NE-45 Lamp Resistance

I'm also interested in building a relaxation oscillator with two NE-2 lamps, running on 240 volt AC. What would I need for this?

Dave,

Did you try Google? Here is one hit that may help, there were many others:

Richard

others:

Yeah, I found that right after I posted the first message.

others:

So this NE-45 lamp only requires about 82k resistance for 240 volts?

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others:

What contact is the internal resistor tied to, the side contact or the bottom contact? Also not too sure if 82k is right for 240 volt operation.

others:

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What difference does it make?

others:

Doesn't matter which side the internal resistor is on (unless you are doing something strange with the bulb, in which case you will have to determine that experimentally).

The data sheet says to use an 82K for 240 volts, so it seems that that is an appropriate value. However, if the lamp really draws 2 mA, 82 K will drop 164 volts, which seems a bit much to me, so I might reduce the resistor to 68K or so (to drop around 120V) if I needed a bit more light.

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others:

The only neons I've ever dealt with before are the common NE-2 which require a 270k resistor for 240 volt operation, I thought all neon indicators did. Guess you learn something new each day. The reason I wanted to know what contact the internal resistor is connected to is so I can put it in series with an external resistor.

--- Since the NE-2 and the NE-45 both have maintaining voltages of about

55V, then the series resistance can be determined from the source voltage and the rated lamp current.

For instance, an NE-2 has a rated current of about 0.5mA, so with a

240VRMS source the resistor must drop the difference between the source voltage and the lamp's maintaining voltage with the rated current going through the lamp:

Vs - Vm 240V - 55V R = --------- = ------------ = 3.7E5 ohms = 370k ohms. Il 5E-4A

Using 270k will result in an RMS current of 685µA through the lamp, which will cause a brighter glow, but will decrease the life of the lamp somewhat.

An NE-45 is rated for 2mA at 120VRMS, so its internal resistor must be:

Vs - Vm 120V - 55V R = --------- = ------------ = 3.25E4 ohms = 32.5k ohms. Il 2E-3A

With a 240V source the total resistance needed would be:

240V - 55V R = ------------ = 9.25E4 ohms = 92.5k ohms, 2E-3A

so the external resistor would be the difference between the two:

Rext = Rt - Rint = 92.5k ohms - 32.5k ohms = 60k ohms

The closest standard 5% value is 62k which would be fine and would extend the life of the lamp.

It (the resistor) would dissipate:

P = I²R = 2E-3A² * 6.2E4R = 2.48E-1 watts ~ 0.25 watt,

so you should use a 62kohm +/- 5%, 1/2 watt resistor.

---

--- :-)

---

--- Since the internal resistor is already in series with the lamp, connecting an external resistor in series with either the bottom contact or the side contact will put it in series with the lamp - internal resistor circuit: View in Courier:

. BOTTOM CONTACT . / .VIN>---[REXT]--->>---[RINT]---+ . | . [LAMP] . | .VIN>------------>>------------+ . \\ . SIDE CONTACT . . . BOTTOM CONTACT . / .VIN>------------>>---[RINT]---+ . | . [LAMP] . | .VIN>---[REXT]--->>------------+ . \\ . SIDE CONTACT

. . BOTTOM CONTACT . / .VIN>---[REXT]--->>------------+ . | . [LAMP] . | .VIN>------------>>---[RINT]---+ . \\ . SIDE CONTACT . . . BOTTOM CONTACT . / .VIN>------------>>------------+ . | . [LAMP] . | .VIN>---[REXT]--->>---[RINT]---+ . \\ . SIDE CONTACT

-- JF

Thanks for that information. Got to swap out the old resistors for my NE-2 lamps with higher rated ones.