Go Down

Topic: [SOLVED] Leakage current in a driver IC; how to compensate? (Read 2 times) previous topic - next topic

Risen


No that is impossible. You are missunderstanding something.

Perhaps you could explain further?

joseph_m

I don't think the resistor is the correct item to control the 70 volt problem.  I would use a transistor to turn on the lights.  Just wire up the transistor as a switch.  The ground of the Nixie has no voltage until the light is lit.  Use this ground voltage of the nixie to turn on, or open, the transistor to the other light.  Then when the nixies are off all lights will be off.  And when the nixies are on there is voltage going to ground that turns on the transistor and the other lights turn on.  I hope that is clear enough.  Lets see if I can do some ascii art here.

Nixie ground pin
0------|-----------------------ground
        |
        |----------------------base of transistor

Figure out the rest.


Now I have never worked with nixie tubes.  So there might be enough voltage flowing to ground when they are off to open the transistor.  I just don't know.  But a little work with a VOM will tell you if this idea will work.

Grumpy_Mike

Quote
Quote from: Grumpy_Mike on April 27, 2012, 07:11:15 PM
No that is impossible. You are missunderstanding something.
Perhaps you could explain further?

A parallel resistor will take current away from the lamp. In any parallel circuit current is split between the two parallel components. Therefore with less current through it the lamp can not be brighter.

dc42


Your divider solution that sounds like it has some promise. Would you mind explaining it a bit further with how you calculated it?


If R1 is the resistor to +170v and R2 is the resistor to ground, then the voltage divider is equivalent to a supply of 170 * R1/(R1 + R2) volts with a series resistor of value R1 * R2/(R1 + R2). So if you can decide what is the ideal voltage supply for the neon lamp supply (I suggest you aim for 120v) and what series resistor you would want to use with that supply, then you can work out the values of R1 and R2 to use by solving those as simultaneous equations.

I have to admit that I didn't do that, instead I reduced R1 a bit (probably not enough) from the value of 200K to allow for some current through R2, and then chose R2 to give a voltage in the range 100 to 120v at the junction.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

Risen

Grumpy: so I was supposed to leave the other resistors already in series at the same time? Nobody had mentioned that, so I took them out when I tried the 1M parallel.

dc42: Thanks a bunch! That was very helpful. Since the Arduinix board doesn't have much room for expansion and I'm not confident in building an additional power supply, I think I will use your solution. I did some calculations for the resistor values I have on hand (exactly 180K and 390K aren't among them) and found some values that work similarly.

Most importantly, I then tested it and it worked out great!

Thanks to everyone, especially dc42.

Go Up