# Ultra bright LED driver circuit

#### MLW

Joined Jun 28, 2011
4
I need help with an ultra bright LED driver circuit. My goal is to have a round PCB with 2, 4 or 6 LEDs of the 5mm size on one board. This board will be placed in the base of a 6 volt spot light reflector. A second board assembly will have the components placed on the other side of the reflector with wires passing through the old bulb mounting hole.

My knowledge of electronics is electro/mechanical design 30+ years. I have laid out PCBs by hand and CAD, and can bread board solder per IPC-610, class 3. I can use a meter to test and or adjust components, but my understanding of the workings of components is very basic.

I need someone to review the LED specification below and suggest and low cost circuit I can build. The LED is green @ 70,000MCD the battery in 6 VDC. As far as the qty. of LED to design in, the more the better until the battery dose not last or the LED burn out.

THANK YOU in advance for any time you spend helping me on this.

LED  5mm Super power utilize a built-in heat sink construction that enables them to be driven up to 100mA.

Color  Green
DC Forward Current  mA 100
DC Forward Voltage V 3.0  4.2
DC Reverse Current uA 10
DC Reverse Voltage  V 5
Power Dissipation mW 500
Dom. Wavelenght nm 525
Operating Temp.  EC -30 - +80

#### #12

Joined Nov 30, 2010
18,224
If you simply put a 30 ohm, 1/2 watt resistor in series with each LED, it should work. One uncertainty is the 3v to 4.2 volt specification. If the voltage needed for the LED goes up to 4.2 volts as the current increases to 100ma, then the answer is 18 ohms, 1/2 watt.

Leftover volts divided by current is the resistance needed.
Wattage of the resistor is (leftover volts squared) divided by resistance.

A constant current circuit would get the best brightness, but it's hard to fit one into 1.8 volts of operating room.

Then there are pulsed constant current supplies.

How fancy do you want to get?

edit: This is a page about pulsed current supplies that can be modified to this purpose, I think.

http://www.romanblack.com/smps/a05.htm

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#### iONic

Joined Nov 16, 2007
1,662
The battery specs are missing for a better assessment of the longevity of on-time. What sort of battery are you intending to use or are you considering a wall adapter.

#### MLW

Joined Jun 28, 2011
4
The spotlight I want to modify has a 6 Volt 3Ah SLA rechargeable battery. Don't know how to tell how long the battery will last using the above resistors. Using the resistors seems so simple.

#### ErnieM

Joined Apr 24, 2011
8,378
I agree that a constant current drive (or sink) would be the way to go; 1.8 volts should be plenty of headroom to squeeze one in. You don't want too much headroom cause you have to burn off all that excess voltage as heat.

The longest run time would require a switching regulator to control the constant current.

#### Wendy

Joined Mar 24, 2008
23,460
There have been several SMPS current regulator designs on this site, though I am not sure how they have performed. They are by far the most efficient way to power LEDs, since the conversion efficiency beats a simple resistor hands down.

Qty 4 LEDs, Vf ≈ 3.0 to 4.2V (you need to pin this down by measuring the actual Vf).

Power supply, 6V @ 100ma per LED.

#### iONic

Joined Nov 16, 2007
1,662
The spotlight I want to modify has a 6 Volt 3Ah SLA rechargeable battery. Don't know how to tell how long the battery will last using the above resistors. Using the resistors seems so simple.
With this battery and you LED's you will get aprox.

6V 3AH = ideally 1A for 3 hours, .5A for 6 hours, and .4A (your circuit) for 7.5 Hours, but I would not count on that... perhaps 6.5H if the battery is brand new.

#### MLW

Joined Jun 28, 2011
4
I looked at the link provided by #12 and I will not have a problem building that circuit on a board for my project. Do remember I am not electric I and mechanical therefore I need to ask about the circuit. The circuit uses a 12 v battery and drives 2 LEDs, it must be to simple to just change the battery voltage and number of LEDs. What changes will need to be made?

#### SgtWookie

Joined Jul 17, 2007
22,230
The Roman Black circuit is a buck-type switcher. As designed, it can only go from higher voltages to lower voltages.

What you really need is more of a boost-type switching constant current supply.

Have a look at the attached 555 timer based boost constant current supply and simulation.

The cyan trace on the plot shows simulated battery voltage from 7v down to 5.7v, which would be your battery voltages over a range of being on a charger, down to where it is completely discharged. The green trace is the current through the LEDs. Note that it remains fairly constant despite the decrease in battery voltage. The red trace shows the control voltage input to the 555 timer; as the current through the LEDs (and therefore R4) decreases, the CV increases, which effectively changes the duty cycle and frequency of the 555 timer.

R1, R2, and C1 control the base frequency and duty cycle of the 555 timer. With the values shown, it's operating in the 120kHz range.

L1, Q1, R3, and D1 are the basic elements of the boost converter. When the output of the 555 timer goes high, Q1 sinks current from the inductor L1. After a period of time, Q1 turns off; but current is still flowing through L1. With Q1 turned off, D1 is the only path for that current, so the charge goes into C3. At some point when C3 is charged up enough, the LEDs start conducting.

The inductor is used in discontinuous mode; that is, the current in L1 builds for a time while Q1 is conducting, and then falls to zero when Q1 is cut off. C3 causes the current flow through the LEDs to be continuous rather than discontinuous.

C2's function is not obvious, but it provides for a "soft start" condition. Without C2, the initial current through the LEDs would be very high, and likely damage them.

R4 is the key to current regulation. Q2 will start conducting when current starts flowing through the base; this occurs when its' base-emitter voltage (typically abbreviated as Vbe) reaches somewhere around 0.62v, or 620mV. Current flowing through R4 causes a voltage to develop across R4 (a voltage drop, if you will); 100mA flowing through a 6.2 Ohm resistor will cause a voltage drop of 0.62 Volts (see Ohm's Law for how this works).
At that point, current flowing via R5 to Q2's base causes Q2 to start conducting, decreasing the CV input on the 555, which decreases the current flow through L1.

The LEDs I used in this simulation have a nominal Vf of 3v; so a total of ~9v. The circuit as shown probably will not supply enough current to operate more than two of your LEDs in series, and definitely not three in series - the major limiting factor being Q1. Replacing Q1 with a ZTX849 NPN transistor would allow operation of more LEDs in series, as the ZTX849 transistor is capable of far more collector current and has much higher gain than the 2N4401.

As shown, the circuit is roughly 76% efficient which isn't great, but is much more efficient than using simple current limiting resistors.

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#### Wendy

Joined Mar 24, 2008
23,460
That is pretty close to what I was thinking about. How much current out of the battery do you see Wook?

#### SgtWookie

Joined Jul 17, 2007
22,230
Current from the battery depends upon the battery voltage, circuit efficiency (which I established at ~76%) and Watts used in the LEDs.

So for example, if there are 3 LEDs in series with a Vf of 4v and 100mA current flowing through them, and the battery has discharged to 6v, efficiency is 76%, then:
Ibattery ~= (4v*3*100mA)/76%/6v = (12v*0.1)/.76/6 = 263mA. As the battery voltage decreases, the battery current will increase.

#### Audioguru

Joined Dec 20, 2007
11,248
The LEDs have a very high brightness rating that is much higher than ordinary bright LEDs operating at 20mA so I betcha those ultra-bright LEDs have a very narrow focussed beam something like a laser pointer. You might not see them if they don't shine directly at you. But if they do shine at you then they will blind you.