Author Topic: Thought : converting 240v 50hz Switch Start fixture to 120v 60hz the easy way ?  (Read 1371 times)
Ash
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Thought : converting 240v 50hz Switch Start fixture to 120v 60hz the easy way ? « on: February 02, 2013, 01:00:45 PM » Author: Ash
This came up in a chat with Funkybulb. We were talking about a 240v 20w Switch Start ballast and how to use it on 120v. Then i thouht if there is any possibility of an overseen simple solution.....

The 240v 50hz balast is of too high impedance for 120v 60hz. So what if we cancel out some of its impedance with a series capacitor ?

I calculated it all the way, but came to unclear results. Considering the error margin in such calculations (due to the lamp not being exactly a resistor), the error margin in the result is so high that it might even include the resonance ondition.....

So here is the calculation done, is it correct ? Is the concept possible ?



We start with a Eltam L18.620 ballast, which is rated at 230v 50hz 18/20w 0.37A PF0.35, and a 20w T12 lamp which IIRC is 57v arc

Lets find the "real power" vdrop :
230v * 0.37a * 0.35 = 29.78w
29.78w - 20w = 9.78w (losses)
9.78w / 0.37a = 26.43v
57v + 26.43v = 83.43v

So we are finding out hat is the inductance of the perfect inductor ballast, which is running an imaginay lamp with Varc = 83.43v

Lets find the "reactive power" vdrop :
sqrt (230v^2 - 83.43v^2) = 214.33v

Lets find Xl :
214.33v / 0.37A = 579 ohm

Lets find l :
579 ohm = 2 * pi * 50hz * l
l = 579ohm / 2 / pi / 50hz = 1.84 H

Now lets wishLets find the "reactive power" vdrop :
sqrt (120v^2 - 83.43v^2) = 86.25v

The impedance we wish :
86.25v / 0.37A = 233.1 ohm

The impedance we have :
Xl = 2 * pi * 60hz * 1.84H = 693.66 ohm

Now we can have 2 options :
Capacitive circuit : Xc = 233.1 ohm + 693.66 ohm
Inductive circuit : Xc = 693.66 ohm - 233.1 ohm

Capacitive option :
Xc = 926.76 ohm
c = 1/(2* pi * 60hz * 926.76ohm) = 2.86 uF

Inductive option :
Xc = 460.56 ohm
c = 1/(2* pi * 60hz * 460.56ohm) = 5.76 uF

So far so good, right ?

The issue is, lets assume we were a bit wrong with estimating the ballast's L. I dont know how to evaluate the error precisely and i dont have an L meter, so lets just say it can be within 25% error so about 1.5-2.5H range instead of 1.84H we found. So if Xl for 1.84H at 60hz is 693.66 ohm. lets see what it'll be :
Xl = 2 * pi * 60hz * 1.5H = 565.48 ohm
Xl = 2 * pi * 60hz * 2.5H = 942.48 ohm

Wait WHAT ?
In both options, the value at which there is a resonance (Xl=Xc) is WITHIN our error margin.....Does not look good at all.... (even if we dont hit it, the difference between a "close" Xl and a "far" Xl would be 100's % of one another....) So it looks like we have a problem



What you think ?
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Re: Thought : converting 240v 50hz Switch Start fixture to 120v 60hz the easy way ? « Reply #1 on: February 02, 2013, 02:24:58 PM » Author: Medved
First you forget to couunt with the less than unity lamp power factor because of the voltage (nearly rectangular) vs current (nearly sinewave) shape mismatch sqrt(2)*2/Pi()=~0.9, but that does is, indeed,  not the main problem there.
You are correct with the sensitivity calculations: More you want to compensate by the capacitor, more it become sensitive to ANY tolerances in the L and/or C.
If you make the circuit on the capacitive side, one effect would play a significant role in helping you: As the core get higher flux, more and more of it's parts saturate (first some edges, then touch points,...), so the inductance goes down as the current increase.
So what happen in your circuit tuned for the capacitive side operation?
Let's assume the circuit is exactly in resonance when there is small current:
Xc=926Ohm, Xl=926Ohm, so the total Z=Xc-Xl=0, so when the voltage is applied, the current start to build up, as the energyu is pumped into the resonator.
But as the current build up, the inductance goes down, so does the inductive reactance. But as the total impedance is Xc-Xl and the Xc stay the same, it mean the overal reactance start to rise, so eventually would limit the current in the circuit to an equilibrum, where the inductance is just reduced so, the impedance yield the corresponding current.
If the circuit condition change so the current would rise with the same impedance, the increase of the current cause the inductance to drop further, so the overall reactance to increase, so reduce the current rise.
You see, than this LC ballast then provide a kind of current stabilizing role - with changing external conditions (mains of lamp arc voltage) it changes it's impedance so the current change only little.
The same mechanism then compensate the component tolerances: Higher current cause lower inductance, so compensate for anything, what changed the higher current and vice versa.

So for real design, I would calculate the circuit for the mean values of all parameters and let the stabilization effect take care of all the variations.

But for the current stabilizing effect have one adverse effect for fluorescents: The current remain pretty the same even during preheat by the starter, while the lamps are usually designed to be preheated by somewhat higher current, expecting the ballast slight saturation cause the current to increase. So the preheating would take longer, so the system would become more "blink happy" with ordinary starters.


And there is yet another effect from the series LC operated below it's resonance (so with dominant capacitance): Lets assume the lamp light. With this, the nominal current flow through the circuit.
Now assume (for the sake of this explanation, for real circuit it does not matter) the circuit is arranged so, the capacitor is just between the phase and inductor, the lamp being between the inductor and Neutral. If you observe the voltage on the middle point between the capacitor and inductor, you would see something like 270VAC. So replacing the 120V mains with the capacitor by the 270V AC source should not change anything for the inductor and lamp, right?

Now what happen, when the current reaches zero cross: Lamp arc disappear, so the circuit become open (for the short instant just after the zero cross). That mean the voltage across the lamp would be the sum of the capacitor and mains voltage, so the peak of the 270V, so about 380V.

But wait a minute, we have 120V mains, but for lamp reignition we have 380V peak voltage available. Now there is flying around "The mains voltage should be twice the lamp arc in order to allow reliable zero cross reignition", isn't the 270V then the equivalent "mains" for the reignition? Well, it in fact is. And it is this effect, what allow to operate lamps with higher reignition spikes, so arc voltages. Well, as the arc voltage rise, the voltage available for reignition drop a bit, but still it remain quite high. Of course the lamp arc voltage should be still low enough, so the ballast could deliver the required current.

Taking into account the current stabilization ability, this arrangement allow to operate lamps with arc voltage up to ~80..85% of the mains voltage, so for 120V mains it would be up to 100V lamps. Well, F30T8, FC12T9,... - all these could be operated without the transformer.
But what remain is the problem with starting: Before any current flow, there is just the mains voltage across the lamp, available to trigger the starter. Only after the starter close, the higher voltage build up. But that mean the starter trigger voltage have the room only between the lamp arc voltage (100V, with 200V reignition spikes should not trip the starter) and the mains (120V, so ~160V peak should reliably make the starter closed). So it is this reason, what keep the usable lamp arc voltage below 60..70% of the mains. In order to go really up to the 85% limit, the starting system should respond to something else than the voltage value across the lamp, so e.g. the tube light (manual start) or circuit current (4 pin thermal starter).
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Re: Thought : converting 240v 50hz Switch Start fixture to 120v 60hz the easy way ? « Reply #2 on: February 03, 2013, 11:16:04 AM » Author: Ash
Is this expected to work in reality ?

Which side you would choose ? The capacitive ?

How close tolerance would be required for the capacitor ? for example using 3uF instead of 2.86uF, resulting in Xc = 884 ohm ok or need better precision ?

Do i need to apply the 0.9 factor ? Didn't i allready apply it implicily when i took 0.37A and 57v for the lamp current and voltage values (which give 0.37 * 57 = 21.09) ?

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Re: Thought : converting 240v 50hz Switch Start fixture to 120v 60hz the easy way ? « Reply #3 on: February 03, 2013, 02:20:27 PM » Author: Medved
Is this expected to work in reality ?

It worked with 230V mains, 0.37A "18W" ballast, 2.7uF/450VAC capacitor and 180V 80W double-circular fluorescent lamp. The current during preheat (so ~5V across the lamp) was about 0.4A, when lit, so at ~180V across the lamp it was about 0.37A... But only with electronic starter, but it didn't restart after brief power cut (the starter is in "fault mode" during normal operation of this setup, so need longer time to reset)

Which side you would choose ? The capacitive ?

The capacitive, just because I want the saturation to work for and not against me.


How close tolerance would be required for the capacitor ? for example using 3uF instead of 2.86uF, resulting in Xc = 884 ohm ok or need better precision ?

I think it would work. You have to make some experiments, the regulation ability of the circuit is not 100%.

Do i need to apply the 0.9 factor ? Didn't i allready apply it implicily when i took 0.37A and 57v for the lamp current and voltage values (which give 0.37 * 57 = 21.09) ?

If the 57V was from the lamp datasheet, the power factor is already included.
If you want to derive the voltage from the lamp rated power and current, of course you have to use it, if you don't want to be 10% off with the expected arc voltage...
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