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Neon indicator across capacitor bank, resistor question?

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Alex M

Thu May 10 2012, 04:36PM

Registered Member #3943 Joined: Sun Jun 12 2011, 05:24PM
Location: The Shire, UK
Posts: 552

If I was to put a neon indicator across a 330v capacitor bank, what resistance resistor would I need to use in series with the neon indicator?

I would like to make it light up when the bank is in the region of 310v and it should not draw too much energy from the bank.

Thanks.

Sulaiman

Thu May 10 2012, 05:57PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

Small neon lamps/bulbs vary quite a bit in specifications but for this application assume about 70v and 0.25 to 2.5 mA depending on size of bulb.
I would use 1.0 to 2.2 MegOhm

If you put a small cap (e.g.100nF) across the neon it will flash
giving a more noticeable indication.

Alex M

Thu May 10 2012, 06:16PM

Registered Member #3943 Joined: Sun Jun 12 2011, 05:24PM
Location: The Shire, UK
Posts: 552

Sulaiman wrote ...

Small neon lamps/bulbs vary quite a bit in specifications but for this application assume about 70v and 0.25 to 2.5 mA depending on size of bulb.
I would use 1.0 to 2.2 MegOhm

If you put a small cap (e.g.100nF) across the neon it will flash
giving a more noticeable indication.

Thanks Sulaiman. I was thinking somewhere around 1megohm would do it but wasn't too sure if that would stop it from lighting up.

The neon is just one out of an old disposable camera, so its not very big.

It doesn't have to be bright since I have my multimeter across it measuring the voltage at all times and I physically turn the charger off when I get to around 310-320v. But the neon indicator will be a nice little "extra".

Sulaiman

Thu May 10 2012, 07:13PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

IN disposable camera flash units there is a film capacitor used with the trigger transformer, put that across the neon for the 'flashing' effect.
You must still use the series 1 MOhm resistor

Proud Mary

Thu May 10 2012, 07:17PM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

DC striking voltage always higher than AC.

Example: Neon indicator bulb typically strike at ~90V AC 50Hz. Once it strikes, voltage across across neon falls to ~60 VAC. When DC is used, striking voltage may be 120 - 130 VDC, and maintaining voltage 90 VDC.

Until moment neon strike, almost no current is drawn, so almost no voltage drop across resistor. So neon will start to conduct when voltage across your capacitors reaches 120V - 130V almost regardless of the size of the series resistor.

As voltage across capacitor climb above strike voltage, neon will conduct more, and more voltage will be dropped across resistor.

Note that the glow discharge is not a simple ON/OFF type of action. Where current is limited in a DC circuit, ionization will start on a small area of the cathode once the striking voltage is reached. This glow discharge will spread as the current is increased, until whole cathode surrounded! Beyond a certain current flow, metal ions from electrodes will start to participate in glow discharge, arc will form, and failure follow soon!

Circuit very non-linear! Negative resistance characteristic after neon strikes!

Note: striking voltage affected by light photons! More daylight = lower striking voltage!

Forty

Thu May 10 2012, 07:37PM

Registered Member #3888 Joined: Sun May 15 2011, 09:50PM
Location: Erie, PA
Posts: 649

How about a voltage divider plus an LED?
Then the current should always be there and
once the voltage passes the turn on voltage of the diode
it should light up, right?
or would it have the problem similar to the neon bulb where it would always be on and just gain brightness as the voltage went up

Steve Conner

Thu May 10 2012, 07:47PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Yup, if you want the neon to strike at 310V you'll need a voltage divider. A simple resistor won't do.

I'd use 220k.

Alex M

Thu May 10 2012, 08:33PM

Registered Member #3943 Joined: Sun Jun 12 2011, 05:24PM
Location: The Shire, UK
Posts: 552

Steve Conner wrote ...

Yup, if you want the neon to strike at 310V you'll need a voltage divider. A simple resistor won't do.

I'd use 220k.

@Steve, something like this?

Edit: sorry the other end of the neon should go to ground (I am bit tired right now).

Proud Mary

Thu May 10 2012, 09:49PM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

In a simple voltage divider we connect two resistors in series across the voltage to be divided, and take the voltage that we want from the junction between the two.

In this case, we want a voltage of 130V (to strike the neon) to appear when you apply 330V across the voltage divider.

We know that these small neons need a current usually between 0.5 mA and 2 mA to fully ignite. So we take a nominal current of the same order to work out the total resistance of our voltage divider. If we decide that 1 mA must flow through the two series resistors then the total of Resistor 1 and Resistor 2 must be... (R = V/I) ... 330 kÎ©. So now we have the two series resistors totaling 330 kÎ© with 1 mA going through them. (In a series string of resistors - no matter how many there are - the current must be the same in all of them.)

So now we have a current - our 1 mA - and a chosen voltage - 130V - to strike the neon to work out the value of the bottom resistor of the two. Again R = V/I, so we have the desired resistance R equal to the voltage we want - 130V - divided by the current we have flowing through the two resistors that form the voltage divider - which is 130/0.001 = 130,000 Î© or 130 kÎ©.

The top resistor has to drop 200V when a current of 1 mA flows through it, so we know that it must have a value of 200 kÎ©.

Now, when your neon strikes, its resistance will fall to between about 10 and 250Î©. This low resistance is now in parallel with your 130 kÎ© resistor, and for all practical purposes shorts it out (because when two or more resistors are connected in parallel, the total resistance must always be less than the value of the lowest resistor in the parallel arrangement).

So when the neon ignites, you can consider the circuit as simply the 200 kÎ© resistor in series with your neon. Will this 200 kÎ© be sufficient to stop the neon burning up?

The resistance of the ignited neon is so small that we can forget about it and just look at what happens when we connect a 200 kÎ© resistor directly across 330V.
I = V/R so the current will be 330V divided by 200,000 Î©, which is 1.65 mA.

What power rating must the resistor have? W = Isq*R so we have a dissipation of 1.65mA squared times 200,000 Î© = 0.5545 W.

In practice, then, we would use a 1 W resistor of 200 kÎ© so we have a good margin of safety.

I have simplified the explanation of the way the neon performs to make it easier to understand and calculate, but the values I have given will do perfectly well in practice.

Bored Chemist

Sat May 12 2012, 01:44PM

Registered Member #193 Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022

This "voltage divider" is a bit odd.
Until the neon strikes it's resistance is practically infinite so all the capacitor voltage will be across it. After it strikes the voltage across the neon will be rather low so practically all the voltage will be across the resistor.

If you get a source of high voltage and connect a high value resistor to it, you get a (rather rough) constant current source.
With 330 V and 200 kÎ© you get about 1.5mA and that's what ends up flowing through the neon .(it's bit lower due to the drop across the neon but trying to treat a negative incremental resistance as part of a voltage divider is an odd way to look at it).

It would be interesting to measure the current through the resistor/ neon combination.

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