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-   -   MFJ259 conversion help (https://www.radiobanter.com/antenna/199501-mfj259-conversion-help.html)

John S December 5th 13 08:42 PM

MFJ259 conversion help
 
On 12/5/2013 1:49 PM, John S wrote:
On 12/5/2013 11:31 AM, Helmut Wabnig wrote:
On Thu, 05 Dec 2013 10:26:33 -0600, amdx wrote:

On 12/4/2013 9:14 AM, John S wrote:
On 12/4/2013 8:48 AM, amdx wrote:
On 12/4/2013 4:50 AM, John S wrote:
On 12/3/2013 9:07 PM, amdx wrote:
I may be asking for something that doesn't have an answer.

I connected a voltmeter to the R meter of my MFJ259.
I checked a bunch of resistors and recorded the voltages.
Now I have all these voltage readings vs. Resistance and
don't know how to relate them except for a conversion graph.

Is there a math function that relates these voltages to Resistance?

Load Resistance Voltage reading

7.5 ohms 0.0388 volts
15 0.0444 volts
25.5 0.0478 volts
39 0.0577 volts
50 0.0614 volts
100 0.0807 volts
140 0.0891 volts
174 0.0935 volts
221 0.0980 volts
249 0.1000 volts
365 0.1060 volts
498 0.1090 volts

If you can figure this out, I'll reward you with the
SWR chart :-)

Thanks, Mikek


V = 0.0191*ln(R) - 0.0077

You're welcome.

Thanks, I'll learn about that and see if I can make it work for
me.

Mikek

Good. I can tell you how I accomplished it if you are interested.

Hi John S,
Please don't let me work you if your not interested, this is now
probably more of a curiosity than a way to make the MFJ259 more useful.
I ran all the numbers with the formula, it is not accurate enough to
me usable. I don't know if that is a calibration error, resistor error,
(I used 1% resistors) a me error, or the wrong formula.


I don't think it correlates very well with my measurements.
Might need to use fixed font to read this.

My Measurements.

Real R Calculated R Measured Voltage


7.5 ohms 11.4 0.0388 volts
15 15.3 0.0444 volts
25.5 18.3 0.0478 volts
39 30.7 0.0577 volts
50 37.3 0.0614 volts
100 102 0.0807 volts
140 158.9 0.0891 volts
174 200 0.0935 volts
221 253 0.0980 volts
249 284 0.1000 volts
365 384.6 0.1060 volts
498 457.4 0.1090 volts

This is really only usable at 100 ohms. Only 2% error.
52% error low end and 8.1% top end
Is there a better formula.
I'm trying to use a digital meter in place of basically a hand drawn
dial that is very nonlinear.
I'm wondering if I calibrated 50 ohms to read a bit lower
or higher voltage it might help

Anyone's helpful thoughts,
Thanks, Mike



Well I did it using an old dirty trick:
adding one fake data point at the end to bend the curve.

http://img560.imageshack.us/img560/1845/yi2.gif

You must then not use the values
which are above the last valid data point.

You may program this formula into an Arduino .-)

w.


Yes, the trend line is very close. But, please perform a calculation on
the 498R value using the equation and let me know if it agrees. If yours
does, then something is wrong here.

Thanks,
John


Helmut -

Please see this. I think Excel is screwed up.

https://imageshack.com/i/mrg919p

Helmut Wabnig[_2_] December 5th 13 10:38 PM

MFJ259 conversion help
 
On Thu, 05 Dec 2013 14:42:02 -0600, John S
wrote:

On 12/5/2013 1:49 PM, John S wrote:


Well I did it using an old dirty trick:
adding one fake data point at the end to bend the curve.

http://img560.imageshack.us/img560/1845/yi2.gif

You must then not use the values
which are above the last valid data point.

You may program this formula into an Arduino .-)

w.


Yes, the trend line is very close. But, please perform a calculation on
the 498R value using the equation and let me know if it agrees. If yours
does, then something is wrong here.

Thanks,
John


Helmut -

Please see this. I think Excel is screwed up.

https://imageshack.com/i/mrg919p



Yes, you are right.
Although EXCEL calculated the curve correctly in its own diagram,
it outputs false (rounded) parameters for the curve fitting polynom.
Compare with the Graphmatica plot, they are ident.
Have to find out tomorrow how to get the polynom factors without
rounding errors, if possible.
http://img834.imageshack.us/img834/8177/altv.gif

w.

amdx[_3_] December 5th 13 10:57 PM

MFJ259 conversion help
 
On 12/5/2013 11:31 AM, Helmut Wabnig wrote:
On Thu, 05 Dec 2013 10:26:33 -0600, amdx wrote:

On 12/4/2013 9:14 AM, John S wrote:
On 12/4/2013 8:48 AM, amdx wrote:
On 12/4/2013 4:50 AM, John S wrote:
On 12/3/2013 9:07 PM, amdx wrote:
I may be asking for something that doesn't have an answer.

I connected a voltmeter to the R meter of my MFJ259.
I checked a bunch of resistors and recorded the voltages.
Now I have all these voltage readings vs. Resistance and
don't know how to relate them except for a conversion graph.

Is there a math function that relates these voltages to Resistance?

Load Resistance Voltage reading

7.5 ohms 0.0388 volts
15 0.0444 volts
25.5 0.0478 volts
39 0.0577 volts
50 0.0614 volts
100 0.0807 volts
140 0.0891 volts
174 0.0935 volts
221 0.0980 volts
249 0.1000 volts
365 0.1060 volts
498 0.1090 volts

If you can figure this out, I'll reward you with the
SWR chart :-)

Thanks, Mikek


V = 0.0191*ln(R) - 0.0077

You're welcome.

Thanks, I'll learn about that and see if I can make it work for me.

Mikek

Good. I can tell you how I accomplished it if you are interested.


Hi John S,
Please don't let me work you if your not interested, this is now
probably more of a curiosity than a way to make the MFJ259 more useful.
I ran all the numbers with the formula, it is not accurate enough to
me usable. I don't know if that is a calibration error, resistor error,
(I used 1% resistors) a me error, or the wrong formula.


I don't think it correlates very well with my measurements.
Might need to use fixed font to read this.

My Measurements.

Real R Calculated R Measured Voltage


7.5 ohms 11.4 0.0388 volts
15 15.3 0.0444 volts
25.5 18.3 0.0478 volts
39 30.7 0.0577 volts
50 37.3 0.0614 volts
100 102 0.0807 volts
140 158.9 0.0891 volts
174 200 0.0935 volts
221 253 0.0980 volts
249 284 0.1000 volts
365 384.6 0.1060 volts
498 457.4 0.1090 volts

This is really only usable at 100 ohms. Only 2% error.
52% error low end and 8.1% top end
Is there a better formula.
I'm trying to use a digital meter in place of basically a hand drawn
dial that is very nonlinear.
I'm wondering if I calibrated 50 ohms to read a bit lower
or higher voltage it might help

Anyone's helpful thoughts,
Thanks, Mike



Well I did it using an old dirty trick:
adding one fake data point at the end to bend the curve.

http://img560.imageshack.us/img560/1845/yi2.gif

You must then not use the values
which are above the last valid data point.

You may program this formula into an Arduino .-)

w.


FWIW, I measured a 549 ohm resistor and
the voltage was. 0.1105mv

Mikek

amdx[_3_] December 5th 13 11:16 PM

MFJ259 conversion help
 
On 12/4/2013 9:48 AM, Ralph Mowery wrote:
"amdx" wrote in message
...
Is that because you think MFJ is junk, or because
there is no easy relationship?


No it is not that I think the MFJ is junk, just there is no easy
relationship.

You did not mention if the volts were AC or DC or RF that I recall.


I'm reading DC volts across a DC meter.


If DC there is no relationship at all. You do not use DC to measure the
inpedance of an antenna. If AC that most voltmeters will show, the
frequency range is too low to get a meaningful showing. If RF, you most
likely loose too much in the length of the leads.


I'm using the MFJ259, you might want to get a little information about it.

What you are doing is sort of like sticking the probes in the side of a
tree and trying to see how tall that tree is. You see something on the
meter, but it does not help to tell how tall that tree is.

What you are doing while a learning experiance , is just a negative one.
One that does not work for anything. Like Tom Edison and the light bulb.
When he had tried about 60 differant things for the filiment of the bulb, he
said I now know 60 things that do not work.


Again, look up the MFJ259

Thanks, Mikek


David Platt December 6th 13 01:13 AM

MFJ259 conversion help
 
In article ,
Ralph Mowery wrote:

So you are using a DC meter . How do you expect to get any thing meaningful
out of it ?


If I understand correctly, the original poster has lightly modified an
MFJ259, and is "tapping out" the DC voltage which drives its internal
meter circuitry. This voltage is a function of the impedance being
calculated by the MFJ's RF-impedance-measurement circuit.

The original poster is *not* trying to use a DC meter to measure the
RF coming out of the MFJ's SO-239 port.



Ralph Mowery December 6th 13 02:56 AM

MFJ259 conversion help
 

"amdx" wrote in message
...
You still don't understand. Do you want to?
I'll post exactly what I'm doing if you want to know.
The simple answer, this antenna analyzer has to DC meters
that display R and SWR. I'm reading the voltage across one of those meters
with my meter set on DC 200mV scale.
Might help if Googled MFJ259 Schematic, don't get the 259B, not the
same.
Mikek


OK, It is my fault for not understanding where you were placing the meter.
I see now that you are just hooking across one of the meters on the MFJ to
get a more accurate measurment.

I thought you were placing it across the actual resistor hooked to the MFJ.



amdx[_3_] December 6th 13 03:26 AM

MFJ259 conversion help
 
On 12/5/2013 8:56 PM, Ralph Mowery wrote:
"amdx" wrote in message
...
You still don't understand. Do you want to?
I'll post exactly what I'm doing if you want to know.
The simple answer, this antenna analyzer has to DC meters
that display R and SWR. I'm reading the voltage across one of those meters
with my meter set on DC 200mV scale.
Might help if Googled MFJ259 Schematic, don't get the 259B, not the
same.
Mikek


OK, It is my fault for not understanding where you were placing the meter.
I see now that you are just hooking across one of the meters on the MFJ to
get a more accurate measurement.

I thought you were placing it across the actual resistor hooked to the MFJ.



Ahh, no problem, sometimes hard to explain things when details you
see as obvious may not be seen that way by others. Glad you got it,
but I'm still a bit lost. The Formula John S gave me was close but not
close enough. They are into a problem with excell now, don't know if
they will get back to me or not :-).

John S December 6th 13 09:57 AM

MFJ259 conversion help
 
On 12/5/2013 4:38 PM, Helmut Wabnig wrote:
On Thu, 05 Dec 2013 14:42:02 -0600, John S
wrote:

On 12/5/2013 1:49 PM, John S wrote:


Well I did it using an old dirty trick:
adding one fake data point at the end to bend the curve.

http://img560.imageshack.us/img560/1845/yi2.gif

You must then not use the values
which are above the last valid data point.

You may program this formula into an Arduino .-)

w.

Yes, the trend line is very close. But, please perform a calculation on
the 498R value using the equation and let me know if it agrees. If yours
does, then something is wrong here.

Thanks,
John


Helmut -

Please see this. I think Excel is screwed up.

https://imageshack.com/i/mrg919p



Yes, you are right.
Although EXCEL calculated the curve correctly in its own diagram,
it outputs false (rounded) parameters for the curve fitting polynom.
Compare with the Graphmatica plot, they are ident.
Have to find out tomorrow how to get the polynom factors without
rounding errors, if possible.
http://img834.imageshack.us/img834/8177/altv.gif

w.


I found these coefficients at a site called ZunZun dot com:

y = a + bx + cx2 + dx3 + fx4 + gx5

Fitting target of lowest sum of squared absolute error =
7.0645972802275931E-06

a = 3.3214807109861584E-02
b = 7.2539508790925885E-04
c = -3.2830626216867792E-06
d = 7.9094355769986563E-09
f = -9.4574857953126017E-12
g = 4.3679252922923517E-15

Worst case error is .202% at 50.

John

Helmut Wabnig[_2_] December 6th 13 10:27 AM

MFJ259 conversion help
 
On Fri, 06 Dec 2013 03:57:40 -0600, John S
wrote:

On 12/5/2013 4:38 PM, Helmut Wabnig wrote:
On Thu, 05 Dec 2013 14:42:02 -0600, John S
wrote:

On 12/5/2013 1:49 PM, John S wrote:


Well I did it using an old dirty trick:
adding one fake data point at the end to bend the curve.

http://img560.imageshack.us/img560/1845/yi2.gif

You must then not use the values
which are above the last valid data point.

You may program this formula into an Arduino .-)

w.

Yes, the trend line is very close. But, please perform a calculation on
the 498R value using the equation and let me know if it agrees. If yours
does, then something is wrong here.

Thanks,
John


Helmut -

Please see this. I think Excel is screwed up.

https://imageshack.com/i/mrg919p



Yes, you are right.
Although EXCEL calculated the curve correctly in its own diagram,
it outputs false (rounded) parameters for the curve fitting polynom.
Compare with the Graphmatica plot, they are ident.
Have to find out tomorrow how to get the polynom factors without
rounding errors, if possible.
http://img834.imageshack.us/img834/8177/altv.gif

w.


I found these coefficients at a site called ZunZun dot com:

y = a + bx + cx2 + dx3 + fx4 + gx5

Fitting target of lowest sum of squared absolute error =
7.0645972802275931E-06

a = 3.3214807109861584E-02
b = 7.2539508790925885E-04
c = -3.2830626216867792E-06
d = 7.9094355769986563E-09
f = -9.4574857953126017E-12
g = 4.3679252922923517E-15

Worst case error is .202% at 50.

John



Yes, that's better

neither EXCEL nor OPEN OFFICE can do it correctly.
Excel generates the correct formula, but outputs only truncated or
rounded coefficients.
I do not know how to access the internal correct coefficients
in Excel.

Then I found this site:
http://www.xuru.org/rt/PR.asp#CopyPaste
Which outputs the following:

http://img43.imageshack.us/img43/6194/jmph.jpg


Which gives the correct curve when inserted into EXCEL
with some editing.
=4,368089718E-15*(B4)^5-9,457445532E-12*(B4)^4+0,000000007909410217*(B4)^3-0,000003283056669*(B4)^2+0,0007253920186*(B4)+0,03 321499593
http://img547.imageshack.us/img547/1012/o8io.gif

To improve the curve fitting I suggest to take additional measurements
in the upper range.

w.

John S December 6th 13 10:40 AM

MFJ259 conversion help
 
On 12/6/2013 4:27 AM, Helmut Wabnig wrote:
On Fri, 06 Dec 2013 03:57:40 -0600, John S
wrote:

On 12/5/2013 4:38 PM, Helmut Wabnig wrote:
On Thu, 05 Dec 2013 14:42:02 -0600, John S
wrote:

On 12/5/2013 1:49 PM, John S wrote:

Well I did it using an old dirty trick:
adding one fake data point at the end to bend the curve.

http://img560.imageshack.us/img560/1845/yi2.gif

You must then not use the values
which are above the last valid data point.

You may program this formula into an Arduino .-)

w.

Yes, the trend line is very close. But, please perform a calculation on
the 498R value using the equation and let me know if it agrees. If yours
does, then something is wrong here.

Thanks,
John


Helmut -

Please see this. I think Excel is screwed up.

https://imageshack.com/i/mrg919p


Yes, you are right.
Although EXCEL calculated the curve correctly in its own diagram,
it outputs false (rounded) parameters for the curve fitting polynom.
Compare with the Graphmatica plot, they are ident.
Have to find out tomorrow how to get the polynom factors without
rounding errors, if possible.
http://img834.imageshack.us/img834/8177/altv.gif

w.


I found these coefficients at a site called ZunZun dot com:

y = a + bx + cx2 + dx3 + fx4 + gx5

Fitting target of lowest sum of squared absolute error =
7.0645972802275931E-06

a = 3.3214807109861584E-02
b = 7.2539508790925885E-04
c = -3.2830626216867792E-06
d = 7.9094355769986563E-09
f = -9.4574857953126017E-12
g = 4.3679252922923517E-15

Worst case error is .202% at 50.

John



Yes, that's better

neither EXCEL nor OPEN OFFICE can do it correctly.
Excel generates the correct formula, but outputs only truncated or
rounded coefficients.
I do not know how to access the internal correct coefficients
in Excel.

Then I found this site:
http://www.xuru.org/rt/PR.asp#CopyPaste
Which outputs the following:

http://img43.imageshack.us/img43/6194/jmph.jpg


Which gives the correct curve when inserted into EXCEL
with some editing.
=4,368089718E-15*(B4)^5-9,457445532E-12*(B4)^4+0,000000007909410217*(B4)^3-0,000003283056669*(B4)^2+0,0007253920186*(B4)+0,03 321499593
http://img547.imageshack.us/img547/1012/o8io.gif

To improve the curve fitting I suggest to take additional measurements
in the upper range.

w.


Together, I think we got a reasonable answer. Thanks for your
information. I did not know about adding an additional data point to
help the curve. Nice work.


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