Correct modeling of parasitic elements

[Paul Burridge]

Hi all,

Returning to this subject with the specific beef I have now. This
example comes from Chris Bowick's book, RF Circuit Design.
See the LT schematic file posted to a.b.s.e. under the same subject
title as this.
In the example Bowick gives, this 10K metal film resistor loses most
of its resistance at 200Mhz; in fact it drops to just 1890 Ohms at
that frequency. The parasitic inductances and capacitance are shown
linked to the resistor (R1) for simulation purposes. I have two
problems here. Firstly, Bowick gives the resitor's lead lengths as
12.7mm each, giving rise to 8.7nH for the type of wire the leads are
made from. Since in this example the leads are so long, aren't the
elements shown associated with R1 the wrong arrangement? Surely the
capacitance should be Directly across R1? Also, the PD across R2 if
Bowick is right should be half the total applied voltage (due to
Thevenin) because the equivalent resistance of R1 at 200Mhz and that
of R2 will be equal. My simulations disagree, showing equality is only
obtained when R2 is made equal to 1750 Ohms. So there's a discrepancy
there, too. Anyone shed any light on this?

Thanks,

p.
--

"What is now proved was once only imagin'd." - William Blake, 1793.

 

[Chuck Harris]

Paul Burridge wrote:

Hi all,

Returning to this subject with the specific beef I have now. This
example comes from Chris Bowick's book, RF Circuit Design.
See the LT schematic file posted to a.b.s.e. under the same subject
title as this.
In the example Bowick gives, this 10K metal film resistor loses most
of its resistance at 200Mhz; in fact it drops to just 1890 Ohms at
that frequency. The parasitic inductances and capacitance are shown
linked to the resistor (R1) for simulation purposes. I have two
problems here. Firstly, Bowick gives the resitor's lead lengths as
12.7mm each, giving rise to 8.7nH for the type of wire the leads are
made from. Since in this example the leads are so long, aren't the
elements shown associated with R1 the wrong arrangement? Surely the
capacitance should be Directly across R1? Also, the PD across R2 if
Bowick is right should be half the total applied voltage (due to
Thevenin) because the equivalent resistance of R1 at 200Mhz and that
of R2 will be equal. My simulations disagree, showing equality is only
obtained when R2 is made equal to 1750 Ohms. So there's a discrepancy
there, too. Anyone shed any light on this?

Thanks,

p.

Hi Paul,

I don't agree with this model at all! Consider that the usual metalfilm
resistor is a glass rod flashed with a metal film, and then laser cut
in a spiral. In other words, it is an air (glass) core solenoid wound
with resistive wire.

If I could ignore any capacitive coupling from the resistor to other
elements (I can't really), I would model the resistor more like this:


O---L(lead)---+--L(resistor)--(R)--+--L(lead)---O
| |
+---C(resistor)------+

C(resistor) is going to be quite significant because it is the
combination of the capacitance due to the size and separation of
the resistor's end caps, and the distributed capacitance due to
the separation between the laser cut turns of the resistor, and
the capacitance between the turns of the coils across the diameter
of the coil.

-Chuck Harris