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