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I'm looking into using a set of IN-12B Nixie tubes in a clock design I've been working on.

If I tie my anode to the 170V striking voltage and use ~16kOhm to limit the digit current to 2.5mA then we're fine. However, I don't know how I am supposed to achieve the 0.3mA current limit for the comma cathode.

I'd like to think that I can just add another resistor in series with my cathode in order to meet the current spec but I've been having trouble finding any examples of people using the IN-12B without ignoring the comma cathode entirely.

What is the right way to go about this?

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2.5 mA × 16 kΩ = 40 V drop, so the sustaining voltage is about 130 V. You could measure this to verify.

If we assume the comma has similar voltage characteristics at the lower current, then you need a 40 V drop @ 0.3 mA, or 130 kΩ.

Or just use your resistance substitution box (start at 1 MΩ) and adjust it until it looks about right.

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  • \$\begingroup\$ Yes, I measured the 130V sustaining voltage. My question is to whether I can add resistance on the cathode in order to meet the current limit spec. \$\endgroup\$ – VBwhatnow Jul 2 '17 at 15:50
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    \$\begingroup\$ Oh, I see what you mean. Yes, putting the extra resistance on the cathode should work. \$\endgroup\$ – Dave Tweed Jul 2 '17 at 15:53
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    \$\begingroup\$ Unfortunately putting and additional resistor in the cathode of the comma is likely to end up with a flashing/off comma and even digit. The Anode is common to all digits so with a single common 16k Anode resistor the required digit strikes and the voltage on the anode drops (below that needed to strike the comma). End result is usually the comma does not strike. .... The right way to cope with the comma is to have no anode resistor for this digit and include a resistor in each cathode line. .... or use transistor/FET as CC drivers in each cathode line. \$\endgroup\$ – Jack Creasey Jul 3 '17 at 2:29
  • \$\begingroup\$ @JackCreasey: I see your point, but can't this be addressed by lighting the comma first? The anode only drops by 4.8 V, which ​should still leave enough to strike the digit. \$\endgroup\$ – Dave Tweed Jul 3 '17 at 4:03
  • \$\begingroup\$ It depends on the struck voltage required for the digit and the comma. Have you ever tried to light two neon lights with a single resistor ...it is not reliable. \$\endgroup\$ – Jack Creasey Jul 3 '17 at 4:28

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