Joule thief

wayneh

Joined Sep 9, 2010
17,496
If I use a12v 2a dc power supply would that be sufficient to power the circuit of a Joule thief? According to my calculations the 120 leds are drawing a total of 12w, so a 12v 2a output should easily cover it. I just thought that the Joule thief would be more economical.
A 24W rated supply should be fine for a 12W load, even allowing for a little loss in the converter.

Chances are you don't need a Joule thief per se if you use a 12V supply. A Joule thief is for producing spikes in voltage to flash an LED (faster than you can see) when the power supply would not otherwise be able to light the LED. So a 3.5V LED can run off a single AA battery, for instance.

In your case, you need to match up your power supply to whatever the LEDs need to operate. That's a job for a DC-DC converter. Buck to reduce voltage (12V to 6V), boost to increase voltage. Combined buck/boost to allow for a wide range of input and outputs. Aa Joule thief is a type of boost converter but usually limited to small loads like a single LED.
 

ian field

Joined Oct 27, 2012
6,536
Sorry, I'm not making myself clear. I need a Joule thief circuit so I can utilize the leds. If I can't run it off a rundown battery then I want to run it off a transformer/power supply at 4.5 volts dc or anything up to 12 volts dc.
The term; "joule thief" into Google should have got you numerous examples of simple blocking oscillator/inverters.
 

takao21203

Joined Apr 28, 2012
3,702
its not even clear what voltage the LEDs need, or whats the circuit. Joule Thief is normally just a very simple circuit, the correct name is blocking oscillators, I have an old PDF where they are discussed, they were used in the 1960s. And still in use for small fluoro lamps or CCFL inverters, altough its often more kind of a flyback inverter with transformer. Often self-driven.

Yes a joule thief will work quite a long while from a discarded D cell, no longer fit for its intended use.

A lead battery with 4.5v is just faulty and wont hold much charge.
 

MikeML

Joined Oct 2, 2009
5,444
Joule thief: a gadget to get 98% of the usable energy out of a primary (non-rechargeable) disposable battery, instead of 90 to 95%...

Not to be used with a rechargeable battery, which has a depth-of-discharge limit such that a Joule thief is useless...

A Joule thief will permanently damage a rechargeable battery!
 
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Thread Starter

boatsman

Joined Jan 17, 2008
187
A 24W rated supply should be fine for a 12W load, even allowing for a little loss in the converter.

Chances are you don't need a Joule thief per se if you use a 12V supply. A Joule thief is for producing spikes in voltage to flash an LED (faster than you can see) when the power supply would not otherwise be able to light the LED. So a 3.5V LED can run off a single AA battery, for instance.

In your case, you need to match up your power supply to whatever the LEDs need to operate. That's a job for a DC-DC converter. Buck to reduce voltage (12V to 6V), boost to increase voltage. Combined buck/boost to allow for a wide range of input and outputs. Aa Joule thief is a type of boost converter but usually limited to small loads like a single LED.
Thanks wayneh for the information. Is there a simple way I can build a DC-DC converter to change the 12v to 6v?
 

wayneh

Joined Sep 9, 2010
17,496
Is there a simple way I can build a DC-DC converter to change the 12v to 6v?
Simple is a relative term. I would say no, and for sure it's cheaper and easier to just buy one. A voltage regulator is simple, such an LM317 based circuit, but that is inefficient.

Will your LEDs run off 5V? You can buy those 12V to USB adapters very cheaply and that would be perfect. Only 5V though.
 

takao21203

Joined Apr 28, 2012
3,702
a dead lead battery wont light a thing, everybody who experienced it knows it.

If it has 4.5v its dead.

Verdict?

Refilling a maintainence free lead battery with water

 

Thread Starter

boatsman

Joined Jan 17, 2008
187
Yes. Don't buy a 12 volt supply when you want a 6 volt supply.

Is this Billy Mayo?
I already have a power supply; one source is the plug in power supply for an external hard drive (12VDC 2A) the other is a small electronic power supply rated at 12 dc 20 -50W . So how can I alter the voltage to 6VDC without losing any power. Should I just place resistors in series? It just seems a waste unless I can find a more practical approach
 

wayneh

Joined Sep 9, 2010
17,496
So how can I alter the voltage to 6VDC without losing any power. Should I just place resistors in series?
Resistors lose power, as will any "linear" voltage regulator, which is essentially a smart resistor. You need a modern DC-DC converter as noted. I'd buy one - they're very cheap. As I said before, I'd try 5V first. If that works, there are many more options open.
 

Thread Starter

boatsman

Joined Jan 17, 2008
187
Sorry, I haven't explained myself clearly. From what I've learned in this forum it's not feasible to build a substitute power supply for the leds. My mention of using a 12vDC 2A power supply was to use the led assembly as a lamp run directly via a power supply from 220v mains, completely cutting out the middleman - the battery charging circuit. I want to connect the power supply directly to the leds via a 2 or three way switch so as to obtain various degrees of illumination. Are there any snags to my idea?
 

ian field

Joined Oct 27, 2012
6,536
its not cclear how many volts your LEDs need or what current
Since the OP stated the LEDs are in an emergency lighting unit, its probably safe to assume white LEDs, voltage in the vicinity of 3.4V. Bog-standard 5mm white LEDs start at 25mA just like a lot of other LEDs, but higher power types are starting to show up in ever increasing numbers.
 

ian field

Joined Oct 27, 2012
6,536
Sorry, I haven't explained myself clearly. From what I've learned in this forum it's not feasible to build a substitute power supply for the leds. My mention of using a 12vDC 2A power supply was to use the led assembly as a lamp run directly via a power supply from 220v mains, completely cutting out the middleman - the battery charging circuit. I want to connect the power supply directly to the leds via a 2 or three way switch so as to obtain various degrees of illumination. Are there any snags to my idea?
The absolute simplest way is to use a "wattles dropper" - that is a capacitor in series with the mains, it relies on Xc being very high relative to Rl so current doesn't change so much when other parameters vary.

You can't use just any old capacitor - it has to be the type rated for connection directly across the mains, such as the mains input filters in a typical SMPSU.

Secondly; you get large current peaks around the zero crossing region of the AC waveform, you have to include a series resistor to limit these current peaks.

The last emergency lighting unit I stripped for parts had a less than ideal layout for the LEDs - they were grouped in series pairs, and all the series pairs were in parallel - you need to take things like that into account when working out the capacitor.

A guide to capacitance value; years ago, a British TV manufacturer introduced the "wattles dropper" for the valve heater chain - the current required was 300mA and the capacitor used was 4.3uF, you can scale this to suit whatever current your LEDs need.
 

wayneh

Joined Sep 9, 2010
17,496
You can't use just any old capacitor - it has to be the type rated for connection directly across the mains, such as the mains input filters in a typical SMPSU.
Can you point to an example at, say, Mouser? I have some big, high voltage ceramics pulled from an old monitor or TV, and I was thinking of using one for this type of application. What spec do I need to look at?
 

bug13

Joined Feb 13, 2012
2,002
So from what I can understand is:

  • OP has a 12V 2A power supply
  • He want to run some white LEDs of the 12V supply.

How about re-range the white LEDs in to 3 in in series, then parallel them up to (in theory) 100 set in parallel. That should be a lot safer (and cheaper?) than messing around with main.

Well, unless you want to mess around with main, and you know what you are doing.
(Assuming 3.5V Vdrop for whilte LEDs)

edit: the placement of 20mA is not right, it should be 20mA for one single parallel only.
Capture.PNG
 
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Wendy

Joined Mar 24, 2008
23,415
I will second MrChips comment. Wattless droppers, which is what Ian Field was refering to, is very much against our terms of service.
 
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