LED and wall adapters

[airbrush]

okay...just to satisfy my curiousity. How should i calculate for a non-regulated adapter. Say for example that i am hooking up 20 LED 3.3v in the provided layout.

 

[David Bridgen]

Hi airbrush,

there was a similar thread, running a few weeks/months back, about series l.e.d.s.

This circuit will assure a constant current through three l.e.d.s despite variations in voltage. Three is the maximum number you can use if their Vf is 3.3 and the minimum supply voltage is 12.

If you want to switch them on and of with, for instance, a 555, connect the 10k to its output instead of directly to the supply.

 

[airbrush]

so would running a series one LED with the correct resistor for 12 volts still work then with a non-regulated adapter as per the schematic?

 

[Morgen]

Yes. If you plan on having 20 LEDs connected, multiply 20 x 20mA to get 400mA. A 12V 500mA adapter will provide enough power with the 555 circuit included. I suggest 470 ohm resistors in series with the LEDs.

If you decide to use a voltage regulator, it can be made with a variable output, making the 555 brightness control circuit is unnecessary.

 

[David Bridgen]

so would running a series one LED with the correct resistor for 12 volts still work then with a non-regulated adapter as per the schematic?

The circuit is a constant-current source. The current through the l.e.d.s will be about 18mA irrespective of variations in the supply.

But perhaps this isn't necessary. Although the output voltage of an unregulated adapter will depend its load, and consequently on the current drawn from it, this wouldn't be a problem if the load is constant.
With a little trial and error you would be able arrive at a suitable value of resistance and the complication of the (many pairs of) transistors would be eliminated.

 

[airbrush]

Yes. If you plan on having 20 LEDs connected, multiply 20 x 20mA to get 400mA. A 12V 500mA adapter will provide enough power with the 555 circuit included. I suggest 470 ohm resistors in series with the LEDs.

If you decide to use a voltage regulator, it can be made with a variable output, making the 555 brightness control circuit is unnecessary.

Okay...i'll explain a little more in detail of what i'm after.
I am wanting to make a table lamp with three rows of lights...each different color and each with its own dimmer control. two of the rows will consist of 10leds each (accent colors) and one row will be either 10 or 20 white led...not sure yet? have to see how bright it is with only 10.

I've checked out the regulated adapters and I really dont want to use them...they are much larger in size than the non regulated.

All of the resistor calculators i've seen dont ask for the mA...just the voltage being used. So how would it be calculated to do as the diagram shows....one led with a resistor for each or is it just much easier to do as you have said...take the mA of the led and mulitply for the number of leds...so i'm confused how you got the resistor value for this. So in your example(20leds at 20mA) you would run 20 leds in a series with only one resistor then..is this correct?

Once again..this forum has been very helpful for a newbie such as myself. Appreciate everyones input.

 

[Dr.EM]

I would have thought it easiest to use an unregulated adapter and fit a suitable regulator IC inside the lamp? Something like these:

**broken link removed**

In this case, buy an adapter with more rated voltage than you need, and suitable current, the regulator will take care of it (don't go over 24v input though). You might need a higher current one for all those LEDs, but even they are inexpensive.

 

[airbrush]

ahh..okay...i didnt catch on to that. Yah that seems like the best route to go....that is the same thing i did for my leds on my motorcycles leds so the current was constant.

So i could just get a 12v dc adapter at 500mA....use a 7810 regulator IC. branch my power off of that to each dimmer and row of leds. So i would need to add some diodes in there as well so it would function properly....does this sound right?

So would i calculate my leds as i normally would....i didnt calculate for mA(load) for bike...just the voltage. ex. 5 2v led @ 20mA running off 10v and get the resistor value for that.
:?

 

[Morgen]

If you are going to use 30 or 40 parallel LEDs each drawing 20mA, then your DC adapter must be able to provide 600 or 800 milliamps. I would look for a cheap 9VDC or 12VDC 1 amp unregulated adapter.
Add a 470uF or more capacitor across the adapter output. Then use three variable regulators (LM317T) so that each row of LEDs is dimmable. The few components needed for the regulators can be soldered directly to the regulator leads, so a circuit board isn't needed. Just screw the TO-220 case somewhere convenient.
The resistor values are determined by subtracting the voltage across the LED (usually about 3.5V for a white LED) from the maximum voltage from the regulator. Then divide the result by the amount of LED current desired (usually 20mA). In my previous post, I divided 9 volts by 20mA to get 450 ohms. 470 ohms is the nearest standard value of 5% resistor.

 

[airbrush]

So what i want can simplified quite a bit then from the original schematic.
So the LM317T would hook up to the pots?
The TO-220 you are referring to, this is a heat sink?

If you could help me out with a schematic for this, that would be great. Until i see it all put together it doesnt make much sense to me. My electrical knowledge is pretty limited. :?

 

[airbrush]

Okay...here is a link that looks like what you are talking about:
**broken link removed**

So this is pretty much what i need then?or does it need to be adjusted for my application?
So how would the whole setup look then for the schematic?

 

[Morgen]

Simplified, yes. I think the schematic you have is for occasions when a constant voltage must be maintained. You aren't restricted by that.
The LM317T will connect to the pots, and has a tab for heatsink purposes. I included a graphic of what this part looks like. I cannot provide a complete schematic without knowing what parts you'll be using, but here is the main section:
**broken link removed**
Where you see C1 in the diagram, that can be the 470uF capacitor, if you place it within 6 inches of the regulators. If you can't place it there, then each regulator will need the 0.1uF cap at its input.
C2 is not required to power your LEDs. It can be left out. Each series LED-resistor combination will be connected as C2 is shown. Just parallel them across the output.
Three 240 ohm resistors can be soldered to pins 1 and 2 of each regulator.
The three pots will be R2 in the diagram.
Other resistors will be placed just above and/or below R2 so that the output voltage can vary between two values. In this way, the circuit will not output an overvoltage that burns up the LEDs, nor will it output an undervoltage that causes the LEDs to turn off. In between will be the dimming range.
It's best to put a switch and 1 amp slow-blow fuse on the primary of the transformer.
I think the other details depend on the specific adapter you use.

 

[airbrush]

Simplified, yes. I think the schematic you have is for occasions when a constant voltage must be maintained. You aren't restricted by that.

are you referring to the original schematic i posted or the link i just provided?
so in this link....
**broken link removed**
could i use this and just adjust the resistors accordingly? so i have a range from say 1v to 3v (assuming my leds are 3v).

 

[Morgen]

Sorry, I meant the original 555 circuit.
The circuit from your posted link is similar to what I described. Could you use it with just some resistor adjustments? Probably. Specific component values can't be determined until the adapter specs are known and the high-low LED current is known.
The output will not be anything like 1-3 volts. If the LED is shown to be 3 volts, then that is what it will be and cannot be changed. It is actually the amount of current through the LEDs that will be adjusted using the potentiometer. Do you understand why that is?
If you know your adapter output in volts and amps, and the LED voltage and LED minimum and maximum current, then that info can be used to design the regulator circuit.

You need the 470uF at the input of the regulators to smooth the DC from the adapter. The linked circuit doesn't include that. The value of the LED resistors will be determined from the maximum voltage output of the regulators. Also, the regulators might need extra heatsinking.
The 680 ohm resistor in the linked circuit determines the minimum output voltage from the regulator. When you build your circuit, leave that resistor out and just use the potentiometer. Adjust the potentiometer until the LEDs go out. Then turn it back slightly so the LEDs are dimly on. Turn off the power to the circuit and measure the resistance across the potentiometer. This will give you the value of resistor that goes in place of the 680 ohm.

 

[airbrush]

Okay..

adapter is 12vdc @ 500mA

LEDs are 3.3v max mA 30
and
1.7v max mA 30
manufacturer didnt list a min mA...does that matter? i want to be able to have them totally dim off.

the potentiometer would be a 500k logarithmic taper if that makes a difference at all?

and i would also do as you mentioned an on/off switch for the whole deal.

appreciate your help Morgen. Alot of this stuff is greek to me, but i'm slowly catching on.

 

[Morgen]

Okay... I would target a maximum LED forward current of 25mA. Easier on the LEDs, and the 5mA difference in candlepower isn't significant IMO. It also gives us a margin to work with.
With a 12V 500ma adapter, and 25mA per LED, that max number of LEDs to connect is 20. There are a few ways to power more LEDs than 20. Get a bigger adapter, use more than one adapter, or connect 2 LEDs in series with each resistor. This third choice will affect operation though, because the LEDs will turn off at a higher voltage (~6.6V instead of 3.3). And if one LED goes bad, the other goes out too, like xmas light strings.
If you want the pot to totally dim off the LEDs, you don't need a resistor where the 680 ohm is located in the linked circuit. 500k is rather large for this circuit, and I don't really know what effect that may have on the regulator. It might end up giving a lot of "play" in the pot's action on the output.
The 12V adapter and 470uF capacitor will provide about 17 volts to the input of the regulator. Figure the maximum regulated output will then be 15 volts. Using the formula from before, with 1 LED and resistor in series, the resistor needs to be about 470 ohms. For 2 LEDs and a resistor in series, the resistor needs to be about 360 ohms.
The regulators will need to dissipate 5W or more. A small heatsink for each would certainly help. They don't need to be store-bought. Large washers and/or "wings" cut from a beer can would probably work fine.

 

[airbrush]

Okay...hmm...you said they will turn "off" at a higher voltage if i create a 2led series. Dont you mean "on"..or am i missing something here? I would probably do as you said...a couple leds in series.

If bigger adapter might be in order, an 800mA...do i need to change anything if i decide to use that instead? I am trying to get the smallest size wallwart that i can, demension wise, that is why i said a 500mA one.

So i can go by the schematic on that link....remove the one resistor...what about the other resistor shown there? do i get rid of the capcitor shown there and put the one mentioned at the input of the regulator instead?

I've never used pots before..you say 500k may be alot of play...that could be fine...if I have 2 or 3 full rotations on it for dimming that is fine.

So the power would be running to three of these setups, would i need to add diodes in there somewhere as well?

 

[Morgen]

Okay...hmm...you said they will turn "off" at a higher voltage if i create a 2led series. Dont you mean "on"..or am i missing something here? I would probably do as you said...a couple leds in series.

They turn off at a higher voltage, they turn on at a higher voltage. If one LED has a forward voltage rating of 3.3 volts, it needs that amount of voltage before it will conduct current and turn on. Any less and it stays off. If 2 of those LEDs are in series, the forward voltage required becomes 6.6 volts. >6.6V is on, <6.6 is off.

If bigger adapter might be in order, an 800mA...do i need to change anything if i decide to use that instead? I am trying to get the smallest size wallwart that i can, demension wise, that is why i said a 500mA one.

Nothing needs to be changed. More amperage from the adapter means you won't need to connect LEDs in series, which is generally a poor circuit design. You could get by with two in series in a noncritical application though.

So i can go by the schematic on that link....remove the one resistor...what about the other resistor shown there? do i get rid of the capcitor shown there and put the one mentioned at the input of the regulator instead?

I suggest you use the datasheet circuit and comments that I posted, and just use the other schematic as a reference.

I've never used pots before..you say 500k may be alot of play...that could be fine...if I have 2 or 3 full rotations on it for dimming that is fine.

Multi-turn pots? Most common pots only have a rotation of 270 degrees or so.

So the power would be running to three of these setups, would i need to add diodes in there somewhere as well?

Yes, three regulator circuits, each row of LEDs with its own dimmer. No diodes needed.

 

[airbrush]

okay...so based off of your suggestions...it should look something like this then? The hightlighted resistor..so this controls the max output then? how is this calculated? Am I missing anything from this?
**broken link removed**

thanks again for your patience and help

 

[Morgen]

That's the circuit. Not much to it.
The highlighted resistor is part of a 1.25V voltage reference for the regulator, so it actually sets the minimum output. Since that resistance value is constant and the 1.25V is constant, the current flow is also constant. If you look at the diagram I posted, you'll see that the ouput is taken across that resistor and the potentiometer. So if 1.25V is across R1, then the remaining voltage will be across R2, the pot. The current through the total resistance is constant, so changing the resistance of R2 with a pot results in a varying voltage for the output. The calculation is shown in that diagram. The Iadj current is only about 100uA and can be ignored in nearly every case. So you have Vout = 1.25(1 + R2/R1). Insert 500k for R2 and you see the problem with that value when R1=240. The circuit will never provide that voltage. It might be possible to use a higher value for R1, but whether you could use a 10k or 20k I am unsure. It probably begins to affect the internal circuitry of the regulator. Easier to just get smaller pots if there is a problem.