On Thursday, November 17, 2022 at 8:44:17 PM UTC-5, Lamont Cranston wrote:
> On Thursday, November 17, 2022 at 6:42:12 PM UTC-6, Ricky wrote:
> > On Thursday, November 17, 2022 at 4:56:02 PM UTC-5, Lamont Cranston wrote:
> > > > So if the resonance is series, as you say, the voltage goes toward zero and the current gets large.
> > > What voltage? I don't believe the drive voltage changes, in fact they have a level control circuit to keep it constant in the HP4342A.
> > > ( I think they put it in the wrong place, but they know more than me)
> > I don't know what your drive circuit is doing,
> As I said, it is a constant voltage a 0.02Ω source is hard to swing.
> The voltage can be adjusted, on the 260A the meter is marked with a max 250Q, then another meter monitors the drive voltage. You can set the voltage at 1X and the drive voltage is 20mV, set at 2 and the drive voltage is 10mV and you multiply the meter reading by 2. I externally monitor the drive voltage and set drive voltage at 3.33mV as 6X multiplier and measure Qs up to 1500.
> >so I'll talk in terms of the impedance. At resonance, the voltages of the L and C are equal and opposite (both >90° out of phase from the drive voltage, >one positive and the other negative phase). So the impedance will appear to approach zero. If your drive voltage >is truly like a constant voltage, the >current will ramp up hugely.
> YES! That is what happens.
> Do you know what the drive voltage it?
> I explained that above, the max on the 260A is 20mV. I'm not sure on the 4342A.
This is all much more clear now. It was a bit of work to get out of the parallel resonance mindset. I think you explained this gear a couple of years (or a few) ago when you were looking to improve its sensitivity or something.
> > > >They must be measuring this current.
> > > Indirectly, they measure the voltage drop across the tuning capacitor.
> > Ok, that makes some sense. The current will be max, so the voltage on the capacitor and inductor will be max, just not if you add them together. It's hard to imagine a constant voltage drive given the nature of the circuit, but if it is low enough, that's fine. Remember, the drive voltage will appear across the cap, multiplied by the Q. So 1VAC drive will show up as 1,000VAC if the Q is 1,000.
> Yes, on the 260A to read a Q of 1000, the drive voltage will need to be reduced to 5mv.
> > > You need to look at the schematic, it consists of a 0.02Ω resistor in series with the inductor under test (plus stray series resistances) in series with the tuning capacitor. The Drive Oscillator is across (in parallel with) the 0.02Ω resistor.
>
> > ??? Is the cap connected to the inductor and this 0.02 ohm resistor, making it a three element loop?
> Yes. I know, it looks strange.
> >Does the drive oscillator have feedback to keep the voltage constant?
> The 260 has a knob where the human adjusts the drive voltage to keep it at the at the correct level.
> The 4342A has an ALC that monitors the input the the transformer. ( seems like it should monitor the output!)
I don't know what an ALC is.
> > > >If you place the 10 Mohm resistor across the series L and C, it is an entirely different analysis than in the fully parallel arrangement.
> > > The 10MΩ is only across the tuning Capacitor.
> > Yeah, this makes the equations a bit more complicated. It provides more current to the coil than the capacitor sees. It distorts the phase angles.
> > > > In fact, with the voltage going toward zero, the added parallel resistor will have almost no current and have very little impact on the measurement.
> > > The drive voltage is constant, it does not go to toward zero.
> > > >
> > > > Sorry, but I'm very lost at this point. With a very low impedance power source, and a very large parallel resistor, I don't see how you can be getting the >results you claim.
> > > I'm not the only one, I have at least one on another group that has done several tests and has the same low readings. He has the newer HP4342A.
>
> > Now that I realize the nature of the full circuit, the resistor will show up in the circuit equations in a way, that will modify the phase angle and alter the >frequency of peak response, so that it is no longer at the resonance of the LC. This is not because of a change in the C, I don't think. You would have to >write out the equations which is easiest using the Laplace transform. I've done it before using simple algebra, but the equations get messy, making it not >so "simple".
> >
> > Essentially, what happens is the voltage on the cap and inductor are no longer 180 out of phase with each other, because each one sees a different drive current.
>
> > > > Wait, you said the tuning capacitor is inside the unit,
> > > Yes.
> > > but you are attaching a coil?
> > > Yes.
> > > So the resistor is only across the coil?
> > > Nope, it is across the capacitor.
> > Ok, so they give you points to add the resistor, got it!
> > > That's not the same thing and the resistor will impact the coil Q factor. Not sure how to write the equation, since the coil parasitic resistor is in series with the pure inductance and your added resistor is in parallel. The parasitic resistor capacitance is different as well, since it in parallel with the coil.
> > > It is across the capacitor.
> > It should be pretty negligible anyway.
> > > > In the application circuits, will you be using a parallel or a series tuning method?
> > > It is a series tuned circuit, But, the 10MΩ is called a parallel connection to the measuring circuit.
> > > Please look at the dropbox picture, The parts in red, I added, to show where the inductor and resistor are placed.
> > > https://www.dropbox.com/s/bjcjzdxtyy63jan/HP%20Injection%20circuit%20with%2010M%CE%A9.jpg?dl=0
>
> > I'm not asking about the Boonton,
> That is actually the 4342A.
> >I'm asking about where you will use the coil.
> The coil I'm using is called a Q standard or work coil.
> From the Radio Museum, "
> The Q-Standard Type 513-A is a shielded reference inductor which has accurately-measured and highly-stable inductance and Q characteristics. Specifically designed for use with Q-Meters Type 260-A and 160-A, the Q-Standard is particularly useful as a check on the overall operation and accuracy of these instruments, as well as for providing precisely-known supplementary Q-circuit inductance desirable for many impedance measurements by the parallel method. (Mikek here, as in measuring resistors)
> The Q-Standard consists of a specially-designed, high-Q coil of Litz wire, wound on a low-loss Steatite form. The coil is hermetically sealed inside a copper shield can which is filled with inert gas under pressure. The desired Q-versus-frequency characteristics are provided by a carbon film resistor shunted across the coil. Two replaceable banana plug connectors mounted on the base serve to connect the unit to the Q-meter circuit.
> The Q-Standard is supplied in a convenient wooden carrying and storage case. Each unit is individually calibrated and marked with its true inductance (L), distributed capacity (Cd), effective Qe and indicated Qi at 0.5, 1.0 and 1.5 Mc, respectively. Tolerance: L ±1%, Cd ±2%, Qe ±3%, measured at 73°F. Any instrument deviating more than ± 7% from the marked value is not operating in accordance with original specifications.
> I'm wondering how similar to your application this test circuit is.
> I'm using it as the work coil, it is part of the LC to do other measurements, like a resistor.
> I don't have an application for the coil, it is a tool for checking calibration and to be used
> as a work coil for other parallel and series external component measurements.
> You need to add the work coil to setup a resonant LC to measure resistors, capacitors,
> and you can even do dielectric constant measurements. This requires a fixture.
>
> The main purpose of a Q meter is to measure Q of inductors, however, it has found other uses.
You are using it as an ohm meter???
--
Rick C.
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