <Jim-T@analog_innovations.com.invalid> wrote in message

news:oe729ukk0tdnbnrs1...@4ax.com...
> 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 :-)
>

Kevin Aylward

unread,
Mar 15, 2002, 2:46:13 AM3/15/02
to
"Bob Penoyer" <rpen...@NOSPAMieee.org> wrote in message
news:01t29u0sevihjf74o...@4ax.com...
> On Thu, 14 Mar 2002 22:48:04 +0100, "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
> > |
> > ---
>
> 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

unread,
Mar 15, 2002, 2:48:05 AM3/15/02
to
Making another check mark against the details category.

"Charles Johnson" <qru...@oneimage.com> wrote in message
news:a6s84e$fa7$1...@suaar1aa.prod.compuserve.com...

Tony Williams

unread,
Mar 15, 2002, 3:40:23 AM3/15/02
to
In article <3C9162EF...@sandia.gov>,

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

unread,
Mar 15, 2002, 3:59:21 AM3/15/02
to
The renowned Tony Williams <to...@ledelec.demon.co.uk> wrote:

> We could patent that. Need a snappy name for it
> though. How about "2-resistor_plus_centred-pot
> differential amplifier"?

"Williams bridge" ;-)

Brad Albing

unread,
Mar 15, 2002, 8:27:55 AM3/15/02
to
Looks like with all this simple, straightforward thinking, we've scared off the
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

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Mar 15, 2002, 9:15:54 AM3/15/02
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In article <JJik8.44$5I...@news2.bloor.is>,
Spehro Pefhany <sp...@interlog.com> wrote:

> "Williams bridge" ;-)

Now, now Speff..... That's going a "Bridge Too Far".

--
Tony Williams.

Frank Bemelman

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Mar 15, 2002, 12:14:22 PM3/15/02
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"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

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Mar 15, 2002, 12:19:43 PM3/15/02