[ARC5] BC-453B dial calibration saga.

[Leslie Smith]

  Hello All,

  My earlier comment, that the inductance must be less not more depends
  on being able to tune the LO at the higher frequency, by adjusting the
  trimmer, of course.   But the basic point was this:  If the dial
  reading is higher than the known frequency, you must take out
  inductance, not add it.  It's counter-intuitive.   Follow the maths.  
  Mathematics and the laws of nature are consistently reliable. 
  Sometimes you THINK the mathematics is unreliable - but no.  The
  problem is in always the mind, never in the maths.

You must (of course) follow the trick suggested and align the dial to
the signal generator at the both the higher and lower end of the band. 
That is given or assumed.

  73 de Les Smith
   vk2bcu at operamail.com

On Fri, Jul 28, 2017, at 15:23, Leslie Smith wrote:
> Hello Skip,
> 
> I think there is a swap in your thinking between "standard" and "dial".
> This is because the problem is counter-intuitive.
> Your "standard" (a signal on 210kHz) is LOW compared with your dial.
> 
> Now I will illustrate what is going on with a "demo" calculation.
> 
> Assume, for the discussion that your total tuning "C" is 155pF at
> 210kHz.
> You HEAR the signal at 210 (it's coming from your signal generator, and
> you know that IS the frequency because you measured it with a counter.) 
>  But the dial reads 215kHz (say).   The signal is your STANDARD.  The
> dial is only an indicator.
> 
> Now I calculate:
> The inductance (L) of your coil is therefore 25330.3/(0.210*0.210*155)
> uH
> This is:  3705.97uH.
> 
> Say you increase the inductance of the coil by 2% - that by 74.12uH 
> (say you used ferrite).
> 
> You haven't touched the dial (the capacitor) but now your set tunes 
> sqrt(25330.3/4789.09 * 155)  = 184.7kHz
> 
> You ADDED inductance and the error increased!!   You need to REMOVE
> inductance to move your receiver toward the correct point on the dial. 
> You must think very hard to understand why this works
> counter-intuitively.
> 
> This is why a trial-and-error maths calculation works.
> 
> Les
> 
> You probably wonder where 25330.29 comes from!
> 
> Taking uH and pF into account, 25330.29 equals 1000,000/(4 pi squared).
> And the 4 pi squared comes because resonance happens when X(l) equals
> X(c)
> and 2 x PI x F x L equals 1000,000/(2 x PI x F x C).
> So - solve for L or C or F and you have the answer.  - Les
> 
> 
> 
> 
> 
>                                           o-o-o  end  o-o-o
> 
> On Fri, Jul 28, 2017, at 14:10, Waldo Magnuson wrote:
> > Hi, 
> >    I'm still trying to get the dial reading to be a little more accurate.
> >     The closest I can get on the 210 kHz alignment is about 216 with C-9
> >    fully unmeshed.  So I thought I would add a little more inductance to
> >    L-5 so I could align to 210 kHz.  Now the closest I can get is 230 or
> >    so.  I increased L-5 by adding a small ferrite core above the slug.  
> >    Am I thinking wrong?  Comments?  Thanks.
> > Skip Magnuson
> > Spokane, WA
> > ______________________________________________________________
> > ARC5 mailing list
> > Home: http://mailman.qth.net/mailman/listinfo/arc5
> > Help: http://mailman.qth.net/mmfaq.htm
> > Post: mailto:ARC5 at mailman.qth.net
> > 
> > This list hosted by: http://www.qsl.net
> > Please help support this email list: http://www.qsl.net/donate.html