More Math for the "SETUP"

[Fred Abse]

On Sat, 15 Feb 2014 13:58:20 -0600, amdx wrote:

> Near as I can figure the Boonton is measuring with about 4 microamps.

According to Page 18 of the Boonton manual, an indicated Q of 250
corresponds to 5 volts across the capacitor. At resonance, therefore, with
an 247uH inductor, at 450kHz, and an indicated Q of 550, that extrapolates
to 10.25 volts across 698 ohms, hence a current of 15 milliamps.

FEMM run at 15 mA:
Total current = 0.015 Amps
Voltage Drop = 0.0453953+I*28.1652 Volts
Flux Linkage = 9.4971e-06-I*1.49318e-08 Webers
Flux/Current = 0.00063314-I*9.95455e-07 Henries
Voltage/Current = 3.02635+I*1877.68 Ohms
Real Power = 0.000340465 Watts
Reactive Power = 0.211239 VAr
Apparent Power = 0.211239 VA

The manual for the Marconi TF1245 is on BAMA. Get it. It includes a good
treatment of errors arising from instrument strays in Q meters.

So does the Boonton manual.

Not in as great a detail.

I don't think we are far of on Q if you put in the tuning capacitor
losses. I have not found a number for that in the Boonton Manual.
I will keep looking.

AFAICS, it isn't there. I'll post a copy of the Marconi capacitor's curves
to A.B.S.E.

At 500pF, and 1MHz, the Marconi capacitor has a stated Q just south of
20,000, rising to 80,000 at 5MHz, then falling off.

PS. see if you can figure anything about our 250uh/621uh discrepancy.
The rod is out of an AM radio, I don't know of any that used other
than #61 material, but?

ISTR lots of manufacturers offering antenna rods, back in the day, from
all around the world. Japanese, Taiwanese, the whole nine yards. Some
better than others.

Oh, I also learned something about the connection points. I'm using
banana pin connection on my coil, but the threaded connectors still need
to be tight or there is a loss. After tightening my Q=550 increased to
Q=610 I'll even try removing the banana plugs and see if the Q goes over
610.

Please do that. Banana plugs aren't very good. The silver-plated "Z" plugs
are better, but not as good as clamping.

The Marconi has gold-plated brass terminals, and shorting straps, BTW. It
actually has two capacitors, and two sets of terminals, covering right up
to 300MHz.



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

 

[amdx]

On 2/16/2014 9:48 AM, Fred Abse wrote:

On Sat, 15 Feb 2014 18:57:04 -0600, amdx wrote:

On 2/15/2014 1:58 PM, amdx wrote:
On 2/15/2014 10:50 AM, Fred Abse wrote:
On Fri, 14 Feb 2014 17:14:29 -0600, amdx wrote:

I went through 7 of what I thought were my “good” caps.
A description of how I setup to measure using the 3db method.
Drawing of layout is here.
http://tinyurl.com/n95t3zl
I have a 4” dia air core coil connected to my frequency generator,
I also had a 50 ohm resistor in parallel with the coil.
This coil was 17” from the inductor under test, on axis.

17 "somethings". My newsreader doesn't support those silly Windows
characters. Please either use ", and ', or "in", "ft", "mm", "m", etc. I
can't tell whether it was inches or feet ;-)

17 inches.

I then attached a capacitor
across the inductor.
This was all sat on a piece of 4” styrofoam and away from lossy
items.
I have a high impedance, low capacitance amplifier, like the second amp
design here.
www.crystal-radio.eu/fetamp/enfetamp.htm
I connected that amp across the LC circuit and connected my scope to the
output. I set the Sig Gen to about 475kHz and adjusted the cap for
maximum amplitude on the scope. I recorded the frequency. Of course
there was some back and forth to find the resonant frequency.

For best accuracy, fine tune the frequency, not the capacitor. I did a
proof thirty years ago, when I liked calculus more than I do today.

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.

That's why Marconi's Dielectric Test Set, a huge beast, with a 2 foot
scale mirror galvanometer, glass insulators, and micrometer capacitors,
used parallel, not series,resonance. Cost as much as a car did, back then.


I then
adjusted the Sig Gen output level so the scope reading was 7 units. Then
I adjusted the frequency up until I had 5 units on the scope,

5/7 isn't quite -3dB. How much bandwidth error you'll get depends on how
sharp, or flat the peak response is.

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

I recorded
the frequency, then lowered the frequency until I had 5 units again, and
recorded the frequency. 7 units x 0.707 = 4.949 or 5 units.
I did three measurements on all seven caps and calculated an average.
Most measurements were close to each other.

That's a rather Byzantine way of going about things.

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

Why not use a parallel connection method on the Boonton? Measure at
several frequencies, and tabulate the results.

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.

Cap 4 had the widest range of measurements 499, 524, and 528.

With Cap 1 average Q = 721

With Cap 2 average Q = 630

With Cap 3 average Q = 701 First 5 @ 475kHz

With Cap 4 average Q = 517

With Cap 5 average Q = 672

With Cap 6 average Q = 689 @ 500kHz

With Cap 7 average Q = 713 @ 600kHz

You don't say what capacitance the capacitors were, or the frequency.

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)

After removing the banana plugs from my inductor and connecting with the
compression thumb nuts,
The Q measured on the Boonton. Q = 618. It needs some work, I would
think it equal or better my 3db testing. My first thought is the wiper
connection on the capacitor.

Lab grade capacitors have multiple wipers, usually bearing on the vane
edges.


So it was a worthwhile exercise, I have marked my caps and know which
are best.

Your thoughts, Mikek

Firstly, without capacitor values, the figures aren't much use. I will say
that they suck as measurement capacitors.

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.



> Boonton (carefully?) don't state parameters for their main capacitor.

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

Marconi, however, did. I'll post

their curves to A.B.S.E. Suffice it to say here that, at 500pF, and 1Mhz,
Q is greater than 20000.

Do the capacitor testing on the Boonton, in parallel with the main
capacitor, measure the C, and Q at various settings, and tabulate the
results. I can do curves with gnuplot, from tables, if you don't have
plotting software.

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 refer you to my last reply, where I wrote about Boonton test current.
Using 15 milliamps, calculated from Boonton's voltmeter figures, FEMM
gives Q at 620. Close enuf?

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.

Inductance given by FEMM is still high, at 633uH. I still suspect that
your ferrite isn't what you think it is. Might it be #67?

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

I'm beginning to believe your Boonton ;-)

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

 

[amdx]

On 2/16/2014 9:48 AM, Fred Abse wrote:

On Sat, 15 Feb 2014 13:58:20 -0600, amdx wrote:

Near as I can figure the Boonton is measuring with about 4 microamps.

According to Page 18 of the Boonton manual, an indicated Q of 250
corresponds to 5 volts across the capacitor. At resonance, therefore, with
an 247uH inductor, at 450kHz, and an indicated Q of 550, that extrapolates
to 10.25 volts across 698 ohms, hence a current of 15 milliamps.

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

FEMM run at 15 mA:
Total current = 0.015 Amps
Voltage Drop = 0.0453953+I*28.1652 Volts
Flux Linkage = 9.4971e-06-I*1.49318e-08 Webers
Flux/Current = 0.00063314-I*9.95455e-07 Henries
Voltage/Current = 3.02635+I*1877.68 Ohms
Real Power = 0.000340465 Watts
Reactive Power = 0.211239 VAr
Apparent Power = 0.211239 VA


The manual for the Marconi TF1245 is on BAMA. Get it. It includes a good
treatment of errors arising from instrument strays in Q meters.

So does the Boonton manual.

Not in as great a detail.

Ok.

I don't think we are far of on Q if you put in the tuning capacitor
losses. I have not found a number for that in the Boonton Manual.
I will keep looking.

AFAICS, it isn't there. I'll post a copy of the Marconi capacitor's curves
to A.B.S.E.

At 500pF, and 1MHz, the Marconi capacitor has a stated Q just south of
20,000, rising to 80,000 at 5MHz, then falling off.


PS. see if you can figure anything about our 250uh/621uh discrepancy.
The rod is out of an AM radio, I don't know of any that used other
than #61 material, but?

ISTR lots of manufacturers offering antenna rods, back in the day, from
all around the world. Japanese, Taiwanese, the whole nine yards. Some
better than others.


Oh, I also learned something about the connection points. I'm using
banana pin connection on my coil, but the threaded connectors still need
to be tight or there is a loss. After tightening my Q=550 increased to
Q=610 I'll even try removing the banana plugs and see if the Q goes over
610.

Please do that. Banana plugs aren't very good. The silver-plated "Z" plugs
are better, but not as good as clamping.

The Marconi has gold-plated brass terminals, and shorting straps, BTW. It
actually has two capacitors, and two sets of terminals, covering right up
to 300MHz.

 

[Fred Abse]

On Sun, 16 Feb 2014 11:05:07 -0600, amdx wrote:

That's a rather Byzantine way of going about things.

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

Overly complicated.

No, I'm lacking there, my cap meter has died, I'm in the market for a
new one. Any suggestions?

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.

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)

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.

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. 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)

 

[amdx]

On 2/16/2014 12:54 PM, Fred Abse wrote:

On Sun, 16 Feb 2014 11:05:07 -0600, amdx wrote:

That's a rather Byzantine way of going about things.

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

Overly complicated.

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.

No, I'm lacking there, my cap meter has died, I'm in the market for a
new one. Any suggestions?

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.

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)

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.

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?

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.

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. 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.

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.

 

[Fred Abse]

On Sun, 16 Feb 2014 14:55:14 -0600, amdx wrote:

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.

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

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.

No, that's equivalent *series* resistance.

0.00265 ohms parallel resistance is called a short circuit

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?

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)

 

[amdx]

On 2/17/2014 1:07 PM, Fred Abse wrote:

On Sun, 16 Feb 2014 14:55:14 -0600, amdx wrote:

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.


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


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.

No, that's equivalent *series* resistance.

Yes, I was going for the equivalent parallel resistance.

0.00265 ohms parallel resistance is called a short circuit

That's pretty short.

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?

0.00265ohms in series with 2.48pF, at 475kHz looks like 6.89 * 10^12 ohms,
in parallel with 2.48pF

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.

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

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.
http://i395.photobucket.com/albums/pp37/Qmavam/Kleijerampinbox.jpg
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

 

[RobertMacy]

On Mon, 17 Feb 2014 14:38:49 -0700, amdx <nojunk@knology.net> wrote:

On 2/17/2014 1:07 PM, Fred Abse wrote:
...snip...
0.00265ohms in series with 2.48pF, at 475kHz looks like 6.89 * 10^12
ohms,
in parallel with 2.48pF

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.

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

 

[amdx]

On 2/17/2014 4:37 PM, RobertMacy wrote:

On Mon, 17 Feb 2014 14:38:49 -0700, amdx <nojunk@knology.net> wrote:

On 2/17/2014 1:07 PM, Fred Abse wrote:
...snip...
0.00265ohms in series with 2.48pF, at 475kHz looks like 6.89 * 10^12
ohms,
in parallel with 2.48pF

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.


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

 

[amdx]

On 2/17/2014 3:38 PM, amdx wrote:

On 2/17/2014 1:07 PM, Fred Abse wrote:

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.

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.
http://i395.photobucket.com/albums/pp37/Qmavam/Kleijerampinbox.jpg
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

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.

http://i395.photobucket.com/albums/pp37/Qmavam/PCBCAPinKleijerAMP_zpse155433e.jpg

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

 

[Fred Abse]

On Mon, 17 Feb 2014 15:38:49 -0600, amdx wrote:

If that conversion is not to involved, I would like you to post it, I'd
like to be able to do it myself.

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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[Fred Abse]

On Mon, 17 Feb 2014 15:38:49 -0600, amdx wrote:

> Here's the input to my amp, is 6.89 trillion ohms possible?

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

(6.89 * 10^12)
Ahh, I don't have the latest build on photobucket, here's a picture of
something previous.
http://i395.photobucket.com/albums/pp37/Qmavam/Kleijerampinbox.jpg

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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[Fred Abse]

On Mon, 17 Feb 2014 11:07:00 -0800, Fred Abse wrote:

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

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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Where does an Australian go to get an avatar?

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[amdx]

On 2/18/2014 1:53 PM, Fred Abse wrote:

On Mon, 17 Feb 2014 15:38:49 -0600, amdx wrote:

Here's the input to my amp, is 6.89 trillion ohms possible?

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

(6.89 * 10^12)
Ahh, I don't have the latest build on photobucket, here's a picture of
something previous.
http://i395.photobucket.com/albums/pp37/Qmavam/Kleijerampinbox.jpg


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.

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

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.

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.

 

[amdx]

On 2/18/2014 1:53 PM, Fred Abse wrote:

On Mon, 17 Feb 2014 11:07:00 -0800, Fred Abse wrote:

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

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.

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

 

[Fred Abse]

On Tue, 18 Feb 2014 16:51:33 -0600, amdx wrote:

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.

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.

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?

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)

 

[Fred Abse]

On Wed, 19 Feb 2014 00:40:33 -0800, Fred Abse wrote:

What you have is the Cgs, and Cgd of the FET, in parallel, making about
1pF

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

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

 

[amdx]

On 2/19/2014 2:40 AM, Fred Abse wrote:

On Tue, 18 Feb 2014 16:51:33 -0600, amdx wrote:

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.

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?

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

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)

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.

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?

You measured the input impedance as 0.00265 ohms

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.

Ignoring experimental error, that's where it comes from. The input
impedance is almost wholly capacitive. As a measuring amplifier, it sucks.

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.

> I'll post the curves, if you like.

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.

 

[amdx]

On 2/19/2014 3:19 AM, Fred Abse wrote:

On Wed, 19 Feb 2014 00:40:33 -0800, Fred Abse wrote:

What you have is the Cgs, and Cgd of the FET, in parallel, making about
1pF

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

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.

 

[Fred Abse]

On Tue, 18 Feb 2014 16:57:39 -0600, amdx wrote:

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.

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

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

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?

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)