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

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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.

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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?




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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.

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Self-serving platitudes.

Facts about microwave resistors. Look it up.

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

It's an _Electronics_Discussion_Group!


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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.

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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