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AC -> 12v DC, 1000mA Output -> Halogen Lamp = Nothing.
DC -> 12v DC, 10 AA Batteries -> Halogen Lamp = Works fine.

Why does the Halogen Lamp not accept the current?
@ 6v, 5W (0.83 Amps, right? Just under the AC output..)

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what are the power requirements of the lamp?

6 volts & 5 watts.

To save a post, I'll ask you, although the question is not limited to anyone:
In accordance to.. 10101001010020984828920's reply, could I possibly use a capacitor and/or transistor to achieve some sort of push-start?

The old "starter's for the older florescent lamps were more like a neon bulb then a capacitor. I do know they use capacitors to start some motors.

You do realize that 1,000 mA = one amp ?

Amperage is calculated from the voltage and the circuit's resistance.

Yes, I do realize that. The Halogen Bulbs take 0.83 amps, as calculated from the V and W -- correct me if I am wrong. The fact that the bulbs are getting an extra 170mA just adds to my head scratching as to why they don't work with the AC Tx. I will mess around with them more and post back if I get something. Thanks for your help.

motors need ac phase shift to start (and some to work too) the fluorescent lamp needs higher voltage to start halogen needs higher current to start try to connect a big capacitor (10 K uF) in parallel with the tx and then connect the lamp to it. it may work. or start it from batteries and flip the switch to the tx what are yo doing with 6 V lamp on 12 V ?? you may want to connect batteries in parallel (6 V higher current) to get it working better

It is the only Tx I have come across that has an acceptable Amperage. I have another @ 9v/100mA and another one @ 12v/500mA.

The biggest caps I have are 2x 670uF and 2x 480uF. Will connecting a few of these in parallel/series be sufficient enough?
-- Now that I think of it, capacitors store charge.. Volts is charge, right? So how will this solution solve the problem if what it needs is more Amperage?
--- Well I do recall you or Goodhart mentioning how if you increase the V then proportionally it will also increase the Amperage.. but bleh it's still a little WooOOooOooOo to me...

problem is that if you manage to light up your halogen lamp youll most likely burn it right away capacitors store charge. how it appears to you depends on how you connect them if you connect the capacitors in series their voltages sum up if you connect in paralle their voltage together is same as of each. but when you load the circuit the charge from the capacitors helps to provide the needed high current in the first moment thats on very surface level. for more complete answer keep on asking !

Thank you very much. Examples with Reasoning behind them helps me understand things much easier. I'll try to connect a capacitor in parallel to one of the Halogen lamps in a few mins, and let you know how it went.

plz get a 6 V power supply before we continue. the 12 V supply is very likely to burn your lamp if we manage to light it somehow (what happens with the batteries is that the load of the lamp rerduces their actual voltage - probably to well below 12 V. this usually does not happen with txformers and more powerfull sources) a computer power supply can give 5 V which is ok for 6 V lamp. if you have one (or can get for free broken one from computer shop and replace the blown capacitors) its great

I've aquired 5 {5v} power supplies, all with 1A.

If he uses your switch idea, it may be that the switch needs to be a MBB (make before break) kind, lest he lose connection in the switching process.

i think its more important to not alow the battery and txformer to get connected together the filament in halogens has quite big thermal capacity. i dont think its going to cool down that fast

A few diodes discretely placed would prevent any type of feedback to the battery, but I see your point which is why I wrote "may" :-)

with diodes you dnt need to do anything fast --> no need a switch at all

but here we anyways have a problem of not enough power from the supply - do you really want to take away even more power with the diodes ? (0.7 V * whatever A)

Well, if we used a Schottky diode the drop would only be 0.22V still I am just poking around writing out loud, so to speak (helps clear my head).

It MAY be that there is too much ripple with the ac to dc supply ? Do they operate on strictly DC ? (I am just guessing because I am not familiar with halogen bulbs, except I have replaced bulbs in my auto.

they are like any incandescent but with more severe dependency between resistance and temperature. it should work ok on 50 hz or higher ac

Do you have a VOM? Verify the current draw (instead of guessing.)

From this source:
Halogen bulbs should work normally at voltages as low as 90 percent of what they were designed for. If the bulb is in an enclosure that conserves heat and a "soft-start" device is used, it will probably work well at even lower voltages, such as 80 percent or possibly 70 percent of its rated voltage.

Like any incandescent bulb, they don't like over-voltage (perhaps even more touchy about that than normal bulbs.) So if rated to 6V, why connect to 12V? Or 15V? (10 batteries...) Have you tried the batteries again? (maybe it died acting as a fuse...)

Or is there some other circuitry connected here we don't know about? We're all thinking "bulb," but you wrote "lamp..."

I first tested with batteries to verify the bulb was still in contact. It lit.
Then I connected the (working) 12v, 1A Tx to the bulb, and it had no result.

I did have another post asking how exactly to lower voltage but keep Amperage, but I don't think I got a straight answer. That's why it's connected to 12v instead of it's rated 6v;
(I understand the difference between Voltage and Amperage, but still have the hardest time understanding why this or why that. I am a very logical person.)

After it didn't work, I tried again with the batteries (they measure 13v), and the bulb worked.

I did have another post asking how exactly to lower voltage but keep Amperage, but I don't think I got a straight answer. That's why it's connected to 12v instead of it's rated 6v;
(I understand the difference between Voltage and Amperage, but still have the hardest time understanding why this or why that. I am a very logical person.)

It's a little tough to get a handle on the concepts when starting out. See if this "light" explanation helps:

-- Amperage is the amount of current a device draws. It's an inherent characteristic of the thing itself.

Different devices (say, a small light bulb and a large DC motor) will draw a different amount of current when connected to the same power source.

-- So the amperage rating of the power source is the amount of current it can potentially supply, and is unimportant unless that limit is exceeded.

-- There's a little thing called "Ohms Law," and it has some interesting consequences:

1) In a strict sense, you cannot lower voltage but keep amperage, as you wrote.

The two are interrelated: I = E / R (current = voltage / resistance). If the voltage drops, then the current drops also (assuming the resistance is constant.)

2) Doubling the voltage doubles the current, also.

Given that, it's very surprising that your lamp has not burned out. Fresh alkaline AA batteries can easily supply 0.83 mA for a short time (although with a slight voltage drop.) You have far exceeded the ratings.

If the batteries read 13V no load, then they might be largely depleted...

You can see why we feel there's some missing information here...

@ 6v, 5W (0.83 Amps, right? Just under the AC output..)

Does this mean that the lamp originally used an adapter with 6V AC output? If so, there may be additional circuitry in the lamp housing itself...

Please repost your topic w/ info on battery arrangement ( are they parallel series) (parallel) or straight series. A halogen Light generally requires a hot filament to vaporize the gas, your 12 volt adapter outputs 1 amp, (12 watts) 5 sets of 2 batteries parallel and then connected in series could output 7.5 volts at 4 to 6 amps, 30 - 45 watts. @ 6v, 5W (0.83 Amps, right? Just under the AC output..) I don't understand this? If this is a joke I will HA-HA once and move on

This meant that on the Halogen Lamp, it specifies it requires 6 volts and 5 watts, which I calculated to equal 0.83 amps. I had it in parenthases with a question mark as a way of asking for verification from whoever could correct me. I then stated that it was just under the output of the mentioned AC adapter.

. Electrically, there's no difference between halogen bulbs and the light bulbs you find in your home. There is no starter or anything else "special" about them. They will light with AC as well as DC. It's just an incandescent bulb. See Halogen lamp at Wikipedia.
. My guess is that your wallwart is bad. Or, as with all incandescent bulbs, there is a high inrush current (resistance is lower when the filament is cold) that may be causing problems for the supply.

Do halogens require correct polarity? I don't think so... huh, maybe the halogen needs a larger staryt up current than the wallwart+cap can handle

no. they dont mind polarity / ac / dc (except maybe small difference in lamp life with longer life on ac) probably thats the point. in halogens its more severe than just incandescents resistance of cool lamp is very low. it appears allmost as a short to the power source if source can supply enough current to the cool lamp so it begins to heat - its resistance rises and it changes quickly to its normal working mode. from now on its easier to maintain the lamp hot if source too weak the lamp is stuck forever in the low resistance mode and overloads the source (allmost a short) to see if that whats happening start the lamp on batteries and then switch quicly to txformer with spdt switch - so itll not cool down while switching