I tried again, this time with four 22 megohm in series with 10
megohm for a nominal 98 megohms total.
1700 kHz, 32.3 pF, coil-alone Q 990; with resistors Q 850.? 990 x
850 / 2? pi? 1.7e6? 32.3e-12? 140? = 17.42 megohms!??? Much much
worse than before. Obviously there is something else going on
here, assuming Mike's formula as stated is correct.
Steve
On 11/16/2022 11:07 AM, Steve Ratzlaff
via groups.io wrote:
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I tried it with my HP4342A Q meter, and a standard 5% 10 megohm
resistor which measures about 10.97 megohms on my Fluke 87V.
(The reading keeps creeping up, never seems to finally stop....)
1700 kHz, 32.5 pF, coil-alone Q 980. With resistor across coil,
Q 635. (32.1 pF)
So plugging into Mike's formula: 980 x 635/ 2?? pi ? 1.7e6 ?
32.5e-12? 345? =? 5.196 megohms----far from what it should
be....
Steve AA7U in AZ
On 11/16/2022 8:35 AM, Mikek wrote:
On Sat, Nov 12, 2022 at 07:42 AM, Mikek wrote:
Impact of 100M ohm
A simple work around is to measure Q of a coil then to add
100M ohms in parallel then retune and take a second Q reading
then calculate out the 100M ohm first reading. However most?
measurements don't need such a process because the 100M ohm
loading is <1% Q reduction.? If you have an excellent 100uH
inductor? Xl=628 ohms at 1MHz, so for a Q=2000? the parallel
loss will be 1.26M ohm . Adding 100M ohm in parallel to 1.26M
produces 1.2443, a 1.26% drop from Q=2,000 to 1975.? In
reality a large coil near to the metalwork of a Q meter will
read low Q because of eddy currents, in my experience the Q
drop due to eddy currents is often far greater than 1%!
Would some perform this experiment on their Q meter?
?I continuously get a about a 19%? low error. I have used two
different Q meters. I'm attaching thin wires to an 0805
resistor.
But please use what you have.
How to:
?Put an inductor on your meter and measure the Q, and find the
capacitance that was used to resonate.
Record both.
Now put a 10Meg¦¸ across the inductor (retune if needed) and
measure the new lower Q.
Record.
Use this formula to find the value of the measured value of the
10M¦¸.
Rp = Q1 x Q2 / 2 x pi x f x capacitance x delta Q.
Capacitance is in Farads, i.e, 150pf = 0.000000000150.
The delta Q is first Q measured minus the second Q measured.
?
I want to know why I don't get an accurate answer.
This method of measuring high value resistors
is in both the Boonton 260A manual and the HP4342A manual.
?
?? Thanks for any incite you can add to this.? Mikek
