Using the LM-13 Frequency Meter As A VFO
by Greg Latta, AA8V
Replacing the Neon Lamps with a 0A2 Regulator Tube Page

0A2 Regulator Tube Without Shield
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Using the LM-13 Frequency Meter As A VFO Pages
Grayed out links will become active as I complete those pages.
 Using the LM-13 Frequency Meter As A VFO - Main Page  Replacing the Neon Lamps with a 0A2 Regulator Tube
 Power Supply  Grid Blocking the Oscillator During Receive
 2 Transistor Buffer Amplifier and 1 to 2 Voltage Step Up Transformer Page  Schematic Diagram and Circuit Descriptions
 How to Read a Vernier Scale  Making an Aluminum Case for the LM-13
 Exterior Photos  Interior Photos
 Resources and Manuals  

The voltage applied to the oscillator in the LM-13 must be regulated if the oscillator is to be stable. Though a conventional gaseous voltage regulator (VR) tube could be used to provide the regulation, the oscillator draws only a couple of mA of current, and a conventional VR tube would have been much larger than needed. At at the beginning of WWII there were no miniature 7-pin regulator tubes such as the 0A2, and the use of one of the then available VR tubes would have substantially increased the size of the LM-13. To save valuable space, neon bulbs were used as regulators instead. The nominal voltage drop across a single neon bulb (without an internal dropping resistor) is 60 V to 70 V. Thus two in series would give a total of 120 V to 140 V, ideal for regulating the voltage on the oscillator. In the image below, taken from page 36 of the manual, the neon regulator bulbs are V-104 and V-105. The bulb sockets are labeled X-104 and X-105

Neon Lamps - Manual Page 36
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The neon regulator bulbs and their sockets are labeled V-104/X-104 and V-105/X-105 in this picture from page 36 of the LM-13 manual.

Replacing The Neon Lamps With An 0A2:
It is unfortunate that in many LM-13s the neon bulbs are now missing, and they are now nearly impossible to find. It is very easy to replace the neon bulbs with a 0A2 regulator tube. The platform holding the bulb sockets must be removed and replaced with a new platform holding a 7-pin tube socket as can be seen in the photo below:

0A2 Regulator Tube Without Shield
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In this view taken from the back of the LM-13 the platform holding the neon bulbs has been replaced with a new one made from bakelite.
The new platform holds a 7-pin tube socket and 0A2 regulator tube. Although this tube socket takes a tube shield, a shield is not necessary.

0A2 With Shield
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In this view, the tube shield has been installed on the 0A2.

The tube is wired so that the anode (pin 1 or 5) is connected to R-116 and the cathode (pin 2, 4, or 7) is connected to ground. See the schematic below:

LM-13 Schematic with 0A2
The tube is wired so that the anode (pin 1 or 5) is connected to R-116 and the cathode (pin 2, 4, or 7) is connected to ground.

You can click here for a 0A2 Tube Data Sheet

Choosing the Proper Value and Power Rating of the Dropping Resistor:
Once the 0A2 is installed, it is necessary to determine the proper value for the dropping resistor. When the LM-13 is used as a VFO, the proper value guarantees enough current through the VR tube to keep it stable and lit, and enough current for the oscillator tube for proper oscillation. I have found that a total current through the dropping resistor of 5 mA to 6 mA seems to work just fine. More than this just produces excess heat and loads down the power supply, and less results in unstable operation of the VR tube.

The proper value of the dropping resistor is easy to calculate. First, find the needed voltage drop by subtracting the VR tube voltage (150 V) from the value of your B+:
VDROP = Your B+ - 150 V

Next, divide the voltage drop by 0.005 A to get the required resistance:
RDROP = VDROP / 0.005 A

Finally, to get the power dissipated in the dropping resistor, multiply the voltage drop by 0.005 A:
Double or triple this value as a safety factor to get the power rating of the dropping resistor.

Let's do an example. My power supply puts out 245 V so:
VDROP = 245 V - 150 V=95 V

We divide this by 0.005 A to get the resistance:
RDROP = VDROP / 0.005 A = 95 V / 0.005 A = 19000 ohm or 19 kohm

The power dissipated is VDROP x 0.005A:
PDISSIPATED = 95 V x 0.005 A = 0.48 W
Tripling this gives 3 x 0.48 W = 1.43 W. So a 19 kohm 2 W resistor would be perfect.

In actuallity, I used a 17.5 kohm 7 W resistor I found in my junk box that was close in physical size to the original R-103. This gives a current of 95 V / 17500 ohm = 5.4 mA which is close enough to the ideal 5 mA.

Installing The New Dropping Resistor:
The easiest way to get the required dropping resistance is to first set the link on the link panel so that R-116 is not in the circuit as shown on the schematic diagram above. R-116 is hard to get to and is probably best left alone. The link panel and R-116 can be seen in the photo below. The link panel is half way down on the left, and R-116 is very near the center of the photo.

Manual Page 39
In this photo, taken from page 39 of the LM-13 manual, the link panel can be seen at left center,
R-116 is near the center of the photo, and R-103 can be seen at top right.

With the link set to bypass R-116, all of the drop resistance can be placed at the location of R-103. R-103 is very easy to get to and is located at the upper right in the photo above. The original resistance of R-103 is 25 kohm. If the required dropping resistance is less than 25 kohm, you can add resistance in parallel with R-103 to lower the effective resistance to the new value. If the required resistance is larger than 25 kohm, you can unsolder one end of R-103 and add resistance to bring the effective resistance up to the required value. You can also simply replace R-103 with a new resistor if you like, but if possible I suggest that you keep the original R-103 for historical reasons.

Once you have everything installed you can power up the LM-13 and check to see if everything is working OK. You should have 150 V across the 0A2 tube and the tube should be lit. (It may be hard to see if it is lit, so darken the room if necessary.) You can verify the current through the dropping resistance by measuring the voltage across it (be careful, both sides are above ground!) and dividing by the total resistance to get the current. It should be around 5 mA.

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All images, designs, and materials on these web pages are the property of Gregory P. Latta and are ©2017 by Gregory P. Latta. You may use them for personal purposes and for educational purposes, but any commercial or other use is strictly prohibited unless written permission is obtained from the author.

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