I have a High Q coil, it's Q measures 1111.

I put a second very high impedance test probe across the coil and the Q
drops to 1090.

 Frequency shift is minuscule.

 What is the parallel resistance of the second very high impedance probe?

Formula please, I have a few more data points to calculate.

                                   Thank you, Mikek

--
This email has been checked for viruses by Avast antivirus software.
https://www.avast.com/antivirus

On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
> I have a High Q coil, it's Q measures 1111.
>
> I put a second very high impedance test probe across the coil and the Q
> drops to 1090.
>
> Frequency shift is minuscule.
>
> What is the parallel resistance of the second very high impedance probe?
>
> Formula please, I have a few more data points to calculate.
>
> Thank you, Mikek

https://www.electronics-tutorials.ws/inductor/parallel-inductors.html

On 8/18/2021 4:31 PM, Ed Lee wrote:
> On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
>> I have a High Q coil, it's Q measures 1111.
>>
>> I put a second very high impedance test probe across the coil and the Q
>> drops to 1090.
>>
>> Frequency shift is minuscule.
>>
>> What is the parallel resistance of the second very high impedance probe?
>>
>> Formula please, I have a few more data points to calculate.
>>
>> Thank you, Mikek
> https://www.electronics-tutorials.ws/inductor/parallel-inductors.html

Thanks, but that doesn't show me how to recalculate Q after adding a
parallel resistor.

                                       Mikek

--
This email has been checked for viruses by Avast antivirus software.
https://www.avast.com/antivirus

On Wednesday, August 18, 2021 at 2:56:19 PM UTC-7, amdx wrote:
> On 8/18/2021 4:31 PM, Ed Lee wrote:
> > On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
> >> I have a High Q coil, it's Q measures 1111.
> >>
> >> I put a second very high impedance test probe across the coil and the Q
> >> drops to 1090.
> >>
> >> Frequency shift is minuscule.
> >>
> >> What is the parallel resistance of the second very high impedance probe?
> >>
> >> Formula please, I have a few more data points to calculate.
> >>
> >> Thank you, Mikek
> > https://www.electronics-tutorials.ws/inductor/parallel-inductors.html
> Thanks, but that doesn't show me how to recalculate Q after adding a
> parallel resistor.
> Mikek

Impedance is fixed for resistor, but function of freq for inductor. It still follow the same equation:

1090 = (1111 * R) / (1111 + R) at the test freq.

On Wednesday, August 18, 2021 at 3:06:01 PM UTC-7, Ed Lee wrote:
> On Wednesday, August 18, 2021 at 2:56:19 PM UTC-7, amdx wrote:
> > On 8/18/2021 4:31 PM, Ed Lee wrote:
> > > On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
> > >> I have a High Q coil, it's Q measures 1111.
> > >>
> > >> I put a second very high impedance test probe across the coil and the Q
> > >> drops to 1090.
> > >>
> > >> Frequency shift is minuscule.
> > >>
> > >> What is the parallel resistance of the second very high impedance probe?
> > >>
> > >> Formula please, I have a few more data points to calculate.
> > >>
> > >> Thank you, Mikek
> > > https://www.electronics-tutorials.ws/inductor/parallel-inductors.html
> > Thanks, but that doesn't show me how to recalculate Q after adding a
> > parallel resistor.
> > Mikek
> Impedance is fixed for resistor, but function of freq for inductor. It still follow the same equation:
>
> 1090 = (1111 * R) / (1111 + R) at the test freq.

You might also need to convert Q to impedance with ( 2 pi w L ) / Q

On 8/18/2021 5:37 PM, Ed Lee wrote:
> On Wednesday, August 18, 2021 at 3:06:01 PM UTC-7, Ed Lee wrote:
>> On Wednesday, August 18, 2021 at 2:56:19 PM UTC-7, amdx wrote:
>>> On 8/18/2021 4:31 PM, Ed Lee wrote:
>>>> On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
>>>>> I have a High Q coil, it's Q measures 1111.
>>>>>
>>>>> I put a second very high impedance test probe across the coil and the Q
>>>>> drops to 1090.
>>>>>
>>>>> Frequency shift is minuscule.
>>>>>
>>>>> What is the parallel resistance of the second very high impedance probe?
>>>>>
>>>>> Formula please, I have a few more data points to calculate.
>>>>>
>>>>> Thank you, Mikek
>>>> https://www.electronics-tutorials.ws/inductor/parallel-inductors.html
>>> Thanks, but that doesn't show me how to recalculate Q after adding a
>>> parallel resistor.
>>> Mikek
>> Impedance is fixed for resistor, but function of freq for inductor. It still follow the same equation:
>>
>> 1090 = (1111 * R) / (1111 + R) at the test freq.
> You might also need to convert Q to impedance with ( 2 pi w L ) / Q
>
OK, the first formula got me an answer of about 60kΩ. I was expecting
2MΩ to 5 MΩ.

 I think I see how I can work with the second formula.

                                Mikek

--
This email has been checked for viruses by Avast antivirus software.
https://www.avast.com/antivirus

On 8/18/2021 5:50 PM, amdx wrote:
> On 8/18/2021 5:37 PM, Ed Lee wrote:
>> On Wednesday, August 18, 2021 at 3:06:01 PM UTC-7, Ed Lee wrote:
>>> On Wednesday, August 18, 2021 at 2:56:19 PM UTC-7, amdx wrote:
>>>> On 8/18/2021 4:31 PM, Ed Lee wrote:
>>>>> On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
>>>>>> I have a High Q coil, it's Q measures 1111.
>>>>>>
>>>>>> I put a second very high impedance test probe across the coil and
>>>>>> the Q
>>>>>> drops to 1090.
>>>>>>
>>>>>> Frequency shift is minuscule.
>>>>>>
>>>>>> What is the parallel resistance of the second very high impedance
>>>>>> probe?
>>>>>>
>>>>>> Formula please, I have a few more data points to calculate.
>>>>>>
>>>>>> Thank you, Mikek
>>>>> https://www.electronics-tutorials.ws/inductor/parallel-inductors.html
>>>> Thanks, but that doesn't show me how to recalculate Q after adding a
>>>> parallel resistor.
>>>> Mikek
>>> Impedance is fixed for resistor, but function of freq for inductor.
>>> It still follow the same equation:
>>>
>>> 1090 = (1111 * R) / (1111 + R) at the test freq.
>> You might also need to convert Q to impedance with ( 2 pi w L ) / Q
>>
> OK, the first formula got me an answer of about 60kΩ. I was expecting
> 2MΩ to 5 MΩ.
>
>  I think I see how I can work with the second formula.
>
>                                 Mikek

 I got the same 60kΩ.

If you are interested here is the website to a high input impedance amp.
He built one and then a second better one.

 He uses the first to measure Q, 3db method, then adds the second high
input impedance amp as a load and recalculates

the Q. Q drops from 1111 to 1090 with the second amp amp added. He
calculates the R of the amp at 5.8MΩ but says in reality

it will be lower, but I don't think 60kΩ.

The chart with numbers is after the next to last picture. I wonder if
I'm still calculating wrong or if the author is wrong.

> http://www.crystal-radio.eu/fetamp/enfetamp.htm
                                       Mikek

--
This email has been checked for viruses by Avast antivirus software.
https://www.avast.com/antivirus

On Wednesday, August 18, 2021 at 4:21:06 PM UTC-7, amdx wrote:
> On 8/18/2021 5:50 PM, amdx wrote:
> > On 8/18/2021 5:37 PM, Ed Lee wrote:
> >> On Wednesday, August 18, 2021 at 3:06:01 PM UTC-7, Ed Lee wrote:
> >>> On Wednesday, August 18, 2021 at 2:56:19 PM UTC-7, amdx wrote:
> >>>> On 8/18/2021 4:31 PM, Ed Lee wrote:
> >>>>> On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
> >>>>>> I have a High Q coil, it's Q measures 1111.
> >>>>>>
> >>>>>> I put a second very high impedance test probe across the coil and
> >>>>>> the Q
> >>>>>> drops to 1090.
> >>>>>>
> >>>>>> Frequency shift is minuscule.
> >>>>>>
> >>>>>> What is the parallel resistance of the second very high impedance
> >>>>>> probe?
> >>>>>>
> >>>>>> Formula please, I have a few more data points to calculate.
> >>>>>>
> >>>>>> Thank you, Mikek
> >>>>> https://www.electronics-tutorials.ws/inductor/parallel-inductors.html
> >>>> Thanks, but that doesn't show me how to recalculate Q after adding a
> >>>> parallel resistor.
> >>>> Mikek
> >>> Impedance is fixed for resistor, but function of freq for inductor.
> >>> It still follow the same equation:
> >>>
> >>> 1090 = (1111 * R) / (1111 + R) at the test freq.
> >> You might also need to convert Q to impedance with ( 2 pi w L ) / Q
> >>
> > OK, the first formula got me an answer of about 60kΩ. I was expecting
> > 2MΩ to 5 MΩ.
> >
> > I think I see how I can work with the second formula.
> >
> > Mikek
> I got the same 60kΩ.
>
> If you are interested here is the website to a high input impedance amp.
> He built one and then a second better one.
>
> He uses the first to measure Q, 3db method, then adds the second high
> input impedance amp as a load and recalculates
>
> the Q. Q drops from 1111 to 1090 with the second amp amp added. He
> calculates the R of the amp at 5.8MΩ but says in reality
>
> it will be lower, but I don't think 60kΩ.
>
> The chart with numbers is after the next to last picture. I wonder if
> I'm still calculating wrong or if the author is wrong.

He said:

"With the formula Z= 2.pi.f.L Q the impedance of the unloaded circuit60 can be calculated (the inductance of the coil is 230 uH):
Z= 1.1 M.Ohm at 600 kHz
Z= 1.57 M.Ohm at 900 kHz
Z= 1.93 M.Ohm at 1200 kHz
Z= 1.97 M.Ohm at 1500 kH"

If adding R drop Z from Z1 to Z2, then Z = Z1 * Z2 / (Z2 - Z1)

Z2 = 1.11M
Z1 = 1.09M
Z1 = (Z2 * R) / (Z2 + R)
Z1 Z2 + Z1 R = Z2 R
Z = Z1 Z2 / (Z2 - Z1) = 60M

On Wednesday, August 18, 2021 at 5:04:38 PM UTC-7, Ed Lee wrote:
> On Wednesday, August 18, 2021 at 4:21:06 PM UTC-7, amdx wrote:
> > On 8/18/2021 5:50 PM, amdx wrote:
> > > On 8/18/2021 5:37 PM, Ed Lee wrote:
> > >> On Wednesday, August 18, 2021 at 3:06:01 PM UTC-7, Ed Lee wrote:
> > >>> On Wednesday, August 18, 2021 at 2:56:19 PM UTC-7, amdx wrote:
> > >>>> On 8/18/2021 4:31 PM, Ed Lee wrote:
> > >>>>> On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
> > >>>>>> I have a High Q coil, it's Q measures 1111.
> > >>>>>>
> > >>>>>> I put a second very high impedance test probe across the coil and
> > >>>>>> the Q
> > >>>>>> drops to 1090.
> > >>>>>>
> > >>>>>> Frequency shift is minuscule.
> > >>>>>>
> > >>>>>> What is the parallel resistance of the second very high impedance
> > >>>>>> probe?
> > >>>>>>
> > >>>>>> Formula please, I have a few more data points to calculate.
> > >>>>>>
> > >>>>>> Thank you, Mikek
> > >>>>> https://www.electronics-tutorials.ws/inductor/parallel-inductors.html
> > >>>> Thanks, but that doesn't show me how to recalculate Q after adding a
> > >>>> parallel resistor.
> > >>>> Mikek
> > >>> Impedance is fixed for resistor, but function of freq for inductor.
> > >>> It still follow the same equation:
> > >>>
> > >>> 1090 = (1111 * R) / (1111 + R) at the test freq.
> > >> You might also need to convert Q to impedance with ( 2 pi w L ) / Q
> > >>
> > > OK, the first formula got me an answer of about 60kΩ. I was expecting
> > > 2MΩ to 5 MΩ.
> > >
> > > I think I see how I can work with the second formula.
> > >
> > > Mikek
> > I got the same 60kΩ.
> >
> > If you are interested here is the website to a high input impedance amp..
> > He built one and then a second better one.
> >
> > He uses the first to measure Q, 3db method, then adds the second high
> > input impedance amp as a load and recalculates
> >
> > the Q. Q drops from 1111 to 1090 with the second amp amp added. He
> > calculates the R of the amp at 5.8MΩ but says in reality
> >
> > it will be lower, but I don't think 60kΩ.
> >
> > The chart with numbers is after the next to last picture. I wonder if
> > I'm still calculating wrong or if the author is wrong.
> He said:
>
> "With the formula Z= 2.pi.f.L Q the impedance of the unloaded circuit60 can be calculated (the inductance of the coil is 230 uH):
> Z= 1.1 M.Ohm at 600 kHz
> Z= 1.57 M.Ohm at 900 kHz
> Z= 1.93 M.Ohm at 1200 kHz
> Z= 1.97 M.Ohm at 1500 kH"
>

Corrections:
> If adding R drop Z1 to Z2, then R = Z1 * Z2 / (Z2 - Z1)
>
> Z1 = 1.11M
> Z2 = 1.09M
> Z1 = (Z2 * R) / (Z2 + R)
> Z1 * Z2 + Z1 R = Z2 R
> R = Z1 * Z2 / (Z2 - Z1) = 60M

On Wednesday, August 18, 2021 at 5:13:05 PM UTC-7, Ed Lee wrote:
> On Wednesday, August 18, 2021 at 5:04:38 PM UTC-7, Ed Lee wrote:
> > On Wednesday, August 18, 2021 at 4:21:06 PM UTC-7, amdx wrote:
> > > On 8/18/2021 5:50 PM, amdx wrote:
> > > > On 8/18/2021 5:37 PM, Ed Lee wrote:
> > > >> On Wednesday, August 18, 2021 at 3:06:01 PM UTC-7, Ed Lee wrote:
> > > >>> On Wednesday, August 18, 2021 at 2:56:19 PM UTC-7, amdx wrote:
> > > >>>> On 8/18/2021 4:31 PM, Ed Lee wrote:
> > > >>>>> On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
> > > >>>>>> I have a High Q coil, it's Q measures 1111.
> > > >>>>>>
> > > >>>>>> I put a second very high impedance test probe across the coil and
> > > >>>>>> the Q
> > > >>>>>> drops to 1090.
> > > >>>>>>
> > > >>>>>> Frequency shift is minuscule.
> > > >>>>>>
> > > >>>>>> What is the parallel resistance of the second very high impedance
> > > >>>>>> probe?
> > > >>>>>>
> > > >>>>>> Formula please, I have a few more data points to calculate.
> > > >>>>>>
> > > >>>>>> Thank you, Mikek
> > > >>>>> https://www.electronics-tutorials.ws/inductor/parallel-inductors.html
> > > >>>> Thanks, but that doesn't show me how to recalculate Q after adding a
> > > >>>> parallel resistor.
> > > >>>> Mikek
> > > >>> Impedance is fixed for resistor, but function of freq for inductor.
> > > >>> It still follow the same equation:
> > > >>>
> > > >>> 1090 = (1111 * R) / (1111 + R) at the test freq.
> > > >> You might also need to convert Q to impedance with ( 2 pi w L ) / Q
> > > >>
> > > > OK, the first formula got me an answer of about 60kΩ. I was expecting
> > > > 2MΩ to 5 MΩ.
> > > >
> > > > I think I see how I can work with the second formula.
> > > >
> > > > Mikek
> > > I got the same 60kΩ.
> > >
> > > If you are interested here is the website to a high input impedance amp.
> > > He built one and then a second better one.
> > >
> > > He uses the first to measure Q, 3db method, then adds the second high
> > > input impedance amp as a load and recalculates
> > >
> > > the Q. Q drops from 1111 to 1090 with the second amp amp added. He
> > > calculates the R of the amp at 5.8MΩ but says in reality
> > >
> > > it will be lower, but I don't think 60kΩ.
> > >
> > > The chart with numbers is after the next to last picture. I wonder if
> > > I'm still calculating wrong or if the author is wrong.
> > He said:
> >
> > "With the formula Z= 2.pi.f.L Q the impedance of the unloaded circuit60 can be calculated (the inductance of the coil is 230 uH):
> > Z= 1.1 M.Ohm at 600 kHz
> > Z= 1.57 M.Ohm at 900 kHz
> > Z= 1.93 M.Ohm at 1200 kHz
> > Z= 1.97 M.Ohm at 1500 kH"
> >
> Corrections:
Again
> > If adding R drop Z1 to Z2, then R = Z2 * Z1 / (Z1 - Z2)
> >
> > Z1 = 1.11M
> > Z2 = 1.09M
> > Z2 = (Z1 * R) / (Z1 + R)
> > Z2 * Z1 + Z2 * R = Z1 * R
> > R = Z2 * Z1 / (Z1 - Z2) = 60M

On Wed, 18 Aug 2021 15:34:26 -0500, amdx <amdx@knology.net> wrote:

>I have a High Q coil, it's Q measures 1111.
>
>I put a second very high impedance test probe across the coil and the Q
>drops to 1090.
>
>  Frequency shift is minuscule.
>
>  What is the parallel resistance of the second very high impedance probe?
>
>Formula please, I have a few more data points to calculate.
>
>                                    Thank you, Mikek

Calculate the effective resistance in both cases.

Qn = wL/Rn

so Rn=wL/Qn

Then with two R values, do the reverse parallel resistor thing to get
the probe resistance.

--

Father Brown's figure remained quite dark and still;
but in that instant he had lost his head. His head was
always most valuable when he had lost it.

On 8/18/2021 7:13 PM, Ed Lee wrote:
> On Wednesday, August 18, 2021 at 5:04:38 PM UTC-7, Ed Lee wrote:
>> On Wednesday, August 18, 2021 at 4:21:06 PM UTC-7, amdx wrote:
>>> On 8/18/2021 5:50 PM, amdx wrote:
>>>> On 8/18/2021 5:37 PM, Ed Lee wrote:
>>>>> On Wednesday, August 18, 2021 at 3:06:01 PM UTC-7, Ed Lee wrote:
>>>>>> On Wednesday, August 18, 2021 at 2:56:19 PM UTC-7, amdx wrote:
>>>>>>> On 8/18/2021 4:31 PM, Ed Lee wrote:
>>>>>>>> On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
>>>>>>>>> I have a High Q coil, it's Q measures 1111.
>>>>>>>>>
>>>>>>>>> I put a second very high impedance test probe across the coil and
>>>>>>>>> the Q
>>>>>>>>> drops to 1090.
>>>>>>>>>
>>>>>>>>> Frequency shift is minuscule.
>>>>>>>>>
>>>>>>>>> What is the parallel resistance of the second very high impedance
>>>>>>>>> probe?
>>>>>>>>>
>>>>>>>>> Formula please, I have a few more data points to calculate.
>>>>>>>>>
>>>>>>>>> Thank you, Mikek
>>>>>>>> https://www.electronics-tutorials.ws/inductor/parallel-inductors.html
>>>>>>> Thanks, but that doesn't show me how to recalculate Q after adding a
>>>>>>> parallel resistor.
>>>>>>> Mikek
>>>>>> Impedance is fixed for resistor, but function of freq for inductor.
>>>>>> It still follow the same equation:
>>>>>>
>>>>>> 1090 = (1111 * R) / (1111 + R) at the test freq.
>>>>> You might also need to convert Q to impedance with ( 2 pi w L ) / Q
>>>>>
>>>> OK, the first formula got me an answer of about 60kΩ. I was expecting
>>>> 2MΩ to 5 MΩ.
>>>>
>>>> I think I see how I can work with the second formula.
>>>>
>>>> Mikek
>>> I got the same 60kΩ.
>>>
>>> If you are interested here is the website to a high input impedance amp.
>>> He built one and then a second better one.
>>>
>>> He uses the first to measure Q, 3db method, then adds the second high
>>> input impedance amp as a load and recalculates
>>>
>>> the Q. Q drops from 1111 to 1090 with the second amp amp added. He
>>> calculates the R of the amp at 5.8MΩ but says in reality
>>>
>>> it will be lower, but I don't think 60kΩ.
>>>
>>> The chart with numbers is after the next to last picture. I wonder if
>>> I'm still calculating wrong or if the author is wrong.
>> He said:
>>
>> "With the formula Z= 2.pi.f.L Q the impedance of the unloaded circuit60 can be calculated (the inductance of the coil is 230 uH):
>> Z= 1.1 M.Ohm at 600 kHz
>> Z= 1.57 M.Ohm at 900 kHz
>> Z= 1.93 M.Ohm at 1200 kHz
>> Z= 1.97 M.Ohm at 1500 kH"
>>
> Corrections:
>> If adding R drop Z1 to Z2, then R = Z1 * Z2 / (Z2 - Z1)
>>
>> Z1 = 1.11M
>> Z2 = 1.09M
>> Z1 = (Z2 * R) / (Z2 + R)
>> Z1 * Z2 + Z1 R = Z2 R
>> R = Z1 * Z2 / (Z2 - Z1) = 60M

OK, more than ten times what the builder calculated and extremely high
input resistance by most standards.

Thanks for doing the math, I'll have to so where I went wrong, but it
was only 3 magnitudes! ;-)

I guessed 240uh, did you see 230uh? I looked didn't see anything.

 Anyway, the numbers may calculate that way, but, I doubt the amp
impedance is that high.

Not sure where the discrepancy is. Measuring High Qs inductors and
getting consistent answers is difficult.

 But, I don't think that is where the discrepancy is.

                                Thanks, Mikek

--
This email has been checked for viruses by Avast antivirus software.
https://www.avast.com/antivirus

On Wednesday, August 18, 2021 at 5:34:49 PM UTC-7, amdx wrote:
> On 8/18/2021 7:13 PM, Ed Lee wrote:
> > On Wednesday, August 18, 2021 at 5:04:38 PM UTC-7, Ed Lee wrote:
> >> On Wednesday, August 18, 2021 at 4:21:06 PM UTC-7, amdx wrote:
> >>> On 8/18/2021 5:50 PM, amdx wrote:
> >>>> On 8/18/2021 5:37 PM, Ed Lee wrote:
> >>>>> On Wednesday, August 18, 2021 at 3:06:01 PM UTC-7, Ed Lee wrote:
> >>>>>> On Wednesday, August 18, 2021 at 2:56:19 PM UTC-7, amdx wrote:
> >>>>>>> On 8/18/2021 4:31 PM, Ed Lee wrote:
> >>>>>>>> On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
> >>>>>>>>> I have a High Q coil, it's Q measures 1111.
> >>>>>>>>>
> >>>>>>>>> I put a second very high impedance test probe across the coil and
> >>>>>>>>> the Q
> >>>>>>>>> drops to 1090.
> >>>>>>>>>
> >>>>>>>>> Frequency shift is minuscule.
> >>>>>>>>>
> >>>>>>>>> What is the parallel resistance of the second very high impedance
> >>>>>>>>> probe?
> >>>>>>>>>
> >>>>>>>>> Formula please, I have a few more data points to calculate.
> >>>>>>>>>
> >>>>>>>>> Thank you, Mikek
> >>>>>>>> https://www.electronics-tutorials.ws/inductor/parallel-inductors..html
> >>>>>>> Thanks, but that doesn't show me how to recalculate Q after adding a
> >>>>>>> parallel resistor.
> >>>>>>> Mikek
> >>>>>> Impedance is fixed for resistor, but function of freq for inductor..
> >>>>>> It still follow the same equation:
> >>>>>>
> >>>>>> 1090 = (1111 * R) / (1111 + R) at the test freq.
> >>>>> You might also need to convert Q to impedance with ( 2 pi w L ) / Q
> >>>>>
> >>>> OK, the first formula got me an answer of about 60kΩ. I was expecting
> >>>> 2MΩ to 5 MΩ.
> >>>>
> >>>> I think I see how I can work with the second formula.
> >>>>
> >>>> Mikek
> >>> I got the same 60kΩ.
> >>>
> >>> If you are interested here is the website to a high input impedance amp.
> >>> He built one and then a second better one.
> >>>
> >>> He uses the first to measure Q, 3db method, then adds the second high
> >>> input impedance amp as a load and recalculates
> >>>
> >>> the Q. Q drops from 1111 to 1090 with the second amp amp added. He
> >>> calculates the R of the amp at 5.8MΩ but says in reality
> >>>
> >>> it will be lower, but I don't think 60kΩ.
> >>>
> >>> The chart with numbers is after the next to last picture. I wonder if
> >>> I'm still calculating wrong or if the author is wrong.
> >> He said:
> >>
> >> "With the formula Z= 2.pi.f.L Q the impedance of the unloaded circuit60 can be calculated (the inductance of the coil is 230 uH):
> >> Z= 1.1 M.Ohm at 600 kHz
> >> Z= 1.57 M.Ohm at 900 kHz
> >> Z= 1.93 M.Ohm at 1200 kHz
> >> Z= 1.97 M.Ohm at 1500 kH"
> >>
> > Corrections:
> >> If adding R drop Z1 to Z2, then R = Z1 * Z2 / (Z2 - Z1)
> >>
> >> Z1 = 1.11M
> >> Z2 = 1.09M
> >> Z1 = (Z2 * R) / (Z2 + R)
> >> Z1 * Z2 + Z1 R = Z2 R
> >> R = Z1 * Z2 / (Z2 - Z1) = 60M
> OK, more than ten times what the builder calculated and extremely high
> input resistance by most standards.
>
> Thanks for doing the math, I'll have to so where I went wrong, but it
> was only 3 magnitudes! ;-)
>
> I guessed 240uh, did you see 230uh? I looked didn't see anything.
>
> Anyway, the numbers may calculate that way, but, I doubt the amp
> impedance is that high.
>
> Not sure where the discrepancy is. Measuring High Qs inductors and
> getting consistent answers is difficult.
>
> But, I don't think that is where the discrepancy is.

It's been a while seen college. But dropping Z by 1.8% should mean R to be 60x. I just assume his impedance calc is correct from the link:

http://www.crystal-radio.eu/endetectorunit1.htm

On 8/18/2021 5:05 PM, Ed Lee wrote:
> On Wednesday, August 18, 2021 at 2:56:19 PM UTC-7, amdx wrote:
>> On 8/18/2021 4:31 PM, Ed Lee wrote:
>>> On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
>>>> I have a High Q coil, it's Q measures 1111.
>>>>
>>>> I put a second very high impedance test probe across the coil and the Q
>>>> drops to 1090.
>>>>
>>>> Frequency shift is minuscule.
>>>>
>>>> What is the parallel resistance of the second very high impedance probe?
>>>>
>>>> Formula please, I have a few more data points to calculate.
>>>>
>>>> Thank you, Mikek
>>> https://www.electronics-tutorials.ws/inductor/parallel-inductors.html
>> Thanks, but that doesn't show me how to recalculate Q after adding a
>> parallel resistor.
>> Mikek
>
> Impedance is fixed for resistor, but function of freq for inductor. It still follow the same equation:
>
> 1090 = (1111 * R) / (1111 + R) at the test freq.
>
>

So R = 57,666 ohms? What R is this?

On 8/20/2021 6:01 AM, John S wrote:
> On 8/18/2021 5:05 PM, Ed Lee wrote:
>> On Wednesday, August 18, 2021 at 2:56:19 PM UTC-7, amdx wrote:
>>> On 8/18/2021 4:31 PM, Ed Lee wrote:
>>>> On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
>>>>> I have a High Q coil, it's Q measures 1111.
>>>>>
>>>>> I put a second very high impedance test probe across the coil and
>>>>> the Q
>>>>> drops to 1090.
>>>>>
>>>>> Frequency shift is minuscule.
>>>>>
>>>>> What is the parallel resistance of the second very high impedance
>>>>> probe?
>>>>>
>>>>> Formula please, I have a few more data points to calculate.
>>>>>
>>>>> Thank you, Mikek
>>>> https://www.electronics-tutorials.ws/inductor/parallel-inductors.html
>>> Thanks, but that doesn't show me how to recalculate Q after adding a
>>> parallel resistor.
>>> Mikek
>>
>> Impedance is fixed for resistor, but function of freq for inductor. 
>> It still follow the same equation:
>>
>> 1090 = (1111 * R) / (1111 + R) at the test freq.
>>
>
> So R = 57,666 ohms? What R is this?

I'm really not sure what R is yet. Here is the scenario. We are using
the 3db method to measure very high Q inductors .

Any test instrument used to measure, is a load on the high Q inductor
and lowers the Q measurement.

A high input impedance amplifier was developed and Q was measured at
1111. ( the amp output drives a scope or volt meter)

Now a second amp was built (trying to improve on the first). With the
first amp still measuring

the Q, the second amp was attached and now Q was remeasured and it
dropped to 1090.

What is the R of the second amp?   I.E. what is the loading of the
second amp?

 Here's the data on the amps.

> http://www.crystal-radio.eu/fetamp/enfetamp.htm
Additional info, 230uh coil, 600kHz measurement.

                                             Mikek

--
This email has been checked for viruses by Avast antivirus software.
https://www.avast.com/antivirus

On 8/20/2021 7:12 AM, amdx wrote:
> On 8/20/2021 6:01 AM, John S wrote:
>> On 8/18/2021 5:05 PM, Ed Lee wrote:
>>> On Wednesday, August 18, 2021 at 2:56:19 PM UTC-7, amdx wrote:
>>>> On 8/18/2021 4:31 PM, Ed Lee wrote:
>>>>> On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
>>>>>> I have a High Q coil, it's Q measures 1111.
>>>>>>
>>>>>> I put a second very high impedance test probe across the coil and
>>>>>> the Q
>>>>>> drops to 1090.
>>>>>>
>>>>>> Frequency shift is minuscule.
>>>>>>
>>>>>> What is the parallel resistance of the second very high impedance
>>>>>> probe?
>>>>>>
>>>>>> Formula please, I have a few more data points to calculate.
>>>>>>
>>>>>> Thank you, Mikek
>>>>> https://www.electronics-tutorials.ws/inductor/parallel-inductors.html
>>>> Thanks, but that doesn't show me how to recalculate Q after adding a
>>>> parallel resistor.
>>>> Mikek
>>>
>>> Impedance is fixed for resistor, but function of freq for inductor.
>>> It still follow the same equation:
>>>
>>> 1090 = (1111 * R) / (1111 + R) at the test freq.
>>>
>>
>> So R = 57,666 ohms? What R is this?
>
> I'm really not sure what R is yet. Here is the scenario. We are using
> the 3db method to measure very high Q inductors .
>
> Any test instrument used to measure, is a load on the high Q inductor
> and lowers the Q measurement.
>
> A high input impedance amplifier was developed and Q was measured at
> 1111. ( the amp output drives a scope or volt meter)
>
> Now a second amp was built (trying to improve on the first). With the
> first amp still measuring
>
> the Q, the second amp was attached and now Q was remeasured and it
> dropped to 1090.
>
> What is the R of the second amp?   I.E. what is the loading of the
> second amp?
>
>  Here's the data on the amps.
>
>> http://www.crystal-radio.eu/fetamp/enfetamp.htm
> Additional info, 230uh coil, 600kHz measurement.
>
>                                              Mikek

What is the inductor's resistance?

On Friday, August 20, 2021 at 12:14:08 PM UTC-4, John S wrote:
> On 8/20/2021 7:12 AM, amdx wrote:
> > On 8/20/2021 6:01 AM, John S wrote:
> >> On 8/18/2021 5:05 PM, Ed Lee wrote:
> >>> On Wednesday, August 18, 2021 at 2:56:19 PM UTC-7, amdx wrote:
> >>>> On 8/18/2021 4:31 PM, Ed Lee wrote:
> >>>>> On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
> >>>>>> I have a High Q coil, it's Q measures 1111.
> >>>>>>
> >>>>>> I put a second very high impedance test probe across the coil and
> >>>>>> the Q
> >>>>>> drops to 1090.
> >>>>>>
> >>>>>> Frequency shift is minuscule.
> >>>>>>
> >>>>>> What is the parallel resistance of the second very high impedance
> >>>>>> probe?
> >>>>>>
> >>>>>> Formula please, I have a few more data points to calculate.
> >>>>>>
> >>>>>> Thank you, Mikek
> >>>>> https://www.electronics-tutorials.ws/inductor/parallel-inductors.html
> >>>> Thanks, but that doesn't show me how to recalculate Q after adding a
> >>>> parallel resistor.
> >>>> Mikek
> >>>
> >>> Impedance is fixed for resistor, but function of freq for inductor.
> >>> It still follow the same equation:
> >>>
> >>> 1090 = (1111 * R) / (1111 + R) at the test freq.
> >>>
> >>
> >> So R = 57,666 ohms? What R is this?
> >
> > I'm really not sure what R is yet. Here is the scenario. We are using
> > the 3db method to measure very high Q inductors .
> >
> > Any test instrument used to measure, is a load on the high Q inductor
> > and lowers the Q measurement.
> >
> > A high input impedance amplifier was developed and Q was measured at
> > 1111. ( the amp output drives a scope or volt meter)
> >
> > Now a second amp was built (trying to improve on the first). With the
> > first amp still measuring
> >
> > the Q, the second amp was attached and now Q was remeasured and it
> > dropped to 1090.
> >
> > What is the R of the second amp? I.E. what is the loading of the
> > second amp?
> >
> > Here's the data on the amps.
> >
> >> http://www.crystal-radio.eu/fetamp/enfetamp.htm
> > Additional info, 230uh coil, 600kHz measurement.
> >
> > Mikek
> What is the inductor's resistance?

You would need another measurement to get that. The issue is not DC resistance since these are AC measurements which are sensitive to skin effect. So you can't measure it with a conventional resistance meter. Once the resistance of this second amp is known, it can be used to measure the coil giving a reading of the coil combined with the second amp that can be used to get all three values, amp 1, amp 2 and coil.

--

Rick C.

- Get 1,000 miles of free Supercharging
- Tesla referral code - https://ts.la/richard11209

On 8/20/2021 11:14 AM, John S wrote:
> On 8/20/2021 7:12 AM, amdx wrote:
>> On 8/20/2021 6:01 AM, John S wrote:
>>> On 8/18/2021 5:05 PM, Ed Lee wrote:
>>>> On Wednesday, August 18, 2021 at 2:56:19 PM UTC-7, amdx wrote:
>>>>> On 8/18/2021 4:31 PM, Ed Lee wrote:
>>>>>> On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
>>>>>>> I have a High Q coil, it's Q measures 1111.
>>>>>>>
>>>>>>> I put a second very high impedance test probe across the coil
>>>>>>> and the Q
>>>>>>> drops to 1090.
>>>>>>>
>>>>>>> Frequency shift is minuscule.
>>>>>>>
>>>>>>> What is the parallel resistance of the second very high
>>>>>>> impedance probe?
>>>>>>>
>>>>>>> Formula please, I have a few more data points to calculate.
>>>>>>>
>>>>>>> Thank you, Mikek
>>>>>> https://www.electronics-tutorials.ws/inductor/parallel-inductors.html
>>>>>>
>>>>> Thanks, but that doesn't show me how to recalculate Q after adding a
>>>>> parallel resistor.
>>>>> Mikek
>>>>
>>>> Impedance is fixed for resistor, but function of freq for inductor.
>>>> It still follow the same equation:
>>>>
>>>> 1090 = (1111 * R) / (1111 + R) at the test freq.
>>>>
>>>
>>> So R = 57,666 ohms? What R is this?
>>
>> I'm really not sure what R is yet. Here is the scenario. We are using
>> the 3db method to measure very high Q inductors .
>>
>> Any test instrument used to measure, is a load on the high Q inductor
>> and lowers the Q measurement.
>>
>> A high input impedance amplifier was developed and Q was measured at
>> 1111. ( the amp output drives a scope or volt meter)
>>
>> Now a second amp was built (trying to improve on the first). With the
>> first amp still measuring
>>
>> the Q, the second amp was attached and now Q was remeasured and it
>> dropped to 1090.
>>
>> What is the R of the second amp?   I.E. what is the loading of the
>> second amp?
>>
>>   Here's the data on the amps.
>>
>>> http://www.crystal-radio.eu/fetamp/enfetamp.htm
>> Additional info, 230uh coil, 600kHz measurement.
>>
>>                                               Mikek
>
> What is the inductor's resistance?
>
As Rick says, you really want to ask, What is Rac, I know how to
calculate Rac from Q and inductive reactance of the coil.

Rac = XL/Q and XL is 867Ω at 600kHz. Q is 1111.  Therefore Rac is
867/1111 = 0.78Ω.

Here's another shot at my question, with both amps attached the Series
Rac is 867/1090 = 0.795Ω

So the second amp adds 0.015Ω of series resistance.

What is the parallel equivalent of 0.015Ω?

                                Thanks, Mikek

--
This email has been checked for viruses by Avast antivirus software.
https://www.avast.com/antivirus

On 8/20/2021 5:12 PM, amdx wrote:
> On 8/20/2021 11:14 AM, John S wrote:
>> On 8/20/2021 7:12 AM, amdx wrote:
>>> On 8/20/2021 6:01 AM, John S wrote:
>>>> On 8/18/2021 5:05 PM, Ed Lee wrote:
>>>>> On Wednesday, August 18, 2021 at 2:56:19 PM UTC-7, amdx wrote:
>>>>>> On 8/18/2021 4:31 PM, Ed Lee wrote:
>>>>>>> On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
>>>>>>>> I have a High Q coil, it's Q measures 1111.
>>>>>>>>
>>>>>>>> I put a second very high impedance test probe across the coil
>>>>>>>> and the Q
>>>>>>>> drops to 1090.
>>>>>>>>
>>>>>>>> Frequency shift is minuscule.
>>>>>>>>
>>>>>>>> What is the parallel resistance of the second very high
>>>>>>>> impedance probe?
>>>>>>>>
>>>>>>>> Formula please, I have a few more data points to calculate.
>>>>>>>>
>>>>>>>> Thank you, Mikek
>>>>>>> https://www.electronics-tutorials.ws/inductor/parallel-inductors.html
>>>>>>>
>>>>>> Thanks, but that doesn't show me how to recalculate Q after adding a
>>>>>> parallel resistor.
>>>>>> Mikek
>>>>>
>>>>> Impedance is fixed for resistor, but function of freq for
>>>>> inductor. It still follow the same equation:
>>>>>
>>>>> 1090 = (1111 * R) / (1111 + R) at the test freq.
>>>>>
>>>>
>>>> So R = 57,666 ohms? What R is this?
>>>
>>> I'm really not sure what R is yet. Here is the scenario. We are
>>> using the 3db method to measure very high Q inductors .
>>>
>>> Any test instrument used to measure, is a load on the high Q
>>> inductor and lowers the Q measurement.
>>>
>>> A high input impedance amplifier was developed and Q was measured at
>>> 1111. ( the amp output drives a scope or volt meter)
>>>
>>> Now a second amp was built (trying to improve on the first). With
>>> the first amp still measuring
>>>
>>> the Q, the second amp was attached and now Q was remeasured and it
>>> dropped to 1090.
>>>
>>> What is the R of the second amp?   I.E. what is the loading of the
>>> second amp?
>>>
>>>   Here's the data on the amps.
>>>
>>>> http://www.crystal-radio.eu/fetamp/enfetamp.htm
>>> Additional info, 230uh coil, 600kHz measurement.
>>>
>>>                                               Mikek
>>
>> What is the inductor's resistance?
>>
> As Rick says, you really want to ask, What is Rac, I know how to
> calculate Rac from Q and inductive reactance of the coil.
>
> Rac = XL/Q and XL is 867Ω at 600kHz. Q is 1111.  Therefore Rac is
> 867/1111 = 0.78Ω.
>
> Here's another shot at my question, with both amps attached the Series
> Rac is 867/1090 = 0.795Ω
>
> So the second amp adds 0.015Ω of series resistance.
>
> What is the parallel equivalent of 0.015Ω?
>
>                                 Thanks, Mikek
>
>
I found, "for a series to parallel conversion, Rp = (Rs^2 + Xs^2) / Rs"

So, Rp = (0.015^2 + 867^2) / Rs

Rp = (0.000225 + 751689) / Rs

Rp = 751689 / 0.015

 Rp = 50,112,600 Ω

This also works if I find Rp of both 0.78Ω = 963,704Ω and 0.795Ω =
945,046Ω, I then find a

50MΩ across 963,704 = 945,046.

That is 10 times what the author calculated.

Off to do more calculations to verify this works the way it seems.

                                Mikek

--
This email has been checked for viruses by Avast antivirus software.
https://www.avast.com/antivirus

On 8/20/2021 6:04 PM, amdx wrote:
> On 8/20/2021 5:12 PM, amdx wrote:
>> On 8/20/2021 11:14 AM, John S wrote:
>>> On 8/20/2021 7:12 AM, amdx wrote:
>>>> On 8/20/2021 6:01 AM, John S wrote:
>>>>> On 8/18/2021 5:05 PM, Ed Lee wrote:
>>>>>> On Wednesday, August 18, 2021 at 2:56:19 PM UTC-7, amdx wrote:
>>>>>>> On 8/18/2021 4:31 PM, Ed Lee wrote:
>>>>>>>> On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
>>>>>>>>> I have a High Q coil, it's Q measures 1111.
>>>>>>>>>
>>>>>>>>> I put a second very high impedance test probe across the coil
>>>>>>>>> and the Q
>>>>>>>>> drops to 1090.
>>>>>>>>>
>>>>>>>>> Frequency shift is minuscule.
>>>>>>>>>
>>>>>>>>> What is the parallel resistance of the second very high
>>>>>>>>> impedance probe?
>>>>>>>>>
>>>>>>>>> Formula please, I have a few more data points to calculate.
>>>>>>>>>
>>>>>>>>> Thank you, Mikek
>>>>>>>> https://www.electronics-tutorials.ws/inductor/parallel-inductors.html
>>>>>>>>
>>>>>>> Thanks, but that doesn't show me how to recalculate Q after
>>>>>>> adding a
>>>>>>> parallel resistor.
>>>>>>> Mikek
>>>>>>
>>>>>> Impedance is fixed for resistor, but function of freq for
>>>>>> inductor. It still follow the same equation:
>>>>>>
>>>>>> 1090 = (1111 * R) / (1111 + R) at the test freq.
>>>>>>
>>>>>
>>>>> So R = 57,666 ohms? What R is this?
>>>>
>>>> I'm really not sure what R is yet. Here is the scenario. We are
>>>> using the 3db method to measure very high Q inductors .
>>>>
>>>> Any test instrument used to measure, is a load on the high Q
>>>> inductor and lowers the Q measurement.
>>>>
>>>> A high input impedance amplifier was developed and Q was measured
>>>> at 1111. ( the amp output drives a scope or volt meter)
>>>>
>>>> Now a second amp was built (trying to improve on the first). With
>>>> the first amp still measuring
>>>>
>>>> the Q, the second amp was attached and now Q was remeasured and it
>>>> dropped to 1090.
>>>>
>>>> What is the R of the second amp?   I.E. what is the loading of the
>>>> second amp?
>>>>
>>>>   Here's the data on the amps.
>>>>
>>>>> http://www.crystal-radio.eu/fetamp/enfetamp.htm
>>>> Additional info, 230uh coil, 600kHz measurement.
>>>>
>>>>                                               Mikek
>>>
>>> What is the inductor's resistance?
>>>
>> As Rick says, you really want to ask, What is Rac, I know how to
>> calculate Rac from Q and inductive reactance of the coil.
>>
>> Rac = XL/Q and XL is 867Ω at 600kHz. Q is 1111.  Therefore Rac is
>> 867/1111 = 0.78Ω.
>>
>> Here's another shot at my question, with both amps attached the
>> Series Rac is 867/1090 = 0.795Ω
>>
>> So the second amp adds 0.015Ω of series resistance.
>>
>> What is the parallel equivalent of 0.015Ω?
>>
>>                                 Thanks, Mikek
>>
>>
> I found, "for a series to parallel conversion, Rp = (Rs^2 + Xs^2) / Rs"
>
> So, Rp = (0.015^2 + 867^2) / Rs
>
> Rp = (0.000225 + 751689) / Rs
>
> Rp = 751689 / 0.015
>
>  Rp = 50,112,600 Ω
>
> This also works if I find Rp of both 0.78Ω = 963,704Ω and 0.795Ω =
> 945,046Ω, I then find a
>
> 50MΩ across 963,704 = 945,046.
>
> That is 10 times what the author calculated.
>
> Off to do more calculations to verify this works the way it seems.
>
>                                 Mike

 I searched the website a little deeper and found the coil used was
175uh not the 230uh we used.

I recalculated everything with 175uh inductor and found the amp was
equivalent to a 27MΩ.

 And I reread his description and where I read 5.780 M. Ohm I should
have seen 5,780 M Ohms, (5.78G)

The author does caution there are a couple more items that will lower that.

 So now, I believe the 27MΩ.

Here's his write up, or you can go to the website.

> http://www.crystal-radio.eu/fetamp/enfetamp.htm

"The 0.3 pF input capacitor is self-made of two copperplates of 1 square
cm at a distance of 3 mm.
By changing the distance between the plates we can adjust the gain of
the amplifier.
The plates must have at least 1 cm distance from the surrounding
grounded box.

The input signal enters the box via a 1mm copperwire, through a 10 mm
hole in the box.
The wire is supported by a piece of polyethene, which is fixed with
nylon screws.
The input amplifier (T1) is screened from the rest of the circuit.

Between the /gate/ (input) of T1 and ground there is a 20 M.Ohm resistor.
But the input resistance of the amplifier is much higher then 20 M.Ohm,
in theorie even 17² times higher  (so, 5780 M.Ohm), this is because over
the 20 M.Ohm resistor is only 1/17th part of the input voltage.
In practice the input resistance will be lower then 5780 M.Ohm because
of dielectric losses e.g. in the gate of the FET."

--
This email has been checked for viruses by Avast antivirus software.
https://www.avast.com/antivirus

On 8/20/2021 5:12 PM, amdx wrote:
> On 8/20/2021 11:14 AM, John S wrote:
>> On 8/20/2021 7:12 AM, amdx wrote:
>>> On 8/20/2021 6:01 AM, John S wrote:
>>>> On 8/18/2021 5:05 PM, Ed Lee wrote:
>>>>> On Wednesday, August 18, 2021 at 2:56:19 PM UTC-7, amdx wrote:
>>>>>> On 8/18/2021 4:31 PM, Ed Lee wrote:
>>>>>>> On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
>>>>>>>> I have a High Q coil, it's Q measures 1111.
>>>>>>>>
>>>>>>>> I put a second very high impedance test probe across the coil
>>>>>>>> and the Q
>>>>>>>> drops to 1090.
>>>>>>>>
>>>>>>>> Frequency shift is minuscule.
>>>>>>>>
>>>>>>>> What is the parallel resistance of the second very high
>>>>>>>> impedance probe?
>>>>>>>>
>>>>>>>> Formula please, I have a few more data points to calculate.
>>>>>>>>
>>>>>>>> Thank you, Mikek
>>>>>>> https://www.electronics-tutorials.ws/inductor/parallel-inductors.html
>>>>>>>
>>>>>> Thanks, but that doesn't show me how to recalculate Q after adding a
>>>>>> parallel resistor.
>>>>>> Mikek
>>>>>
>>>>> Impedance is fixed for resistor, but function of freq for inductor.
>>>>> It still follow the same equation:
>>>>>
>>>>> 1090 = (1111 * R) / (1111 + R) at the test freq.
>>>>>
>>>>
>>>> So R = 57,666 ohms? What R is this?
>>>
>>> I'm really not sure what R is yet. Here is the scenario. We are using
>>> the 3db method to measure very high Q inductors .
>>>
>>> Any test instrument used to measure, is a load on the high Q inductor
>>> and lowers the Q measurement.
>>>
>>> A high input impedance amplifier was developed and Q was measured at
>>> 1111. ( the amp output drives a scope or volt meter)
>>>
>>> Now a second amp was built (trying to improve on the first). With the
>>> first amp still measuring
>>>
>>> the Q, the second amp was attached and now Q was remeasured and it
>>> dropped to 1090.
>>>
>>> What is the R of the second amp?   I.E. what is the loading of the
>>> second amp?
>>>
>>>   Here's the data on the amps.
>>>
>>>> http://www.crystal-radio.eu/fetamp/enfetamp.htm
>>> Additional info, 230uh coil, 600kHz measurement.
>>>
>>>                                               Mikek
>>
>> What is the inductor's resistance?
>>
> As Rick says, you really want to ask, What is Rac, I know how to
> calculate Rac from Q and inductive reactance of the coil.
>
> Rac = XL/Q and XL is 867Ω at 600kHz. Q is 1111.  Therefore Rac is
> 867/1111 = 0.78Ω.
>
> Here's another shot at my question, with both amps attached the Series
> Rac is 867/1090 = 0.795Ω
>
> So the second amp adds 0.015Ω of series resistance.
>
> What is the parallel equivalent of 0.015Ω?
>
>                                 Thanks, Mikek

Use Rp = QX. This is the Q for a resistor in parallel with the
reactance. So you have 1111*867 = Rp = 963k. The parallel resistance for
a Q of 1090 is R = 1090 * 867 = 945k. So, how much parallel resistance
do you need to reduce the total resistance from 963k down to 945k? That
would be R = 1/(1/945k - 1/963k) or 50 Megs!

As an aside, how much more "AC" resistance will the inductor have than
the DC value?

On 8/20/2021 6:34 PM, John S wrote:
> On 8/20/2021 5:12 PM, amdx wrote:
>> On 8/20/2021 11:14 AM, John S wrote:
>>> On 8/20/2021 7:12 AM, amdx wrote:
>>>> On 8/20/2021 6:01 AM, John S wrote:
>>>>> On 8/18/2021 5:05 PM, Ed Lee wrote:
>>>>>> On Wednesday, August 18, 2021 at 2:56:19 PM UTC-7, amdx wrote:
>>>>>>> On 8/18/2021 4:31 PM, Ed Lee wrote:
>>>>>>>> On Wednesday, August 18, 2021 at 1:34:35 PM UTC-7, amdx wrote:
>>>>>>>>> I have a High Q coil, it's Q measures 1111.
>>>>>>>>>
>>>>>>>>> I put a second very high impedance test probe across the coil
>>>>>>>>> and the Q
>>>>>>>>> drops to 1090.
>>>>>>>>>
>>>>>>>>> Frequency shift is minuscule.
>>>>>>>>>
>>>>>>>>> What is the parallel resistance of the second very high
>>>>>>>>> impedance probe?
>>>>>>>>>
>>>>>>>>> Formula please, I have a few more data points to calculate.
>>>>>>>>>
>>>>>>>>> Thank you, Mikek
>>>>>>>> https://www.electronics-tutorials.ws/inductor/parallel-inductors.html
>>>>>>>>
>>>>>>> Thanks, but that doesn't show me how to recalculate Q after
>>>>>>> adding a
>>>>>>> parallel resistor.
>>>>>>> Mikek
>>>>>>
>>>>>> Impedance is fixed for resistor, but function of freq for
>>>>>> inductor. It still follow the same equation:
>>>>>>
>>>>>> 1090 = (1111 * R) / (1111 + R) at the test freq.
>>>>>>
>>>>>
>>>>> So R = 57,666 ohms? What R is this?
>>>>
>>>> I'm really not sure what R is yet. Here is the scenario. We are
>>>> using the 3db method to measure very high Q inductors .
>>>>
>>>> Any test instrument used to measure, is a load on the high Q
>>>> inductor and lowers the Q measurement.
>>>>
>>>> A high input impedance amplifier was developed and Q was measured
>>>> at 1111. ( the amp output drives a scope or volt meter)
>>>>
>>>> Now a second amp was built (trying to improve on the first). With
>>>> the first amp still measuring
>>>>
>>>> the Q, the second amp was attached and now Q was remeasured and it
>>>> dropped to 1090.
>>>>
>>>> What is the R of the second amp?   I.E. what is the loading of the
>>>> second amp?
>>>>
>>>>   Here's the data on the amps.
>>>>
>>>>> http://www.crystal-radio.eu/fetamp/enfetamp.htm
>>>> Additional info, 230uh coil, 600kHz measurement.
>>>>
>>>>                                               Mikek
>>>
>>> What is the inductor's resistance?
>>>
>> As Rick says, you really want to ask, What is Rac, I know how to
>> calculate Rac from Q and inductive reactance of the coil.
>>
>> Rac = XL/Q and XL is 867Ω at 600kHz. Q is 1111.  Therefore Rac is
>> 867/1111 = 0.78Ω.
>>
>> Here's another shot at my question, with both amps attached the
>> Series Rac is 867/1090 = 0.795Ω
>>
>> So the second amp adds 0.015Ω of series resistance.
>>
>> What is the parallel equivalent of 0.015Ω?
>>
>>                                  Thanks, Mikek
>
> Use Rp = QX. This is the Q for a resistor in parallel with the
> reactance. So you have 1111*867 = Rp = 963k. The parallel resistance
> for a Q of 1090 is R = 1090 * 867 = 945k. So, how much parallel
> resistance do you need to reduce the total resistance from 963k down
> to 945k? That would be R = 1/(1/945k - 1/963k) or 50 Megs!
>
> As an aside, how much more "AC" resistance will the inductor have than
> the DC value?
>
That's wire dependent, frequency dependent and dependent on spacing of
the turns on the coil.

 Properly chosen Litz wire will reduce frequency dependence and spacing
will reduce the effect

that I can't remember the name of, but it's magnetic crowding when two
wires on a coil are close to each other.

I'd like to think I will remember as soon as I click send!

OK, I pulled out Terman's 1943 Radio Engineers Handbook, the term is
'Proximity Effect"

 He even has a formula for Rac/Rdc = H+/k/ (nds/do)^2 G, but you need
to pull numbers from a table after you do a calculation

using diameter of the wire and frequency and get/x, /then go across and
find two more terms. So I can't give you a formula.

And then it is still an approximation.

                       Mikek

--
This email has been checked for viruses by Avast antivirus software.
https://www.avast.com/antivirus

On Friday, August 20, 2021 at 6:13:00 PM UTC-4, amdx wrote:
> On 8/20/2021 11:14 AM, John S wrote:
> > On 8/20/2021 7:12 AM, amdx wrote:
> >>
> >> I'm really not sure what R is yet. Here is the scenario. We are using
> >> the 3db method to measure very high Q inductors .
> >>
> >> Any test instrument used to measure, is a load on the high Q inductor
> >> and lowers the Q measurement.
> >>
> >> A high input impedance amplifier was developed and Q was measured at
> >> 1111. ( the amp output drives a scope or volt meter)
> >>
> >> Now a second amp was built (trying to improve on the first). With the
> >> first amp still measuring
> >>
> >> the Q, the second amp was attached and now Q was remeasured and it
> >> dropped to 1090.
> >>
> >> What is the R of the second amp? I.E. what is the loading of the
> >> second amp?
> >>
> >> Here's the data on the amps.
> >>
> >>> http://www.crystal-radio.eu/fetamp/enfetamp.htm
> >> Additional info, 230uh coil, 600kHz measurement.
> >>
> >> Mikek
> >
> > What is the inductor's resistance?
> >
> As Rick says, you really want to ask, What is Rac, I know how to
> calculate Rac from Q and inductive reactance of the coil.
>
> Rac = XL/Q and XL is 867Ω at 600kHz. Q is 1111. Therefore Rac is
> 867/1111 = 0.78Ω.

I thought that Q number was for the coil with amplifier A attached? So that would not give you the resistance of the coil until you account for the amplifier resistance, no?

How did you measure XL?

--

Rick C.

+ Get 1,000 miles of free Supercharging
+ Tesla referral code - https://ts.la/richard11209

On 8/20/2021 7:13 PM, Rick C wrote:
> On Friday, August 20, 2021 at 6:13:00 PM UTC-4, amdx wrote:
>> On 8/20/2021 11:14 AM, John S wrote:
>>> On 8/20/2021 7:12 AM, amdx wrote:
>>>> I'm really not sure what R is yet. Here is the scenario. We are using
>>>> the 3db method to measure very high Q inductors .
>>>>
>>>> Any test instrument used to measure, is a load on the high Q inductor
>>>> and lowers the Q measurement.
>>>>
>>>> A high input impedance amplifier was developed and Q was measured at
>>>> 1111. ( the amp output drives a scope or volt meter)
>>>>
>>>> Now a second amp was built (trying to improve on the first). With the
>>>> first amp still measuring
>>>>
>>>> the Q, the second amp was attached and now Q was remeasured and it
>>>> dropped to 1090.
>>>>
>>>> What is the R of the second amp? I.E. what is the loading of the
>>>> second amp?
>>>>
>>>> Here's the data on the amps.
>>>>
>>>>> http://www.crystal-radio.eu/fetamp/enfetamp.htm
>>>> Additional info, 230uh coil, 600kHz measurement.
>>>>
>>>> Mikek
>>> What is the inductor's resistance?
>>>
>> As Rick says, you really want to ask, What is Rac, I know how to
>> calculate Rac from Q and inductive reactance of the coil.
>>
>> Rac = XL/Q and XL is 867Ω at 600kHz. Q is 1111. Therefore Rac is
>> 867/1111 = 0.78Ω.
> I thought that Q number was for the coil with amplifier A attached? So that would not give you the resistance of the coil until you account for the amplifier resistance, no?

Yes, that would seem to be true. All I can say is we always speak of
measured Q, so Q always includes the R of the test instrument.

I think sense we now have two instruments we could work out what true Q is

> How did you measure XL?
>
  XL is calculated,  2pifL and 175uh at 600kHz. Caution first iteration
used a 230uh Inductor and 600kHz.

                                                      Mikek

--
This email has been checked for viruses by Avast antivirus software.
https://www.avast.com/antivirus

On Friday, August 20, 2021 at 8:21:57 PM UTC-4, amdx wrote:
> On 8/20/2021 7:13 PM, Rick C wrote:
> > On Friday, August 20, 2021 at 6:13:00 PM UTC-4, amdx wrote:
> >>
> >> As Rick says, you really want to ask, What is Rac, I know how to
> >> calculate Rac from Q and inductive reactance of the coil.
> >>
> >> Rac = XL/Q and XL is 867Ω at 600kHz. Q is 1111. Therefore Rac is
> >> 867/1111 = 0.78Ω.
> > I thought that Q number was for the coil with amplifier A attached? So that would not give you the resistance of the coil until you account for the amplifier resistance, no?
> Yes, that would seem to be true. All I can say is we always speak of
> measured Q, so Q always includes the R of the test instrument.
>
> I think sense we now have two instruments we could work out what true Q is
> > How did you measure XL?
> >
> XL is calculated, 2pifL and 175uh at 600kHz. Caution first iteration
> used a 230uh Inductor and 600kHz.

I thought you were measuring an unknown coil. How do you know the coil is 175 uH? Even if that is what it is supposed to be, they have tolerances.

Yes, there are three resistances, one each for the amps and one in the coil.. Three measurements of different combinations allows you to calculate each of the three resistances and Q values. Otherwise the Q and resistance will vary with the amp.

--

Rick C.

-- Get 1,000 miles of free Supercharging
-- Tesla referral code - https://ts.la/richard11209