Question on Arduinix and INS-1

[William Lee]

Hi all-

This is my first posting to the group.  I'm still relatively new to electronics and to the wonderful world of Nixies.  I've built a couple of Nixie clocks from kits and stuff like the Ogi Lumen and Arduinix platform.  I'm looking forward to eventually building my own clock from scratch, but have a question on the Arduinix if someone would be willing to give it a shot.

 

For those unfamiliar with it, the arduinix is an Arduino shield that uses an inductor based set up to provide HV for the nixies and allow them to be controlled by the Arduino.  It's a very cool setup.  It also provides PCBs to mount IN-17s.  I have one of them that allows 4 IN-17s with two INS-1s as the divider/colon on the display.  The cathodes on the PCB each have a resistor for each nixie and INS-1.  

 

My question is related to the INS-1s.  I want to ensure that they don't burn out by getting the full nixie voltage from the arduinix (approximately 180V).  Is this just as simple as picking a resistor with the correct value and power rating? 

 

Sorry for what I'm sure is a very basic question!

[John Rehwinkel]

> My question is related to the INS-1s. I want to ensure that they don't burn out by getting the full nixie voltage from the arduinix (approximately 180V). Is this just as simple as picking a resistor with the correct value and power rating?

You've got it right. The INS-1 runs on the same voltage as the nixies, just less current. And the resistor controls the current.

> Sorry for what I'm sure is a very basic question!

It's a good question, and that's why we're here. And it can be less basic than it seems. To calculate the resistor size, you have to subtract the tube's maintaining voltage from your power supply voltage, and then divide by the desired current.

Welcome!

- John

[Dylan Distasio]

Thanks, John.  I'm a little confused though.  I thought that the firing voltage on the INS-1 was 65V and the current was .5mA.  Do I not need to worry about reducing the voltage going to the INS-1 down from ~160-180V as long as I address the current issue?  Also, is "maintaining" voltage in your reply the same thing as firing voltage.  

In other words, let's say my PSU voltage is 180V.  Would my calculation be:

(180V-65V) / 0.0005A = 230k resistor?    

Also, in terms of the Power rating needed on the resistor, can you tell me if this is the right calculation.  I want to make sure I am not missing anything now to avoid burning out a component later.  I wasn't sure if I should be using the current that I am looking for (.5mA) or the current that is flowing into the resistor from the PSU for this calculation.

P= I^2 x R = .0005A^2 x 230000 ohms = 0.0575 watts.  Assuming this is correct, does that mean I can use anything 1/8W (.125) or better on the resistor.

Thanks for the welcome!

Best,

Dylan

--
You received this message because you are subscribed to the Google Groups "neonixie-l" group.

To post to this group, send an email to neoni...@googlegroups.com.
To unsubscribe from this group, send email to neonixie-l+...@googlegroups.com.
For more options, visit this group at http://groups.google.com/group/neonixie-l?hl=en-GB.

[John Rehwinkel]

> Thanks, John. I'm a little confused though. I thought that the firing voltage on the INS-1 was 65V and the current was .5mA. Do I not need to worry about reducing the voltage going to the INS-1 down from ~160-180V as long as I address the current issue? Also, is "maintaining" voltage in your reply the same thing as firing voltage.

Right, if you calculate your resistor for the correct current, it will drop the required voltage (Ohm's law and all that).

Firing (or "striking") voltage is generally higher than the maintaining voltage. The firing voltage
is the voltage required to initiate the discharge, after which the voltage drops some, to the "maintaining" voltage.

For the INS-1, the striking voltage is specified as not less than 65V and not more than 90V. The maintaining voltage is specified as not more than 55V.

> In other words, let's say my PSU voltage is 180V. Would my calculation be:
>
> (180V-65V) / 0.0005A = 230k resistor?

You have the right idea, but you want to use the 55V figure, as the steady state value should be that or less.

(180V - 55V) / 0.0005A = 250k resistor.

With the 230k resistor calculated from the striking voltage, you'd get 0.54mA (if the tube's maintaining voltage is 55V), which is probably just fine too. It's not critical.

> Also, in terms of the Power rating needed on the resistor, can you tell me if this is the right calculation. I want to make sure I am not missing anything now to avoid burning out a component later. I wasn't sure if I should be using the current that I am looking for (.5mA) or the current that is flowing into the resistor from the PSU for this calculation.
>
> P= I^2 x R = .0005A^2 x 230000 ohms = 0.0575 watts. Assuming this is correct, does that mean I can use anything 1/8W (.125) or better on the resistor.

The same current flows through the resistor and the indicator, the 0.5mA you mentioned. Even though the power supply is capable of supplying more current, the resistor limits it to this value.

Your calculation is correct, any reasonable resistor will have sufficient power dissipation capability. However, there's another subtle factor that's often overlooked, and that is resistor
voltage capability. Some small resistors can't hold off a lot of voltage, so you might want to use a quarter watt or half watt unit for its voltage capability, even though you won't be dissipating
much power.

You can also solve the calculation the other way, volts times amps:

(180V - 55V) x 0.0005A = 0.0625 watts (a larger figure due to the difference between striking and maintaining voltages).

- Cheers,
John

[Imbanon]

If in doubt with your calculation, try this online calculator
http://www.mcamafia.de/nixie/ncp_en/ncp.htm
Very simple, and does the job quickly