lm317 as current regulator

No, but it's easily fixable, and it won't fry any components as it stands.

The fix is simple, and can be added later. Just add a

Ed

I bypassed responding to the rest of your note - there's nothing on topic that would be helpful to others. But the misinformation below is so completely wrong and dangerous to the poor LM317 that response is required.

this

That is complete and utter rubbish.

The 300 volt transformer you proposed will yield ~424 volts DC, applied to Vin of the LM317. The LED string is 40 LEDs in series each with a 3.5V drop for a total drop of 140 volts, and is connected to the junction of the adj pin and the 50 ohm resistor. The 50 ohm resistor, connected from Vout to Adj, has a 1.25 volt drop and sets the current to .025 mA. The LM317 sees Vin to Vout, which will be 282.75 volts (424-1.25- 140)

I assume you can read an ascii schematic: (View in fixed font)

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Do you understand it now?

Ed

but

Yes. I was wrong. I could put it down to a thirty hour session on other problems, but it doesn't mattaer. Wrong is wrong.

The spec doesn't say that.

It says it can handle 1.5A. It says it can handle 37V. It doesn't say that it can do both at the same time.

There's a spec for maximum junction temperature, and a spec for thermal resistance of the package. Given those specs, the ambient temperature, the thermal resistance of the heat sink, and the airflow, you can determine the maximum power dissipation that can be handled without exceeding the maximum junction temperature.

Does this chip include overtemp shutdown?

On Sun, 25 Jun 2006 00:50:05 -0400, "James Thompson" put finger to keyboard and composed:

If you settle for the series resistor solution, then you don't even need to rectify the supply. You could just wire the LEDs as strings of anti-parallel pairs with a limiting resistor (or capacitor).

|--|>|--| |--|>|--| +--| |--- / ---| |--- R ---+ | |--||--| | +--| |--- / ---| |--- R ---+ |--|

Thank you frank. I believe you hit at the heart of the matter. What you said about using a regulator per series string is along the lines of my original thinking. Especially that the led's are sensitive to current changes.

Ok here it is as best I can come up with now, help me if I make a mistake.

Now for the supply end and isolation from mains, 2 identical 120 to 18 volt transformers, 18 volt winding together for isolation. To the 120 out of second transformer is a bridge rectifier with Capacitor. This would develope about 160 volt dc +/- a few volts. From this dc, can the lm317 handle the regulation being 48 volt difference of 160 and 112, and would this be a workable solution? I could sellect a 120 to 18, then a 120 to 24 volt transformer for the isolation part to drop some volts on the ac side. As long as each transfomer handles at least 1 amp. Thanks again. JT

No. However, your solution below with the 18 and 24 volt transformers will work, but with much beefier transformers. See that response below.

The LM317 has *MAXIMUM* of 40 volts difference Vin to Vout. The LM317HV has a *MAXIMUM* of 60 volts difference. Your 48 volt figure is wrong, and does not consider worst case.

At 120 v AC, you won't get 160 volts, you'll get ~ 167 at the cap. (1.414 * (120-2)) = 166.852. Your design range will be 145.6 to 181 volts input to the regulator. Your maximum difference is 181-112 or 69 volts - you *must* reduce the difference, even if you use the higher voltage LM317HV.

That will work with 34 strings of 30 LEDs per string. The maximum Vin-Vout, worst case, will be 20 volts, and the minimum will be 3.5, so the LM317s will get enough overhead at the minimum, and Vin-Vout won't be exceeded at the maximum.

The problem here is the 1 amp transformer rating. That's

Frankly, this solution is not very good. While it will protect the LED's and ensure no fluctuation in the light output due to line voltage variation, you'll need beefy, expensive transformers, and 34 LM317's .

Ed

Snip..

Thank you ed, you are a genuine help here! I have adjusted the the led count to 40 per so the led drop will be 140 volt. With the remainder on the lm317 and under the 37 volt i/o maximum. Would you suggest not using transformers for isolation as like you said considering the wattage ( 1280 led = 40 per string X 32 rows )( .0875 watt per led X 1280 = 112 watt + about 16 watt (.5 per lm317) on the 32 lm317 =

It's even worse than the numbers above show. The transformer supplies current through diodes to the capacitor and to the load. The total power consumed must be provided by the transformer, but the diode and the charge on the capacitor prevents the transformer from supplying power for some portion of the cycle. Consider: +---Load---+ | | xformer A--->|---B---+---cap----+---gnd

When the xformer output at A is more positive than B, the diode conducts and sends current into the load AND the cap. So more current is needed than just the current that the load draws.

When the voltage at A is below the voltage at B, the load draws current from the cap. The transformer can't provide current, because the diode blocks it. The transformer will provide current once again when the voltage at A is higher than the voltage at B. The net effect is that the current is drawn in spurts, and the peak current exceeds the current that the leds will use.

Regarding the "best" configuration: "best" is a subjective term. Lowest cost? Least amount of work? Highest reliability? Least flicker? Most stable brightness? Safest?

If you don't mind investing a lot of work, the best solution in my opinion is a 10 dollar power supply and 341 strings of 3 LEDs each with a resistor in series with each string. The last led can be handled alone, with a different resistor. This solution is safe, cheap etc etc - the downside is a lot of work, and to a minor degree, differences in current through the LEDs. The power supply is cat# 15625PS for $9.95 at

It will provide 12 volts at 10 amps (plus other voltages we don't care about). Three LEDs at 3.5 volts will need a dropping resistor: R = (12-(3*3.5))/I For example, say you want 25 mA: R = 1.5/.025 or 60 ohms. You could use a 59 ohm or a 60.4 ohm - they are standard values, and are available for 2 cents each in lots of 200 from Mouser. The single LED will need a

To turn the power supply on, connect a switch from the green wire to any black wire. +12 volts is available on the yellow wire.

Ed