On Thu, 04 Aug 2011 14:53:33 -0700, John Larkin
wrote:

On Thu, 04 Aug 2011 15:07:00 -0500, John Fields
wrote:

On Thu, 04 Aug 2011 09:17:41 -0700, John Larkin
wrote:

On Thu, 04 Aug 2011 10:17:00 -0500, John Fields
wrote:

On Wed, 03 Aug 2011 20:56:57 -0700, John Larkin
wrote:

On Wed, 03 Aug 2011 22:35:35 -0500, "
wrote:

On Wed, 03 Aug 2011 19:25:05 -0700, John Larkin
wrote:

On Wed, 03 Aug 2011 19:22:36 -0500, "
wrote:

On Wed, 03 Aug 2011 17:06:22 -0700, John Larkin
m wrote:

On Wed, 03 Aug 2011 18:56:30 -0500, "
wrote:

On Wed, 03 Aug 2011 16:46:11 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART. com wrote:

On Wed, 03 Aug 2011 15:07:32 -0400, Archon
wrote:

On 8/3/2011 11:39 AM, Lanny wrote:

but this resistence have a polarity?
---
No.

--
JF

They have a particular name these little resistance?

Regards


They do when I drop one and can't find it
JC


They work better at high frequencies if you install them upside down.
Less inductance.

Less inductance or higher capacitance (resistive element closer to the plane)?

What I see in a TDR setup is less series inductance. The context is a
coplanar waveguide PCB trace with a gap that's bridged by the
resistor.

ftp://jjlarkin.lmi.net/0805_res_fix.JPG

The impedance bump at cm 3.5 is inductive...

ftp://jjlarkin.lmi.net/0805_normal.JPG

and if you flip it over, it's a lot smaller

ftp://jjlarkin.lmi.net/0805_flipped.JPG

But that's telling you that the sqrt(L/C) is closer to the transmission line
Z0, no? C will certainly be higher with the resistor closer to the plane.


I'd have to think about that. The normal way, you'd have the substrate
(alumina, Er around 10) down, and then the PCB, Er more like 4.6
maybe, and air on top. Inverted, the resistance element sees the PCB
looking down, and alumina+air looking up. Too complex for my tiny
brain, especially after the day I've had.

---
Excuses?
---

But the alumina is the same in both orientations. Down adds the PCB plane
capacitance.

But the TDR sure looks inductive in both cases. Effective bandwidth is
around 12 GHz. 1-cent 0805 resistors are pretty good way up into the
GHz.

Less inductive or more capacitive (canceling a constant inductance) = higher
impedance?

I'm an engineer, not a philosopher. What I see here is inductance. I
assume that the issue is loop area, and less area is less L.

---
Since loop area doesn't change, then a decrease in inductance can't be
linked to that.

Of course it changes. Think about it for Pete's sake.

---
Maybe we're talking apples and oranges; what do you mean by loop area
and what's the thickness of the FR4 on the PCB?

If the resistor is mounted upside-down, the current flow is close to
the PCB.

If it's mounted the normal way, the current has to climb up one end
cap, cross over the top, and go back down the other cap. That makes an
arch, which encloses loop area. That makes inductance.

The FR4 thickness doesn't affect this inductance.

---
But it does affect the capacitance, so if the resistive element is
located closer to the ground plane, then the capacitance will
increase.
---

test board was chopped out of this

ftp://jjlarkin.lmi.net/Z250A.jpg

which is 0.062 FR4 with a ground plane 20 mils down from the top.

---
Thank you. That 20 mils of thickness between the resistive element and
the ground plane was what I was looking for.
---

I do test circuits like this now and then and toss in little adapters,
filter layouts, anything that might be handy for experimenting.

---
As do most of the rest of us, according to our disciplines, so why do
you find it necessary to single yourself out and exalt yourself as
unique by patting yourself on the back?
---

Super microwave resistors have no end caps at all, just the resistive
element and two solderable end zones, all planar. You mount these
element down, of course.

---
Self-serving platitudes.

The issue here is whether the decrease in inductive reactance
indicated by your TDR is due to the lowering of inductance due to the
shortening of the connections to the end caps, or is due to the
increase of capacitance due to the resistive element moving closer to
the ground plane.

Got some numbers?






--
JF

On Thu, 04 Aug 2011 21:34:19 -0700, John Larkin
wrote:

On Thu, 04 Aug 2011 20:56:09 -0500, Jeffrey Angus
wrote:

On 8/4/2011 5:49 PM, John Larkin wrote:

Can't afford scroll bars?

Shouldn't have to scroll down 100 lines of crap to read
a 2 line response.

What part of that don't you understand?


Design any interesting electronics lately?

---
Standard line for you when you've been sussed, huh?

--
JF

On Sat, 06 Aug 2011 17:56:03 -0500, John Fields
wrote:

On Thu, 04 Aug 2011 21:34:19 -0700, John Larkin
wrote:

On Thu, 04 Aug 2011 20:56:09 -0500, Jeffrey Angus
wrote:

On 8/4/2011 5:49 PM, John Larkin wrote:

Can't afford scroll bars?

Shouldn't have to scroll down 100 lines of crap to read
a 2 line response.

What part of that don't you understand?


Design any interesting electronics lately?

---
Standard line for you when you've been sussed, huh?

It's a good question to put to whiney lurkers. In an electronics
newsgroup, credibility comes from electronics.

And speaking of whiney lurkers, design any interesting electronics
lately?


John

On Sat, 06 Aug 2011 17:45:03 -0500, John Fields
wrote:

On Thu, 04 Aug 2011 14:53:33 -0700, John Larkin
wrote:

On Thu, 04 Aug 2011 15:07:00 -0500, John Fields
wrote:

On Thu, 04 Aug 2011 09:17:41 -0700, John Larkin
wrote:

On Thu, 04 Aug 2011 10:17:00 -0500, John Fields
wrote:

On Wed, 03 Aug 2011 20:56:57 -0700, John Larkin
wrote:

On Wed, 03 Aug 2011 22:35:35 -0500, "
wrote:

On Wed, 03 Aug 2011 19:25:05 -0700, John Larkin
wrote:

On Wed, 03 Aug 2011 19:22:36 -0500, "
wrote:

On Wed, 03 Aug 2011 17:06:22 -0700, John Larkin
om wrote:

On Wed, 03 Aug 2011 18:56:30 -0500, "
wrote:

On Wed, 03 Aug 2011 16:46:11 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART .com wrote:

On Wed, 03 Aug 2011 15:07:32 -0400, Archon
wrote:

On 8/3/2011 11:39 AM, Lanny wrote:

but this resistence have a polarity?
---
No.

--
JF

They have a particular name these little resistance?

Regards


They do when I drop one and can't find it
JC


They work better at high frequencies if you install them upside down.
Less inductance.

Less inductance or higher capacitance (resistive element closer to the plane)?

What I see in a TDR setup is less series inductance. The context is a
coplanar waveguide PCB trace with a gap that's bridged by the
resistor.

ftp://jjlarkin.lmi.net/0805_res_fix.JPG

The impedance bump at cm 3.5 is inductive...

ftp://jjlarkin.lmi.net/0805_normal.JPG

and if you flip it over, it's a lot smaller

ftp://jjlarkin.lmi.net/0805_flipped.JPG

But that's telling you that the sqrt(L/C) is closer to the transmission line
Z0, no? C will certainly be higher with the resistor closer to the plane.


I'd have to think about that. The normal way, you'd have the substrate
(alumina, Er around 10) down, and then the PCB, Er more like 4.6
maybe, and air on top. Inverted, the resistance element sees the PCB
looking down, and alumina+air looking up. Too complex for my tiny
brain, especially after the day I've had.

---
Excuses?
---

But the alumina is the same in both orientations. Down adds the PCB plane
capacitance.

But the TDR sure looks inductive in both cases. Effective bandwidth is
around 12 GHz. 1-cent 0805 resistors are pretty good way up into the
GHz.

Less inductive or more capacitive (canceling a constant inductance) = higher
impedance?

I'm an engineer, not a philosopher. What I see here is inductance. I
assume that the issue is loop area, and less area is less L.

---
Since loop area doesn't change, then a decrease in inductance can't be
linked to that.

Of course it changes. Think about it for Pete's sake.

---
Maybe we're talking apples and oranges; what do you mean by loop area
and what's the thickness of the FR4 on the PCB?

If the resistor is mounted upside-down, the current flow is close to
the PCB.

If it's mounted the normal way, the current has to climb up one end
cap, cross over the top, and go back down the other cap. That makes an
arch, which encloses loop area. That makes inductance.

The FR4 thickness doesn't affect this inductance.

---
But it does affect the capacitance, so if the resistive element is
located closer to the ground plane, then the capacitance will
increase.
---

test board was chopped out of this

ftp://jjlarkin.lmi.net/Z250A.jpg

which is 0.062 FR4 with a ground plane 20 mils down from the top.

---
Thank you. That 20 mils of thickness between the resistive element and
the ground plane was what I was looking for.
---

I do test circuits like this now and then and toss in little adapters,
filter layouts, anything that might be handy for experimenting.

---
As do most of the rest of us, according to our disciplines, so why do
you find it necessary to single yourself out and exalt yourself as
unique by patting yourself on the back?

Show us some of the PC boards you've designed lately.


---

Super microwave resistors have no end caps at all, just the resistive
element and two solderable end zones, all planar. You mount these
element down, of course.

---
Self-serving platitudes.

Facts about microwave resistors. Look it up.


The issue here is whether the decrease in inductive reactance
indicated by your TDR is due to the lowering of inductance due to the
shortening of the connections to the end caps, or is due to the
increase of capacitance due to the resistive element moving closer to
the ground plane.

Got some numbers?

I already did a rough calculation on the capacitance. The resistor is
soldered across a small gap in the trace, so in fact most of the
capacitance of the resistor element is to the trace, not to the PCB.
And there's a sizable air gap between the resistor element and the
pcb, caused by the solder.

John

On Sat, 06 Aug 2011 22:20:14 -0700, John Larkin
wrote:

On Sat, 06 Aug 2011 17:56:03 -0500, John Fields
wrote:

On Thu, 04 Aug 2011 21:34:19 -0700, John Larkin
wrote:

On Thu, 04 Aug 2011 20:56:09 -0500, Jeffrey Angus
wrote:

On 8/4/2011 5:49 PM, John Larkin wrote:

Can't afford scroll bars?

Shouldn't have to scroll down 100 lines of crap to read
a 2 line response.

What part of that don't you understand?


Design any interesting electronics lately?

---
Standard line for you when you've been sussed, huh?

It's a good question to put to whiney lurkers. In an electronics
newsgroup, credibility comes from electronics.

---
I'd suspect your credibility to be seriously damaged, then, by virtue
of the amount of non-electronic garbage you post.
---

And speaking of whiney lurkers, design any interesting electronics
lately?

---
I hardly think that slapping you around can be referred to as "whiney
lurking" and the answer to your question is "yes".

--
JF

On Sat, 06 Aug 2011 22:28:07 -0700, John Larkin
wrote:

On Sat, 06 Aug 2011 17:45:03 -0500, John Fields
wrote:

On Thu, 04 Aug 2011 14:53:33 -0700, John Larkin
wrote:
..
..
..

I do test circuits like this now and then and toss in little adapters,
filter layouts, anything that might be handy for experimenting.

---
As do most of the rest of us, according to our disciplines, so why do
you find it necessary to single yourself out and exalt yourself as
unique by patting yourself on the back?

Show us some of the PC boards you've designed lately.

---
PCB???

Test circuits have to be on PCB's?
---

Super microwave resistors have no end caps at all, just the resistive
element and two solderable end zones, all planar. You mount these
element down, of course.

---
Self-serving platitudes.

Facts about microwave resistors. Look it up.

---
Sure, but what's the point other than for you to flap your gums?
---

The issue here is whether the decrease in inductive reactance
indicated by your TDR is due to the lowering of inductance due to the
shortening of the connections to the end caps, or is due to the
increase of capacitance due to the resistive element moving closer to
the ground plane.

Got some numbers?

I already did a rough calculation on the capacitance. The resistor is
soldered across a small gap in the trace, so in fact most of the
capacitance of the resistor element is to the trace, not to the PCB.
And there's a sizable air gap between the resistor element and the
pcb, caused by the solder.

---
OK, if you want to play that way, then everything else being equal
means that with the element up, the alumina substrate will be forcing
the element farther away from the cut trace than it would be with the
element down.

Such being the case means that the capacitance between the element and
the cut trace will be greater when the element is down than when it's
up, neglecting the Er of the alumina with R up.

So, even if the inductance is lowered by R down, C will be increased,
making Z closer to the line's impedance and, therefore, lowering the
bump on the TDR.

So, all that's needed is to resolve the difference between C up and
C down WRT to the alumina substrate with R up.

Wanna take a crack at it?

--
JF

On Sun, 07 Aug 2011 07:28:06 -0500, John Fields
wrote:

On Sat, 06 Aug 2011 22:20:14 -0700, John Larkin
wrote:

On Sat, 06 Aug 2011 17:56:03 -0500, John Fields
wrote:

On Thu, 04 Aug 2011 21:34:19 -0700, John Larkin
wrote:

On Thu, 04 Aug 2011 20:56:09 -0500, Jeffrey Angus
wrote:

On 8/4/2011 5:49 PM, John Larkin wrote:

Can't afford scroll bars?

Shouldn't have to scroll down 100 lines of crap to read
a 2 line response.

What part of that don't you understand?


Design any interesting electronics lately?

---
Standard line for you when you've been sussed, huh?

It's a good question to put to whiney lurkers. In an electronics
newsgroup, credibility comes from electronics.

---
I'd suspect your credibility to be seriously damaged, then, by virtue
of the amount of non-electronic garbage you post.

My electronic credibility is independent of my opinions on snow, beer,
politics, or drugstore cowboys.

---

And speaking of whiney lurkers, design any interesting electronics
lately?

---
I hardly think that slapping you around can be referred to as "whiney
lurking" and the answer to your question is "yes".

Cool! We are all impressed by that.

John

On Sun, 07 Aug 2011 08:47:52 -0500, John Fields
wrote:

On Sat, 06 Aug 2011 22:28:07 -0700, John Larkin
wrote:

On Sat, 06 Aug 2011 17:45:03 -0500, John Fields
wrote:

On Thu, 04 Aug 2011 14:53:33 -0700, John Larkin
wrote:
.
.
.

I do test circuits like this now and then and toss in little adapters,
filter layouts, anything that might be handy for experimenting.

---
As do most of the rest of us, according to our disciplines, so why do
you find it necessary to single yourself out and exalt yourself as
unique by patting yourself on the back?

Show us some of the PC boards you've designed lately.

---
PCB???

Test circuits have to be on PCB's?

Controlled-impedance ones sure so. And few cool parts are available
any more in thru-hole, so a PCB is necessary to do any serious
prototyping. You can sometimes get away with adapters, like the Bellin
things, and a lot of wire, but a PCB saves a lot of time for
nontrivial stuff. And, as noted, you can add a lot of other useful
gidgets to the layout, saw them off, do fun things with them off to
the side.

You don't design PCBs?




---

Super microwave resistors have no end caps at all, just the resistive
element and two solderable end zones, all planar. You mount these
element down, of course.

---
Self-serving platitudes.

Facts about microwave resistors. Look it up.

---
Sure, but what's the point other than for you to flap your gums?

It's an _Electronics_Discussion_Group!


---

The issue here is whether the decrease in inductive reactance
indicated by your TDR is due to the lowering of inductance due to the
shortening of the connections to the end caps, or is due to the
increase of capacitance due to the resistive element moving closer to
the ground plane.

Got some numbers?

I already did a rough calculation on the capacitance. The resistor is
soldered across a small gap in the trace, so in fact most of the
capacitance of the resistor element is to the trace, not to the PCB.
And there's a sizable air gap between the resistor element and the
pcb, caused by the solder.

---
OK, if you want to play that way, then everything else being equal
means that with the element up, the alumina substrate will be forcing
the element farther away from the cut trace than it would be with the
element down.

Such being the case means that the capacitance between the element and
the cut trace will be greater when the element is down than when it's
up, neglecting the Er of the alumina with R up.

So, even if the inductance is lowered by R down, C will be increased,
making Z closer to the line's impedance and, therefore, lowering the
bump on the TDR.

Possibly so. It's too complex for me to do the numbers, and I don't
own a 3D EM simulator. The experiment is my bottom line for now, and
it shows that the inverted resistor has less HF parasitics.



So, all that's needed is to resolve the difference between C up and
C down WRT to the alumina substrate with R up.

Wanna take a crack at it?

No. I measured the entire net effect, and that's all I need and all I
have time for.

John

On Sun, 07 Aug 2011 07:30:41 -0700, John Larkin
wrote:

On Sun, 07 Aug 2011 07:28:06 -0500, John Fields
wrote:

On Sat, 06 Aug 2011 22:20:14 -0700, John Larkin
wrote:

On Sat, 06 Aug 2011 17:56:03 -0500, John Fields
wrote:

On Thu, 04 Aug 2011 21:34:19 -0700, John Larkin
wrote:

On Thu, 04 Aug 2011 20:56:09 -0500, Jeffrey Angus
wrote:

On 8/4/2011 5:49 PM, John Larkin wrote:

Can't afford scroll bars?

Shouldn't have to scroll down 100 lines of crap to read
a 2 line response.

What part of that don't you understand?


Design any interesting electronics lately?

---
Standard line for you when you've been sussed, huh?

It's a good question to put to whiney lurkers. In an electronics
newsgroup, credibility comes from electronics.

---
I'd suspect your credibility to be seriously damaged, then, by virtue
of the amount of non-electronic garbage you post.

My electronic credibility is independent of my opinions on snow, beer,
politics, or drugstore cowboys.

---
Fair enough, but your electronic credibility certainly suffers when
you refuse to admit to error when it's staring you in the face.

--
JF

On Sun, 07 Aug 2011 07:40:40 -0700, John Larkin
wrote:

On Sun, 07 Aug 2011 08:47:52 -0500, John Fields
wrote:

On Sat, 06 Aug 2011 22:28:07 -0700, John Larkin
wrote:

On Sat, 06 Aug 2011 17:45:03 -0500, John Fields
wrote:

On Thu, 04 Aug 2011 14:53:33 -0700, John Larkin
wrote:
.
.
.

I do test circuits like this now and then and toss in little adapters,
filter layouts, anything that might be handy for experimenting.

---
As do most of the rest of us, according to our disciplines, so why do
you find it necessary to single yourself out and exalt yourself as
unique by patting yourself on the back?

Show us some of the PC boards you've designed lately.

---
PCB???

Test circuits have to be on PCB's?

Controlled-impedance ones sure so. And few cool parts are available
any more in thru-hole, so a PCB is necessary to do any serious
prototyping. You can sometimes get away with adapters, like the Bellin
things, and a lot of wire, but a PCB saves a lot of time for
nontrivial stuff. And, as noted, you can add a lot of other useful
gidgets to the layout, saw them off, do fun things with them off to
the side.

You don't design PCBs?

---
Unless it's trivial, not any more.

I do the dimensioning, and maybe parts placement, but I leave the
non-critical routing to a PCB design outfit here in town.
---

Super microwave resistors have no end caps at all, just the resistive
element and two solderable end zones, all planar. You mount these
element down, of course.

---
Self-serving platitudes.

Facts about microwave resistors. Look it up.

---
Sure, but what's the point other than for you to flap your gums?

It's an _Electronics_Discussion_Group!


---

The issue here is whether the decrease in inductive reactance
indicated by your TDR is due to the lowering of inductance due to the
shortening of the connections to the end caps, or is due to the
increase of capacitance due to the resistive element moving closer to
the ground plane.

Got some numbers?

I already did a rough calculation on the capacitance. The resistor is
soldered across a small gap in the trace, so in fact most of the
capacitance of the resistor element is to the trace, not to the PCB.
And there's a sizable air gap between the resistor element and the
pcb, caused by the solder.

---
OK, if you want to play that way, then everything else being equal
means that with the element up, the alumina substrate will be forcing
the element farther away from the cut trace than it would be with the
element down.

Such being the case means that the capacitance between the element and
the cut trace will be greater when the element is down than when it's
up, neglecting the Er of the alumina with R up.

So, even if the inductance is lowered by R down, C will be increased,
making Z closer to the line's impedance and, therefore, lowering the
bump on the TDR.

Possibly so. It's too complex for me to do the numbers, and I don't
own a 3D EM simulator. The experiment is my bottom line for now, and
it shows that the inverted resistor has less HF parasitics.



So, all that's needed is to resolve the difference between C up and
C down WRT to the alumina substrate with R up.

Wanna take a crack at it?

No. I measured the entire net effect, and that's all I need and all I
have time for.

---
Then you sure have spent an inordinate amount of time denying that the
capacitance changes from R up to R down.

--
JF

On Sun, 07 Aug 2011 11:20:31 -0500, John Fields
wrote:

On Sun, 07 Aug 2011 07:40:40 -0700, John Larkin
wrote:

On Sun, 07 Aug 2011 08:47:52 -0500, John Fields
wrote:

On Sat, 06 Aug 2011 22:28:07 -0700, John Larkin
wrote:

On Sat, 06 Aug 2011 17:45:03 -0500, John Fields
wrote:

On Thu, 04 Aug 2011 14:53:33 -0700, John Larkin
wrote:
.
.
.

I do test circuits like this now and then and toss in little adapters,
filter layouts, anything that might be handy for experimenting.

---
As do most of the rest of us, according to our disciplines, so why do
you find it necessary to single yourself out and exalt yourself as
unique by patting yourself on the back?

Show us some of the PC boards you've designed lately.

---
PCB???

Test circuits have to be on PCB's?

Controlled-impedance ones sure so. And few cool parts are available
any more in thru-hole, so a PCB is necessary to do any serious
prototyping. You can sometimes get away with adapters, like the Bellin
things, and a lot of wire, but a PCB saves a lot of time for
nontrivial stuff. And, as noted, you can add a lot of other useful
gidgets to the layout, saw them off, do fun things with them off to
the side.

You don't design PCBs?

---
Unless it's trivial, not any more.

I do the dimensioning, and maybe parts placement, but I leave the
non-critical routing to a PCB design outfit here in town.
---

Super microwave resistors have no end caps at all, just the resistive
element and two solderable end zones, all planar. You mount these
element down, of course.

---
Self-serving platitudes.

Facts about microwave resistors. Look it up.

---
Sure, but what's the point other than for you to flap your gums?

It's an _Electronics_Discussion_Group!


---

The issue here is whether the decrease in inductive reactance
indicated by your TDR is due to the lowering of inductance due to the
shortening of the connections to the end caps, or is due to the
increase of capacitance due to the resistive element moving closer to
the ground plane.

Got some numbers?

I already did a rough calculation on the capacitance. The resistor is
soldered across a small gap in the trace, so in fact most of the
capacitance of the resistor element is to the trace, not to the PCB.
And there's a sizable air gap between the resistor element and the
pcb, caused by the solder.

---
OK, if you want to play that way, then everything else being equal
means that with the element up, the alumina substrate will be forcing
the element farther away from the cut trace than it would be with the
element down.

Such being the case means that the capacitance between the element and
the cut trace will be greater when the element is down than when it's
up, neglecting the Er of the alumina with R up.

So, even if the inductance is lowered by R down, C will be increased,
making Z closer to the line's impedance and, therefore, lowering the
bump on the TDR.

Possibly so. It's too complex for me to do the numbers, and I don't
own a 3D EM simulator. The experiment is my bottom line for now, and
it shows that the inverted resistor has less HF parasitics.



So, all that's needed is to resolve the difference between C up and
C down WRT to the alumina substrate with R up.

Wanna take a crack at it?

No. I measured the entire net effect, and that's all I need and all I
have time for.

---
Then you sure have spent an inordinate amount of time denying that the
capacitance changes from R up to R down.

I never denied that it changes. I said that it's too complex to solve
analytically, and I don't have the EM simulation software to evaluate
it. My estimate is that most of the impedance change can be explained
by the increased loop area, hence inductance, of the "standard"
mounting method. My bottom line is the reality of the impedance
difference, which I measured and posted.

All you've done, all you ever do, is whine.

John

All you've done, all you ever do, is whine.

John

Do you possibly think you and JF could meet somwhere away from us mere
mortals and continue your absolutly OFF TOPIC discussion in private,
Please :'(

--
John G.

On Mon, 08 Aug 2011 15:17:23 +1000, John G
wrote:

All you've done, all you ever do, is whine.

John

Do you possibly think you and JF could meet somwhere away from us mere
mortals and continue your absolutly OFF TOPIC discussion in private,
Please :'(

What have you contributed to this thread?

If you don't like what we post, why are you reading our posts?

And what's OFF TOPIC about discussing the behavior of resistors in an
electronics newsgroup?

John

John Larkin formulated on Monday :
On Mon, 08 Aug 2011 15:17:23 +1000, John G
wrote:

All you've done, all you ever do, is whine.

John

Do you possibly think you and JF could meet somwhere away from us mere
mortals and continue your absolutly OFF TOPIC discussion in private,
Please :'(

What have you contributed to this thread?

If you don't like what we post, why are you reading our posts?

And what's OFF TOPIC about discussing the behavior of resistors in an
electronics newsgroup?

John.

Well your diatribe has nothing to do with the OP's original question.

--
John G.

On Sun, 07 Aug 2011 22:06:54 -0700, John Larkin
wrote:


All you've done, all you ever do, is whine.

---
A liar, having lied once, finds lying easier and easier with each lie
and a lie, no matter how often repeated, remains a lie.

Was it once upon a time Liarkin and you decided to drop the 'i'?

--
JF

tuinkabouter wrote:

[snip]

Pedantic nit to pick:

100 = 10 × 100 ohm = 10 ohms

should be written as:

100 = 10 × 10^0 ohm = 10 ohms

--
Paul Hovnanian
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When I was in high school, I remember boys and girls slept
together all the time. We called it algebra class. -- Jay Leno