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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
Chipbee40_SpamNo@yahoo.com> 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?
What I see in a TDR setup is less series inductance. The context is aOn Wed, 03 Aug 2011 16:46:11 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 03 Aug 2011 15:07:32 -0400, Archon
Chipbee40_SpamNo@yahoo.com> 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)?
But that's telling you that the sqrt(L/C) is closer to the transmission lineOn Wed, 03 Aug 2011 18:56:30 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> 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
Chipbee40_SpamNo@yahoo.com> 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.
I'd have to think about that. The normal way, you'd have the substrateOn Wed, 03 Aug 2011 17:06:22 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 03 Aug 2011 18:56:30 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> 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
Chipbee40_SpamNo@yahoo.com> 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.
But the alumina is the same in both orientations. Down adds the PCB planeOn Wed, 03 Aug 2011 19:22:36 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 17:06:22 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 03 Aug 2011 18:56:30 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> 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
Chipbee40_SpamNo@yahoo.com> 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.
Less inductive or more capacitive (canceling a constant inductance) => higherBut 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.
I'm an engineer, not a philosopher. What I see here is inductance. IOn Wed, 03 Aug 2011 19:25:05 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Wed, 03 Aug 2011 19:22:36 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 17:06:22 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 03 Aug 2011 18:56:30 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> 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
Chipbee40_SpamNo@yahoo.com> 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?
The mechanics matter. Understanding the physics matters.On Wed, 03 Aug 2011 22:35:35 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 19:25:05 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Wed, 03 Aug 2011 19:22:36 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 17:06:22 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 03 Aug 2011 18:56:30 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> 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
Chipbee40_SpamNo@yahoo.com> 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.
"One measurement is worth a thousand expert opinions."On Wed, 03 Aug 2011 20:56:57 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Wed, 03 Aug 2011 22:35:35 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 19:25:05 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Wed, 03 Aug 2011 19:22:36 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 17:06:22 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 03 Aug 2011 18:56:30 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> 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
Chipbee40_SpamNo@yahoo.com> 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 22:35:35 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 19:25:05 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Wed, 03 Aug 2011 19:22:36 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 17:06:22 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 03 Aug 2011 18:56:30 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> 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
Chipbee40_SpamNo@yahoo.com> 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.
---On Wed, 03 Aug 2011 23:38:45 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 20:56:57 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Wed, 03 Aug 2011 22:35:35 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 19:25:05 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Wed, 03 Aug 2011 19:22:36 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 17:06:22 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 03 Aug 2011 18:56:30 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> 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
Chipbee40_SpamNo@yahoo.com> 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.
Of course it changes. Think about it for Pete's sake.On Wed, 03 Aug 2011 20:56:57 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Wed, 03 Aug 2011 22:35:35 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 19:25:05 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Wed, 03 Aug 2011 19:22:36 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 17:06:22 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 03 Aug 2011 18:56:30 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> 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
Chipbee40_SpamNo@yahoo.com> 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.
---On Thu, 04 Aug 2011 10:17:00 -0500, John Fields
jfields@austininstruments.com> wrote:
On Wed, 03 Aug 2011 20:56:57 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Wed, 03 Aug 2011 22:35:35 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 19:25:05 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Wed, 03 Aug 2011 19:22:36 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 17:06:22 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 03 Aug 2011 18:56:30 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> 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
Chipbee40_SpamNo@yahoo.com> 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
If the resistor is mounted upside-down, the current flow is close toOn Thu, 04 Aug 2011 09:17:41 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Thu, 04 Aug 2011 10:17:00 -0500, John Fields
jfields@austininstruments.com> wrote:
On Wed, 03 Aug 2011 20:56:57 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Wed, 03 Aug 2011 22:35:35 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 19:25:05 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Wed, 03 Aug 2011 19:22:36 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 17:06:22 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 03 Aug 2011 18:56:30 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> 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
Chipbee40_SpamNo@yahoo.com> 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?
Can't afford scroll bars?While I really do enjoy the engineering pissing contest
here, could you two take the time to EDIT your fucking
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
Measurements, alone, don't lead to understanding.On Wed, 03 Aug 2011 23:38:45 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 20:56:57 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Wed, 03 Aug 2011 22:35:35 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 19:25:05 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Wed, 03 Aug 2011 19:22:36 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 17:06:22 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 03 Aug 2011 18:56:30 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> 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
Chipbee40_SpamNo@yahoo.com> 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."
But reality exists. The resistors behave the way I measured them. NoOn Thu, 04 Aug 2011 07:16:49 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 03 Aug 2011 23:38:45 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 20:56:57 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Wed, 03 Aug 2011 22:35:35 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 19:25:05 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Wed, 03 Aug 2011 19:22:36 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 17:06:22 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 03 Aug 2011 18:56:30 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> 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
Chipbee40_SpamNo@yahoo.com> 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.
For that particular geometry, no. I don't get any generally usefulOn Thu, 04 Aug 2011 18:21:58 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Thu, 04 Aug 2011 07:16:49 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 03 Aug 2011 23:38:45 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 20:56:57 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Wed, 03 Aug 2011 22:35:35 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 19:25:05 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:
On Wed, 03 Aug 2011 19:22:36 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 03 Aug 2011 17:06:22 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 03 Aug 2011 18:56:30 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> 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
Chipbee40_SpamNo@yahoo.com> 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.
I disagree. It's useful to understand the physics to generalize theIt 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.
One less in that list is good.Strictly speaking, I don't need to understand it. I use lots of things
I don't understand.
Shouldn't have to scroll down 100 lines of crap to readCan't afford scroll bars?
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?