# Design recommendation with LED Array Design

I am designing a LED array. I am planning to use 12 White LEDs of 0.5W each. It has 150mA forward current and Forward voltage of 3.6v.

I have a power supply of 12V,1A

Here is my Schematic for it.

Here is the Forward Current vs Forward Voltage Graph for the LED

Is supply enough for glowing the LEDs to its fullest? Is supply current and voltage is enough for the array design? Resistance value is enough to drop voltage ?

• Comments are not for extended discussion; this conversation has been moved to chat. – Dave Tweed Oct 28 '16 at 10:06

Your LED datasheet shows a figure (which I take as representative only) of $3.01\:\textrm{V}$ when operating at $150\:\textrm{mA}$. It also shows that the manufacturer bins their LEDs for voltage ratings, with a maximum $V_{fwd}$ in tenths of a volt from $2.7-3.2\:\textrm{V}$. If you are buying these from an authorized supplier (or directly), it is likely you'll be asked to tell them which bin you want and there may be a price difference from one to another. If these are coming from a supplier who isn't asking you about the bin, then it may be the case that these are unbinned and may span anywhere within the range of $2.6-3.2\:\textrm{V}$, in operation at the same current. (The $2.6\:\textrm{V}$ comes from the minimum voltage of the minimum voltage LED bin.) Looking at the curve of $V_{fwd}$ vs current, I would estimate a local effective resistance of about $1.8\:\Omega$. (Just by estimating $R=\tfrac{\Delta V}{\Delta I}$.) There are therefore two bracketing models for your LEDs:

\begin{align*} V_D&=2.33\:\textrm{V}+1.8\:\Omega\cdot I_D \\ \\ V_D&=2.93\:\textrm{V}+1.8\:\Omega\cdot I_D \end{align*}

Now just by way of example, let's use your $8.2\:\Omega$ current limiting resistor and your $12\pm 5\%\:\textrm{V}$ power supply and see what range of behavior you could expect in the more unlucky cases.

On one hand, you might get three LEDs all with $2.6\:\textrm{V}$ @ $150\:\textrm{mA}$. So in this case you have:

\begin{align*} 12\:\textrm{V}&=3\cdot\left(2.33\:\textrm{V}+1.8\:\Omega\cdot I_D\right)+8.2\:\Omega\cdot I_D \\ \\ I_D&\approx 375\:\textrm{mA} \end{align*}

On another hand, you might get three LEDs all with $3.2\:\textrm{V}$ @ $150\:\textrm{mA}$. So in this case you have:

\begin{align*} 12\:\textrm{V}&=3\cdot\left(2.93\:\textrm{V}+1.8\:\Omega\cdot I_D\right)+8.2\:\Omega\cdot I_D \\ \\ I_D&\approx 240\:\textrm{mA} \end{align*}

Plus a little for power supply variation.

One main thing that comes from this is that you can easily see just how much variation in current you might see in the worst case situation. And you most certainly will see differences in the lighting and differences in the dissipation between these two circumstances. Plus, if you should get all of them with the lowest voltage rating bin, you could expect 4 strings to sum to currents as bad as $1.5\:\textrm{A}$, which exceeds your power supply rating. Finally, all of these currents are way in excess of your supposedly designed value of $150\:\textrm{mA}$. Which means your resistor was also not calculated correctly. (That came from your use of $3.6\:\textrm{V}$, of course.)

Resistors don't make good current regulators when they have very little voltage headroom to work with. That's one conclusion.

Another is that you may actually want to consider inserting a simple, active current regulator in line with each string to improve the situation. Something like this:

simulate this circuit – Schematic created using CircuitLab

Now this isn't perfect, either. But it is better. One immediate problem I need to call out is that the 2N3906 isn't appropriate here. It might need to dissipate $750\:\textrm{mW}$ (or a lot less), so you need to select something in a larger package than a TO-92. A TO-220 would be lots better. And there are a lot of choices there. I'll leave that for your own consideration if you decide to even go this way. But keep it in mind.

The circuit I've shown sets the current with $R_1$. In this case, I'm assuming a $V_{BE}$ drop of about $800\:\textrm{mV}$, so the current should be about $140\:\textrm{mA}$ with the lower voltage LED types and about $120\:\textrm{mA}$ with the higher voltage LED types. That's a lot better regulation than with just a resistor. The reason this works is $R_2$ and $Q_1$'s collector current. The PNP BJT dumps any excess current through its collector into ground. In this way, the LEDs get a relatively stable current supply and the BJT tosses aside any excess that might otherwise have gone into them. I've set $R_2$ so that there is barely any excess current that has to be dumped by the BJT if the LED voltages are on the maximum high side. Regulation would be better still if I could set that excess current to a higher level. But if I did, you'd exceed your power supply rating with all four LED strings applied. So I am shaving things close here.

The resistors should be at least a quarter watt in size, too. But this is one possible approach to improving things if you don't know what you are getting.

There will also be a variation in the load current seen by your power supply. But I set this up so that the worst case is about $230\:\textrm{mA}$ per string. So with four strings this means it still is less than the power supply's rating.

In effect, I've added much better current regulation for each LED string. There's still some variation possible. But with the likely mix of LED values, I'm betting that you may not be able to notice the difference.

• good analysis, but without better tolerance and "overall" requirements , just a start – Tony Stewart Sunnyskyguy EE75 Oct 27 '16 at 20:34
• @TonyStewart.EEsince'75 It's all I can do for now. – jonk Oct 27 '16 at 20:35
• agreed , we need a "How to Specify ( so you can design optimum requirements) then design is perfect... webpage based on many factors cost, performance, temp rise , consistency, Ohm's law , tolerance , Rth etc. This excellent part has many options ( unspecified in question ) – Tony Stewart Sunnyskyguy EE75 Oct 27 '16 at 20:36
• @TonyStewart.EEsince'75 You do and I consider that to be wonderful! I also have sustained a full 300 hours per year as a volunteer for developmentally delayed/disabled adults and children, plus contribute tens of thousands of dollars to those charities where I supply my time. It's part of giving back for the benefits of having been born here in the US, I think. Most of what I have gained personally was by luck. People around the world work harder, are smarter, etc., than I am. So it's mostly luck... (plus some skill and talent.) I owe back almost everything I am for that luck. – jonk Oct 27 '16 at 21:01
• nice.. I was never brilliant like my colleagues, I just had to work longer hours. – Tony Stewart Sunnyskyguy EE75 Oct 27 '16 at 21:02

Let's do the math!

Vf = 3* 3,6V = 10.8V

Vr = 12V - 10.8V = 1.2V

So the voltage on the resistor will be 1.2V.

1.2V = 8.2R * i

i = 146mA

iTotal = 4*146mA = 584mA

Pled = 3.6V * 146mA = 0.5256W

Presistor = 1.2V*146mA = 0.1752W

So, going to the questions:

Is supply enough for glowing the LEDs to its fullest? Is supply current and voltage is enough for the array design?

Well, 0.584A < 1A by a good margin, so yes. (Assuming the LEDs fullest is 0.5W and they are properly cooled.)

Resistance value is enough to drop voltage?

Calculated current is 146mA, close enough to intended value. So it is an ok value. The resistor has to be at least 1/2W rated or more. With given care to dissipate heat you should be fine.

Let's check:

You want the LED current to be 0.15A and you supply the whole string with 12VDC.

First, if I were you, I would check $V_F$ vs $I_{LED}$ graphs and find $V_F$ for an LED current of 0.15A. Let's assume that it's 3.6VDC.

With 12VDC supply, of course it's best to connect 3 LEDs plus a limiter resistor in series. With $V_F=3.6V$ and $I_{LED}=0.15A$, $R_L = \frac {12 - (3 \cdot 3.6)}{0.15}=8 \Omega$. Power dissipation of $R_L$ will be $P_{R_L}=0.15^2 \cdot 8 = 0.18W$. Seems like 0.25W resistors would be enough but I personally recommend you to connect two $16\Omega/0.25W$ resistors in parallel.

Total load current will be $I_{TOT}=4\cdot 0.15 = 0.6A=600mA$, please make sure that your power supply is sufficient.

• I have a query, I will be using PWM to dim the Leds. In data sheet of LED, it has mentioned the Pulse current of 200mA. What does it mean? Should I care about it? – Embedded Geek Oct 27 '16 at 19:08
• Actually no. Since you're driving with PWM, pulse currents can be higher than forward current if you keep the on-time of PWM short enough. For this design, it's not a problem. – Rohat Kılıç Oct 27 '16 at 19:30

Let's do some better math. on choice of Vf bins and If and Pd with better focus on specs and overall requirements. (\$, deg C rise, % error and ambient temperature range n.b.!!)

• This is a good quality 500 mW White LED.

Although there are still a few unknowns such as;

• Vf bin for LED part number, (p.n.)
• heatsink & interface Thermal Resistance , these LEDs are Rjc=20'C/W
• choice of transistor
• I would assume a good choice of heatsink is < 10'C/W
• thus 30'C/W *1.5W=45'C rise suitable for 40'C ambient max.
• Design of Constant Current,
• since V drop is 1 LED or 3V the traditional 2 transistor series regulator works pretty well
• CC was tested with a supply of 10% tolerance for a few reasons
• 5% due to source or 11.4V min
• 5% due to load voltage margin on CC driver for LED's
• LED's are avail from 2.7 to 3.2 @150mA in 0.1V bins
• thus 3S string is 8.1 to 9.6V leaving 1.8Vmin for CC drop (Vce+Vbe)
• 11.4Vcc - 9.6Vled = 1.8Vmin where Vbe=0.7 and Vce=1.1
• choice of NPN requires low Vce(sat) @150mA
• since Vce(sat) has Ic/Ib=10,
• Vce(sat) must be <<0.4V to get hFE=100 @1.1V ,
• ideally Vce=2Vmin for Q2 @ 150mA for good hFE, Q1 is low current.
• choose lower Vf max selection on part numbers or tighter %Vcc

JAVA Sim.link Feel free to change values (right mouse in new tab)

LEDs= 2.9V Zener model with ESR=2 Ohms

From datasheet computed in Excel ( actually Open Office ccalc.exe )