Thanks Tom,
Any idea what it would be at a MHz or 10?
I might set it up tomorrow and measure it at RF, but, if it's less then an ohm, I'm not sure how I would do that. :-)
??????????????????????????????????????? Mikek
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Hi Mikek, Selecting Zo is an implementation choice. I picked 50ohm as
all of my bench patch cables are 50ohm.
Added to: is "SimSmith Example" which
demonstrates operation. G is oscillator equivalent with output of 100mV
(0.0002W). A is 50:1 ideal transformer. It's not perfect, as there is
some loading loss as shown with VSWR at about 1.02. Resulting with a
measured Q=196.7 from a Q=200 Inductor DUT.
John KN5L
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On 9/16/22 7:51 PM, Mikek wrote: I'm not sure I agree, the schematic I posted shows a 75¦¸ resistor across
the transformer as Zo.
And I'm waiting for an analysis of the impedance converter amp to see if
we can find it's output impedance.
I'm speculating that it is low, but the 220¦¸ resistor has me questioning.
????????????????????? Mikek
|
It¡¯s milliohms at dc. Take the resistance seen to the left of Q1¡¯s base and divide by the product of betas. If dc betas are 100 each, you get about 5 milliohms, again at dc.
Tom Sent from my iThing; please forgive the typos and brevity
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On Sep 16, 2022, at 17:51, Mikek <amdx@...> wrote:
?I'm not sure I agree, the schematic I posted shows a 75¦¸ resistor across the transformer as Zo.
And I'm waiting for an analysis of the impedance converter amp to see if we can find it's output impedance.
I'm speculating that it is low, but the 220¦¸ resistor has me questioning.
????????????????????? Mikek
|
Yes, I keep confusing the 260A circuit driving the 0.02¦¸ and the HP 4342A not.
??????????????????????????????? Mikek
|
I'm not sure I agree, the schematic I posted shows a 75¦¸ resistor across the transformer as Zo.
And I'm waiting for an analysis of the impedance converter amp to see if we can find it's output impedance.
I'm speculating that it is low, but the 220¦¸ resistor has me questioning.
????????????????????? Mikek
|
The operating impedance for this example is 75 ohm. 75 Ohm R1 in tuning
capacitor ass'y is matching load for impedance converter.
The secondary of T1 has little to no change in load to the impedance
converter.
John KN5L
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On 9/16/22 7:44 PM, Mikek wrote: Someone sent me a private messaged and corrected me on the output
impedance of the 50 to 1 transformer,
it is 0.001¦¸ or 1 milli¦¸. So, impedance ratio of 2500 x -.001 = 2.5¦¸
primary impedance. I'm not sure I know how to figure the output
impedance of the
power amp in the HP4342A, I would have thought 220¦¸.
I'm posting the schematic of the impedance converter the drives the
injection transformer primary and asking,
What is the output impedance of the impedance converter?
????????????????? Thanks, Mikek
|
Someone sent me a private messaged and corrected me on the output impedance of the 50 to 1 transformer,
it is 0.001¦¸ or 1 milli¦¸. So, impedance ratio of 2500 x -.001 = 2.5¦¸ primary impedance. I'm not sure I know how to figure the output impedance of the
power amp in the HP4342A, I would have thought 220¦¸.
I'm posting the schematic of the impedance converter the drives the injection transformer primary and asking,
What is the output impedance of the impedance converter?
????????????????? Thanks, Mikek
|
Hi Mikek, Added to: is example HP-400E
application.
John KN5L
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On 9/16/22 7:23 PM, John KN5L wrote: Hi Mikek, In the figure, Zo is output impedance of the oscillator,
normally 50 ohms. The Zo resistor is the impedance match for the
oscillator. The "Injection Transformer" supplies 1/50 of the voltage
from the oscillator.
I'd try a FT50-43 for the core. An example in:
John KN5L
On 9/16/22 5:57 PM, Mikek wrote:
John, now that I have thought about your graph a little bit, it seems
clear the primary needs more inductance.
To be flat down to 200kHz, I 'think' it would need about 160uh.
That would be 168 turns on that core.
?Was the point of your graph to show that the transformer needs more
work? Maybe a different core, maybe ferrite instead of iron?
Does anyone have a picture of the HP 4342A transformer?
?I added a schematic picture of the section about the HP 4342A
transformer. 50 to 1, has a 1 turn, 0.1¦¸ secondary output impedance, and
is flat from 10kHz to 70 MHz.
?My understanding is a 50 to 1 turns ratio is a 2500 to 1 impedance
ratio, so 0.1 x 2500 = 250¦¸ input impedance. Is this at 10kHz???
The resistor across the primary is 75 ohms.? Yes, I'm confused by it all.
????????????????????????????????????????????????? Mikek
|
Hi Mikek, In the figure, Zo is output impedance of the oscillator,
normally 50 ohms. The Zo resistor is the impedance match for the
oscillator. The "Injection Transformer" supplies 1/50 of the voltage
from the oscillator.
I'd try a FT50-43 for the core. An example in:
John KN5L
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On 9/16/22 5:57 PM, Mikek wrote: John, now that I have thought about your graph a little bit, it seems
clear the primary needs more inductance.
To be flat down to 200kHz, I 'think' it would need about 160uh.
That would be 168 turns on that core.
?Was the point of your graph to show that the transformer needs more
work? Maybe a different core, maybe ferrite instead of iron?
Does anyone have a picture of the HP 4342A transformer?
?I added a schematic picture of the section about the HP 4342A
transformer. 50 to 1, has a 1 turn, 0.1¦¸ secondary output impedance, and
is flat from 10kHz to 70 MHz.
?My understanding is a 50 to 1 turns ratio is a 2500 to 1 impedance
ratio, so 0.1 x 2500 = 250¦¸ input impedance. Is this at 10kHz???
The resistor across the primary is 75 ohms.? Yes, I'm confused by it all.
????????????????????????????????????????????????? Mikek
|
John, now that I have thought about your graph a little bit, it seems clear the primary needs more inductance.
To be flat down to 200kHz, I 'think' it would need about 160uh.
That would be 168 turns on that core.
?Was the point of your graph to show that the transformer needs more work? Maybe a different core, maybe ferrite instead of iron?
Does anyone have a picture of the HP 4342A transformer?
?I added a schematic picture of the section about the HP 4342A transformer. 50 to 1, has a 1 turn, 0.1¦¸ secondary output impedance, and is flat from 10kHz to 70 MHz.
?My understanding is a 50 to 1 turns ratio is a 2500 to 1 impedance ratio, so 0.1 x 2500 = 250¦¸ input impedance. Is this at 10kHz???
The resistor across the primary is 75 ohms.? Yes, I'm confused by it all.
????????????????????????????????????????????????? Mikek
|
Here's two Q measurement methods, using S11 and S21 bandwidth. Measuring
a reference Micrometals device for comparison.
John KN5L
|
Hello George,
?
You could have said the same about the bible ¨C and I would have
just deleted ¨C or I might have said ¡°Merely another assertion¡±.
?
As soon as anyone tells me that something or somebody¡¯s work
covers ¡®every¡¯ anything, I go into scientist mode. There is nothing known
nowadays in physics or even its offspring, electronics, that covers everything.
Scientists and engineers will find new ¡®things¡¯. That is the essence of
science.
?
Now, had you defended Grover on the grounds that his formulae
are based on specific, but as yet irrefuted, laws of physics, I might have paid
more attention. But you made no mention of Faraday, Gauss, Hertz or even Lenz,
some of the forefathers of modern-day electro-magnetics.
?
73 de Brian, VK2GCE
?
?
Hi Brian,
?
It has formulas to calculate inductance for every conceivable
shape and size conductor and type of coil.
?
?
?
Hello George,
As the bible is full of contradictions, what is so valuable
about Grover?
73 de Brian, VK2GCE
?
?
The bible for inductance calculations is:
?
Inductance Calculations by FREDERICK-W-GROVER
?
|
Oops, I really should remember to read all my incoming emails to see where the topic is going before making comments
Johm
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On 9/15/2022 7:06 PM, John Kolb wrote: A comparison to a 1" long square wire 0.3" x 0.3"? would be a more equivalent comparison.
?John
On 9/15/2022 3:43 PM, Mikek wrote:
On another note: How does inductance change from say, a 1" long 0.30" diameter wire vs a 1" x 1"? square 0.30" thick?<
Found a calculator, 1" wire 21.1nH, 1" square plate 7.1nH.
_._,_._,_
------------------------------------------------------------------------
|
30 ga wire is 0.01" diameter. Your inductance is correct, 26.7 nH. I've handled LOTS of 30 ga wire :)
John
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On 9/15/2022 7:30 PM, Mikek wrote: I'm sorry, I got the numbers messed up.
Corrected below.
?How does inductance change from say, a 1" long 0.10"
diameter wire (30 Gauge wire) vs a 1" x 1"? square 1.4mil thick? (1oz copper pcb)
Found a calculator, 1" of 30 gauge wire, 26.7nH, 1" square copper pcb 7.19nH.
??????????????????????????? Mikek
|
Hi Brian, ? It has formulas to calculate inductance for every conceivable shape and size conductor and type of coil. ? ?
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From: [email protected] [mailto:[email protected]] On Behalf Of Brian
Sent: Friday, 16 September 2022 12:03 PM
To: [email protected]
Subject: Re: [Test Equipment Design & Construction] Making a Q-meter /? Hello George, As the bible is full of contradictions, what is so valuable about Grover? 73 de Brian, VK2GCE ? ? The bible for inductance calculations is: ? Inductance Calculations by FREDERICK-W-GROVER ? ? ? I'm curious where you found that calculator. I would've guessed something a bit over 10nH for the plate, based on scaling the value for the wire. Not a huge difference, but I'm curious to see what formula that calculator is using for the square plate.
-- Cheers,
Tom -- Prof. Thomas H. Lee Allen Ctr., Rm. 205 350 Jane Stanford Way Stanford University Stanford, CA 94305-4070 On 9/15/2022 15:43, Mikek wrote: >On another note: How does inductance change from say, a 1" long 0.30" diameter wire vs a 1" x 1"? square 0.30" thick?<
Found a calculator, 1" wire 21.1nH, 1" square plate 7.1nH.
?
|
>The 0.02 ohm resistor is in series with the Inductor being measures so affects the Q measurement. I don't understand why the inductance of the thermocouple is meaningful, considering it's in series with the inductance of the connecting cable and wiring, switch contacts, and the oscillator coil secondary winding?<
?>
The 0.02 ohm resistor is in series with the Inductor being measures so affects the Q measurement. I don't understand why the inductance of the thermocouple is meaningful, considering it's in series with the inductance of the connecting cable and wiring, switch contacts, and the oscillator coil secondary winding?<
OK. I'll go with that.
Did anyone look at the Vishay surface mount resistors??
The have very tiny values, I can't find the data but, say we took two 0603 smd .04¦¸ resistors and put them in parallel and reduced the
inductance by half. That could be .25nH. (if indeed the smallest resistor has the smallest inductance)
A thought for the current sensor, pcb mount a 0603 smd resistor (small value of your choice. Then take a 1210
smd thermistor and using silver filled epoxy, attach it too the 0603 resistor at a 90* angle. The 1210 is big enough that it would span the
0603 resistor then wire could attach the thermistor to tracks on the pcb. Flow one amp through the 0603 resistor and measure the thermistor.
???????????????????????????????????????? Mikek
|
I'm sorry, I got the numbers messed up.
Corrected below.
?How does inductance change from say, a 1" long 0.10"
diameter wire (30 Gauge wire) vs a 1" x 1"? square 1.4mil thick? (1oz copper pcb)
Found a calculator, 1" of 30 gauge wire, 26.7nH, 1" square copper pcb 7.19nH.
??????????????????????????? Mikek
|
The 0.02 ohm resistor is in series with the Inductor being measures so affects the Q measurement. I don't understand why the inductance of the thermocouple is meaningful, considering it's in series with the inductance of the connecting cable and wiring, switch contacts, and the oscillator coil secondary winding?
John
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On 9/15/2022 4:10 PM, Mikek wrote: That brings up the question, "What is the inductance of the precision resistor and the nichrome wire?
My thinking is they had to have a lot of heat to develop the current required to drive the meter.
See diagram.
We have opamps to give us all the gain we need. I did some figuring, a 10*C increase in temp will cause a 400mv change in a type K thermocouple.
A 0.125" long by 11.6 mill trace of 1 oz copper will rise 10*C from ambient with 1 amp flowing.
A 1" by 1", 1oz copper pcb will have about 7nH (I think that's high compared to the inductance of the precision resistor and the nichrome wire, but I don't know?)
If you etch it to have a 0.125 x 11.6 mill trace in the middle, the inductance will increase. the concept would be to put the thermocouple in the middle of that 0.125" trace,
spot weld as in the prc68 webpage in my last post.
?Here is a drawing of my thinking, ( I drew most of it so I would understand what the circuit is)
|
And is that the inductance from one side of the square to the other, a 0.30" length, or from one end o the other, a 1" length.
John
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On 9/15/2022 4:03 PM, Tom Lee wrote: I'm curious where you found that calculator. I would've guessed something a bit over 10nH for the plate, based on scaling the value for the wire. Not a huge difference, but I'm curious to see what formula that calculator is using for the square plate.
-- Cheers,
Tom
--
Prof. Thomas H. Lee
Allen Ctr., Rm. 205
350 Jane Stanford Way
Stanford University
Stanford, CA 94305-4070
On 9/15/2022 15:43, Mikek wrote:
On another note: How does inductance change from say, a 1" long 0.30" diameter wire vs a 1" x 1"? square 0.30" thick?<
Found a calculator, 1" wire 21.1nH, 1" square plate 7.1nH.
_._,_._,_
------------------------------------------------------------------------
|
A comparison to a 1" long square wire 0.3" x 0.3" would be a more equivalent comparison.
John
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On 9/15/2022 3:43 PM, Mikek wrote: On another note: How does inductance change from say, a 1" long 0.30" diameter wire vs a 1" x 1"? square 0.30" thick?<
Found a calculator, 1" wire 21.1nH, 1" square plate 7.1nH.
_._,_._,_
------------------------------------------------------------------------
|
Brian
John Kolb