Am I right?

Maybe I missed something, but why did you need the 555 output to be optically coupled to the LED driver transistor? Why not just connect the ouput of the 555 to a transistor that can handle 300 mA? A 2N4401 (NPN) will handle 600mA. Or find an appropiate FET to do the job. It seems to me that controlling the AC side of the power supply makes things more complicated than needed. HTH

-Dave Pollum

I want the optocoupler because the control pair (ground and LED cathode) will be in an "industrial" environment.

Rather safe than sorry, I suppose.

I just thought having the transistor in the DC path would potentially be more risky.

One example.... not all 2n4401's are alike? ...what about that and the influence on the "constant current" requirements of the supply...or any possible changes in the characteristics of the transistor over time?

If I dealt with the AC side, that would be eliminated because as long as the transformer secondary stayed within a certain window the 317 wouldn't care?

Use a PIC.

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OK, so I understand why you want the optoisolator. So the output of the 555 drives the opto's LED, which turns on the opto's transistor. That transistor then controls a driver transistor (i.e. 2N4401) that handles the 300 mA current of the LEDs. According to a graph in the Fairchild 2N4401 data sheet

2N4401.pdf), when it's saturated its Vce is 0.25 volts. That graph shows that the current gain is only 10, so the base current would need to be 30 mA. These values differ from what is stated in the tables, so some experimanting would be required to determine what you need. And as far as I know, all 2N4401's are the same. I suggested a 2N4401 because I knew it could handle the current, it's a common part, and it's cheap. There may be other transistors that are more suitable for your project. HTH

-Dave Pollum

Don, It seems like the cure for everything except the common cold is either a 555 timer or a PIC ;)

-Dave Pollum

SORRY you guys, I didn't think this through before posting.

I liked the idea of controlling the AC side and keeping the series path in the power LEDS clear but I would need the power supply ON all the time for the 555 etc.

Wasn't thinking.

I'll breadboard a transistor to open/close the DC path....an FET would be better, for lower on resistance?

Also, an FET would have less current drive requirement, right?

I'm going to try and switch 300ma....that's quite a bit of current.......maybe they make an opto with a "second side" that will do the trick.

That's not an ungodly amount of current to switch.

Are you running the 317 as a current regulator? That way you won't need dropping resistors (assuming 300 mA is the right forward current for the LED).

Then, a low-side switch can be anything you want that can handle the volts and mA. ; you might want a bleeder resistor directly to ground from the 317 output - when you open it, it will go as close to the supply V as it can, trying to push 300 mA through an open. =:-O

There might be a more sophisticated circuit, like constant current with a voltage limit, but you'd have to look that up.

Good Luck! Rich

series.

this,

opto

The cure for a 555 timer is a PIC.

Is there a compelling reason why you don't use the opto output to pull down the reference pin of the 317?

Sarcastic f*c^er :)

--
Gibbo

This email address isn\'t real.

--
Sounds almost like a product...

Why make things difficult by adding another pass element, when you already have a perfectly good one in an LM317?

To easily control the current, since the LM317 is there anyway, why not just use it? Then the control current to switch things on and off is quite low. All you have to do is run the output of the LM317 through a resistor R1 to the string of LEDs. R1 = 1.25V/LED_current, or about 4 ohms for 300mA. Then put, say, a 1k resistor between the junction of R1 and the top LED cathode, and the LM317 control pin. Now if you pull down the LED control pin, the LM317 just delivers a bit over a milliamp to the pull-down device. If the voltages are appropriate, you can just use a 555 timer, or you could use some other arrangement of transistors or whatever. You can even drive the control point on the LM317 with an opto isolator output directly. Surely you don't have to worry about the LED current with three in series across about 1.25 volts, in the "off" state. You do have to account for the LM317's 50uA nom., 100uA max control pin current, which will drop through the 1k resistor when the LEDs are on. It would make a max. difference of less than 10%, assuming that 1k resistor, if you just ignore it.

Cheers, Tom

The cure for a PIC is a 555.

-- Service to my country? Been there, Done that, and I've got my DD214 to prove it. Member of DAV #85.

Michael A. Terrell Central Florida

Funny how you can get away with that. ;-)

series.

this,

an

opto

Programmed to look like a 555?

OK, that works for me. ;)

--
JF

Yeah, I'm going to look at that approach later and breadboard something.

I've never really tried to fiddle with the control pin (adj?) pin on the 317 but it has occured to me in the past that something could be done with it.

Even in it's current regulated setup where the output is taken from the input pin, what you're saying is I can pull the adj pin low and effectively turn off the voltage, right?

Make sure there is some resistance there (1k like you say).....you wouldn't want it see a straight shot to ground?

Just take the opto output (npn out).....connect the emitter to ground and the collector through 1 k to the adj pin.

Sounds like the way to go.

I have the opto trigerring a 4401 right now and all is fine....put +5 DC on the opto emitter and using a 560 ohm on the collector to drive its base....switching the 300 ma and not even warm.

Oh, and I love Picbasic but I'll take the 23 cent 555 for this !

I'm using 2 optos at once, one for the dc switch and the other npn out to reset the 555 on each control closure.

I meant to say output from adj pin...not input.

output...power resistor....to adjust

and adjust = current reg output

Um, the 1k resistor I suggested goes between the top of the LED string and the LM317 adj pin. You yank down directly on the adj pin. The LM317 output will follow 1.25V above that. So if you manage to fully ground the adj pin, the output voltage will be 1.25V. 1.25V across a series string of 3 LEDs will cause a current of perhaps nanoamps to flow in the LEDs, depending on what kind they are. But it will be at most a tiny current, even it they are pretty hot.

When the adj pin is allowed to "float", it will be held by the 1k resistor to very nearly the voltage at the top of the LED string. The LM317 output pin will be 1.25V above that. So to get 300mA current in the LEDs, you put a resistor between the output pin and the top of the LED string: 1.25V/300mA is about 4 ohms.

When you pull the adj pin low, the thing which pulls it low must handle the current out of the adj pin of the LM317, at most 100uA (from the data sheet), plus the current in the 1k resistor. Since the LED current is very low now, with the adj pin held low, there will be essentially no drop in the 4-ohm resistor, so 1.25V drops across the

1k resistor, causing 1.25mA current. So the net current in the "low- puller" (whatever it may be) is 1.35mA at most. You do want to make sure you only pull DOWN on the adj pin to turn things off. Do NOT connect it directly to the output (pin 3) of a 555, that can pull it UP. That would try to turn on the LEDs harder than you intend. Instead, put a diode in (anode to the adj pin) so you can only pull down if you have any doubts about your circuit. If the LEDs you're driving have even a moderately high voltage drop, you can safely use just about any silicon signal diode. You don't need a diode if you're driving the adj pin low using an open-collector or open-drain transistor.

You shouldn't need any capacitors on the output end of the LM317 for all this to work.

Cheers, Tom

It can't work the way I understand your description - I'm obviously not getting it. A schematic would help. Here's a possibility for yanking down - feel free to modify it to show what you have in mind. The 1K resistor from the top of the LED string to the adj pin escapes me, so it's not in the diagram, but it should work as drawn.

Ed I -> ~300 mA ----- Vcc + ----Vin|LM317|Vout-----+