LED strings problem

[polashd]

Dear all

I made LED strings (09 strings) for my aquarium lighting.

I used a transformer (220v AC to 12vX2 center taped 1Amp rating).

After rectification and filtering (with 1000uf 50v cap) I get 33vDC.

It is then input to 9 strings each of Lm317 (constant current of around 22ma using 56 ohm resistor), array of 8 to 10 LEDs (26v to 27v @22ma), 0.33uf ceramic (or..) cap at input pin, output cap 10uf(electrolytic) 50v.


Individually each string works fine. But as I keep adding strings parallely, current flow through each strings droops (as min as 10ma). Even the input DC also drops several volts from initial 33v.

Transformer capacity = 12v x 1amp=24v x 0.5amp=12watt
Each strings power consumption=33v x 0.022ma=0.726watt
09 string power draw: 0.726watt x9=6.534watt (about half of transformer capacity)
09 string current draw: 22ma x 9=0.198ma

Then what is the problem, where am I making mistake.

Please help.

 

[Pommie]

The 33V is the unloaded peak to peak voltage. When you draw 1A expect that to fall significantly. If you reduce the number of LEDs in each string it should work better.

Mike.

 

[polashd]

Dear mike
thnks for your reply
I did that, reduced 1 to 3 LEDs (shorted) and average amp increased slightly (from 12/13 to 15/16ma).
inst. of 33v DC what is the workable voltage I should expect?
As the Power rating of input transformer is much higher then required (drawn by all strings) why its not supplying it?
would u please explain.
thanks

 

[KeepItSimpleStupid]

LED's are current mode type of devices. They don't all have the same Vf or forward voltage drop.

As the Power rating of input transformer is much higher then required (drawn by all strings) why its not supplying it?

Not sure why this comes up from time to time, but think of it this way:
A car battery can supply hundreds of Amps, why doesn't a 12 Watt bulb draw 400 Amps?
Your house might have 200 Amps of 240 available, and your 1000 W toaster takes what it needs; 1000 W.

The load determines the current draw. The source (transformer) must be rated > the load or must be able to supply the load.

 

[Pommie]

Connect all your strings up and see what voltage you get instead of 33V and work from that. Note that when you increase the current the voltage will drop even further. As you have power to spare then make your string voltage much lower so you get the right current. Note also that the LM317 and resistor can drop nearly 4V at times.

Mike.

 

[audioguru]

The transformer provides 24V at 1A = 24VA.
The LEDs operate from the rectified supply of (24V x 1.414)= 33V.
Then the current available for the LEDs is 24VA/33V= 0.73A.

Nobody said what is the color or voltage of each LED. 3.5V white ones? 1.8V red ones?
Eight LEDs at 3.5V each total 28V. The LM317 needs about 2V plus 1.25V for its current-setting resistor. The rectifier bridge reduces the voltage available to 31V so the circuit is on the edge of not having enough voltage. It certainly will not work with ten 3.5V LEDs.

 

[Diver300]

The transformer is rated at 24 V at full current, when the load is resistive. The peak voltage is only at 1.4142 times that for a really short time. In fact, the voltage is only more than 30 V for around 1/3 of the time, so if you try to take all your power only when the voltage is more than 30 V, you will get much more voltage drop and the peak voltage will be reduced.

Of course, a lot of modern electronic appliances work like that, and take most of their current at the voltage peak. Almost any appliance that runs on DC internally will take little current unless the voltage is near the peak, although there are now some rules to make larger electronic appliances do that to a lesser extent.

The result is that the mains voltage already has the peak voltage reduced, so the 24 V output will have a lower peak than you would calculate. You then need to subtract:-

Diode drop
LM317 minimum voltage difference
LM317 reference voltage
LED tolerances

and the one that seems to have been forgotten, capacitor ripple voltage.

You said you are using 1000 μF, and are trying to run 9 strings at 22 mA so 198 mA. If the mains is at 50 Hz (which I am assuming as it is 220 V) then the capacitor has to support the LEDs for around half a cycle or 10 ms. That means that the charge is 0.00198 C, so the capacitor which has a capacitance of 0.001 F will reduce by 1.98 V during that time

If we add that lot up, we get:-

30 V peak (assuming we are going to be happy with current only flowing 1/3rd of the time)
1.6 V drop across 2 diodes
2 V minimum LM317 difference
1.25 V LM317 reference
1 V LED tolerance (10 at 0.1 V each, just a guess)
2 V capacitor ripple

which comes to 22.15 V.

So it is no surprise that your 26 - 27 V string isn't working as you want it to.

Now I haven't measured anything before posting this, apart from noticing that the mains is tending to be clipped. You might well find that a few of these items aren't as bad as I am suggesting. However, you need to do calculations like this to get a circuit to work in the real world.

There is also the supply voltage tolerance. Mains in Europe is generally 230 V ±6%, so can be 216 V, so that takes another 2% off the 30 V for worst-case conditions.

You can reduce the diode drop by using Schottky diodes. A larger capacitor will give less ripple. There are other current regulator circuits that have lower voltage drops.

Also, a 2 x 12 V, 1 A transformer would be expected to run at around 13.8 V with no load (https://www.farnell.com/datasheets/1829278.pdf) so I would expect the open circuit voltage to be around 38 V, as the diode drop will only be around 1 V with no load. If you are getting 33 V at no load, the transformer voltage will almost certainly be quite a bit lower than 12 V per winding when loaded, and the transformer is not meeting specification.