Hi

This resistence, R26 and R27:
http://www.123rf.com/photo_552275_si...on-simbol.html


What kind of resistance they are, and where can buy?


Regards

Lanny wrote in message
...

Hi

This resistence, R26 and R27:

http://www.123rf.com/photo_552275_si...lose-up-select
ive-focus-on-simbol.html


What kind of resistance they are, and where can buy?


Regards




So what do the numbers on the tops read as ?
x10 or more magnification required

So what do the numbers on the tops read as ?
x10 or more magnification required


Perhaps similar to the R47


Regards

On 8/2/2011 11:41 AM, Lanny wrote:
Hi

This resistence, R26 and R27:
http://www.123rf.com/photo_552275_si...on-simbol.html


What kind of resistance they are, and where can buy?

This type of resistors are SMD (Surface Mount Device) resistors.

You can buy these in any good electronic parts shop or online.
Digikey, Farnell


From the wiki: https://secure.wikimedia.org/wikipedia/en/wiki/Resistor

Surface mounted resistors are printed with numerical values in a code
related to that used on axial resistors. Standard-tolerance
surface-mount technology (SMT) resistors are marked with a three-digit
code, in which the first two digits are the first two significant digits
of the value and the third digit is the power of ten (the number of
zeroes). For example:
334 = 33 × 104 ohms = 330 kilohms
222 = 22 × 102 ohms = 2.2 kilohms
473 = 47 × 103 ohms = 47 kilohms
105 = 10 × 105 ohms = 1.0 megohm

Resistances less than 100 ohms are written: 100, 220, 470. The final
zero represents ten to the power zero, which is 1. For example:
100 = 10 × 100 ohm = 10 ohms
220 = 22 × 100 ohm = 22 ohms

Sometimes these values are marked as 10 or 22 to prevent a mistake.

Resistances less than 10 ohms have 'R' to indicate the position of the
decimal point (radix point). For example:
4R7 = 4.7 ohms
R300 = 0.30 ohms
0R22 = 0.22 ohms
0R01 = 0.01 ohms

Precision resistors are marked with a four-digit code, in which the
first three digits are the significant figures and the fourth is the
power of ten. For example:
1001 = 100 × 101 ohms = 1.00 kilohm
4992 = 499 × 102 ohms = 49.9 kilohm
1000 = 100 × 100 ohm = 100 ohms

000 and 0000 sometimes appear as values on surface-mount zero-ohm links,
since these have (approximately) zero resistance.

More recent surface-mount resistors are too small, physically, to permit
practical markings to be applied.

On Tue, 02 Aug 2011 14:12:10 +0200, tuinkabouter
wrote:

On 8/2/2011 11:41 AM, Lanny wrote:
Hi

This resistence, R26 and R27:
http://www.123rf.com/photo_552275_si...on-simbol.html


What kind of resistance they are, and where can buy?

This type of resistors are SMD (Surface Mount Device) resistors.

You can buy these in any good electronic parts shop or online.
Digikey, Farnell


From the wiki: https://secure.wikimedia.org/wikipedia/en/wiki/Resistor

Surface mounted resistors are printed with numerical values in a code
related to that used on axial resistors. Standard-tolerance
surface-mount technology (SMT) resistors are marked with a three-digit
code, in which the first two digits are the first two significant digits
of the value and the third digit is the power of ten (the number of
zeroes).

There is also an SMD resistor marking scheme that uses a mix of numbers
and letters. E.g. 25C = 17.8K 1% and C25 is 10K 5%. A decoder for those
is available free at http://www.schematica.com/

--
Rich Webb Norfolk, VA

On 8/2/2011 3:22 PM, Rich Webb wrote:
On Tue, 02 Aug 2011 14:12:10 +0200, tuinkabouter
wrote:

On 8/2/2011 11:41 AM, Lanny wrote:
Hi

This resistence, R26 and R27:
http://www.123rf.com/photo_552275_si...on-simbol.html


What kind of resistance they are, and where can buy?

This type of resistors are SMD (Surface Mount Device) resistors.

You can buy these in any good electronic parts shop or online.
Digikey, Farnell


From the wiki:https://secure.wikimedia.org/wikipedia/en/wiki/Resistor

Surface mounted resistors are printed with numerical values in a code
related to that used on axial resistors. Standard-tolerance
surface-mount technology (SMT) resistors are marked with a three-digit
code, in which the first two digits are the first two significant digits
of the value and the third digit is the power of ten (the number of
zeroes).

There is also an SMD resistor marking scheme that uses a mix of numbers
and letters. E.g. 25C = 17.8K 1% and C25 is 10K 5%. A decoder for those
is available free at http://www.schematica.com

Perhaps you can update the wiki.

"Lanny" wrote in message
...

Hi

This resistence, R26 and R27:
http://www.123rf.com/photo_552275_si...on-simbol.html


What kind of resistance they are, and where can buy?


Regards




As a matter of interest, why do you feel that these devices need replacement
?

Arfa

In article ,
"Arfa Daily" wrote:

"Lanny" wrote in message
...

Hi

This resistence, R26 and R27:
http://www.123rf.com/photo_552275_si...lose-up-select
ive-focus-on-simbol.html


What kind of resistance they are, and where can buy?


Regards




As a matter of interest, why do you feel that these devices need replacement
?

Arfa

Seems to be two camps of respondents. But the picture seems to be a
generic stock photo, so I'm in the camp that believes the OP was just
inquiring in general about SMD.

What kind of resistance they are, and where can buy?


Regards




As a matter of interest, why do you feel that these devices need
replacement
?

Arfa

Seems to be two camps of respondents. But the picture seems to be a
generic stock photo, so I'm in the camp that believes the OP was just
inquiring in general about SMD.

but this resistence have a polarity?

Regards

On Wed, 3 Aug 2011 10:08:11 +0200, "Lanny" wrote:

What kind of resistance they are, and where can buy?


Regards




As a matter of interest, why do you feel that these devices need
replacement
?

Arfa

Seems to be two camps of respondents. But the picture seems to be a
generic stock photo, so I'm in the camp that believes the OP was just
inquiring in general about SMD.

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

--
JF

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

--
JF

They have a particular name these little resistance?

Regards

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

On Wed, 3 Aug 2011 17:39:31 +0200, "Lanny" wrote:



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

--
JF

They have a particular name these little resistance?

---
"Surface mount resistors"

--
JF

"John Fields" ha scritto nel messaggio
...
On Wed, 3 Aug 2011 17:39:31 +0200, "Lanny" wrote:



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

--
JF

They have a particular name these little resistance?

---
"Surface mount resistors"

still answering to that guy?
Do you realize hi's kidding everybody?

On 8/3/2011 3:08 AM, Lanny wrote:
but this resistence have a polarity?

Regards

No you're thinking of electrolytic resistors.
They usually have to polarity clearly marked on them.

Jeff



--
"Everything from Crackers to Coffins"

On Wed, 3 Aug 2011 22:20:18 +0200, "Vale" wrote:


"John Fields" ha scritto nel messaggio
.. .
On Wed, 3 Aug 2011 17:39:31 +0200, "Lanny" wrote:



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

--
JF

They have a particular name these little resistance?

---
"Surface mount resistors"

still answering to that guy?

---
No, I'm just answering him, not answering to him.
---

Do you realize hi's kidding everybody?

---
Don't spoil my troll.

--
JF

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.

That's the way they always land on the bench anyhow.


John

On Wed, 03 Aug 2011 16:46:29 -0500, Jeffrey Angus
wrote:

On 8/3/2011 3:08 AM, Lanny wrote:
but this resistence have a polarity?

Regards

No you're thinking of electrolytic resistors.
They usually have to polarity clearly marked on them.

Jeff

There actually were electrolytic resistors, actually electrolytic
rheostats, but they were usually used at AC.

I once saw a big multi-megavolt Marx generator that used liquid
resistors, long clear hoses filled with water and something blue,
copper sulfate or some such. Looked cool.


John

On Wed, 03 Aug 2011 16:46:11 -0700, John Larkin
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)?

That's the way they always land on the bench anyhow.

Yours hit the bench?

" wrote:

On Wed, 03 Aug 2011 16:46:11 -0700, John Larkin
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)?

That's the way they always land on the bench anyhow.

Yours hit the bench?


Did you forget that he can only 'Pitch a bitch'? ;-)


--
It's easy to think outside the box, when you have a cutting torch.

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

On Wed, 03 Aug 2011 16:46:11 -0700, John Larkin
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


The capacitive bump around cm 1 is the SMA connector transition.

It really helps to flip them.

John

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

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

On Wed, 03 Aug 2011 16:46:11 -0700, John Larkin
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.

The capacitive bump around cm 1 is the SMA connector transition.

It really helps to flip them.

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

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

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

On Wed, 03 Aug 2011 16:46:11 -0700, John Larkin
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.

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.

John

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
wrote:

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

On Wed, 03 Aug 2011 16:46:11 -0700, John Larkin
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.

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?

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
wrote:

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

On Wed, 03 Aug 2011 16:46:11 -0700, John Larkin
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.

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.

John

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
wrote:

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

On Wed, 03 Aug 2011 16:46:11 -0700, John Larkin
m 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.

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.

The mechanics matter. Understanding the physics matters.

On Wed, 03 Aug 2011 23:38:45 -0500, "
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
wrote:

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

On Wed, 03 Aug 2011 16:46:11 -0700, John Larkin
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.

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.

The mechanics matter. Understanding the physics matters.

"One measurement is worth a thousand expert opinions."

Werner von Braun, I think.

John

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
wrote:

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

On Wed, 03 Aug 2011 16:46:11 -0700, John Larkin
m 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.

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.

Since the difference between the two configurations is that in the
upside-down case the resistive element is closer to the PCB plane by
the thickness of the alumina substrate, the capacitance between the
resistive element and the PCB plane must increase.

That is, as long as the Er of the alumina doesn't overcome the
increased distance between the resistive element and the PCB plane
with the resistor right side up, there will be an increase in
capacitance with the resistor upside-down and, as KRW noted, that
flattening of the bump shows that some of the L is being cancelled by
that C, causing the impedance discontinuity to diminish.

How thick is the FR4 in your fixture?

--
JF

On Thu, 04 Aug 2011 07:16:49 -0700, John Larkin
wrote:

On Wed, 03 Aug 2011 23:38:45 -0500, "
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
m 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.

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.

The mechanics matter. Understanding the physics matters.

"One measurement is worth a thousand expert opinions."

Werner von Braun, I think.

---
How typically Larkinesque!

You've painted yourself into a corner and instead of admitting to it
you're desperately trying to change the subject a la "scattergun"
style.

--
JF

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
wrote:

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

On Wed, 03 Aug 2011 16:46:11 -0700, John Larkin
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.

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.

But the bottom line ramains: by actual measurement in a DC-12 GHz
bandwidth, the normal resistor has a bunch of inductance (I'll
calculate how much) and the inverted resistor has a lot less.

An 0805 resistor has a pretty small footprint; 0.004 square inches.
About half of that is end cap, so figure the element is 0.002. The
resistor element will have about 0.03 pF of capacitance to the PCB,
probably less because of the air gap between the resistor and the
board. My 12 GHz TDR wouldn't see 0.03 pF; the time constant with 50
ohms is only 1.5 picoseconds.

So it's really inductance.

John

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
m 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.

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?
---

But the bottom line ramains: by actual measurement in a DC-12 GHz
bandwidth, the normal resistor has a bunch of inductance (I'll
calculate how much) and the inverted resistor has a lot less.

An 0805 resistor has a pretty small footprint; 0.004 square inches.
About half of that is end cap, so figure the element is 0.002. The
resistor element will have about 0.03 pF of capacitance to the PCB,
probably less because of the air gap between the resistor and the
board. My 12 GHz TDR wouldn't see 0.03 pF; the time constant with 50
ohms is only 1.5 picoseconds.

So it's really inductance.

John


--
JF

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
wrote:

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

On Wed, 03 Aug 2011 16:46:11 -0700, John Larkin
om 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.

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. The little resistor
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.

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

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.

John

While I really do enjoy the engineering ****ing contest
here, could you two take the time to EDIT your ****ing
replies?

There's NO reason other than sloth for having to wade
through 2-3 pages of repeatedly quoted replies to see
the next step in your comments.

Jeff

On Thu, 04 Aug 2011 17:46:44 -0500, Jeffrey Angus
wrote:

While I really do enjoy the engineering ****ing contest
here, could you two take the time to EDIT your ****ing
replies?

There's NO reason other than sloth for having to wade
through 2-3 pages of repeatedly quoted replies to see
the next step in your comments.

Jeff

Can't afford scroll bars?

Feel free to not read my posts.

John

On Thu, 04 Aug 2011 07:16:49 -0700, John Larkin
wrote:

On Wed, 03 Aug 2011 23:38:45 -0500, "
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
m 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.

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.

The mechanics matter. Understanding the physics matters.

"One measurement is worth a thousand expert opinions."

Measurements, alone, don't lead to understanding.

On Thu, 04 Aug 2011 18:21:58 -0500, "
wrote:

On Thu, 04 Aug 2011 07:16:49 -0700, John Larkin
wrote:

On Wed, 03 Aug 2011 23:38:45 -0500, "
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
wrote:

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

On Wed, 03 Aug 2011 16:46:11 -0700, John Larkin
om 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.

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.

The mechanics matter. Understanding the physics matters.

"One measurement is worth a thousand expert opinions."

Measurements, alone, don't lead to understanding.

But reality exists. The resistors behave the way I measured them. No
amount of theorizing is going to make them behave any different.

It wouldn't take a lot of math to reconcile the dimensions with the
amounts of capacitance and inductance to explain the things I've
measured. That's interesting, but as an engineer I now know that
mounting the resistors upside-down makes them more ohmic at high
frequencies, and that's useful. People's theorizing about resistors
here wasn't especially useful, in the sense of being predictive to
behavior.

Strictly speaking, I don't need to understand it. I use lots of things
I don't understand.

John

On Thu, 04 Aug 2011 17:18:38 -0700, John Larkin
wrote:

On Thu, 04 Aug 2011 18:21:58 -0500, "
wrote:

On Thu, 04 Aug 2011 07:16:49 -0700, John Larkin
wrote:

On Wed, 03 Aug 2011 23:38:45 -0500, "
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.

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.

The mechanics matter. Understanding the physics matters.

"One measurement is worth a thousand expert opinions."

Measurements, alone, don't lead to understanding.

But reality exists. The resistors behave the way I measured them. No
amount of theorizing is going to make them behave any different.

For that particular geometry, no. I don't get any generally useful
information without understanding, though.

It wouldn't take a lot of math to reconcile the dimensions with the
amounts of capacitance and inductance to explain the things I've
measured. That's interesting, but as an engineer I now know that
mounting the resistors upside-down makes them more ohmic at high
frequencies, and that's useful. People's theorizing about resistors
here wasn't especially useful, in the sense of being predictive to
behavior.

I disagree. It's useful to understand the physics to generalize the
information.

Strictly speaking, I don't need to understand it. I use lots of things
I don't understand.

One less in that list is good.

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?

Jeff

--
"Everything from Crackers to Coffins"

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?

John

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?

Oh, he understands just fine - he's just being a big poopyhead. ;-)

Usually, when he carries on those long threads, he's just troll-baiting;
I think he's achieved his Master Baiter rating by now. ;-D

Cheers!
Rich