On the idea perhaps the resonating capacitance was too small
compared with any (tiny) inductance of the resistor, I resonated
at the other end, using 450 pF, using the nominal 98 Mohm.
1447 x 1423 / 2? pi? 491,437 Hz? 450 e-12? 24? =? 61.75 megohms.
Much better than before but still far from nominal 98 megohms.
On 11/16/2022 12:41 PM, Steve Ratzlaff
via groups.io wrote:
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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:
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
