Project: Fluoro Lamp Power Inverter (12v)

[hsab]

hi for all members.

no comment :roll:

 

[williB]

hi hsab,
what does it do..?

 

[hsab]

hi

this sheme works with 12v batteries

we can use it at home when light cut off; in vehicule to repair wheel inthe dark far way home........................

i build it and it works ok .

by

 

[drrogla]

what does this circut draw from the power suply (i.e. car battery)?
and how powerful can the fluorescent be?

 

[hsab]

hi

use 12 v batteries

power lamp fluo from 18watts to 36 watts

@+

 

[satishkikani]

nice

this is nice[/i][/b]

 

[stanleycarl2]

Hello,
how are u doing today?iam stanley from Nigeria.iam a student,i need ur help.I was told in school to present a project topic for my final year.I need ur help ok.
my email address is stanleycarl2@yahoo.com
i will like us to communicate via my email address cos i have time checking it everyday.
thankz
stanley

 

[etes_oroq]

Hi hsab:

hope you're having a nice day today.

We had a similar project way back in my college days. Just want to ask if the it's possible to use a 1/2-inch (12.5mm) diameter AM radio antenna ferrite rod instead of the "philips tv psu ferrite" used in this project ?

By the way, I'm from the Philippines, so i have to find the JIS (Japanese) equivalents of BC series transistors. That I can google my way from here.

jun balista [etes_oroq]

 

[audioguru]

we can use it at home when light cut off

Why did your lights get cut off? Did you forget to pay your electricity bill?

 

[Rolf]

hi

this sheme works with 12v batteries

we can use it at home when light cut off; in vehicule to repair wheel inthe dark far way home........................

i build it and it works ok .

by

This is OLD technology, it is much more efficient to use the battery power to drive LED's directly. If I remember correctly, white high efficiency LED's produce around 130 lumen per watt versus about 80 for fluorescents.

 

[Hero999]

Flourescent are more efficient than LEDs, well for producing white light anyway.

The problem with this circuit it it drives the tube of pulsed DC not AC so it won't be as birght and it won't last as long. You need a push-pull driver and centre tapped transformer primary to get an AC output and drive the lamp properly.

 

[fonze]

Found **broken link removed** haven't tried it yet, I am in the proces of building it.

 

[Hero999]

That isn't a very good way of doing this, please refer to my post linked above.

 

[Oznog]

Flourescent are more efficient than LEDs, well for producing white light anyway.

The problem with this circuit it it drives the tube of pulsed DC not AC so it won't be as birght and it won't last as long. You need a push-pull driver and centre tapped transformer primary to get an AC output and drive the lamp properly.

The output of the transformer shown is still AC! However that type of drive is indeed inefficient and makes poor use of the magnetic core. The output may be a bit "spikey" too.

Rolf, what Hero999 says is accurate- good flourescents are more efficient than LEDs.

Starting and driving a flourescent is a bit tricky, and requirements vary from tube to tube. Of particular interest is that the impedance of the tube changes as it ages or just warms up. A fixed voltage may either not be enough to drive the tube with enough current OR it will provide the tube with too much current. I'm not big on home-grown solutions myself because it's such a complicated task to do right.

 

[Hero999]

The output of the transformer shown is still AC!

I believe we've seen this discussion on this forum before. The transformer's primary is driven by one transistor connected to a positive supply. When the transistor shuts off the back-emf generate by the field around the primary collapsing induces a large negitive pulse in the secondary, this repeates continiously. Hence it is generating pulsed DC not AC. The pulses travel in one direction only, from the cathode to the anode causing the catode to be brighter and excess electrode sputter around it, this isn't good. One solution to this problem is to simply add a capacitor in series with the tube, this will effectively block the DC level (remember pulsed DC is simply an AC voltage plus half it's peak value DC offset).

However that type of drive is indeed inefficient and makes poor use of the magnetic core.

True, because the pulses are only one polarity the same is true for the field, in effect only half the magnetising potential of the core is being used.

The output may be a bit "spikey" too.

That's an understatement, there's a heave peek at the start of each pulse followed by ringing (which may be damped when the tube ignites).

Rolf, what Hero999 says is accurate- good flourescents are more efficient than LEDs.

To date this is correct, but while fluorescent tube technology is mature LEDs are continuing to improve, it's only a matter of time before they beat fluorescents.

Starting and driving a flourescent is a bit tricky, and requirements vary from tube to tube. Of particular interest is that the impedance of the tube changes as it ages or just warms up.

It starts off open circuit before ignition and drops one struck, then continiues to drop as the current increases until either there's no more gas to ionise or the wiring resistance or fuse limits the current.

A fixed voltage may either not be enough to drive the tube with enough current OR it will provide the tube with too much current.

A constant current source is idea but this is normally approximated by a large inductor, at mains frequencies, or with a high frequency inverter a capacitor can be used (which will also solve the pulsed DC problem). Often the transformer is specially designed to have a very high leakage inductance to limit the current, this is true bot for some mains ballasts and inverters.

 

[Oznog]

I believe we've seen this discussion on this forum before. The transformer's primary is driven by one transistor connected to a positive supply. When the transistor shuts off the back-emf generate by the field around the primary collapsing induces a large negitive pulse in the secondary, this repeates continiously. Hence it is generating pulsed DC not AC. The pulses travel in one direction only, from the cathode to the anode causing the catode to be brighter and excess electrode sputter around it, this isn't good.

Certainly not. No transformer can ever create DC without a rectifier, no matter what you do with the primary.

Current in induced in the secondary winding when the mag flux changes. The mag flux increases when the primary transistor is on and creates current in the secondary. When the transistor is off, it decreases and makes current in the other direction. The sum of current over time must always be zero. The shape of the waveform is typically undesirable- when the transistor is on, flux increases rapidly making a high current spike on the output. Then it decreases at a more moderate rate when the transistor is off.

However, the core flux is always in the same direction. It increases and decreases but does not reverse, so the core is essentially biased at all times like an electromagnet. This reduces the capacity of the core. Furthermore, if you do not allow enough time for the flux to be drained by the load during the "off" period, current will increase even higher the next "on" period and so on until the core flux reaches a saturation point. At that time the current during the "on" period is limited only by the total resistance in the primary circuit and will typically draw enough current to smoke.

It is not a very appropriate design. He needs a split primary pretty badly.

 

[Hero999]

Certainly not. No transformer can ever create DC without a rectifier, no matter what you do with the primary.

Alright I'll be a bit more acurate doesn't create steady DC, it makes very small positive pulses and very large negitive pulses.

Try connecting the primary of a small mains transformer to an AA battey in series with a switch, connect a neon lamp to the secondary. When you hold the button the neon might flash very dimly as the field builds but it will flash even brighter when you release the button as the field collapses and notice the corrona is only generated around one electrode.

Current in induced in the secondary winding when the mag flux changes. The mag flux increases when the primary transistor is on and creates current in the secondary.

True but not much current is generated as the field builds slowly.

When the transistor is off, it decreases and makes current in the other direction.

When the transistor suddenly shuts off, the current in the primary decay more rapidly and generates a huge back EMF, it explains why you can generate 1kV from a 1:20 transformer with only 12V at the primary.

when the transistor is on, flux increases rapidly making a high current spike on the output. Then it decreases at a more moderate rate when the transistor is off.

You've got that the wrong way round, the current increases slowly when the transistor turns on, remember inductors don't like sudden change in current and it's for this reason that a huge spike is generated when the transistor turns of; it's like turning a relay off.

However, the core flux is always in the same direction. It increases and decreases but does not reverse

That's true in this case and it's why large negitive spikes are generated.

, so the core is essentially biased at all times like an electromagnet.

It does turn off (minus the small current in the feedback).

This reduces the capacity of the core.

No, it's because the core is only ever being magnetised with one polarity.

Furthermore, if you do not allow enough time for the flux to be drained by the load during the "off" period, current will increase even higher the next "on" period and so on until the core flux reaches a saturation point. At that time the current during the "on" period is limited only by the total resistance in the primary circuit and will typically draw enough current to smoke.

That will only happen if the tube is disconnected.

It is not a very appropriate design. He needs a split primary pretty badly.

Either way I agree.

But it does use pulsed DC, read the link I posted before. I've bought cheep DC fluorescent tube fittings before and one end of the tube always goes black and it doesn't illuminate uniformly

 

[Nigel Goodwin]

But it does use pulsed DC, read the link I posted before. I've bought cheep DC fluorescent tube fittings before and one end of the tube always goes black and it doesn't illuminate uniformly

There's too much rubbish on these forums about 'pulsed DC', there's not really any such thing - it's AC, as simple as that!.

The reason one end blackens is because the AC used isn't symmetrical, nothing to do with 'pulsed DC'. It's simply because the inverters used are crude, and waveshape isn't of much concern.

 

[Oznog]

Running a tube with too little current will also blacken the ends in short order. I assume doing it with too much current will do the same thing.

Waveforms which heavily deviate from a sine wave are a problem, but so is failure to regulate tube current. IIRC those great compact flourescents actually require a bit of starting cycle algorithm to get long life. Take one apart sometime, the 110v ones- even the cheap ones- still have like two dozen components inside.

There are some quality 12V DC devices out there:
**broken link removed**
And these are some excellent 12v DC lamp bulbs. They're really good. I've used them.
**broken link removed**

They're certainly expensive for a flourescent, but those are not the "crude" solutions mentioned.

The simple answer is to simply use a power inverter.

 

[Hero999]

The reason one end blackens is because the AC used isn't symmetrical, nothing to do with 'pulsed DC'. It's simply because the inverters used are crude, and waveshape isn't of much concern.

I did explain this a couple of posts a go.

One solution to this problem is to simply add a capacitor in series with the tube, this will effectively block the DC level (remember pulsed DC is simply an AC voltage plus half it's peak value DC offset).

None the less you do have a point, but how else should I explain it?

How about this:?

This inverter is no good because the AC waveform it produces is very asymmetrical (mostly negitive). Current only flows in one direction through the tube the electrons are only emitted from one end of the tube, this causes one end to be brighter than the other and excess electrode splutter also occurs at one end cause it to go black.