# [Boatanchors] Problem solved...probably. FW: Question for Weston 269 Meter

David Rossetti drossetti at comcast.net
Sun Dec 30 14:41:50 EST 2012

```William,

I was intrigued from the get-go by your meter challenge. A quick Internet
image search revealed the very cool and unique fan layout for the Weston 269
meter.

As a tube tester collector, a key component to any of these devices is the
meter. It has been my experience that meters from the same manufacturer with
the same model number often have been internally shunted to provide
different operating characteristics. Just as you may find various drawings
of vintage gear schematics showing alterations to the circuit layouts yet
exhibiting the same drawing number, such was the state of configuration
control/management in those days.

A cursory search on the Internet has uncovered various full scale (f.s.)
characteristics for the Weston 269. The lowest f.s. deflection I found was 1
ma. I would have to agree with Ron. Your candidate substitute meter, with a
f.s. deflection of 50 ma, probably has been internally shunted.

Alan Douglas addresses meter substitution in his tube tester 'bible'; "Tube
Testers and Classic Electronic Test Gear". The donor meter has to have a
sensitivity equal to or greater than the meter you are replacing.
Sensitivity is based on both the f.s. deflection amperage and the equivalent
f.s. voltage.

Tube tester schematics generally show both the f.s. amperage and the
internal meter resistance for the main meter. The meter internal resistance
plays into how the meter itself affects the entire circuit it is integrated
into. The equivalent f.s. voltage is derived using ohms law (V = IR). The
equivalent f.s. voltage (V) is equal to the f.s. amperage (I) multiplied by
the internal meter resistance (R). As an example, the schematic for the
Hickok 539A tube tester shows the main meter has a 97 uA f.s. deflection,
and a 250 ohm internal meter resistance. I normalize everything to volts,
amps, and ohms to keep the decimal places straight. Doing the math, we can
determine the f.s equivalent voltage:

V = IR
V = I(97 uA or .000097 A) times R (250 ohms)
V = .02425 volts or 24.25 mV

Just to keep it clear in my mind, a donor meter with ‘greater sensitivity’
means both the f.s. amperage and the equivalent voltage are lower than the
identified values (in this case, less than 97 uA and less than 24.25 mV). To
understand why this is so becomes clear as you look at the method used to

The process for meter substitution is actually pretty straight forward. The
resulting meter must exhibit both the same f.s. amperage characteristics,
and the same internal meter resistance characteristics, as the original
meter. If the donor meter has the same f.s. amperage characteristic, we are
good to go for that element. If not, the only choice we have is to find a
meter with ‘greater sensitivity’ (a lower f.s. amperage characteristic) and
shunt it in parallel with the proper resistor (more on this later). Once
this is done, we add a series resistor to bring the total internal meter
resistance back to spec. Remove the meter scale from the original meter,
scan it, print it on card stock, and fit it into your new meter, and you are
good to go. In your case, since you are dealing with the same meter model,
the scale can just be carefully substituted (as you indicated).

The challenge then is what is the f.s. deflection and internal resistance
requirement for the Weston 269 meter in your application? Based on what you
stated, these characteristics can't be derived from your original meter, in
that it was blown. Perhaps the schematic will show you. If the f.s.
deflection requirement is 50 ma, you are at least good for that element. If
the requirement is greater than 50 ma, you can use your newly purchased
meter as a base, and shunt it to achieve the required f.s. deflection.
Unfortunately, if your required f.s. deflection is less than 50 ma, it won't
work without internal meter modification (that being to remove or replace
the internal shunt, which can get very tricky). Shunting any meter that is
already less sensitive than the meter you are trying to replace will just
make it even less sensitive.

Determining the shunt resistor to be used is where the equivalent f.s.
voltage comes into play. As an example, and as discussed earlier, using ohms
law (V = IR), the equivalent f.s. voltage (V) for the 539A meter is equal to
the f.s. amperage (I = 97 uA) multiplied by the internal meter resistance (R
= 250 ohms), or 24.25 mV. This is the voltage that will cause the original
meter to measure full scale. The replacement meter must be shunted to read
full scale when the same voltage is applied. If we were to use a 50 uA meter
as a substitute, we have a difference of 47 uA that must be accommodated by
a shunt resistor to keep the full scale equivalent voltage constant. The

V (24.25 mV) = I(47 uA) x R
R = 24.25 mV / 47 uA = 516 ohms

So for this example, a 50 uA f.s. meter can be shunted using a 516 ohm
resistor in parallel to the meter to replicate a 97 uA f.s. meter. Now all
we have to do is place another resistor in series with this shunted meter to
give it the required internal meter resistance. As you may recall from
Electricity 101 class, the calculation for determining the total resistance
of a set of resistors in parallel is found by adding up the reciprocals (or
inverse) of the resistance values, and then taking the reciprocal of the
total. Another way of stating this is: one over the sum of the inverse of
the ohm values of the resistors. If, by example, let's say my substitute 50
um meter has an internal meter resistance of 5000 ohms. The inverse of that
value is 1/5000 or .0002. The shunt resistor which is in parallel with the
meter has a value of 516 ohms. The inverse of that is 1/516 or .0019. The
sum of these two values is .0021. The reciprocal of that value is 1/.0021,
which equals 468 ohms.

As long as the internal resistance requirement for this sample is equal to
or greater than 468 ohms, we can add another resistor in series to get us to
the required internal resistance. However, if it turns out the internal
resistance requirement is less than 468 ohms, we have a problem. There is no
resistor that can be added in series to 468 ohms to get us to anything less
than this value. In this example, the shunted meter has too much internal
resistance to be used as a substitute.

In the example for the 539A tube tester, and as mentioned earlier, the
substitute meter we select must not only have a sensitivity value of 97 uA
or less, but must also have an f.s. equivalent voltage of 24.25 mV or less.
If we were to use a 50 uA f.s. meter as a substitute for the 539A meter, we
can again use ohms law to determine the maximum internal meter resistance it
can have. The equation works out as so:

V (24.25 mV) = I (50 uA) times R
R = 24.25 mV / 50 uA = 485 ohms

If we shunt a 50 uA f.s. meter that has an internal meter resistance of 485
ohms with a 516 ohm resistor, we get a meter that goes full scale at the
same voltage as the original 97 uA f.s. meter. Let’s do the math. This meter
has an internal meter resistance of 485 ohms. The inverse of that value is
1/485 or .0021. The shunt resistor which is in parallel with the meter has a
value of 516 ohms. The inverse of that is 1/516 or .0019. The sum of these
two values is .004. The reciprocal of that value is 1/.004, which equals 250
ohms.

If the substitute meter has an internal meter resistance of less than 485
ohms, the resulting internal meter resistance of the shunted replacement
meter would be something less than 250 ohms. Whatever that difference is, we
find a precision resistor of that value and place it in series with this new
substitute meter to bring its internal meter resistance up to spec. Replace

Many of the older meters show the f.s. amperage value on the face of the
meter, but have not found any images of the Weston 269 showing the f.s.
value. If not marked, the f.s. equivalent voltage can be determined using a
known DC voltage power source and some known high ohm value resistors, as
you did. Meter resistance can generally be safely measured using a modern
digital ohm meter since it will not use more than 50 uA of power. NOTE: it
is not advisable to attempt to measure internal meter resistance with an
analog ohm meter. The ohm meter itself can put out amperage values that can
destroy the meter being tested. Once you know the f.s. equivalent voltage
and the internal meter resistance of your candidate meter, you can use ohms
law to calculate its f.s. amperage sensitivity.

Enough with the examples. You need to find out what the f.s. deflection and
internal resistance requirements for the Weston 269 meter in your
application is. Then you can determine if the meter you purchased will work
as a substitute.

Dave Rossetti

-----Original Message-----
From: boatanchors-bounces at puck.nether.net
[mailto:boatanchors-bounces at puck.nether.net] On Behalf Of Ron Barlow
Sent: Sunday, December 30, 2012 1:19 PM
To: Bry Carling; William Morton; boatanchor
Subject: Re: [Boatanchors] Problem solved...probably. FW: Question for
Weston 269 Meter

If I understand correctly, a shunt is not missing, but perhaps one has been
Be very careful, William, while performing the scale swap. The meter
movement and/or the pointer can easily be damaged.
GL 73 & HNY de Ron  n4gjv

________________________________
From: Bry Carling <bcarling at cfl.rr.com>
To: William Morton <w_b_morton at hotmail.com>; boatanchor
<boatanchors at puck.nether.net>
Sent: Sunday, December 30, 2012 11:42 AM
Subject: Re: [Boatanchors] Problem solved...probably. FW: Question for
Weston 269 Meter

Perhaps it's missing a shunt?

William Morton <w_b_morton at hotmail.com> wrote:

>
>Hello All,
>I ended up buying a couple meters from that auction site and spent last
>night calibrating the first one that has already arrived.  While the
>information I found on the model 269 indicated 1 mA full scale, a rig
>set up with a 1.5v battery and linear pot gave me something on the
>order of 50 mA full scale.  Hmm... the meter moves smoothly and the
>scale on this particular meter is - surprise! - 50 mA.  Perhaps this
>particular meter was an earlier version or a modified example with a
>different sensitivity.
>In any case, I am able to remove the outer cover and can see the scale
>plate is held by two screws.  Should be a simple matter of swapping out
>the scale plates and moving along with the project.  A learning
>experience, to be sure.
>Happy New Year to All!
>Best Regards,
>William
>From: w_b_morton at hotmail.com
>To: boatanchors at puck.nether.net
>Subject: Question for Weston 269 Meter
>Date: Mon, 24 Dec 2012 14:45:41 +0000
>
>
>
>
>
>Hello All,
>Season's Greetings from Minnesota, where it is currently snowing.
>
>Today's question is about the beautiful Weston 269 meter.  Does anyone
>have any experience in changing/replacing the background scale?  I have
>a pair of meters, each with a DC volt scale ranging from 0 to 500.
>Unfortunately each of them has been blown (I will spare you the sad
>details).
>I see several for sale on eBay but they are scaled for mA or dB so I
>would need to change them somehow.  Luckily I can take my blown meters
>apart for trial purposes but I would be grateful for some advice as to
>the proper method.
>Alternatively, if someone has a pair already scaled to 500 VDC and is
>willing to part with them I would be very interested in arranging a
>purchase.
>Best Regards,
>William _______________________________________________
>Boatanchors mailing list
>Boatanchors at puck.nether.net
>https://puck.nether.net/mailman/listinfo/boatanchors

regards,
Bry Carling

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