Need help making a resistive ballast.

[RayRay1132]

I have started to acquire a stupidly ridiculous amount of old cfl bulbs with bad ballasts and I was looking to make a couple of resistive ballasts for fun (and simplicity). I've used this led current dropping resistor calculator thing, set the forward drop to 0, input voltage to 120, and current draw to 300mA (http://www.circuitous.ca/LEDcalc.html) to try to find out what value resistor I need. According to the calculator I need a 400 ohm resistor for a 36 watt bulb running at 300mA. This sounds correct, however if someone could just confirm this before I wire it up that would be great.
Thanks in advance, Ray, KD2JID

 

[Nigel Goodwin]

Just dump the old bulbs, CCFL bulbs are crap, and resistive ballasts would only make them even less efficient - use the far better LED bulbs.

CCFL usually use electronic inverters, not a 'ballast' at all as such - and normally require a much higher voltage than mains to initially strike. I'm dubious about what just a resistor would do?

 

[RayRay1132]

Just dump the old bulbs, CCFL bulbs are crap, and resistive ballasts would only make them even less efficient - use the far better LED bulbs.

CCFL usually use electronic inverters, not a 'ballast' at all as such - and normally require a much higher voltage than mains to initially strike. I'm dubious about what just a resistor would do?

I just want to have some fun with these bulbs; I know about leds and such. I've noticed on some small fluorescent lights pre 1990s, they had a resistor and glow starter. Just wondering if my math is correct for that resistor.

 

[Les Jones]

If they have a glow starter they will have an inductive ballast. (A choke.) When power is first applied the gas in the starter is ionized. This heats a bimetal strip in the starter which closes a contact connecting the heaters in the tube in series with each other and the choke, The heaters heat up during this time but as the gas is no longer ionized the bimetal cools an opens the contact. If there is enough current through the choke when the contacts open there is high voltage back EMF from the choke which causes the lamp to strike. (The gas in the tube is now ionized.) The flicker when starting is repeated attempts at starting until the starter contacts open at a point on the cycle that generates enough back EMF to strike the tube. Once the tube has struck the voltage between it's ends is not enough to ionize the gas in the starter. (You could then remove the starter and the tube would remain lit.)

Les.

 

[RayRay1132]

If they have a glow starter they will have an inductive ballast. (A choke.) When power is first applied the gas in the starter is ionized. This heats a bimetal strip in the starter which closes a contact connecting the heaters in the tube in series with each other and the choke, The heaters heat up during this time but as the gas is no longer ionized the bimetal cools an opens the contact. If there is enough current through the choke when the contacts open there is high voltage back EMF from the choke which causes the lamp to strike. (The gas in the tube is now ionized.) The flicker when starting is repeated attempts at starting until the starter contacts open at a point on the cycle that generates enough back EMF to strike the tube. Once the tube has struck the voltage between it's ends is not enough to ionize the gas in the starter. (You could then remove the starter and the tube would remain lit.)

Les.

The little fixture I found had both a resistor and a glow bottle starter, no inductive choke ballast

 

[crutschow]

Just wondering if my math is correct for that resistor.

No.
To calculate the proper ballast resistor you need to know the voltage drop across the bulb at its operating current, which you haven't stated.
The resistor value would then be 120V minus the bulb drop divided by the current.
Note that it may require a high power resistor, depending upon the actual bulb voltage drop.

 

[RayRay1132]

If they have a glow starter they will have an inductive ballast. (A choke.) When power is first applied the gas in the starter is ionized. This heats a bimetal strip in the starter which closes a contact connecting the heaters in the tube in series with each other and the choke, The heaters heat up during this time but as the gas is no longer ionized the bimetal cools an opens the contact. If there is enough current through the choke when the contacts open there is high voltage back EMF from the choke which causes the lamp to strike. (The gas in the tube is now ionized.) The flicker when starting is repeated attempts at starting until the starter contacts open at a point on the cycle that generates enough back EMF to strike the tube. Once the tube has struck the voltage between it's ends is not enough to ionize the gas in the starter. (You could then remove the starter and the tube would remain lit.)

Les.

Im pretty sure the bulb was so small that 120 volts could light it once the filaments were hot, I think thats the only reason they added the starter

 

[RayRay1132]

No.
To calculate the proper ballast resistor you need to know the voltage drop across the bulb at its operating current, which you haven't stated.
The resistor value would then be 120V minus the bulb drop divided by the current.
Note that it may require a high power resistor, depending upon the actual bulb voltage drop.

I did state the operating current which was 300mA

 

[RayRay1132]

No.
To calculate the proper ballast resistor you need to know the voltage drop across the bulb at its operating current, which you haven't stated.
The resistor value would then be 120V minus the bulb drop divided by the current.
Note that it may require a high power resistor, depending upon the actual bulb voltage drop.

I'm going to overdrive the bulb slightly and run it at 36 watts. 120volts @ 300mA

 

[rjenkinsgb]

The resistor (or ballast) and lamp form a divider.

300mA through a 400 Ohm resistor means 120V across the resistor and zero volts across the lamp.

 

[RayRay1132]

The resistor (or ballast) and lamp form a divider.

300mA through a 400 Ohm resistor means 120V across the resistor and zero volts across the lamp.

I don't think the lamps really has any resistance though, its a mercury vapor type bulb, which is essentially a short circuit.

 

[audioguru]

Does the lamp produce light when it gets 0V?? You said the lamp draws 300mA from 120VAC so it and its ballast are not a dead short.
The ballast produces a high starting voltage then the ballast limits the AC current while still giving a high voltage to the lamp.
The fluorescent lamp also has a filament at each end.

 

[rjenkinsgb]

I don't think the lamps really has any resistance though, its a mercury vapor type bulb, which is essentially a short circuit.

That would mean zero volts across the lamp. Zero volt at any current is zero power.

The lamp itself [excluding the ballast] must have (volts * amps) = 36W if is a 36W lamp!

 

[RayRay1132]

That would mean zero volts across the lamp. Zero volt at any current is zero power.

The lamp itself [excluding the ballast] must have (volts * amps) = 36W if is a 36W lamp!

ok, I'm starting to get a little confused (I understand the basic concept), what would be the easiest way to let 300mA across the tube while having enough voltage to keep the tube ionized? I've seen it done before with resistors, I just really need some help with the math side of choosing the resistor.

 

[audioguru]

The electronic ballast used in a GE CF bulb in post #12 is as simple and as cheap as can be. If they could do it with only simple resistors then their cost would have been lower causing their profit to be massive. They like lots of profit.

 

[RayRay1132]

Here’s some pictures showing what I mean on a small lamp. It has no magnetic ballast, just a resistor capacitor and glow bottle starter. Even if the mains voltage isn’t enough to start the bulb after heating filaments, I have another way of starting it. Can someone just help me with the math for finding the right resistor pls. Thanks in advance -Ray

 

[crutschow]

I don't think the lamps really has any resistance though, its a mercury vapor type bulb, which is essentially a short circuit.

Yes it has a very low dynamic resistance, but it still has a significant voltage drop when operating, since lamp operating power equals the lamp voltage drop times the current.

I'm starting to get a little confused (I understand the basic concept),

Yes you are, and not quite.

I did state the operating current which was 300mA

Yes you did.
But that's not sufficient to calculate the resistor value.
You seem to think you can calculate the resistor value without knowing the lamp voltage, but you can't, since some of the voltage will be dropped across the lamp, and the rest will be dropped across the resistor.

 

[RayRay1132]

Yes it has a very low dynamic resistance, but it still has a significant voltage drop when operating, since lamp operating power equals the lamp voltage drop times the current.
Yes you are, and not quite.
Yes you did.
But that's not sufficient to calculate the resistor value.
You seem to think you can calculate the resistor value without knowing the lamp voltage, but you can't, since some of the voltage will be dropped across the lamp, and the rest will be dropped across the resistor.

how would i figure out the lamp voltage?

 

[audioguru]

I have never seen such a small fluorescent light tube. The resistor limits the turn on current surge.
The value of the capacitor also limits the light tube's continuous (AC) current.

 

[crutschow]

how would i figure out the lamp voltage?

The best way is to measure it under load.
Otherwise, if you know what gas is in the tube, you might be able to determine the voltage from an internet search.