Help! creating 1 watt LED driver circuit...

[Clarkdale44]

Hello

I want to create led driver circuit as it consumes less power than cfl's, i tried making it few months back and ended up damaging the led, so i wanna do this proper way this time. I want to power it from a 12 v battery. I will be using 4 led's.
I want it simple and cheap...
I found this circuit online, how to modify it for 4 led's? or provide me any better circuit. Also tell me the formula for calculating led's.
**broken link removed**
I appreciate all the help.
Thank you

 

[MikeMl]

Watch this video. I agree with almost everything presented there except when he connects parallel strings of 1W Leds to the driver. I would put a 10Ω resistor in-series with each string...

Come back if you have questions.

 

[AnalogKid]

A normal little red or green LED is much like a zener diode. Within a range of currents, the voltage drop across it is nearly constant. This makes calculating the current limiting resistor fairly easy.

Hig power LEDs are different. The data sheet has voltage drops at different current levels, but the forward voltage varies more with current that for lower power LEDs. The part behaves better if you give it a constant current and let the voltage drop across be whatever it wants.

One approach for multiple LEDs is to look at their nominal forward voltage, stack up two or three in series, and use an LM317 in a constant current design to supply the current.

ak

 

[MikeMl]

...

One approach for multiple LEDs is to look at their nominal forward voltage, stack up two or three in series, and use an LM317 in a constant current design to supply the current.

I dislike the LM317 for this use. It has a minimum differential voltage of 3V plus the 1.25V across the current setting resistor, so it wastes a minimum of 4.25*0.3 = 1.25W to drive a single 1W LED.

 

[Tony Stewart]

Efficient 4' Fluorescent Tubes are now 88 Lumens per Watt. The smaller CFL's are closer to 50 LPW.
Unless you design and choose your LEDs and PS source properly, you will get probably much less.

First you would start with LEDs from 100~130 LPW, then make /buy a PSU that is 80~95% efficient.

Good 1W LEDs are typically 350 mA at 2.9V = 1 Watt
The rest will have higher tolerance on Vf ( e.g. 3.4)

Running a 3V LED from a 12V supply with a series regulator and series resistor is going to be 3/12 or 25% efficient alone.

Using strings of 4 gets you almost 12V with a 2.9V LED. If you have these LEDs and a good 12.0V supply , you can regulate the current with a measured power Resistor that ends up being anywhere from 0 to 0.5 Ohm. Otherwise a regulated LED driver with constant current (CC) source up to 350mA is recommended.

 

[Les Jones]

If you can afford to use 5 LEDs one of the cheap step up regulators that use an LM2577S regulator can be modified as a constant current source. If you could be sure the 12 volts did not go above 12 volts it should work with 4 LEDs You could go up to a maximum of 8 LEDs as the maximum output voltage rating is 35 volts. Here is a link to the modification.
https://lesjhobbies.weebly.com/led-driver-01.html

Les.

 

[MikeMl]

I like this better than using the LM317.

Note the power in one of the LEDs (red trace) as the battery voltage varies from 9.5V to 14.5V (x-axis). It begins regulating when battery voltage >10.2V.

Note the power wasted in M1 (green trace) and R1 (blue trace). M1 will have to be on a heat-sink if it is dissipating more than about 3/4W.

 

[Colin]

The first circuit is wrong.
It is supposed to be a constant current circuit.

 

[Tony Stewart]

MikeL circuit is very stable and well regulated to just about any common device and is a perfect solution for unregulated voltages in this range such as a car, where efficiency is not a major concern, only cost, performance and reliability.

The 0.6V Vbe drop is a small price for current regulation and the additional MOSFET loss used as a linear regulator in this very good classic CC stable regulator design is roughly equal to one 1/2 LED at 14.5V or 67% efficiency.

If maximum efficiency was the requirement, it could only be done with added complexity of a SMPS design, or a stable power source and LED's to match with the low loss solution I outlined in #5

Choice is yours depending on your true requirements.

 

[Colin]

A 7805 in Constant Current mode needs 3.5v plus 5v overhead.
An LM317 needs only 5v. This leaves only 7v and two 3.6v LEDs need 7.2v.

If you have a fixed voltage of 12v, it is best to use a single resistor and 3 LEDs.
All the other ways are far too complex and expensive.

 

[kinarfi]

This is a design I've used many times, it will work with 4 or more LEDs in series,
After looking harder at Post #6 This is virtually the same, but set up for 833 ma, the use of a zener would not be practical, it would have to be able to absorb all the power that the several LEDs had been using. Also, the LT1270 is rated at 8 amps and the LT1270A is rated at 10 amps.

 

[Willen]

I like this better than using the LM317.

Note the power in one of the LEDs (red trace) as the battery voltage varies from 9.5V to 14.5V (x-axis). It begins regulating when battery voltage >10.2V.

Note the power wasted in M1 (green trace) and R1 (blue trace). M1 will have to be on a heat-sink if it is dissipating more than about 3/4W.

View attachment 93571

Hi Mike,
Such the simplest and clever circuit always attracts me! So useful even for charging Ni-Cad or lead-acid batteries from this nice constant current source!

I know that PN junction of the NPN making the base of the transistor regulated at 0.6V so the resistor only will pass 0.6V/1.9R= 0.3A constant current. Here main power dissipation is in the resistor but the transistor junction is making the base voltage regulated always, so doesn't it dissipate huge power same as resistor? The voltage regulation made me amazing! Always I get fail to understand the operation of transistor in such condition.

 

[MikeMl]

Willen, look at the plots in my simulation of post #7. I show the power dissipation in three of the parts. The power dissipation in the NPN is a few uW.

Base current is only nA. Collector current is uA. Vbe is ~0.65V.

 

[Willen]

I think collector current is limited by 100k (12V/100k=120uA) and same the gate current. Otherwise it would dissipate huge power maybe. Nice trick. Using almost no current, the Base is limiting voltage, amazing to me still!

 

[MikeMl]

Here is the dirty little secret of the LED driver I posted in #7: At least it doesn't make the LED(s) go int0 thermal runaway...

 

[Willen]

Here is the dirty little secret of the LED driver I posted in #7: At least it doesn't make the LED(s) go int0 thermal runaway...

View attachment 93581

It seems that if we placed a LED and the NPN touching each another (to transfer heat from LED to transistor). When temperature go high, the base goes low and power limited in synchronous way. But seems not more practical.

 

[Tony Stewart]

I also have a little secret method of preventing thermal runaway with some engineering calculations.

If you designed or know the thermal resistance [°C/W] and know the temperature coefficient of the LED [mV/°C] and the Pd [W] max power rating at some target temp like 85°C , I have developed a method that works for finding the critical minimum series resistance where thermal runaway can be prevented.

It works by adding a small external Rs, which is near the internal LED ESR. It depends on Pd Max, expect Tj rise, thermal resistance of heatsink and and by adding Rs to the string , if current rises due to self heating and reduced junction voltage the external voltage drop on the external series resistance or cable resistance must have at least the same voltage drop as expected from the rise in temp to eliminate the cause of thermal runaway. This added resistance reduces the NTC effect since any current rise would induce more voltage drop in the external series resistance. Normally one sees very large values which are quite inefficient, whereas I have discovered to avoid the critical point of runaway by adding less than an Ohm in series with the string of LEDs.

NTC of White Power LEDs = -3.4 mV/°C

 

[MrOneShotAway]

Here is the dirty little secret of the LED driver I posted in #7: At least it doesn't make the LED(s) go int0 thermal runaway...

View attachment 93581

Can any one explain how this circuit works? When the NPN transistors is on, the 100K ohm resistor kicks in and limits the current to the LED's? Why the mosfet?

 

[Tony Stewart]

The 1.9Ω is the current sense, which is amplified at a threshold of 650mV but does not limit the current.
The collector drives the 100k resistor pullup bias to drive the FET , which limits the series LED current.

Since the Vbe diode characteristic, is that of a negative temp coeffic. or NTC, the threshold is shown here to reduce by -1.5mV/°C which if it were thermally connected to a LED metal clad PCB would serve to reduce the current a little bit as the LED's generate heat.

 

[MikeMl]

Can any one explain how this circuit works?

If the NPN is initially off, the 100K resistor (rather slowly) begins charging the gate capacitance of the NFET. As the gate voltage [V(g) green trace] reaches the threshold voltage of the NFET, it begins turning on. The current through the LEDs, the NFET, and R1 [I(D1) blue trace] starts increasing.
As the LED current increases, the voltage drop across R1 [V(b) violet trace] increases toward the normal Vbe of ~0.6V, at which point the NPN begins turning on, drawing current through the 100K, clamping the gate voltage V(g) to the point where the NFET has substantial voltage from drain to source, thereby regulating the current though the LEDs and R1.

The circuit stabilizes with I=Vbe/R1. In this example, I = 0.6/1.9 ≈ 0.3A