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