OK, seems like enough understand to do the long story.
Here’s the scenario:
I have a 167uH capacitor resonating with a 151pf cap on my Q meter, Q = 1092.
Now I put a resistor across the tuning capacitor, I retune* for resonance and measure the lowered Q. Because of the resistors capacitance I must reduce the capacitance to get back to resonance.
The idea is to measure the resistor I put in parallel with the capacitor by the drop in Q.
I’m having trouble getting the Q meter to measure the resistor as the correct value.
First my Q meter measures LC in parallel resonance. And I think it might be because
I'm resonating the self capacitance of the resistor.
The details
I set up an inductor and measured Q at 1MHz, the Q is 1092, the reactance is 1,053Ω.
So, Rp = Q times X =1092 x 1053 = 1,149,876Ω =Rp (Rp = Parallel resistance at resonance)
Then I put a 10MΩ Metal Film ½ W resistor in parallel with the tuning capacitor.
I had to reduce* the tuning cap by 0.37pf and the Q is reduced to 882. Making Rp = 928,746Ω
Q times X = 882 x 1053 = 928,746Ω
*Meaning the resistor has 0.37pf of capacitance across it.
The parallel combination of a 1,149,876Ω and 10MΩ is 1,031,290Ω.
But, my Q meter gives an Rp of 928,746Ω saying my 10MΩ has a value of,
4,820,000 Ω. i.e. 1,149,876Ω//4,820,000Ω = 928,746Ω
But wait, my resistor has 0.37pf of capacitance. At 1MHz that is 430,128Ω
So, I have 430,128Ω of capacitive reactance in parallel with 10MΩ or 429,751Ω.
That doesn’t jive with my measurement.
What I don’t have a handle on is this, the capacitance of the resistor is tuned out by the inductor, so I would think it is not seen and we have just Rp. (a pure parallel resistance)
But, the self capacitance of the resistor has a Q, if the R loss of the capacitor is magnified by the Q of the capacitor, Then we have a pure resistance across the 10MΩ.
For an example, let's say the Q of the “self capacitance” of the resistor is 21.72, and if Rp=Q times Xc, (that is a question, I’m not sure)
Then 21.72 x 430,128Ω = 9,342,398Ω and 10MΩ//9,342,398Ω = 4,820,000 Ω . This the answer my Q meter gives.
If any of this makes sense, is it possible the “self capacitance” of the resistor has a Q of 21.72?
The reason I'm doing this is, I want to measure the input impedance of high input impedance amplifiers, maybe as high as 100MΩ and a few pf.
Mikek