# I don't get resistors

#### Rocky_circuits

Joined Nov 1, 2011
57
One of the most basic components of circuits, and I don't really understand it. Atleast for my application. I'm using Nixie Tubes, and I am supposed to have a resistor on their common anodes.

Here's two scenarios:
1) I tune my power supply to about 135-139 volts. I put a 1k resistor on the anode of a nixie tube and test it with a multimeter. It pulls 2.5mA which is exactly what I am looking for. But the data sheet says I need 180 (nominal) volts to ignite it. Apparently not. I was told this may make for unreliable striking but here's scenario 2 which confuse me about that.

2) I tune my power supply to 180 volts. I put a 18k resistor on it and the multimeter reads 2.5mA. Great looks good. But I test with my voltmeter the voltage after the resistor and it is 135v, approximately the same as just using 135v PS and the 1k resistor in scenario one.

So at this point I am very confused. Why bother turning the power supply up to 180 volts? I'm thinking that there is a property of resistors that I might not understand or something of that sort.

on a side note I got everything counting and working great on the clock, just need to make sure I'm not going to burn things out if left on at this point and build an enclosure by this thursday

#### DerStrom8

Joined Feb 20, 2011
2,390
It's important to remember that resistors reduce current flow, and also have a voltage drop determined by the value. If you have a 12 volt source connected to two 1K resistors in series, according to ohm's law you will have 6 milliamps flowing through the circuit (12V/(1K+1K)). Then, since you have 6 mA through the resistors, that means you have a voltage drop across one of them of 6mA*1K=6V. When you increase resistor size, you'll reduce the current drawn and increase the voltage drop (reduce the voltage going through the circuit). You'll need to match up the proper resistors to give you what you need.

Sorry if I didn't answer your actual question, but I hope this helps anyway

Regards,
Der Strom

#### Rocky_circuits

Joined Nov 1, 2011
57
It's important to remember that resistors reduce current flow, and also have a voltage drop determined by the value. If you have a 12 volt source connected to two 1K resistors in series, according to ohm's law you will have 6 milliamps flowing through the circuit (12V/(1K+1K)). Then, since you have 6 mA through the resistors, that means you have a voltage drop across one of them of 6mA*1K=6V. When you increase resistor size, you'll reduce the current drawn and increase the voltage drop (reduce the voltage going through the circuit). You'll need to match up the proper resistors to give you what you need.

Sorry if I didn't answer your actual question, but I hope this helps anyway

Regards,
Der Strom
It does help a little bit, but would that not mean that scenario 1 and 2 are roughly exactly the same (excluding efficiency) since they yield the same current and voltage?

Joined Dec 26, 2010
2,148
This is as much concerned with the properties of the tubes as those of resistors. Actually, somewhat simpler but related issues occur with LEDs, and people new to them often display a similar reluctance to use adequate series resistance.

There are at least three problems with working the tubes with a lower supply voltage:-

1. The tube requires a somewhat higher voltage to start than its steady running voltage.
2. The striking and running voltages may be subject to variation caused by ageing, temperature, or individual tolerances.
3. The supply voltage must also be subject to variation.
The use of a larger supply voltage ensures reliable starting, even if things deteriorate a bit. Having a reasonably big resistance, with a significant voltage across it "dilutes" the effects of small voltage changes, so that the effect on the electrode current and brightness will be restrained. Think about it, with a resistor of R ohms in series with the electrode, if there is a change ΔV in the difference between the supply voltage and the tube voltage, the alteration in the current will be ΔV/R. A bigger value for R means less variation.

You are of course at liberty to doubt these explanations - I recollect a very similar discussion on another thread - but really it does work out this way.

#### Rocky_circuits

Joined Nov 1, 2011
57
So a higher voltage and bigger resister minimizes chance for potential error or failure in the future as things are "broken in" or get old.

Although something you say is what peaks my interest I believe. Your #1 if I read it correctly talks about having a higher voltage than the normal running voltage.

Does the resistor not always step the voltage down immediately? With my understanding of it, after the resistor there is only 135 volts and the tube never sees 180 volts even though the power supply is putting that many out.

#### SgtWookie

Joined Jul 17, 2007
22,230
If there is no current through the resistor, then there is a difference of zero volts across the resistor.

So, if 180V is on one end of the resistor, and there is no current flowing through the resistor, then both ends of the resistor measure 180V.

Once the tube starts conducting, then current flows through the resistor; and there is then a voltage drop across the resistor.

#### MrChips

Joined Oct 2, 2009
29,207
The higher voltage gives the Nixie tube a better chance of starting, like giving a runner at the starting blocks a kick in the rear end to get started.The higher series resistance creates a constant current supply so that the Nixie tube gets the same amount of current independent of power fluctuations and prevents the tube from burning out.

Joined Dec 26, 2010
2,148
As a result of Ohm's law, the voltage of V volts dropped across a resistor of R ohms is equal to the current through it I amps multiplied by R: V = I*R. For a perfect resistor this is an instantaneous effect. Real resistors have imperfections such as parasitic inductance, but usually not on a scale which would bother us much when running a Nixie tube.

If the supply to the resistor feeding the tube was turned on suddenly, the tube voltage might get to the supply voltage for a matter of some microseconds (maybe a bit longer?) before the ionisation in the gas built up enough to get the current going. As soon as this happened, the voltage would fall.

See page 5 of this .pdf: it gives a nice explanation. http://www.bader-frankfurt.de/loads/burroughs.pdf

Edit: But note that the detailed numbers may not necessarily line up with those for the devices you are using - stick to the specs you have. What is clear is that the manufacturer of these devices warned against trying to skimp on volts.

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