luxeon led dimmer circuit

Does the BD139 get very hot? I agree that you can use a lower value current sense resistor. But you also have to scale the reference voltage down by half. I think you don't need the first opamp to supply the divider, since it is a constant load, so it can be factored into the zener bias design. Also, the divider can be very much higher value resistors, since the opamp draws almost no current from it. For instance, a 10k pot in series with a 100k resistor (to use with a .5 ohm sense resistor). You could use the opamp to regulate the current to the zener, instead, to make that voltage almost completely independent of the battery.

The idea is to have the opamp output drive the zener through a current limiting resistor, while the zener is also tied to non inverting input. The inverting input is connected to a voltage divider (could also be your output adjust divider) between the opamp output and ground, which sets the voltage gain of the opamp to something a little more than 1. Lets say you use a divider that creates 5.6 volts when the opamp produces 8.4. (say, a 10k resistor to the opamp output and a

20k to ground) To bias the zener with about 5 mA, the resistor from output to zener would have to be about 2.8/.005=560 ohms. you may have to add a 10k pull up resistor to the opamp output to make sure it starts positive when the power is applied. You tap off the lower half volt across the lower part of the 20k resistor to produce the 0 to .5 volt needed for the current reference.

Yes, but not as bad as the other power components, so probably not desperately in need of heat sinking. (The LEDs get hot enough to scald when run at full current, so I'll probably heat sink them with some aluminium sheet. They already have mini heat sinks built in though.)

Yes, I was assuming that.

Thanks for the idea - I'll give it a try.

cheers, andy.

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Another way to control brightness is to use what is called a PWM circuit. Basically, you give your pass device a pulse every once in a while, and control the percentage of time it is on. Since you aren't just burning up energy with a resistor, it is often cooler and more efficient.

A simple circuit for this consists of a cmos 555, an N-channel JFET, a cap, and a pot:

-----------------------o--------------o--Vcc | | | ||--' .--------. | || J1=BF245B .------+Vss Vcc+----' .->||--. | | CN555 | | | | | | | | | | tr & th----------o------o | | | R=470 | | .---+OUT DISC+---o-\\/\\/\\/\\--------o | | | | | ^ | | | '--------' '----' | | | | | | | | '-Gate of logic level NMOS ----- C=1uF | ----- | |

-o------------------------------------o--Gnd Call the current through J1 'I'.

The period is going to be the sum of the charging time

Tc = (1/3 * Vcc) * C / I

and the discharge time is

Td = R*C*ln((2Vcc - 3RI)/(Vcc - 3RI))

The duty cycle, which is what you are interested in, will obviously be

Tc/(Tc+Td)

since OUT is high during Tc.

The odd thing is that if you simplify this, the duty cycle doesn't depend on the size of the capacitor; it only depends on vcc, Id(J1), and R. Thus, you can pick a cap that is small enough so you don't see a flash, but not too so the pass transistor requires too much dynamic current. The duty cycle defined by this monster:

Vcc D = --------------------------------------------- Vcc + 3 R I ln((2 Vcc - 3 R I)/(Vcc - 3 R I))

So, when R is 0, D is 1, and when R is Vcc/3I, D is 0

That means that when R=Vcc/3I, the thing will simply stop, with output low. The reason is that the discharge pin won't be able to pull the trigger pin lower than Vcc/3. When R=0, it won't take any time to discharge the timing node (well, almost no time) so D

i am also working with some leds (lumileds superflux). My project is to control a cicuit of leds that will be a parking light/turn signal. The light while illuminated as parking light mode is set at one brightness, and gets almost twice as bright when turn signal is on. The only diagram i can findf for this particular circuit is this

but being a beginner, im not sure how to construct this complicated of a circuit. i know i should use pc board (or so ive been told) but it is still pretty intimidating. Also, anyone know where i can get a 15v zener diode??

For the 15 v zener Cat #1N4744 from

4 for $1.00

The circuit at the url does not give you the values for 3 of the resistors, nor the part to use for the darlington PNP transistor.

Here's a different way to do it:

------- 1N4001 Tail -----in| LM317 |out---+---->|----+ ------- | | Adj [240R] | | | | +----------+ | | | [POT] 5K | | | Gnd | | ------- | Turn -----in| LM317 |out---+---->|----+--- To led array ------- | Adj [240] | | +----------+ | [POT] 5K | Gnd

Here's the LED array:

+----+----+----+---- Vcc | | | | V V V V --- --- --- --- | | | | V V V V --- --- --- --- | | | | V V V V --- --- --- --- | | | | V V V V --- --- --- --- | | | | [51R][51R][51R][51R] | | | | +----+----+----+----Gnd

All resistors are 1/2 watt. You can adjust the pots for the desired brightness, and then replace them with the next closest standard value resistor.

The 51 ohm value was determined assuming Vf of 2.5 volts and max current of 70 mA for the LEDS - if your LEDS are different, a new value needs to be computed.

Ed

i dont want to sound like a total dumb@$$, but i cant understand that kind of diagram!!! lol i really appreciate the help and the circuit appears to be much less complicated, but where can i get some help so i can better understand that diagram?

Not sure which circuit you're talking about, but the basics are:

- lines mean an electrical connection between one part and another. A circle means a connection between three lines (you never show four lines connecting, so a crossing without a dot means the wires don't touch)

- square boxes are resistors. resistors with 3 connectors and an arrow on the middle one are variable resistors (pots)

- two lines with a space between are capacitors

- a circle with an arrow in it is a diode. If there's a lightning arrow off it, it's a light emitting diode.

- a circle with a bar in it an 3 lines coming off is a transistor (the line with the arrow is the emitter, the line into the middle of the bar is the base, and the other line is the collector)

- triangles are 'operational amplifiers' - high gain amplifiers which you 'program' with external components to make them into reliable amplifiers or control elements. These usually come in Dual In Line IC packages, often with more than one on the same chip.

- a box with several lines coming off is a more complex chip, which could do any number of things, so you have to look up the chip number.

The idea of these diagrams is that they show just the 'logic' of the connections between the components. Given a circuit diagram, you should be able to wire up the real parts to make a working circuit. Or else you can use these diagrams to look at the circuit before you build it and see how it's going to work, or show other people, etc.

any decent electronics book or website should tell you enough to make sense of diagrams like this, at least enough to build the circuit and understand it at a simple level.

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Is it the ascii art that you have trouble viewing? You need to set your newgroup reader to read the messages in a fixed width font such as courier. When you see something like this: -[240R]- it is a

240 ohm resistor. [51R] is a 51 ohm resistor

--->|--- is a diode - so is V --- It was used as an LED (light emitting Diode)

[POT] is a potientiometer and the LM317 is a three terminal voltage regulator whose pins are labeled in, out and Adj

Does that help? Ed

very much so. for some reason i just couldnt wrap my brain around that. its definitely less complicated than the original circuit i found. ill give it a shot.