More Math for the "SETUP"

17 inches.

That's what I did. I never hit 475kHz exactly, I was 100 to 300 Hz off by the time fine tuned the frequency. But much easier than adjusting the capacitor.

Did you calculate the difference? I get 1.03%, that's better than my scope and my eyes even with my magnifying glass.

Could you explain what you mean? Do you mean overkill with 3 readings or do you see some thing I made overly complicated?

This was a test of the Boonton. Looks to me like the Boonton is reading low. Although, when I did the 3db method on the Boonton, it gave the same Q as the normal method. That's why I suspect the internal capacitor may have extra losses. I need to look into to see what kind of connection it has for to the moving vanes.

I need to look into the parallel connection method, I don't think you can get rid of the about 30pf from the internal cap.

No, I'm lacking there, my cap meter has died, I'm in the market for a new one. Any suggestions? But I can tell you it is very close to 450pf. I gave an about number on the frequency. 475kHz for the first 5 caps and 500kHz and 600kHz for the last two. They are smaller caps and would not tune down to 475kHz. If I give you exact resonant frequency, what could you do with it? Not knowing the exact capacitance. (I have the frequency numbers)

Very close top 450pf, for the first 5.

If that's what you think, then you must also think the cap in the Boonton sucks. It shows more losses than what I consider my good caps.

Ya, I find that lacking, it should be there. What were they hiding?

Marconi, however, did. I'll post

What am I going to learn? Can we back out new Qs for the caps? I would like to do that and send you the data for plotting.

I may have made an error. I'm looking at page 9 of the manual. Can I make any measurement between Hi, Lo, and Grd with my inductor at resonance to find the current. Using XL/E. Or do I need to connect to the Q meter. No, across the Q meter won't work because that depends on where XQ is set.

Well if it doesn't give the right inductance for the number of turns, It may be #67, although rods are know to very a lot, depending on where the wire is located on the rod. That was an adjustment at the factory, slide the coil until it tunes the band properly

Hmm, I started having less confidence in the Q on the Boonton after getting higher numbers with the separated 3db method and "good" caps.

Thanks, Mikek

Hmm, one of us got off track there, my measurements were at 475kHz.

I just rechecked my voltage across the inductor @ 475kHz. Using my scope with a x10 probe, I get 9Vpp or 3.2V RMS. This does load the circuit a bit and adds some capacitance, I had to adjust the internal cap to bring it back to resonance at 475kHz. With the scope probe attached the Q dropped to 425. The difference in the capacitance with scope probe and without is

20.2pf. 452.7 - 432.5 = 20.2

Ok.

Overly complicated.

You don't need a cap meter. The Boonton will do everything you need. Just make sure it's on calibration. You can calibrate it with not much more than an LF oscillator and voltmeter.

450pf. I gave an about number on

As long as the capacitor you are measuring is smaller than the maximum of the Q meter's capacitor, all you have to do is find, or make, a suitable inductor, resonate that, then connect the specimen cap across Hi and Gnd, and resonate again,, using only the Boonton capacitor, not altering the frequency. The difference in dial readings is the unknown capacitance, the difference in Q readings can be used to calculate the Q of the unknown capacitor. You needn't worry about your residual 30pF, you're measuring "by difference". Repeat at several widely-spaced frequencies. Record and tabulate all your raw results. I used to do it all the time, before vector impedance meters came along. Far more accurate than most bridges. We even used to measure scope probes that way.

If the specimen capacitor is greater than the Q meter capacitor, you have to connect it in series with the tuned circuit, instead, and calculate differently.

Hell, we only had slide rules, and paper, back then.

In some ways, a Q meter is more useful for measuring capacitance than most high-end LCR meters, which won't measure things which have one end grounded. That includes my HP LCR meter that cost 5 figures new.

No. You know , from page 18, that a reading of 250 on the Q voltmeter corresponds to 5 volts across the resonating capacitor, or at least should be, if it's calibrated correctly. The scale is linear, so you know the voltage from whatever meter reading you have. That voltage,divided by the reactance of the tuning capacitor, which, at resonance, equals the reactance of the coil under test, gives you the current. No need for any other equipment, just make sure that R310 et al. are adjusted correctly.

--
"Design is the reverse of analysis" 
                   (R.D. Middlebrook)

I'm at a loss as to how it could be any easier. Once you find resonance, just shift the frequency down to 70.7% in voltage and and the same thing up in frequency. Please explain if you have another idea.

In fact I did measure my scope probe, and got 20.2 pf. 100Mhz probe. I have 300MHz probe, I'll try that later.

I also measured the low capacitance, high impedance amp I referred to in an earlier thread. When I put the amp across the Boonton internal cap, I had to reduce the Boonton cap by 2.48 pf. The Q dropped from

593.75 to 592.5 for 1.25 reduction in Q. I got 2.48pf as the loading capacitance. The amp added .00265 ohms of equivalent parallel resistance. 250uh 475kHz Is that enough info to calculate the parallel equivalent resistance? I'd like to know about my amp, the input impedance is 2.48 pf in parallel with (what) Resistance?

Yes, I want to take some time to run a calibration, but I want to understand what I'm doing before I start, I don't want to make it worse.

Your injection method is, essentially, a loose transformer. You cannot determine exactly what impedance is reflected into your test circuit by your source.

No, that's equivalent *series* resistance.

0.00265 ohms parallel resistance is called a short circuit :-) 0.00265ohms in series with 2.48pF, at 475kHz looks like 6.89 * 10^12 ohms, in parallel with 2.48pF

That looks a little high - electrometer sort of input resistance.

--
"Design is the reverse of analysis" 
                   (R.D. Middlebrook)

Yes, I was going for the equivalent parallel resistance.

That's pretty short.

Thanks you, that's what I was looking for. If that conversion is not to involved, I would like you to post it, I'd like to be able to do it myself.

Here's the input to my amp, is 6.89 trillion ohms possible? :-) (6.89 * 10^12) Ahh, I don't have the latest build on photobucket, here's a picture of something previous.

Where you see the 2 copper plates, I have replaced that with a 1/8 inch disc of 1/32 inch thick Teflon PCB material (Rogers 5880). Possibly 3/16 inch disc. I'll measure it later. So the R is the leakage through the Teflon in series with the input fet gate resistance and through the polystyrene at the input off the box. I'll post a link to the latest build later. Thank you, Mikek

Equivalent Circuits: where j=sqrt(-1)

at one, and only one frequency, it is possible to make these two circuits equivalent:

Series Rs+1/(j*Xs)

Parallel 1/(1/Rp+j*Xp)

Define Q as Xs/Rs, One fast approximation in a high Q circuit is that the Resistance has little effect on the reactance so simply taking the Q^2 times will get you close. for example if you have 100j ohms of capacitance in series with 1 ohm, it looks about like 100j ohms in parallel with 10k ohm in parallel. See it's still has a ratio of 100:1. it's easy to go back and forth that way.

Equivalency: Rs+j*Xs=1/(1/Rp+1/(j*Xp)) 1/(Rs+j*Xs)=1/Rp+1/(j*Xp) multiply by 1: (Rs-j*Xs)/(Rs^2+Xs^2)=1/Rp+1/(j*Xp) flip left and right side, and equate real to real and imaginary to imaginary real 1/Rp=Rs/(Rs^2+Xs^2) Rp=(Rs^2+Xs^2)/Rs=(1+Xs^2/Rs^2)*Rs, remember approx? Rp=(1+Q^2)*Rs imaginary be my guest

Thanks, I'll print that for the next time I want to figure this out. Mikek

As I said, I would post the latest build of the Low capacitance, high resistance amp input. (Kleijer amp) I forgot I had 20 Meg from gate to ground, so it's the resistance of the small dot of PCB material in series with the 20 Meg in parallel with the fet gate resistance. I'll try washing my finger prints from the pcb cap and see if I can get the 6.89 trillion ohms up a little. :-) I understand it may not be trillions of ohms, but it is really high resistance. >

The amp has a voltage gain of 1.3, when the output 50 ohm output is left unterminated. Mikek

OK, take a resistance and reactance in series, the impedance is:

Z = R + jX

Inductive reactance is positive, capacitive reactance is negative.

Firstly, we convert to an admittance, Y, where Y = 1/Z:

Y = G - jB = 1/(R + jX)

B is called the susceptance. Inductive susceptance is positive, capacitive susceptance is negative.

Resistance and reactance in series go to conductance and susceptance in parallel. (Units are mhos, or siemens - reciprocal ohms)

As previously explained about division of complex quantities:

*** G = R/(R^2 + x^2) and B = X/(R^2 + X^2) ***

*** Then R(parallel) = 1/G, and X(parallel) = 1/B ***

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Emphatically, no. Even Keithley electrometers only do about 10^14 ohms.

I took a look at that guy's web page. There's no way on earth that the input impedance is remotely as high as claimed. A plain JFET source follower, come on!

The input impedance will be dominated by Cgs, and Cds, which for the FET used are in the order of 2 to 2.5pF.

A Spice simulation of the front end shows an input impedance of 2.9k -j1.3meg.

Most of that is due to the "gimmick" capacitor.

Sorry for all your hard work, building it, but the design is a lemon.

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Where does an Australian go to get an avatar? 

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There's another point about that. If you are measuring by the "two voltmeter" method, which is how Q meters do it "natively", injection impedance isn't that important.

If you employ a "center frequency / 3dB bandwidth" method, it becomes vital, since the injection impedance now forms part of the measured Q.

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The gate is in series with the input capacitor, the 0.57pf gimmick.

I'm willing to listen, but, I don't see that. The gimmick capacitor, which is the input capacitor, is a 0.57pf capacitor, will not have 2.9k of leakage. Also, the proof is in the use, when applied across the inductor the Q just drops ever so slightly, I don't recall but it was only 1 or 2 Q points. Where do you see the low R coming from? Thanks, Mikek

PS. You sure are tough :-) I appreciate it.

Wouldn't the injection impedance lower Q? My Q went up. Well, that was with different capacitors, so I can't say that for sure without pulling the Boonton cap out and comparing it to my other "good" caps.

I must say, I thought the light coupling would have a very minor effect, I was 17 inches away with an untuned coil. I had a 50 ohm resistor across the drive coil, would it be better to put a high resistance in series with the drive coil to help raise injection impedance?

Thanks, Mikek

It's not leakage, it's the real part of the input impedance, which is mainly capacitive.

The magnitude of the input impedance at 1kHz is between 500 and 550 megohms. At 500khZ, it's down to 1.3 megohms. Angle varies from about 87 to 90 degrees, from 1kHz, to 500kHz. Almost wholly capacitive.

What you have is the Cgs, and Cgd of the FET, in parallel, making about

1pF, in parallel with 20 megohms, the whole in series with the fabricated capacitor.

You measured the input impedance as 0.00265 ohms in series with 2.48pF. Ignoring experimental error, that's where it comes from. The input impedance is almost wholly capacitive. As a measuring amplifier, it sucks.

I'll post the curves, if you like.

--
"Design is the reverse of analysis" 
                   (R.D. Middlebrook)

Sorry, that should be 5pF, it's a BF256.

--
"Design is the reverse of analysis" 
                   (R.D. Middlebrook)

First let me see if I understand what you are saying. For now I'm going to assume the 2.9k -j1.3meg is at 475kHz.

Are you suggesting, I could take a 0.25pf cap and put it in parallel with 2.9k and this would look like the input to my amp?

And I could take this same circuit and put it across an inductor on my Boonton and that would load the same as my amp?

I plead ignorance. If I had just a capacitor wouldn't it be the capacitance and a very high leakage resistance?

Am I mixing up a series and parallel conversion? (2.9k -j1.3meg)

I got that number by figuring how much R need to be added to reduce Q by the same amount adding the amp to the Boonton caused.

in series with 2.48pF.

This is how much I needed to increase the cap to bring the Boonton back to resonance.

I would think that I want an amplifier input with very low capacitance and very high Resistance. So far when I put it across an L on the Boonton, it acts like a very low capacitance and very high Resistance.

Is your spice model compatible with LTspice? I can make a run on that. Curves won't help until I understand the low R and why it doesn't affect Q on the Boonton. Thanks for playing along with me, Mikek

btw, I did run through the calibration no surprises, except one of the pots is in a different place, but it did what it should. It now measures Q just a little lower. The internal cap is very tarnished, don't know whether it could be improve it in any way. Also I noted he says his cap is 0.3pf, I measured a piece of PCB material with about 0.297 times the area of my gimmick and divided by

0.297 to get the 0.56pf. I thought I had made my cap smaller than his, apparently not.

So, I could lower the capacitance at the input with a different fet, if we come to some agreement that the amp does something worthwhile. :-) Mikek

Btw, not sure I should mention this, I have two more populated boards for the amp. The front end for each amp is not built yet.

Very risky. Unless you have a definite fault, leave it alone.

Look around the hamfests for GR standard capacitors, if you can find any.

No, just ditch the transformer arrangement, and use a known, low-value resistor. Surface mount current sensing resistors of a few milliohms are cheap and plentiful. Getting enough voltage is another matter, as is measuring with a 2% at best scope with 10 to 20pF from the probes...

--
"Design is the reverse of analysis" 
                   (R.D. Middlebrook)