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Re: Making a Q-meter /

 

I was thinking the capacitive divider was to reduce the C loading of the voltmeter on the resonant circuit, but it's to reduce the input resistance. Wouldn't matter if you are measuring 20 ?H coils, but measuring a 1.2H inductor at 25 kHz. XsubL = 188 k¦¸. 10 M¦¸ input R starts to affect accuracy.

As long as the input C is stable, it just adds to the minimum C that can be used for resonance.

I see that Mikek's amplifier is marked 0.25 pF and 100M, in spite of having a 10M FET bias resistor. I guess that bootstrapping increases the R as well as decrease the C.

John KK6IL

On 9/22/2022 11:24 AM, Mikek wrote:
OK, so you are unloading the resonant LC improving (getting closer to) the actual Q.
I thought you were making some improvement to the injection transformer.
?Be aware the HP 400E has, as the manual says, "an input capacitance of 25pf or less on the 1V range."
The 3V range has about 12pf. No info on other ranges.
So that changes your capacitive divider from a 12 to 1 to a 14.5 to 1. Assuming I have any clue, and I might not.
Now we are back to a high input impedance amplifier.
I need to build this, to find the frequency response, but I have two HP 3400A RF meters I need to get working before I start another project.
?It might need a 50 driver on the output.
??????????????????????????????? Mikek
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