HV5622PJ-G High Voltage Shift Register

Other Hardware General Electronics
1

Hello, I've already successfully used some 74hc595 shift register on my Arduino with this tutorialhttps://www.arduino.cc/en/Tutorial/ShiftOut . Now I wanted to use some HV5622PJ-G High Voltage Shift Register to drive my IN-12a nixie tubes. On the datasheet (http://datasheet.octopart.com/HV5622PJ-G-Supertex-datasheet-11741471.pdf) I found that apart from Data Input, Clock and Latch there are two extra pins, Polarity and Blanking. At first I tought to just redirect them to GND, but looking at the datasheet I don't think it would be a good idea. I also couldn't find an explaination of what they are used for specifically and how to drive them with an arduino. Also let the current flow through the Nixie, the shift register must redirect to ground, I don't know if this has anything to do.

2

In that datasheet, you want to look at the Function Table on page 4.

BL needs to be high (otherwise it forces all the outputs on)

POL is used to invert the outputs or not. If you set POL high, then a '1' sent into the shift register will turn that output on.

You can connect BL and POL to the +ve supply of the chip.

Note that this device seems to be designed for 12V, so you'll need to level shift the Clock, Data and Latch lines from the processor with something that can stand 12v.

Yours,
TonyWilk

3

Thank you for the clarification, also I have another question, if I connect a nixie tube to +180vdc, to make it light up, should the shift register output be high or low?

Thank you very much, so with POL on low, if I shift 1 than the output will be LOW, and the nixie will discharge through it, right?

4

AchilleM:
Thank you very much, so with POL on low, if I shift 1 than the output will be LOW, and the nixie will discharge through it, right?

How I read the datasheet is that shifting in a '1' will turn ON the output (to make it go LOW) when POL is HIGH (not inverted) - but it isn't exactly crystal clear.

Doesn't seem much of a problem, if it happens to be the wrong way round, you'd just have to invert the data bits before you shift 'em out.

Yours,
TonyWilk

5

Thank you very much again. I really appreciate your help :slight_smile: