Re: difference in VFO between BITX20 and BITX20A: connection of resistor?

larry <ag8o1@...>
 

COMPUTER CHRASED
I lost info on L1.L2 turns and size wire on bitx17a? larry AG8O

----- Original Message -----
From: Arv Evans
To: BITX20@...
Sent: Friday, May 28, 2010 9:01 AM
Subject: Re: [BITX20] difference in VFO between BITX20 and BITX20A: connection of resistor?



Chris PA2CRX

The BITX20A BFO has been optimized for best quality sine-wave output.
This included taking
output from the tank circuit instead of from the emitter. There is also
the filtering action of
R32 and C91 added to attenuate higher frequency energy. This
arrangement has the effect of
lowering the output voltage and making this voltage somewhat dependent
on impedance
distribution around the feedback loop (L7, C15, C34, C35, C37 and C38).
Some BITX20A
builders have noticed that the VFO output changes as you open or close
C38. This is because
the feedback loop impedance is altered by changing the value of C38.

If one plans to use the BITX20A VFO design at other frequencies, you may
find that the output
voltage might be inadequate if significant change is made to the
impedance of the feedback loop.
If you use Jim Tonne's "BITX_Tune" software to re-design the BITX20A VFO
for other frequencies,
you should always follow through with running the new design in LT-Spice
to show the magnitude
of output voltage with your new design.

Q8 in the BITX20A VFO design is biased to half the supply voltage (9V)
by R33 and R34. This should
then cause the emitter resistor R36 to show ((9/2)-0.6) = 3.9 volts.
With R36 being 2.2K ohms,
then the current through Q8 should be (3.9 / 2200) = 1.8 ma. Q7 is also
biased to 4.5 volts by
R33 and R34, so it's 1K emitter resistor should show a voltage drop of
(4.5-0.6) = 3.9 V, resulting
in an emitter current of 3.9/1000 = 3.9 ma for Q7. Since both Q8 and Q7
use the same regulated
9 volts, the total current drain would be (3.9 + 1.8) = 5.4 ma.
Resistor R35 is 100 ohms, so the
current through it to the 9V Zener diode would be ((12 - 9) /100) = 30
ma. This 30 ma minus the
5.4 ma load from Q8 and Q7 leaves 24.6 ma through the Zener diode, which
is plenty for regulation
purposes. This is 24.6 * 9 = 0.22 watts which explains why that
regulator runs a bit warm.

Remember that you cannot make reliable DC voltage measurements on VFO
circuits with the oscillator
running. What you will see is the average of the AC sine wave and is
dependent on purity of that
waveform. Best way to measure voltages in a VFO is to stop oscillations
by opening the feedback
loop, or swamping it with a 0.1 mfd capacitor...then make the DC
measurements with the stage
operating as an amplifier instead of as an oscillator.

Hope this helps answer some of your questions.

Arv - K7HKL
_._

On 05/28/2010 08:04 AM, vdberghak wrote:
>
> Hi all,
> In my BITX17 I use a VXO. However, I made also the VFO from the
> original schematic. It oscillates directly, however, the voltage on
> the collector is about 6 volt so the zener of 9V1 do not the
> stabilising function... The output is so low, my frequency counter
> does not display the frequency. Enough output for the counter at the
> coil or at the 2K2 resistor.
>
> Then I looked to the schematic of the BITX20A. The 4K7 resistor is not
> connected to the emittor/2K2 resistor but it is connected to the coil!
> (not DC connected at all).
> Is this wrong in the schematic? Or it is wrong in the (original)
> schematic on Farhans site?
>
> If the original schematic is correct, I wonder why the output is so
> low and the current so (relatively) high...
>
> Any comments?
> Thanks,
> Chris, PA3CRX
>
>

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