Opamp resistor magnitude

Phil Hobbs · Apr 28, 2014

On 04/28/2014 01:03 PM, George Herold wrote:

On Monday, April 28, 2014 12:39:54 PM UTC-4, Phil Hobbs wrote:
On 04/28/2014 11:06 AM, Gone Postal wrote:

On Mon, 28 Apr 2014 10:13:46 -0400, Phil Hobbs

snip previous stuff

There's also noise to worry about, of course. Your garden variety op
amp has an input noise voltage of very roughly 10 nV/sqrt(Hz), which is
about the same as the thermal (Johnson) noise of a 6k resistor. So
going too high will cost you voltage noise performance.

The other nice thing about 10 k ohm, is you can put a 1/4W 10k across
the +/- 15V rail and *not* let the magic smoke out.
And 10 k is very close the the "quantum" of conductance.
2*e**2/h =~ 1/12.9k ohm) Coincidence? I don't think so.
I think God also likes ~10k ohm :^)
(please excuse the extreme hubris in the above.)

George H.

Maybe that's the Imperial unit of resistance. One attoparsec equals one
decifoot, to within observational error (at least mine), which proves
once again (if any further proof were needed) that God is an Englishman.

(Of course those sillies have gone all metric on us, but I digress.)

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

Apr 28, 2014

John Fields <jfields@austininstruments.com> wrote:

On Mon, 28 Apr 2014 13:32:10 -0400, Phil Hobbs
pcdhSpamMeSenseless@electrooptical.net> wrote:

Maybe that's the Imperial unit of resistance. One attoparsec equals
one decifoot, to within observational error (at least mine), which
proves once again (if any further proof were needed) that God is an
Englishman.

Well, now that the sun manages to set on the British Empire, maybe
He can take a Saturday off every now and then...

Actually, it still hasn't. Thank the Pitcairn Islands.

http://what-if.xkcd.com/48/

Matt Roberds

John Fields · Apr 29, 2014

On Mon, 28 Apr 2014 12:54:20 -0400, Gone Postal
<gone_postal@it.doesn't.exist> wrote:

John Fields, thank you very, very much for that thoughtful
explanation, it covers exactly what I was wanting to know but didn't
quite have the knowledge to ask correctly.

I had thought I understood op amps thoroughly, having used them in
some light-weight audio level circuitry in the past. I'm trying to
thoroughly understand them as I try to understand everything else,
before I embark on designing anything substantial. The most I've
managed to design and build from the ground up so far was a power
supply with current fold-back, from discrete parts, but it was a seat
of the pants design. I want to understand what's really happening in
a circuit so I can understand more or less what to expect and
understand why I expect it before I ever breadboard anything.

And thank you for treating a basic problem from a first time asker for
what it was.
Gone Postal

---
My pleasure.

BTW, here's a link to a handy opamp design reference:

http://www.ti.com/lit/an/slod006b/slod006b.pdf

John Fields

Jim Thompson · Apr 29, 2014

On Mon, 28 Apr 2014 13:32:10 -0400, Phil Hobbs
<pcdhSpamMeSenseless@electrooptical.net> wrote:

One attoparsec equals one
decifoot, to within observational error (at least mine), which proves
once again (if any further proof were needed) that God is an Englishman.

Sno-o-o-ort >:-}

(Of course those sillies have gone all metric on us, but I digress.)

Cheers

Phil Hobbs

They tried metric on all the highway signs around here, but their use
has gradually faded away... might have been all the bullet holes they
collected >:-}

...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: Contacts Only | |
| Voice

Jim Thompson · Apr 29, 2014

On Mon, 28 Apr 2014 12:55:57 -0400, Gone Postal
<gone_postal@it.doesn't.exist> wrote:

On Mon, 28 Apr 2014 23:25:50 +1000, "Phil Allison" <phil_a@tpg.com.au
wrote:

"Gone Postal = Wanker "

Thank you George. I've been following that as a general rule of thumb
(staying in the huuped singled K and the general 10's of Kohm range),
but I was seeing circuits that had other orders of magnatude and I was
just starting to wonder why so high, in particular with two of three
circuits I saw that just struck me as a wee bid odd..

** FFS - why not post links to them ????

You pompous, boring, illiterate, anonymous fart.

.... Phil

Because they're in Books and you can't link to a book.

GP

Jim Thompson · Apr 29, 2014

On Mon, 28 Apr 2014 12:54:20 -0400, Gone Postal
<gone_postal@it.doesn't.exist> wrote:

On Mon, 28 Apr 2014 10:30:15 -0500, John Fields
jfields@austininstruments.com> wrote:

On Mon, 28 Apr 2014 00:44:23 -0400, Gone Postal
gone_postal@it.doesn't.exist> wrote:

I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved? I've seen different circuits with resistors of, say
thousands, tens or hundered of thousands and even one that used
Megohm-range resistors.
Why the variation in magnitude of the resistors, assuming a fixed
ratio? Does it have anything to do with the overall impedance of the
opamp circuit block?
I thank you for whatever illumination that can be provided.

GP

---
Using a fixed-pitch font to display the basic configurations for
inverting and non-inverting opamps we have, for the supply
connections:

. V+
. |
. --|-\
. | >--
. --|+/
. |
. V-

and for the inverting configuration:

. +--[R2]--+
. | |
.Vin>--[R1]-+--|-\ |
. | >--+-->Vout
.GND/0V>-------|+/

Note that with the non-inverting (+) input at 0V, then Vout must
assume whatever value is required to force the - input to whatever
voltage is on the + input; 0V.

If Vin is at 1V, and if R1 = R2, then we have a voltage divider that
looks like this:

1V
|
[R1]
|
+---0V
|
[R2]
|
-1V

where -1V is the output of the opamp and 0V is the opamp's - input.

Accordingly, as the input voltage swings positive and negative, then
the output voltage will swing with the same magnitude as the input,
but 180 degrees out of phase.

The circuit, then, has a voltage gain of -1.

Assuming, now, that the signal generator driving R1 has an output
impedance of 10kohms and that R1 and R2 are each 10k, our circuit
now looks like this:

. SIGNAL 10k
. GENERATOR +--[R2]--+
. +-----------+ 10k | |
. | +--[Rg]--|--[R1]-+--|-\ |
. | | 10k | | >--+-->Vout
. | OSC | +--|+/
. +-----------+ |
. GND

If the signal generator has an open-circuit output of 1 volt, when
it's connected to R1 (since the end of R1 connected to the minus
input of the opamp will be at 0V) the current through RgR1 will be:

E 1V
I = --------- = ------ = 50 microamps
Rg + R1 20kR

Now, since Rg, R1, and R2 are in series and the current into the -
input of the opamp is miniscule, the current through Rg and R1 also
flows through R2.

Then, since R2 = 10kR and the current through it is 50 microamps,
the voltage across it must be:

E = IR = 50ľA * 10kR = 0.5V

Therefore, even though R1 = R2, the gain of the [entire] circuit
won't be -1 because of the effect of R1 loading the source.

Knowing that, and taking the source impedance into consideration,
the resistances of R1 and R2 can be selected so that the opamp's
output voltage will be whatever's desired.

The resistor selection criteria for the non-inverting amp are
arguably less stringent with, roughly, not starving the opamp -
input for bias current on one end and minimizing the noise into the
opamp on the other.

John Fields

John Fields, thank you very, very much for that thoughtful
explanation, it covers exactly what I was wanting to know but didn't
quite have the knowledge to ask correctly.

I had thought I understood op amps thoroughly, having used them in
some light-weight audio level circuitry in the past. I'm trying to
thoroughly understand them as I try to understand everything else,
before I embark on designing anything substantial. The most I've
managed to design and build from the ground up so far was a power
supply with current fold-back, from discrete parts, but it was a seat
of the pants design. I want to understand what's really happening in
a circuit so I can understand more or less what to expect and
understand why I expect it before I ever breadboard anything.

And thank you for treating a basic problem from a first time asker for
what it was.
Gone Postal

John Fields · Apr 29, 2014

On Mon, 28 Apr 2014 13:32:10 -0400, Phil Hobbs
<pcdhSpamMeSenseless@electrooptical.net> wrote:

Maybe that's the Imperial unit of resistance. One attoparsec equals one
decifoot, to within observational error (at least mine), which proves
once again (if any further proof were needed) that God is an Englishman.

---
Well, now that the sun manages to set on the British Empire, maybe
He can take a Saturday off every now and then...

John Larkin · Apr 29, 2014

On Mon, 28 Apr 2014 00:44:23 -0400, Gone Postal
<gone_postal@it.doesn't.exist> wrote:

I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved? I've seen different circuits with resistors of, say
thousands, tens or hundered of thousands and even one that used
Megohm-range resistors.
Why the variation in magnitude of the resistors, assuming a fixed
ratio? Does it have anything to do with the overall impedance of the
opamp circuit block?
I thank you for whatever illumination that can be provided.

GP

Low value resistors could load the opamp output, or in low-power
applications use too much power. If they self-heat, or cause the opamp
to heat, you can get weird thermal transients, usually only an issue
in super-precision circuits.

If the resistors are too high, you can get...

more Johnson noise

more noise from opamp input noise current

Noise peaking from the pole (RC rolloff) that's created inside the
feedback loop by the resistors and the opamp input capacitance

Ringing/oscillation, same reason

DC errors, from opamp input bias current or PCB leakage.

Current-mode opamps also care about the impedance seen at the
inverting input; it directly affects the opamp gain-bandwidth.

--

John Larkin Highland Technology, Inc

jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com

David Eather · Apr 29, 2014

On Tue, 29 Apr 2014 06:28:51 +1000, David Eather <eather@tpg.com.au> wrote:

On Mon, 28 Apr 2014 23:25:50 +1000, Phil Allison <phil_a@tpg.com.au
wrote:

"Gone Postal = Wanker "

Thank you George. I've been following that as a general rule of thumb
(staying in the huuped singled K and the general 10's of Kohm range),
but I was seeing circuits that had other orders of magnatude and I was
just starting to wonder why so high, in particular with two of three
circuits I saw that just struck me as a wee bid odd..

** FFS - why not post links to them ????

You pompous, boring, illiterate, anonymous fart.

.... Phil

Anonymouse? From O Nil Phallis? Your last significant creation was
what? You will go down unmourned and unloved. Not to mention speedily
forgotten. A well deserved legacy for a POS like you.

David Eather · Apr 29, 2014

On Tue, 29 Apr 2014 06:28:51 +1000, David Eather <eather@tpg.com.au> wrote:

On Mon, 28 Apr 2014 23:20:09 +1000, wrote:

On Mon, 28 Apr 2014 06:04:10 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Monday, April 28, 2014 12:44:23 AM UTC-4, Gone Postal wrote:
I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved? I've seen different circuits with resistors of, say
thousands, tens or hundered of thousands and even one that used
Megohm-range resistors.
Why the variation in magnitude of the resistors, assuming a fixed
ratio? Does it have anything to do with the overall impedance of the
opamp circuit block?
I thank you for whatever illumination that can be provided.

I think 10k ohm resistors are my favorite. (then 1 k ohm)
In general I'd like to choose R's that are smaller.
First this tends to reduce the effects of stray capacitance.
(Some RC corner that move to higher frequency with smaller R)
And then if you care, smaller R has less noise.. though most of the
time that doesn't matter much.
Now of course you can't make 'em too small.
First the opamp mya not have enough poop to drive it.
and second if the R's get very small (a few ohms) you then
start to worry about the resistance of the traces and contacts.

George H.

GP
Thank you George. I've been following that as a general rule of thumb
(staying in the huuped singled K and the general 10's of Kohm range),
but I was seeing circuits that had other orders of magnatude and I was
just starting to wonder why so high, in particular with two of three
circuits I saw that just struck me as a wee bid odd..

GP

Please understand that Phil Allison is a malignant narcissist. He heaps
unwarranted abuse on anyone if he thinks he can get away with it. The
reason he does this is that it makes him feel superior to berate people
and put them down, It makes him feel good. In part it is compensation
for his own lack of achievement - he failed at uni and his personality
won't allow him to work with "normal" people. He repairs toasters and
any other bits of electronics he can get on his kitchen table. Of course
his personality is so toxic that he lives alone. He seems to be a rather
good repair man, but please don't mistake that for anything "special" -
there would be thousands as good or better in Australia and 10's of
thousands in the USA.

NOTE to people of sci.electronics: this is exactly the behavior of PA
that I have been saying is unacceptable. A newbie politely asks a
question - of course he doesn't get the question totally right because
he is still learning - and because the OP is polite or possibly hesitant
PA starts to heap abuse on them. It is not acceptable to do this to
anyone!

Gone Postal · Apr 29, 2014

On Mon, 28 Apr 2014 13:46:25 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Mon, 28 Apr 2014 12:54:20 -0400, Gone Postal
gone_postal@it.doesn't.exist> wrote:

John Fields, thank you very, very much for that thoughtful
explanation, it covers exactly what I was wanting to know but didn't
quite have the knowledge to ask correctly.

I had thought I understood op amps thoroughly, having used them in
some light-weight audio level circuitry in the past. I'm trying to
thoroughly understand them as I try to understand everything else,
before I embark on designing anything substantial. The most I've
managed to design and build from the ground up so far was a power
supply with current fold-back, from discrete parts, but it was a seat
of the pants design. I want to understand what's really happening in
a circuit so I can understand more or less what to expect and
understand why I expect it before I ever breadboard anything.

And thank you for treating a basic problem from a first time asker for
what it was.
Gone Postal

---
My pleasure.

BTW, here's a link to a handy opamp design reference:

http://www.ti.com/lit/an/slod006b/slod006b.pdf

John Fields

Thank you very much for the reference!

GP

Phil Allison · Apr 29, 2014

"Phil Hobbs"

** Hey dil,

got any idea what over answering is?

Or that YOU are feeding a very obvious troll ?

My god you are one dumb shithead.

.... Phil

<pcdhSpamMeSenseless@electrooptical.net> wrote in message
news:VOWdnW1jft6H_8POnZ2dnUVZ_oSdnZ2d@supernews.com...

On 04/28/2014 03:06 AM, Gone Postal wrote:
On 28 Apr 2014 06:13:27 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2014-04-28, Gone Postal <gone_postal@it.doesn't.exist> wrote:
I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved?

mainly the impedance of the input.
the imput will have high resistance but not infinite.

Thanks for the answer, that's what I was generally getting at.

Capacitance is very important as well, both input and feedback.

Op amps always have at least 1.5 pF of input capacitance, and usually more
than that. Say you have a unity gain inverter with 1 meg input and
feedback resistors and Cin = 3pF.

Your feedback network will roll off starting at

f_c = 1/(2 pi (3pF) (500k)) = 106 kHz.

That means that the output swing will start to rise there, to keep the
input in balance. So you wind up with a gain peak. If you had picked 10k
instead of 1 meg, the peak would start 100 times higher in frequency, i.e.
10.6 MHz, which is liable to be outside the amplifier bandwidth, so you
wouldn't see it unless the op amp were reasonably fast.

The shunt capacitance of the feedback resistor is also important, though
it doesn't really matter for unity gain inverters, since the effect is
about the same on each one. It's typically about 0.12 pF for a 1/4 W
axial resistor and 0.05 pF for an 0603. (You also have to watch out for
the capacitance between pads--a nearby ground plane is your friend here.)

However, if you had an amp with a gain of -100, i.e. 1 meg on the input
and 100 meg feedback, that 0.05 pF starts to dominate the feedback at

f_RC = 1/(2 pi (50 fF) 100 meg) = 32 kHz.

Layout problems will make this worse.

Adjusting the feedback capacitance to cancel out the effect of the input
capacitance is a typical way of controlling these problems when you can't
just use lower impedances.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

Phil Allison · Apr 29, 2014

"Phil Hobbs"

** Hey dil,

got any idea what over answering is?

Or that YOU are feeding a very obvious troll ?

My god you are one dumb shithead.

.... Phil

Phil Allison · Apr 29, 2014

"Jim Thompson"

** Hey dil,

got any idea what over answering is?

Or that YOU are feeding a very obvious troll ?

My god you are one dumb shithead.

.... Phil

Phil Allison · Apr 29, 2014

"Gone Postal"
"Phil Allison"

"Gone Postal = Wanker "

Thank you George. I've been following that as a general rule of thumb
(staying in the huuped singled K and the general 10's of Kohm range),
but I was seeing circuits that had other orders of magnatude and I was
just starting to wonder why so high, in particular with two of three
circuits I saw that just struck me as a wee bid odd..

** FFS - why not post links to them ????

You pompous, boring, illiterate, anonymous fart.

Because they're in Books and you can't link to a book.

** FFS wot a wanker !!!!

Find similar examples on the net and link them.

You are such a damn obvious LIAR.

..... Phil

Phil Allison · Apr 29, 2014

"Jim Thompson"

** Drop dead you stinking septic cunt.

<To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote in message
news:7f9tl9l6d7eicchogr0uueqo9tchdlogcv@4ax.com...

On Mon, 28 Apr 2014 12:55:57 -0400, Gone Postal
gone_postal@it.doesn't.exist> wrote:

On Mon, 28 Apr 2014 23:25:50 +1000, "Phil Allison" <phil_a@tpg.com.au
wrote:

"Gone Postal = Wanker "

Thank you George. I've been following that as a general rule of thumb
(staying in the huuped singled K and the general 10's of Kohm range),
but I was seeing circuits that had other orders of magnatude and I was
just starting to wonder why so high, in particular with two of three
circuits I saw that just struck me as a wee bid odd..

** FFS - why not post links to them ????

You pompous, boring, illiterate, anonymous fart.

.... Phil

Because they're in Books and you can't link to a book.

GP

The general rule around here is to just ignore Allison's posts... just
filter him into oblivion.

...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

Phil Allison · Apr 29, 2014

"John Fields"

** Hey dil,

got any idea what over answering is?

Or that YOU are feeding a very obvious troll ?

My god you are one dumb shithead.

.... Phil

<jfields@austininstruments.com> wrote in message
news:uolsl9hcdqo6gmp6d3snmlbsifmaiqv1eh@4ax.com...

On Mon, 28 Apr 2014 00:44:23 -0400, Gone Postal
gone_postal@it.doesn't.exist> wrote:

I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved? I've seen different circuits with resistors of, say
thousands, tens or hundered of thousands and even one that used
Megohm-range resistors.
Why the variation in magnitude of the resistors, assuming a fixed
ratio? Does it have anything to do with the overall impedance of the
opamp circuit block?
I thank you for whatever illumination that can be provided.

GP

---
Using a fixed-pitch font to display the basic configurations for
inverting and non-inverting opamps we have, for the supply
connections:

. V+
. |
. --|-\
. | >--
. --|+/
. |
. V-

and for the inverting configuration:

. +--[R2]--+
. | |
.Vin>--[R1]-+--|-\ |
. | >--+-->Vout
.GND/0V>-------|+/

Note that with the non-inverting (+) input at 0V, then Vout must
assume whatever value is required to force the - input to whatever
voltage is on the + input; 0V.

If Vin is at 1V, and if R1 = R2, then we have a voltage divider that
looks like this:

1V
|
[R1]
|
+---0V
|
[R2]
|
-1V

where -1V is the output of the opamp and 0V is the opamp's - input.

Accordingly, as the input voltage swings positive and negative, then
the output voltage will swing with the same magnitude as the input,
but 180 degrees out of phase.

The circuit, then, has a voltage gain of -1.

Assuming, now, that the signal generator driving R1 has an output
impedance of 10kohms and that R1 and R2 are each 10k, our circuit
now looks like this:

. SIGNAL 10k
. GENERATOR +--[R2]--+
. +-----------+ 10k | |
. | +--[Rg]--|--[R1]-+--|-\ |
. | | 10k | | >--+-->Vout
. | OSC | +--|+/
. +-----------+ |
. GND

If the signal generator has an open-circuit output of 1 volt, when
it's connected to R1 (since the end of R1 connected to the minus
input of the opamp will be at 0V) the current through RgR1 will be:

E 1V
I = --------- = ------ = 50 microamps
Rg + R1 20kR

Now, since Rg, R1, and R2 are in series and the current into the -
input of the opamp is miniscule, the current through Rg and R1 also
flows through R2.

Then, since R2 = 10kR and the current through it is 50 microamps,
the voltage across it must be:

E = IR = 50ľA * 10kR = 0.5V

Therefore, even though R1 = R2, the gain of the [entire] circuit
won't be -1 because of the effect of R1 loading the source.

Knowing that, and taking the source impedance into consideration,
the resistances of R1 and R2 can be selected so that the opamp's
output voltage will be whatever's desired.

The resistor selection criteria for the non-inverting amp are
arguably less stringent with, roughly, not starving the opamp -
input for bias current on one end and minimizing the noise into the
opamp on the other.

John Fields

Phil Allison · Apr 29, 2014

"John Fields"

** Hey dil,

got any idea what over answering is?

Or that YOU are feeding a very obvious troll ?

My god you are one dumb shithead.

.... Phil

Phil Allison · Apr 29, 2014

"Gone Postal"

** Piss OFF you damn TROLL !!!!

John Fields, thank you very, very much for that thoughtful
explanation, it covers exactly what I was wanting to know but didn't
quite have the knowledge to ask correctly.

I had thought I understood op amps thoroughly, having used them in
some light-weight audio level circuitry in the past. I'm trying to
thoroughly understand them as I try to understand everything else,
before I embark on designing anything substantial. The most I've
managed to design and build from the ground up so far was a power
supply with current fold-back, from discrete parts, but it was a seat
of the pants design. I want to understand what's really happening in
a circuit so I can understand more or less what to expect and
understand why I expect it before I ever breadboard anything.

And thank you for treating a basic problem from a first time asker for
what it was.
Gone Postal

---
My pleasure.

BTW, here's a link to a handy opamp design reference:

http://www.ti.com/lit/an/slod006b/slod006b.pdf

John Fields

Thank you very much for the reference!

GP

George Herold · Apr 29, 2014

On Tuesday, April 29, 2014 9:25:17 AM UTC-4, Phil Hobbs wrote:

On 4/29/2014 3:59 AM, Phil Allison wrote:

Trolls need love too, Phil.

Phil (H.), Did you ever read any of the disc-world novels by Terry Pratchet?
He's got that dry British wit similar to Doug Adams. (And nice characters.)
I'm just polishing one off...(that has lovable Trolls in it.) You need to skip the early ones and start in the middle somewhere. And to keep this on topic... one novel was called "Going Postal".

George H.

Cheers

Phil Hobbs

--

Dr Philip C D Hobbs

Principal Consultant

ElectroOptical Innovations LLC

Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203

Briarcliff Manor NY 10510

hobbs at electrooptical dot net

http://electrooptical.net

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