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
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
plane)?
But the alumina is the same in both orientations. Down adds the PCB plane capacitance.
Less inductive or more capacitive (canceling a constant inductance) => higher impedance?
plane)?
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
plane)?
The mechanics matter. Understanding the physics matters.
plane)?
"One measurement is worth a thousand expert opinions."
Werner von Braun, I think.
John
plane)?
-- 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?
plane)?
line
-- 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.
plane)?
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
plane)?
line
-- 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?
plane)?
line
plane.
higher
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
plane)?
line
Measurements, alone, don't lead to understanding.
plane)?
line
plane.
higher
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
plane)?
line
plane.
higher
For that particular geometry, no. I don't get any generally useful information without understanding, though.
I disagree. It's useful to understand the physics to generalize the information.
One less in that list is good.
plane)?
line
plane.
-- Excuses?
the plane)?
line
plane.
higher
Show us some of the PC boards you've designed lately.
Facts about microwave resistors. Look it up.
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
-- 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?
It's an _Electronics_Discussion_Group!
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.
No. I measured the entire net effect, and that's all I need and all I have time for.
John
-- 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.
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