[R-390] R-390 VFO Question

[Barry Scott]

Okay, Larry.  Thanks for the reply.

On further reading, the VFO is a Hartley design and, given that the
frequency formula for a Hartley is an inverse function of the LC values,
presuming that at xx 000, the iron core is "out" of the coil and a
clockwise turn of the KC knob causes the core to be pushed further inside
the coil increasing the L, then the frequency would indeed drop with CW
motion of the knob.  Sorry for the awkward way of stating that but I think
it makes sense to me.

Thanks again,
Barry - N4BUQ
Thanks,
Barry - N4BUQ


On Sun, Sep 15, 2024 at 11:10 PM Larry Haney <larry41gm2 at gmail.com> wrote:

> Hi Barry,  You are absolutely correct in your deductions.  When the KC is
> at its lowest of 000, the vfo is at its highest (3.455 mh).
>
> Regards, Larry
>
> On Sun, Sep 15, 2024 at 7:23 PM Barry Scott <72volkswagon at gmail.com>
> wrote:
>
>> Looking at the simplified schematic for the 3rd mixer (V205), the output
>> from the 2nd mixer (V204) tunes from 3 to 2 MC and the VFO tunes from
>> 3.455
>> MC to 2.455 MC yielding a constant 455 kc mixer product.  Is it correct to
>> interpret those numbers to mean that if the counter starts at XX  000 and
>> the KC control is rotated CW to XX +000 the VFO's output frequency starts
>> at 3.455kc and falls 1000 kc for 10 turns CW on the KC knob?
>>
>> I'm asking because I want to know what the frequency of the VFO is
>> supposed
>> to be when the dial is at XX  000 and I presume it's 3.455 kc but wanting
>> to make sure.
>>
>> Thanks,
>> Barry - N4BUQ
>> ______________________________________________________________
>> R-390 mailing list
>> Home: http://mailman.qth.net/mailman/listinfo/r-390
>> Help: http://mailman.qth.net/mmfaq.htm
>> Post: mailto:R-390 at mailman.qth.net
>>
>> This list hosted by: http://www.qsl.net
>> Please help support this email list: http://www.qsl.net/donate.html
>>
>