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LEDs and powering 'em

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Raul

New Member
I am trying to come to terms with LEDs and power supplies.
Goals:
To learn how to build and to power LED arrays & do it.
To learn about the relationship between power supplies and LED power consumption.
To learn how to build my own power supply.
There is an Wikihow article on turning a PC power supply into a lab supply here:
How to Convert a Computer ATX Power Supply to a Lab Power Supply (with video) - wikiHow


I have several low voltage Track Lighting transformers left over from a renovation. The produce 24VAC. I think I’d rather make my own but the temptation to use these is high.

I'm still trying to figure out how to do what I want.
-============================================

Powering LEDs
Take this:
R = (VS - VL) / I
where
R = the resistor
VS is the cumulative voltage demand of the LEDS in series
I = the LED draw in Millliamps (mA)


Example:
A series of 6 LEDs (assume 2 VDC for each) I can take a 120VAC line voltage rectify it to DC and end up with: 120 - 6 / 0.02 = 5850 Ohms resistance
Does that make sense? I need a resister of 5850 Ohms??

I just know I am only a quarter right and there’s more I need to know.

And I see different way of connecting a resistor to an LED.
Some people are putting the resistor in the Negative terminal and others on the Positive.
Examples:
http://wolfstone.halloweenhost.com/Lighting/litlpo_PoweredByFullWave.png
http://www.theledlight.com/img-tech/series101.jpg
Is one more correct than the other?
Does it matter which terminal is used for the resistor?

-=========================================================
Then on to the power supply.
If I understand what I’ve read correctly, I need no less than the voltage the LEDS will draw down. So a 120VDC (or AC) line will not drive more than 51 LEDs?
Does that make sense?
I can only drive 51 Two Volt LEDs on a 120 Volt supply line?
That makes no sense.

I've seen systems where way more than 50 LEDS are running in a plug and play bank of hundreds of LEDs and the thing is sold to plug into a 120 VAC line. There isn’t the physical space in them for banks of capacitors. So I’m missing something.

What am I doing wrong??

This tells me that the statement about matching power supply to LED consumption is error - - - or at least how I have understood it is error. :
This is 72 white (2 volt) LEDs on a board
<B>LED1188</B><BR>880nm/50 Degree<BR>Assembled And Tested - LED1188
The Description says:
This assembled and tested unit 72-880nm LEDs with a 50 degree viewing angle. The LEDs in this unit are capable of drawing 100mA. each. The Bigger IR ILLUMINATOR kit is designed to operate between 12 and 13.8Vdc, with maximum efficiency at 13.2Vdc. IR Illuminator draws between 600-800mA. Units can be daisy chained. At 24 feet, the beam is approximately 8 feet in diameter. PC board measures 3.7"x1.95".

They says it's tested with this itty bitty $4.00 power supply
<B>PWR1171</B><BR>Wall Wart for IR Illuminator - PWR1171
That tells me I haven’t yet got a handle on LED power consumption or how to feed them.

Input would be appreciated.
 

Diver300

Well-Known Member
Most Helpful Member
I am trying to come to terms with LEDs and power supplies.
Goals:
To learn how to build and to power LED arrays & do it.
To learn about the relationship between power supplies and LED power consumption.
To learn how to build my own power supply.
There is an Wikihow article on turning a PC power supply into a lab supply here:
How to Convert a Computer ATX Power Supply to a Lab Power Supply (with video) - wikiHow


I have several low voltage Track Lighting transformers left over from a renovation. The produce 24VAC. I think I’d rather make my own but the temptation to use these is high.

I'm still trying to figure out how to do what I want.
-============================================

Powering LEDs
Take this:
R = (VS - VL) / I
where
R = the resistor
VS is the cumulative voltage demand of the LEDS in series
I = the LED draw in Millliamps (mA)


Example:
A series of 6 LEDs (assume 2 VDC for each) I can take a 120VAC line voltage rectify it to DC and end up with: 120 - 6 / 0.02 = 5850 Ohms resistance
Does that make sense? I need a resister of 5850 Ohms??

That is the right idea but you have a few details wrong.

120 VAC has a peak voltage of sqrt(2) * 120 = 170 V
You would want a capacitor to smooth the rectified voltage.
6 LEDs in series would be 12 V
The resistor you need for that is (170 - 12)/0.02 = 7900 Ω.

You could use a standard value of 8.2 kΩ (8200 Ω) and get just a bit less current. You need a power rating of over 3 W for the resistor.

However, that is not safe as it uses 120 V mains without any isolation, and inefficient with less than 10% of the power going to the LEDs.

You should use the 24 V output of the transformers that you have. Then the resistor is (24*sqrt(2)-12)/.02 = 1097 ohms, so a 1 kΩ, 1/2 W resistor would do, and give you just over 20 mA, or a 1.2 kΩ 1/2 W resistor would give a bit less than 20 mA. 1 W resistors would be better as they wouldn't get as hot.

I just know I am only a quarter right and there’s more I need to know.

And I see different way of connecting a resistor to an LED.
Some people are putting the resistor in the Negative terminal and others on the Positive.
Examples:
http://wolfstone.halloweenhost.com/Lighting/litlpo_PoweredByFullWave.png
http://www.theledlight.com/img-tech/series101.jpg
Is one more correct than the other?
Does it matter which terminal is used for the resistor?

Both of those are OK. As I said, you should be careful of circuits that use the mains directly.

You can put the resistor on either +ve or -ve.

-=========================================================
Then on to the power supply.
If I understand what I’ve read correctly, I need no less than the voltage the LEDS will draw down. So a 120VDC (or AC) line will not drive more than 51 LEDs?
Does that make sense?
I can only drive 51 Two Volt LEDs on a 120 Volt supply line?
That makes no sense.

I've seen systems where way more than 50 LEDS are running in a plug and play bank of hundreds of LEDs and the thing is sold to plug into a 120 VAC line. There isn’t the physical space in them for banks of capacitors. So I’m missing something.

What am I doing wrong??
You are correct in thinking that there is a limit, but it is a limit for LEDs in series. Because of the sqrt(2) factor it would be more than 51.

However, what happens is that you have several LEDs in series, with a resistor to control the current. When you want more LEDS, you have a separate group with their own resistor, and put that in parallel with the first.

I don't know where you get the idea that lots of capacitors are needed.

This tells me that the statement about matching power supply to LED consumption is error - - - or at least how I have understood it is error. :
This is 72 white (2 volt) LEDs on a board
<B>LED1188</B><BR>880nm/50 Degree<BR>Assembled And Tested - LED1188
The Description says:
This assembled and tested unit 72-880nm LEDs with a 50 degree viewing angle. The LEDs in this unit are capable of drawing 100mA. each. The Bigger IR ILLUMINATOR kit is designed to operate between 12 and 13.8Vdc, with maximum efficiency at 13.2Vdc. IR Illuminator draws between 600-800mA. Units can be daisy chained. At 24 feet, the beam is approximately 8 feet in diameter. PC board measures 3.7"x1.95".

They says it's tested with this itty bitty $4.00 power supply
<B>PWR1171</B><BR>Wall Wart for IR Illuminator - PWR1171
That tells me I haven’t yet got a handle on LED power consumption or how to feed them.

Input would be appreciated.

They have 6 LEDs in series. As they are IR leds, their voltage will be a bit less than 2 V each, so 6 give a bit less than 12 V. There is a resistor for the 6 LEDs. The 6 LEDs take about 50 mA, so they are not running at their maximum, but that's fine.

Then there are 12 such groups of 6 LEDs, in parallel. Each group runs from 12 V, each takes 50 mA, so the parallel combination is still 12 V, and is 12 (groups) * 50 mA = 600 mA.

That power supply looks about right for 12 V and 600 mA
 
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Raul

New Member
Thank you. That was enormously helpful. Got me past a couple of barriers.

As to the Cap. I know about motor start and run Caps but I rather suspect that this will be of a different sort.
 

Sceadwian

Banned
The capacitor (if any) you chose will be based on your own requirements. Aside from smoothing rectified voltage they'll also gobble up small transients if chosen properly. LED's are prone to destruction from even 'low' energy transients. I've personally powered an 80 LED array in anti parallel array right off a wall socket with nothing else but the LED's simply by carefully choosing the number of LEDs to put in each anti parallel series string to meet maximum allowed voltage levels and average current.
 

Raul

New Member
Forward Voltage drop.
I've been trying to get a handle on the expression.
What's it mean by forward?
What is meant by drop?
What if there's more than one LED in the string running off one resistor?


If an LED is rated at 2 volts and there are 6 of them in a parallel the "Forward Voltage Drop" across the 6 LEDs is a constant at the 2 volt factory rating of the LED. REALLY? This can't be. IT has to be some cumulative number subsequent to the drop across all the LEDs in any given parallel string.


I think that it is the voltage delta between anode and cathode of the LED that arises when a current is passed across it.
This means that there is a physical gap in the LED across which voltage must pass. In bridging the gap some voltage is lost (the fairies take it?) and that is the drop.

I found this the forum won't let me use sub and super script

ID = IS (eqVD/NkT -1)
Where
ID is the Diode Current in amps ( my case 15mA - 20mA)
IS Saturation current in Amps (1*10-12 amps)
e = 2.718281828 (Eueler’s constant)
q = the electron charge (1.6*10-19 coulombs)
VD = voltage across the LED in volts
N = non ideality or emission coefficient (a number between 1 & 2)
K = 1.38 * 10-23 (Boltzmann’s constant)
T = junction temp in Kelvin
Here is the JPEG image with the correct script
Voltage Drop formula.png

The formula is more like Greek to me than Greek was in College.

If all I wanted to do was drive a few LEDs I'd lust start wiring and playing with resistors but I want a series of parallel strings of LEDs with totals in the hundreds. I'm looking to build a light source.

arrrgggghhh

And to make matters less clear I found this
https://www.oksolar.com/led/led_color_chart.htm
On the chart there is a column for voltage drop which is reasonably identical to the rated voltage consumed by any given LED.

Am I over thinking this. Is "forward Voltage Drop " really nothing more than the manufacturer's rated voltage for any given LED and the total calculated drop is gotten by simply adding them up?
SIX, 2Volt LEDs = Forward Voltage Drop of 12 ? That make sense?
 
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MikeMl

Well-Known Member
Most Helpful Member
In a series circuit, it matters not what order things are connected; the current is the same in all of them.

If you can put n Leds and a suitable current-limiting resistor in a series-string, you can connect as many strings in parallel as you want...
 

RODALCO

Well-Known Member
Keep it simple.
As long the led current is below 20 mA you can not go wrong.

Supply voltage, subtract voltage across leds / divide by 20 mA yields the series resistor.

Say led (green) 2.2 Volts

24 Volts ac and 6 leds in series.

24-13.2 / 0.02 = 540 ohms. use 620 or 680 R.

On higher voltages 120 or 240 Volts and if 50 or 60 Hz flicker is not an issue then put an 1N4007 diode in series which will reduce the power dissipation in the series R.

120 - 13.2 / 0.02 = 5340 ohms. use 5600 or 6800 ohms.

Don't bother with the peak value as the leds will have an on - off ratio at the frequency and can handle the 22-25 mA for a short period with no problems.

The resistor power rating

P=I²R

0.02²*5600 = 2.24 Watts

Your led current is less especially when the 1 N4007 diode is in the circuit as well.
 

Diver300

Well-Known Member
Most Helpful Member
If an LED is rated at 2 volts and there are 6 of them in a parallel the "Forward Voltage Drop" across the 6 LEDs is a constant at the 2 volt factory rating of the LED. REALLY? This can't be. IT has to be some cumulative number subsequent to the drop across all the LEDs in any given parallel string.

In a parallel circuit, the voltage is the same for each item. The current is the sum of all the currents taken.

House wiring is parallel circuits. Every appliance is at 230 V (120V in USA). The current taken is all of the currents added together.

However, you shouldn't run LEDs in parallel, because their voltages vary. Although they say 2 Volts, you can get some that will be higher, and some that will be less. Also, one thing that equation shows is that the hotter an LED becomes, the bigger the current if the voltage is fixed. So if you fix the voltage at 2 V, across 6 LEDs, and one gets hot, it takes more current, gets hotter, takes even more current and so on. This is called a thermal runaway and ends with a dead LED.

Appliances don't do that. If a lamp gets hotter, it will take less current.

You can run several LEDs in series, and that keeps the current in each one the same. A resistor controls that current. If one LED gets hotter, because the current is fixed, the voltage reduces, so it gets colder, so thermal runaway is impossible.
 

Raul

New Member
Power supply schematic

Searching out a way to cool my beer using Peltier effect I stumbled across this
https://www.electro-tech-online.com/custompdfs/2009/12/uc3638.pdf

Yah I brew my own and find that close control over temperature during the ferment can produce exceptional results even such that I can cause the yeasts to produce citrus like notes of flavor just by controlling temperature.

Anyway This PDF seemed a good give back.
 
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