I recieved some IN-14 nixie tubes for Christmas, and was hoping to implement them in a project i have been considering for some time. The details are unimportant, except that each nixie tube should have its own discrete pcb and driver. After looking at ICs for some time I found the MAXIM 6966, a LED sink driver with some nifty features (though i would have preferred an I2C interface instead of SPI). It seems perfect for my project (driving a single nixie tube) for the following reasons:

  • Tiny 3x3mm QFN package
  • Exactly 10 outputs
  • Open-drain functionality (output pins either low or Hi-z)
  • PWM functionality on outputs

My problem is this... the datasheet says the outputs have up to 7V protection, but i'm not sure if that applies to both Hi-Z and Low output configurations, or just the Low output configuration. With an output set to Hi-Z, the corresponding pin is likely to be biased to a high voltage such as 170V... well above the 7V protection. So that is to say, if the 7V max condition also applies to the Hi-Z pin configuration then I cannot use it.

I hope it will work fine, but I suspect it isn't suitable. Could someone confirm my suspicion before I dismiss this otherwise fantastic IC?

Here is a link to the product page: https://www.maximintegrated.com/en/products/interface/controllers-expanders/MAX6966.html

Thanks :)


2 Answers 2


No, that device is not suitable for driving Nixie tubes. When an output is switched off, it will see the full voltage at the output pin.

The datasheet clearly states that

I/O Port Outputs Are 7V-Rated Open Drain

That falls well short of the 170V you'll be using.

Development of drivers specifically designed for Nixie tubes pretty much reached a dead end with the 74141, but there are some modern parts around that will carry out serial to parallel conversion and drive high voltage open-drain outputs. Look at the Microchip HV5122 or HV5522 for example.

  • \$\begingroup\$ I was hoping that wasn't the case but i suppose its wishful thinking. I was hoping to avoid the HV series because the lack of any 10, 12, 16, or similarly ranged bit parts throws a spanner in my plan of discretising the tube circuits due to their large size (i would need a 32-bit part). \$\endgroup\$
    – Joe
    Commented Dec 29, 2017 at 12:42
  • \$\begingroup\$ You don't have to use all 32, use one per tube and just use the first 10 bits? \$\endgroup\$
    – Finbarr
    Commented Dec 29, 2017 at 12:57
  • \$\begingroup\$ I can yes, but a 32-pin package would likely be too large to implement neatly for the configuration i'm considering. \$\endgroup\$
    – Joe
    Commented Dec 29, 2017 at 13:18
  • \$\begingroup\$ You might want to note that the rating of the 74141 (60V) is also much less than 170V. \$\endgroup\$ Commented Dec 29, 2017 at 13:52
  • \$\begingroup\$ Yes. And i saw a circuit employing the MAX6922, which is rated for 76V on it's inputs. I don't quite understand this... surely the idling cathodes will be biased by the anode voltage and require the I/O connected to the idle cathodes to be rated at the anode voltage (e.g. 180V)? unless the idle cathodes are floating? \$\endgroup\$
    – Joe
    Commented Dec 29, 2017 at 14:42

While the supply voltage vastly exceeds the maximum voltage for the driver IC, it is still possible to use them together if there is a transistor in series with the driver output in order to reduce the voltage below the upper limit.

This method is explored in TI's Application Report SLVA280, "Using TLC5940 With Higher LED Supply Voltages and Series LEDs". Note that while the title implies specific parts, the techniques described can be used with any open-drain driver and non-inductive load.


  • \$\begingroup\$ If i understand it correctly, this would be a good solution if you want to keep a particular driver IC in the design. However, i would prefer not to add 10 discrete mosfets if i can help it. \$\endgroup\$
    – Joe
    Commented Dec 29, 2017 at 13:20
  • 1
    \$\begingroup\$ They may make MOSFET arrays that can withstand the required voltage. \$\endgroup\$ Commented Dec 29, 2017 at 13:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.