Thread: How do I dim my 12v garden lighting?

Mate Thats all i would be doing and when you do get the tranformer if ya pick it up from a wholesaler you can usually by them with a flex and 3 pin plug already on it so it is just a matter of unplugging the existing transformer from the 240v supply diconnect the 12v side of it its only 2 screws reconnect it to the 12v side of the new trany and plug it in the dimmer will cost ya between $30- and $40 bucks you most likely will not get that from Bunnings and make sure when you do get it that it will work with whatever transformer you go with personally i would buy from a wholesaler and not bunnings as some of the people there know sweet Fanny Adams as the trasnformer is under the house just screw it to a bearer I wouldnt go to the trouble of mounting it in an enclosure besides the tranformer needs to "breathe" to dissipate heat..

After having done some research, it appears that your transformer may in fact be electronic, which is why HPM said it would shut down if it was "dimmed". It appears as though the weight of such a device is not a reliable way of telling if it's electronic or magnetic since many electronic "transformers" may use a wire wound transformer to "step down" the voltage. See the attached document. If anybody is interested, I have the full PDF document available. THIS DOCUMENT ONLY REFERS TO TRANSFORMERS THAT SUPPLY HALOGEN LIGHTING. THE DOCUMENT, IN NO WAY, REFERS TO STANDARD "HIGHER QUALITY" POWER TRANSFORMERS AS THEY ARE VERY DIFFERENT IN DESIGN.

At the moment, I'm having trouble downloading the HPM tech data for your transformer (Thailand has a woeful internet bandwidth). Disregarding this, I have managed to find the list price of a Clipsal "leading edge" dimmer, which is between $38 & $48 excl GST. Since HPM have already told you that your transformer is not dimmable, you will obviously need to buy an electronic transformer (make sure it is dimmable) & then procure the correct dimmer for it. Bear in mind that you need a minimum 150VA transformer. This could be costly (more than $30 or $40). Add the cost of the transformer to the cost of the dimmer & what is the total cost compared to buying a 240v to 9v "high quality" step-down transformer?

At the end of the day, what do you want? Do you want a lighting system that will not give you much trouble? Do you want long lamp life? If this is the case, spend the money on an appropriate electronic "transformer" with an appropriate dimmer. Remember, "pay peanuts, get monkeys". I don't believe that your problem will be easily solved by a visit to any "hardware" store because your problem is unique. You simply want to reduce the voltage to your lamps but bear in mind that Halogen lamps last the longest when they get hot enough to use the "Halogen effect".

Disregarding the fact that you have installed the lights incorrectly for your needs, if I was in your situation, I would opt for a reduced output voltage step-down magnetic transformer (the cheapest option). Since I would not be turning the lights on & off very often (I assume they will stay on for hours at a time), a "proper" power transformer at a reduced voltage output would solve my problems. Also, a slightly reduced voltage at the lamps will noticeably increase their life. Since I would not have any electronics involved, I could expect a very long life from the "proper" power transformer (not a transformer designed for use with Halogen lamps).

"Proper" power transformers are nothing like magnetic Halogen lighting transformers because they (the "proper" transformers) are much more efficient & use high quality iron.

As the previous Chief Engineer of 2 hotels in Sydney CBD, I oversaw the construction of one hotel (1998). During the construction period, I required that "standard Halogen lighting transformers" not be used. In their place, I required that each room be supplied with an appropriate "power transformer", which could supply all of the Halogen lamps (all of the rooms only had ELV Halogen lamps). At this time, "electronic" transformers were expensive. The "pros" were: energy savings; less fire risk; almost zero transformer replacements & therefore reduced maintenance costs.
The "cons" were: if the transformer failed; all the lights would fail; the cost of a new transformer.

The "cons" didn't enter the equation because I chose a reputable transformer manufacturer who I had been using for many years prior whilst I was a switchboard design engineer (power & control). BTW, I was a "working" electrical fitter mechanic for about 8 years after I finished my trade.

Thanks elkangorito. You answered most of my (unposted ) questions.

Originally Posted by Wood Borer

I would be tempted to use a suitable rectifier diode in series with the secondary on the transformer which will immediately give you half power without dissipating a whole lot of heat. They are cheap and it would be an easy experiment to try before mucking around changing transformers etc.

Assuming that it is a conventional transformer producing an AC output, I'd agree with Wood Borer. This would be the easiest and simplest approach to try first.

I don't think magnetisation of the core will be an issue as it seems that the unit is considerably oversized already (140W (VA?) to run 30W of bulbs).

Do you have a "VA" rating for the unit? With half-wave rectification, as long as the current is below 0.64 of the secondary rating there shouldn't be any problems. If you have a VA rating, divide it by the output voltage to give the secondary current rating. Multiply this figure by 0.64 to give the half-wave rectified current rating.

Originally Posted by elkangorito

I would opt for a reduced output voltage step-down magnetic transformer (the cheapest option). Since I would not be turning the lights on & off very often (I assume they will stay on for hours at a time), a "proper" power transformer at a reduced voltage output would solve my problems. Also, a slightly reduced voltage at the lamps will noticeably increase their life. Since I would not have any electronics involved, I could expect a very long life from the "proper" power transformer (not a transformer designed for use with Halogen lamps).

Thanks for the extensive research, elkangorito. This is why I love forums.

Sounds good. So if the output voltage would not be adjustable (dimmable), what do I do if that is still too bright?

Jason

Originally Posted by chrisp

Do you have a "VA" rating for the unit? With half-wave rectification, as long as the current is below 0.64 of the secondary rating there shouldn't be any problems. If you have a VA rating, divide it by the output voltage to give the secondary current rating. Multiply this figure by 0.64 to give the half-wave rectified current rating.

The tranny says 150VA. ( 150 / 12 ) x 0.64 = 8.

Originally Posted by orange squishy

The tranny says 150VA. ( 150 / 12 ) x 0.64 = 8.

You are good for 8A. At full voltage a 10W bulb will draw about 0.84A, as the voltage is lowered the current will probably increase a bit but you probably won't have any trouble driving 3 x 10W bulbs.

If I were you, I'd try something like a Dick Smith Z3336 bridge rectifier - but only use one diode of the bridge. i.e. connect an AC leg and, say, the + leg in series with one of the output wires from the transformer. The advantage of using a diode bridge is that it is easy to connect to using spade terminals, and it is also easy to mount using the centre hole. Depending upon the current the bridge may get a bit warm but it is easy to screw it to a metallic heatsink if needed (too hot to hold is too hot).

Why not just wire the bulbs in series.

Originally Posted by orange squishy

Thanks for the extensive research, elkangorito. This is why I love forums.

Sounds good. So if the output voltage would not be adjustable (dimmable), what do I do if that is still too bright?

Jason

Well Jason, I guess that this is your ultimate problem. Of course, the more expensive option of buying a dimmable electronic transformer with an appropriate dimmer unit, will allow you to get your desired results.
If you had the means, you could apply 9v to a 10W 12v lamp & see for yourself if this voltage reduction is adequate for your needs.

The rectifier sounds like a reasonable idea but I still think I wouldn't do it. Halogen lamps rely on heat to give them their longer life. Half wave rectification (DC) has 2 negative affects on Halogen lamps;

1] the large voltage reduction will not allow the lamp to get hot enough, therefore reducing its' life.

2] DC will be applied to the lamp, which can cause "polarisation" of the re-deposited tungsten particles. This will also reduce the life of the lamp.

Also, bear in mind that if you choose to use the rectifier method, the voltage reduction will be greater than 50%. This will not equate to a "light" reduction of 50%. I think that you will find a 25% voltage reduction (9v) to be adequate but as mentioned before, do a "bench test" to make sure (if you can).

Originally Posted by elkangorito

Resistors convert electrical energy into heat. If they didn't do this, they wouldn't work. Also, all of the power losses in a circuit are additive. That is, the power consumed by the resistor plus the power consumed by the lamp equals the total power loss, which is more than if the resistor was not in the circuit.

I'm sorry but I do not agree with you.

Take a simple example 12V DC source with a 12 ohm load. Power dissipated in the load = 12 Watts.

Connect a 2 Ohm resistor in series with the 12 Ohm load. Total power dissipated ~10.29 Watts.

If a 3 Ohm resistor was in series with the 12 Ohm load the total power dissipated would be 9.6 Watts.

Or using algebra total power dissipated = (V^2)/R. V is constant but as R increases the total power decreases.

Originally Posted by Wood Borer

I'm sorry but I do not agree with you.

Take a simple example 12V DC source with a 12 ohm load. Power dissipated in the load = 12 Watts.

Connect a 2 Ohm resistor in series with the 12 Ohm load. Total power dissipated ~10.29 Watts.

If a 3 Ohm resistor was in series with the 12 Ohm load the total power dissipated would be 9.6 Watts.

Or using algebra total power dissipated = (V^2)/R. V is constant but as R increases the total power decreases.

12v DC source supplying one 12 Ohm load;

I = E/R
= 12/12
= 1 Amp

Power (total) = E x I
= 12 x 1
= 12 Watts


12v DC source supplying a 12 Ohm load plus a 2 Ohm load;

Total resistance = 12 Ohms + 2 Ohms
= 14 Ohms

I = E/R
= 12/14
= 0.857 Amps

Power (total) = E x I
= 12 x 0.857
= 10.28 Watts

Power dissipated by 2 Ohm resistor;
P = I squared x R
= 0.857 squared x 2
= 1.469 Watts

Power dissipated by 12 Ohm load;
P = I squared x R
= 0.857 squared x 12
= 8.813 Watts

Total power dissipated = (E squared x Ir) + (E squared x Iload)
= 1.47 Watts + 8.81 Watts

Power total = Power (resistor) + power (load)
= 1.47 + 8.81
= 10.28 Watts.

Please bear in mind that less current is going to the load because the 2 Ohm resistor is in series with the load. This means that the load has a decreased performance with regard to its' requirement. If a series resistor is used, it will limit the current by dissipating the relevant amount of heat as can be seen above.

Also, in a series circuit, current is constant & the voltage drop is relevant to each series resistance. This is Kirchoffs Law.

Perhaps we are grossly misunderstanding each other here.

You wrote

That is, the power consumed by the resistor plus the power consumed by the lamp equals the total power loss, which is more than if the resistor was not in the circuit.

My interpretation of your statement is that more power is dissipated if a resistor was placed in series with the globe than the power dissipated with only the globe in the circuit.

Clearly this is not the case by both my examples and your examples.

There may however be truth in your statement because the resistance of the filament of globe is dependent on the current through the globe. For example, when the filament is cold, the resistance is quite low but as the current is increased the filament increases in temperature and the filament resistance increases which is common for all the filaments I am aware of ie positive temperature coefficient.

I am a bit busy at the moment but perhaps at some time in the future when I get a bit of time I'll analyse such a circuit and post it here.

We are possibly getting away bit from the original question posted though.

Wood Borer,

What I was trying to say was that if a resistor is used, it will be an additive to the total power used, even though the power output has been limited (& decreased) by the resistor. The power that the resistor uses is not productive ie it is a power loss.

Thanks for all the great advice and feedback guys. The job is now done.

elkangorito, I didn't have the means to test 9v to the lights but now that I have finished, 9v would have been too much.

I also did not want to get too complicated with components I was unfamiliar with.

I took the simple path:

1 x Osram ET-REDBACK 60VA/230-240 electronic transformer for Low V Halogens ($15)
1 x Clipsal trailing-edge dimmer ($53)

- I removed the HPM Garden Light transformer
- I wired a male 3 pin plug to the new transformer and the lights to the output
- I mounted the transformer to the bearer with a cement sheet spacer for heat protection (a very weatherproof location under the house)
- Had my sparky mate wire in the dimmer

Done. Awesome. I need to dim it down to about 20% for the right ambience. Looks great.

I went out into the rain this morning and got y'all a photo (attached).

Cheers, thanks a lot,
Jason