resistance values for LED's

Yup I know what you mean thats what the technician in the shop asked me -he said "do you want the big type"

I know now-- Ha!! he means 1/2 watt.

So if I used a 1k Resistor I would be within limits. A 9mm resistor is a big thing to incorperate in to this small instrument plug and the brightness could be reduced ?

 

YUP. See post #3. The other 27 posts are redundant. Interesting reading but irrelevant to your requirement.

1kohm will see 0.14W max on your 12V system so the little 1/4W will handle it. You can also get 1/8W but these are really small. About 4mm long. Maybe these are the ones you lost. They are not much thicker than the wire leads.

Do you have a multi-meter? A useful one costs about $20 here and can give all sorts of insight into matters electrical.

Rick W

 

Aren't we in boating ******** a bunch of pricks - Frosty You better give a good donation to Rick Willoughby I am restrained by "...You must spread some Reputation around before giving it to Rick Willoughby again...." else I would too...

 

Yes mas just signed on to do that. What a good bloke,!! you would'nt think he was an Aussie would you.

 

LED light dimmer

Well it's taken me awhile to get this together, but better late than never, as they say.

Below is the circuit I plan to use in my own boat for light control, in conjunction with the switching regulator circuit from a post above. Every effort has been made to eliminate resistive losses. Assuming that four Cree high-power LEDs are controlled at a voltage that turns them all on at 2.8 amps (3.5 volts) then the transistor doing the switching (the IRLI3803) will dissipate only 0.047 watts, and then only when on. This is because this MOSFET type transistor has an "ON" resistance of only 0.006 ohms. It can handle 76 amps, by the way. So, 2.8 amps passing through 0.006 ohms will have a voltage drop of 0.0168 volts, multiply that times the current (2.8) and we get 0.047 watts, which is lost as heat. This will be controlling 9.75 watts of LED light.

Of course that is way too much light for most purposes! I went through the little exercise in the paragraph above to show how efficient this approach can be, compared to "resistive" approaches. We need to tame this down a bit, though, so let's go through the dimmer circuit below step by step, assuming that it is powered at 3.5 volts by the switching circuit above.

This is a Pulse Width Modulation controller, which means that it contains an oscillator running at a (more or less) fixed frequency but varies the "duty cycle" of the output pulses from "on" very little to on almost the entire cycle. So, these pulses turn on the IRLI3803 output transistor in a range from ON almost all the time within a pulse "window" to almost never on, thus varying the perceived light output.

If you look at the circuit diagram you will see a chip labeled 74AC14. This chip contains six "inverters" which do exactly what they say, they take a "low" signal on the input and turn it to a "high" signal on the output, or vice versa. We will only be using four of these inverters. These four inverters are used to produce two oscillators and buffers for their outputs. One oscillator is the one I described above that varies the duty cycle, and the other is used to power a voltage doubler circuit, which is needed by the output transistor (IRLI3803) to drive the gate fully ON. Let's look at the voltage-doubler gate-drive oscillator first.

To the left of the 74AC14 chip is a 0.1 uF capacitor with one end grounded and the other end going to pin three of the 74AC14 chip, plus to a 10K resistor that is connected to pin four of the same 74AC14 chip. Pin three is the input to one of the inverters on this chip, and pin four is the output.

When power is first applied, the capacitor connected to pin three is discharged, so the inverter at pin three, sensing this, tells the output at pin four to go high. This output is connected to the top of a resistor connected to this capacitor, and since the voltage is now high starts to charge the capacitor. At some point the voltage on this capacitor exceeds a threshold in the inverter and instead of seeing a low input it then sees a high input . . . so the inverter output goes low. Now this same resistor discharges the capacitor, until the entire cycle starts over. The voltage on the capacitor just keeps going up and down, making the inverter output alternately high and low. The circuit is oscillating, in other words. The values shown for the capacitor and resistor will produce an oscillation frequency somewhere around 1 kHz.

The output from this oscillator is connected to another inverter at pin five, thus causing the output at pin six to also oscillate. This second inverter protects the first from any effects from further down the line. That's why it is called a "buffer." This second output is connected to a capacitor. The other end of the capacitor is connected to two diodes (these are Schottky diodes, which have very low forward voltage drops). When the output at pin six is low the uppermost diode conducts and charges up this capacitor. When the output at pin six is high that kicks the charge on the capacitor up, just like putting two batteries in series (the voltage already on the capacitor plus the voltage from the output at pin six going high), and it conducts through the lower diode to charge a 4.7uF capacitor further down the wire. Thus the voltage is (roughly) doubled.

Now look at the 0.47uF capacitor connected to pin 13 of the 74AC14 chip. This capacitor is in a circuit very similar to the oscillator circuit described above, with one important difference. Now, when the capacitor is charged the charge has to go through the "wiper" on a potentiometer, then through whatever is left of the resistor to the left of the wiper, and through a diode to charge the 0.47 uF capacitor. On the discharge part of the cycle the current has to pass through the other diode and the right part of the potentiometer to get back to the wiper and hence to pin twelve, the output of the inverter. So, depending on the position of the wiper on the potentiometer some fraction of the resistance controlling charging (one direction through the potentiometer from the wiper) and discharging (the other direction) is metered out, and that fraction controls the time that it charges or discharges . . . thus modulating the width of the output pulses either on for a great part of the cycle or a small part of the cycle. Note though, that the total time of the cycle remains the same, because as you subtract from one side by moving the potentiometer wiper you add to the other! This oscillator is therefore the Pulse Width Modulation oscillator, and the duty cycle is controlled with the potentiometer shown. The output of the inverter at pins thirteen and twelve is also routed to the next inverter at pin eleven as a "buffer" to further circuitry to control the switching of the power MOSFET (IRLI3803). This PWM oscillator will oscillate at roughly 200 Hz.

Power MOSFETs are wonderful devices; extremely fast, capable of massive currents, and with very low "forward voltage" drops in most applications. They would be perfect switches if it wasn't for one little annoying fact, the gate (which controls the switching) has inherent capacitance. This means that it resists turning on to activate the "switch" and it also resists turning off. In practice this produces annoying time delays if you don't provide sufficient initial current to overcome the inherent capacitance of that gate. Once you do get the thing switched, however, it takes very little current to keep it that way. The sole purpose of the three small MOSFET transistors in between the output from the PWM oscillator and the power output transistor (the IRLI3803) is to provide this fast switching at the gate of the IRLI3803, at the required voltages and currents to overcome the inherent capacitance of that gate.

The output from pin ten of the 74AC14 (the PWM output) goes through a 3.3 ohm resistor to the gate of a ZVN3306A transistor. If this output is HIGH the ZVN3306A will turn ON and pull the 10K resistor attached to it's drain to very near ground, thus also pulling the gate of a TP0610KL transistor down, and turning it ON. (The ZVN3306 transistor is an "N" channel type of MOSFET, just like the IRLI3803, only much smaller and faster. The two TP0610KL transistors are "P" channel types, also small and fast, and these function as a "mirror image" (if you will) of the "N" channel types. Very similar in principle to the difference between NPN and PNP bipolar transistors.)

When the first TP0610KL transistor is turned on it passes current from it's drain through a diode, through a 3.3 ohm resistor and then to the gate of the big IRLI3803 transistor. This is a low impedance path, so something like 6 volts with the charge potential of the 4.7 uF capacitor hits that gate and overcomes the capacitance there very quickly, thus turning ON the IRLI3803. At the same time, this high voltage from the drain of the first TP0610KL raises the gate voltage of a second TP0610KL to near it's source voltage, thus turning this second transistor OFF.

Now when the output from the PWM oscillator goes LOW, first the ZVN3306A turns OFF, which causes the first TP0601KL to turn OFF . . . and allows the 10K resistor attached to the gate of the second TP0601KL to pull the gate LOW, thus turning it ON. This, in turn, provides a low impedance discharge path from ground through the drain and source of this second TP0601KL to the gate of the big IRLI3803, and turns it off.

Simple, right?

OK, maybe not. There are some details that need to be addressed as well. This circuit and the switching power supply above both use rapid switching to achieve their respective goals. If you break down the frequency components of a "square wave" like these devices use, you find that there is the fundamental frequency (in this case nominally 1000 Hz and 200 Hz) and then a series of harmonics of this fundamental. These time-varying currents can transmit from any conductor, just like a radio. Fortunately they are easy to suppress, just build the basic circuit in a metal box and use shielded cables (marine grade, of course) to the LEDs, if they are remotely located.

I just did a quick price check at digikey.com for the major components and got this:

74AC14 0.42
IRLI3803 2.14
TP0610KL 0.86
ZVN3306A 1.17
11DQ06 0.51

All prices in US dollars in quantity one.

If you try this, please let me know how it goes!

BillyDoc

 

BillyDoc,

I just stumbled across some products from Maxim that appear to have some potential for shipboard LEDs:
http://www.maxim-ic.com/solutions/hb_led_drivers/parts.mvp/scpk/2001/pl_pk/0
Constant-current LED drivers, apparently with all the control circuitry on one chip.
They're aimed at the auto industry (which is using them for tail lamps, turn signals, etc.). Prices seem to be in the $1 to $5 range. Any thoughts?

 

Hi Matt,

That looks very interesting indeed! I just took a very quick look at the MAX16812 data sheet and it looks like it does everything needed for yachtie LED use. If I understand the data sheet, and I only gave it a quick skim (sorry!), it looks like a combination buck/boost regulator to set the current level with a PWM stage added on for dimming . . . which is exactly what you would want. And the cost is currently $7.16 for the chip on digikey, which isn't bad at all. I wish I could study it out more thoroughly, but I'm swamped.

BillyDoc

 

I'm not a LED lover, but I recently found some impressive ones.

Usually these white, almost blue led's makes my eyes burn after a short while, but the ones I got are stunning and makes a light that is like sunlight, very eye pleasing. One can also distinguish colours of items very well.

I'm making a few for the boat, on off with dimmer. Expecting the boards any day from month end. The board gets 3 LED's at 1 W each drawing 200mA off 12V. Emmision is 360/180 deg which is nice, I don't want spot lights. The LED's come with heatsinks mounted. See PDF.

Life expectancy is good at 100 000 hours. The previous LED's I got had like a 3000 hour life expectancy and the light colour sucked. They also drawed 1A with less emission and ran hot as hell.

 

led lights

I thought some of you who are into LED's might be interested in this .

A friend of mine has just converted all his wheel house cabin lights to LED using a 12-36v led driver. (his boat runs on 24v) He bought through a reputable supplier here in UK. As he knows I m a lighting engineer, he called me in to show me the effect so I have witnessed this for myself

On turning on the lights, the electrial noise from the drivers wipes the satalite signal out to the chart plotter. and his hand held GPS. the plotter has external aireal so its interfearing at the head end. hand held ---fair enough.

As some one did explain the way the drivers work, I can imagine the pulsing transients with all the eletrical noise cuming from them.

Me -----on my newly aquird wreck,will be fitting nice and quite resisters,
Or even better---- nice oil filled lamps. Much better swinging action than an LED so I can practice my tails of daring-do before going to the club bar

Kim

 

The circuitry to generate the high voltage is the problem. It is used to convert the voltage to a higher than supply voltage so the LED's work in series and not in parallel circuits.

It works almost the same as a switchmode regulator. A power component, usually a FET, switches the supply through an inductor. The inductor charges and when the FET switches off the back EMF power is fed through a schotkey diode to a capacitor. This pumping the cap up through the diode is what enables the higher voltage.

The noise is generated by the inductor that transmits energy into the air similar to what a radio does when it transmits. The frequencies switched goes as high as 500kHz to gain efficiency and so get away with smaller components to save space and cost. It's like having magnets near your compass. Lots of noise emitted.

There is a way your friend can get his boat back. You can buy a screen material from spares shops. If you can wrap the switching circuitry you can contain the emissions.

My circuits are all linear, ie there are no switching done and emission will be almost zero.

 

Hi Kim,

You make a very good point, that transients from rapidly switching anything will cause nasty problems . . . if not properly shielded or suppressed.

But there are definite advantages to using PWM or switch-mode power supplies as well, and you just have to weigh those advantages against the disadvantages, like with most engineering. As I see it, when you are talking about suppressing transients you are talking about a problem that you address and solve "up front" in the initial design and implementation stages of the project, and then you enjoy the results for as long as you use the system. If you just take the easy way and "go with resistors" you then also have to live with the consequences of that decision, which are a much greater power draw and the production of unwanted heat, for the duration of your use of the system.

Frankly, I'm surprised that your friend purchased a commercial system (presumably for a boat) and was not advised about this issue. In any case, please tell him that the fix is to use shielded cables to the lights if they are remotely mounted, only join the shields to ground at one point to prevent ground loops, and to put the main PWM or switch-mode circuitry in a grounded metal box. Clip-on ferrites on the cables will also help. That way, just as with a co-axial cable to an antenna, the transients stay confined and cause no problems. It's several extra steps and an extra expense to be sure, and depending on how precious power is in a particular situation may not be worth the bother.

I like oil lamps quite a bit myself. Except for the mess and the smell.

BillyDoc

 

Billydock, linear does not nessesarily mean high energy losses and lots of heat. The only losses one have in a linear circuit is the heat generated by the components, and this is where a switchmode have an advantage.

There are however ways to design linear suppies with such low losses that the difference between linear and switching is almost neglectable, and without the switching supply's costs, and of course you don't have to deal with emission problems.

 

Somehow managed to hit the post button prematurely! Sorry!

 

Hi Fanie,

I hope you are right and can tell me how it works, because I would drop the switch-mode stuff in a heartbeat if that's true. I don't like the emission problem one bit!

However, my current understanding of linears is somewhat different from yours. In all cases that I am familiar with linear regulators adjust the bias voltage (or current) on the base (or gate) of the main output transistor to adjust it's effective resistance to something very near the resistance of a simple resistor that you could put in it's place if you knew that other variables would remain constant. Many data sheets include circuit diagrams for these regulators (sometimes they are simplified, and sometimes missing completely). I think if you look at one you will find a bipolar transistor that forms the output stage of the regulator. In the case of the 78xx series I know you will.

So, this output resistor is being dynamically "adjusted" to provide exactly (or nearly so) the resistance necessary to limit the input voltage to the desired output voltage. It's a transistor acting like a resistor, in other words, and as such produces heat just like any other resistor. Unlike a resistor, however, it is also using energy for it's control function. A linear regulator will, therefore, always waste more energy than a resistor if the situation is static. It is usually dynamic though, which is why we use linear regulators.

Switch-mode regulators do something entirely different. They exploit Lenz's second law and the resulting time lag of the inrush current when a voltage is applied to an inductor. Let's review a little physics here before going on.

Lenz's second law states that (my paraphrase here): "The magnetic field produced by a current will induce a current in the original conductor in the opposite direction from the first current and this opposing current will have an opposing magnetic field." A great demonstration of this effect is the so-called "repulsion coil." Basically you get a bundle of steel gas welding rods, dip them in varnish to insulate them, and cram as many as you can in a plastic tube about 1 meter long. Then, on one end you wind a bunch of wire around the plastic tube to make a solenoid. If you put DC current through this wire you have simply made a big electromagnet. If you put AC current through this same wire interesting things can be made to happen. The usual experiment it to place a big conductive ring over the long end of the plastic tube with welding rod "core." I've cut aluminum "washers" maybe 10 cm in diameter and 3 mm thick for this purpose. After the washers have been slipped over the tube DO NOT AIM THE TUBE AT ANYTHING YOU CARE ABOUT!!! Because when you power up the coil of wire it will cause a magnetic field which will induce a current in the conductive ring and this second current will have it's own magnetic field which opposes the first. The result is a high-velocity conductive ring that usually puts an impressive dent in the ceiling.

So what's this have to do with switch-mode regulators?

In an inductor, which is basically a bunch of wire wound around something, when you initially put voltage across the ends of the wire a current flows. This current has a magnetic field which induces a second current in opposition to the first and thus delays the buildup of both the magnetic field and the current for the inductor as a whole. At some point the maximum current does flow at a constant rate and there is no "increasing" magnetic field to kick in Lenz's law. But now you have a big steady-state magnetic field. What happens if you remove your initial voltage source? The answer is that the magnetic field gets moving again as it collapses and generates a second current in the wires. The first current as the magnetic field built up generated a current that opposed the first current, but as this magnetic field collapses it again generates a current that has a magnetic field that opposes the collapse . . . but the main and over-riding current that is generated is in the same direction as the original current that caused an increasing field!

This point is key to understanding switch-mode regulators, so let me emphasize it again in a simpler way: You put a voltage across an inductor which causes a current to flow in the usual direction, but with a time lag because of the opposing current. When you remove this current from the inductor the magnetic field that has built up collapses, but is again delayed, and generates another current IN THE SAME DIRECTION.

In terms of the speed of most electrical phenomena the time delays an inductor can induce are very, very long.

So, say you have an awkwardly high voltage you want to make a much lower voltage as efficiently as possible. There is always a resistor, but that has a few problems as we all know. Instead of a resistor you could stick an inductor in the circuit and arrange the circuit with a switch and some diodes so that the inductor is only turned on long enough to build up the voltage across it that you want, then turn it off and let it "discharge" this voltage where you want it. The inductor would be "bucking" the voltage because of it's Lenz Law effect . . . which is exactly what a "buck mode" switching regulator does.

But wait! There's more! As the cheesy ads used to say.

This energy we are playing with is being stored in the form of a magnetic field, and the magnetic field doesn't know anything about voltages and currents. So, if you design the system right you can build up a magnetic field in an inductor using a high current, then switch off the current source and let the rebounding magnetic field generate a higher voltage than the original one! This is "boosting" the voltage and is the basis for the "boost mode" regulators. If you want to get fancy you can even do boost and buck combined regulators.

The point I am trying to make is this: I am not aware of any technologies that can be used for voltage regulation besides resistive approaches (linear regulators) or energy storage and release approaches (switch mode regulators). And as you know, resistors of all types make heat and waste energy. Switch mode regulators make little heat and are very efficient as a result. If you know of another approach PLEASE let me know!

BillyDoc

 

Ok, think about this -

If you have a single LED with say 3V3 drop over it then to work it off 12V the series component has to drop 8.7V, not very efficient. At 100mA you lose 0.87W, and only 0.33W goes to the LED (basic Ohms Law)

If you now series 3 of the same LED's, you get a 9V9 in total on the LED's, you can already see it looks a lot better. If you use a series resistor you have a 0.21W loss at 100mA drawed and 0.99W goes to LED power.

To make the 0.21W or 2V1 less wastefull, you can add a current regulator ie LM317 which will drop about another 1.25V off the 2V1. This also serves a protection purpose when surges on the 12V occur, even rising to 40V will maintain a constant regulated current through the LED's in series for the short surge time.

This leaves us with a residual of 0.9V. To add another enhancement, add a schotkey diode on the supply line for polarity protection. This will roughly drop another 0.2V leaving 0.7V to be lost.

The losses now is around 5%, I can live with that. What you gain from it is simplicity and any one can build it, very common and economic off the shelf components, no noise emission, and a well protected circuit.

For more light you just parallel or duplicate the complete same circuit.

The 100mA used is just for easy calculation, most LED's doesn't have to be driven at max current. At about 70 - 80% of full spec current you can hardly see a change in light emission. Running the LED on lower current will run it cooler and increase expected life considerably.

The loss increase if the 12V is risen to say 14V. The current regulator will however maintain a constant current through the LED's, so light emission remains the same.