VTVM question

[Ken Dupuis Wn2Sqc]

Hi Mark. There was an article in the February 1969 issue of Ham Radio
Magazine that shows how to eliminate the battery in the Heath VTVM units.
Basically a bridge off the 6.3 volt winding through an 18 ohm 2 watt
resistor, the bridge output goes to ground through a 100mfd 15vdc cap,
the other side of the resistor is 1.5 vdc and goes to ground through 2
1N538 diodes. Sounds like the circuit you may have tried.  The article
notes that you must lift the 6.3 winding from ground to use a bridge,
want a copy? Ken

On Sat, 6 Apr 2002 18:30:48 -0500 Mark Graalman WB8JKR <wb8jkr at JUNO.COM>
writes:
>   Hi Stu,
>
>   I was just messing around with robbing a regulated 1.4 volts
> from the filament supply and eliminating the battery. I had
> it working well, but in order to have enough standing
> current in the regulator (two silicon diodes in series) to
> keep the internal resistance of the supply low enough not to
> screw up the accuracy of the ohm meter in the low range
> I decided it placed too much load on the meter's transformer
> so I gave it up and put the battery back in. Rather change a
> battery once in awhile than maybe replacing the transformer.  ;^)
>
> Mark  WB8JKR
>
>
> On Sat, 6 Apr 2002 14:24:11 -0800 "Stu Lyon" <slyon at pacificnet.net>
> writes:
> > Mark,
> > On my V-7A, which *should* be similar to the IM-18,etc, I
> measured
> > 135 mA
> > under the conditions you stated. The VTVM showed 0.6 ohms, which
> is
> > the
> > internal resistance of the multimeter I used.
> >
> > Just out curiosity, why do you need this information?
> >
> > 73, Stu W6CUX
> > Winnetka, CA
> > slyon at pacificnet.net
>
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