Hi,

It works in simulation, but LED1 or LED2 could be an Achilles' heel if either of them opened up. That would cause the rest of the LEDs to get maximum current flow through them; then they would all have to be replaced. If LED3 failed open, all the other LEDs would simply turn off.

LED's failing open is less than 1 in a thousand, the normal failure mode for any diode is short circuit.
LED failure is a extremely rare thing, as long as the current stays within a sensible range.

Using eg. an LM317 all LEDs might kick the bucket if the LM317 failed.

I have had about 3..5 LM317 fail short (i.e. full voltage on the output) over the years, but the only LED's I have killed were by purpose or an *oops, was the PSU set to 30V/10A?*
So I know which circuit I'd personally be using.

 

Soeren,
As an experiment, I re-ran the simulation of your schematic after deleting D13-D15 to see how it might work for our OP.

The circuit entered high frequency oscillations as V+ came up; in the vicinity of 600kHz. Adding a 10nF capacitor from the base of Q1 to ground eliminated the oscillation.

 

Instead of trying to get xxx volts try to get xxx ma as a current source out of the lm317 then let the LED just see the current to turn them on. Your 12 v is probably 13.7v so why bother with voltage and the design is simpler assume 4 lets in series then add a res. 82 Ω from out to adj. simple enough for you. add more strings of 4 the reduce res. accordingly. If you add to many then it is going to get hot because of power dissipation.

 

Instead of trying to get xxx volts try to get xxx ma as a current source out of the lm317 then let the LED just see the current to turn them on.

The LM317 can be used as a constant current source, but when used in this manner, it has a minimum 3v drop across itself.

It would also require an LM317 for each string.

Your 12 v is probably 13.7v so why bother with voltage and the design is simpler assume 4 lets in series then add a res. 82 Ω from out to adj. simple enough for you. add more strings of 4 the reduce res. accordingly. If you add to many then it is going to get hot because of power dissipation.

You are incorrect.

Boaters typically use interior lighting whether the engine is running or not. The electrical system may have anywhere from nearly 15v down to perhaps 11.5v, if the battery is allowed to become fully discharged.

His LEDs have a Vf of 3.2 at 20mA.
With an LM317 used in current limiter mode, the formula for the resistor is:
R=1.25/DesiredCurrent, in this case, 1.25/20mA = 1.25/0.02 = 62.5 Ohms. 62 Ohms is the closest standard value.

The 82 Ohm resistor you suggested would result in only 15.24mA current in a single series string, if there was enough voltage available.
However, four 3.2v 20mA LEDs in series with an LM317 as a current regulator would require 4 * 3.2 + 3 = a minimum of 15.8v; his electrical system is not capable of generating that much output.

If three LEDs were placed in series with an LM317 wired as a current regulator, it would still require a minimum of 3 * 3.2 + 3 = 12.6v. This would work as long as the battery was completely charged, but then the LEDs would begin to dim. So, our OP would be forced to use only two LEDs in series per regulator.

Your suggestion of using a single current regulator for several parallel strings is not a good idea; if one of the strings burned open, the remaining strings would be forced to share the excess current. This would rapidly lead to ALL of the LED strings being burned out.

 

I've proposed this regulator for LED circuits before...

The problem is tolerance, the zener can be too low (±5%, which creates a possible 10.8V to 12.0V output). Besides test selection, is there another way to do this you think?

My thought is, baring the zener tolerance, this has got to be the lowest insertion drop you can achieve with a voltage regulator.
True or False?

BTW, R2 can be eliminated if there are LED circuits there instead.

 

I've proposed this regulator for LED circuits before...

The problem is tolerance, the zener can be too low (±5%, which creates a possible 10.8V to 12.0V output). Besides test selection, is there another way to do this you think?

Use a TL431, which is an "adjustable Zener"

My thought is, baring the Zener tolerance, this has got to be the lowest insertion drop you can achieve with a voltage regulator.
True or False?

False.

 

My thought is, baring the Zener tolerance, this has got to be the lowest insertion drop you can achieve with a voltage regulator.
True or False?

False.

OK, then what is? I thought 0.6V was pretty good. I guess if I could get a germanium transistor I could go for 0.3V.

 

Here's something I just threw together. It's not really a practical circuit as-is. However, there is no fixed dropout voltage like with BJT's; the limiting factor is current vs the Rds(on) of the MOSFET.

V1 is just a simulated voltage source that varies from 10v to 14v.
Q1 is a P-channel power MOSFET.
R1 keeps the MOSFET gate held @ source potential by default.
R2 and D1 form a voltage reference that stays around 700mV or so.
R3 simply represents a load.
R4 sets the output voltage.
R5 provides a bit of hysteresis to keep the comparator from cycling too frequently.
C1 is the transient cap.
What the circuit really needs at the moment is something to limit current besides the MOSFET's Rds(on); an inductor would be a good candidate.
The MOSFET also needs a driver; just using the LM339 to switch the MOSFET's gate on and allowing R1 to pull it back off is really quite pitiful - but I left it that way just to show that it can work, even though not ideal. The MOSFET would spend too much time in the linear region.

 

I'm seeing around .2V insertion loss. Sound about right? I like the germanium transistor better, in terms of parts count, not that they are available. I'm thinking something a noob can build.