Yesterday's post using SimSmith to solve for 11Mohm resistor capacitance
and Q by match Steve's Q measurements. Results were 0.4pF with Q=45.
Xc = 1/(2*Pi*1.7e6*0.4e-12) = 234kohm
Rp = 234k x 45 = 10.53Mohm
11M || 10.53M = 5.38M
On 11/16/22 12:07 PM, Steve Ratzlaff wrote:
So plugging into Mike's formula: 980 x 635/ 2 pi 1.7e6 32.5e-12
345 = 5.196 megohms
Safe to conclude Steve's 11Mohm resistor parallel resistance is between
5.4Mohm and 5.2Mohm at 1.7MHz?
John KN5L
On 11/19/22 4:43 PM, John KN5L wrote:
On 11/19/22 1:47 PM, Jacques Audet wrote:
This resistor Rp comes with a capacitor in parallel Cp, the inter
electrode capacitance of the physical resistor.
This capacitor has a "poor" dielectric:? made by the actual resistance
material.
On 11/16/22 12:07 PM, Steve Ratzlaff wrote:
1700 kHz, 32.5 pF, coil-alone Q 980. With resistor across coil, Q 635.
(32.1 pF)
Attached is a SimSmith model of Steve's 11/16/22 measurement.
Set DUT Q to match measurement without 11M resistor.
Added 11M parallel resistor and adjusted CpC and CpQ for resonance and
Q, as shown in trace, Steve measured.
Modeling suggests 11M resistor has 400fF parallel capacitance with Q of 45.
John KN5L
