Beginner: LM350 as constant current not working

[MagicMatt]

I think I may have just fried some rather expensive LEDs

I rigged up an LM350 as per this page (Constant Current circuit)....
http://diyaudioprojects.com/Technical/Voltage-Regulator/

Tried feeding it a 12V input voltage, used a 2R7 resistor in place of R1
The LED states a forward voltage of 2.1V@350mA to 2.3V@700mA (it's a high intensity power LED obviously)

Result... LED got very hot and I pulled the power quick.

I then tried my test meter across the LED - it was getting over 4.2V!
In series with the LED the current is coming out at 2Amp, which is the max the power supply can deliver! No wonder it got very hot!

I then tried putting a 470ohm resistor in place of R1, thinking this would regulate to a very very small current.... but no, I still get the 2Amp across the LED!

Circuit is obviously doing something, as the 12V has gone down to around 4.2V, but why is my resistor seemingly ignored?

I tried it with NO resistor - same thing.

Now I can't make sense of the calculations. I have tried three different LM350 devices, triple checked the pin config against the data sheet, and tested the resistor values with my multimeter.

Does that circuit just not work with an LM350 - am I being stupid?

I don't know what to test or do now, and I'm also worried I may have blown one of the LEDs. Can anyone offer any advice, or even a sanity check??

Are all LM350s the same? Do the pin configs change? Is it possible I have one with a different pin out? From left to right, with the bulge facing towards me, I believe the pins are Adj, Vout, Vin ..?

 

[davenn]

You are feeding it with 12 V but you only want ~ 2.2V
The reguator is regulating current in that config, NOT voltage .

try dropping the input V to the reg down to closer to the required voltage for the LED's

Dave

 

[MagicMatt]

Tried a 6V 800mA power supply... got 800mA across the LED.

If I didn't connect anything to Adj shouldn't I get no output?

Unless I'm reading the specs wrong, it can regulate happily up to 35V difference, as long as the difference between In and Out is 3V or more, 12V in but 2.2V out should be fine. I was expecting 2.2V only because that's the quoted forward curernt of the LED at the current I want it regulated to.

The rest of the circuit will need 12V, and that's what the power supply. If it's regulating current shouldn't putting the 470R in there have dropped current down to very little?

I wanted to run 12 of these in total... plan was to group them into 4 sets of 3 in series... doing that still throws the full 2amps across them.

 

[(*steve*)]

It sounds like you're wiring up the LM350 incorrectly. (edit: check here)

Make sure you have the pins identified correctly.

The best way to test a constant current power supply is with a resistor. Select a resistor that will drop some sensible fraction of your 12V (say 6V) at the current you believe you're setting it at. Then measure the voltage across this resistor when testing the circuit. If the resistor has a power rating such that it can be left connected across your 12v, then it won't get too hot (it might get hot though)

In your case, 700mA, 6V... ohms law says 8.57 ohms, use 8.2 ohms (it will drop 5.74V at 700mA). Technically you'd want a 20W resistor, but a 10W resistor should be fine for short periods of time with 12V across it.

 

[MagicMatt]

Thanks, it hadn't even crossed my mind to use a dummy load!

Checked the pin out (see original post) - looks the same as I have. Triple checked the wiring - also looks correct.

I think I may also try wiring it up off-board, and just solder the component on to test, in case there's an issue on the bread-board as the legs are pretty fat to go into the small holes - could it be shorting somwehere inside the board?


Resistor wise... 5.74V x 0.7mA = 4Watts.... so why a 20W resistor?

I only have 0.25W and 0.5W in that kind of value, so if I use 3 x 2W 2R7 in series that will also split the power dissipation wouldn't it? I expect them to get hot, but I only need it to run long enough for the multimeter to give a reading.

Wish I didn't have to leave for work now... (8:30am here, finish work 6pm).... will let you know what happens.

 

[MagicMatt]

Ok, resistor across Output and Adj is rated at 2R7 and measured at 3R0

3 resistors in series between Output and Gnd rated at 2R7 measured 8R3

Test meter says current flowing is 0.98A, voltage drop 8.29V


....?


I was expecting 0.42A - less than half of what was measured. I'm totally baffled.

Edit - Doubled the resistance between Output and Adj, and it did NOTHING. Readings identical to before. I have tried all 5 of the LM350 now... could they be faulty??

 

[daddles]

If you don't have a bench DC power supply with constant current/constant voltage, a handy device to put in series with the (expensive) load is a fuse -- this will keep the LED's current to safe levels when experimenting. Of course, the load resistor suggestion is even better.

I recently built a 1 A constant current source using an op amp to drive a TO-220 IRF3205 MOSFET. This worked well because the MOSFET's resistance was low enough to let me not have to bother with a heat sink for it. This was a much better solution than a LM317 constant current source I also built for 1 A -- the LM317 runs from 12 V and gets quite hot under a steady state load, even with a TO-3 heat sink.

 

[ModemHead]

I have tried all 5 of the LM350 now... could they be faulty??

Check the voltage between the Output and Adjust pins of the LM350, it should be 1.25V or thereabouts (as long as there is at least 5 to 10mA of output current). Otherwise the regulator is faulty.

 

[MagicMatt]

Check the voltage between the Output and Adjust pins of the LM350, it should be 1.25V or thereabouts (as long as there is at least 5 to 10mA of output current). Otherwise the regulator is faulty.

0v on all 5 of them.

 

[MagicMatt]

I recently built a 1 A constant current source using an op amp to drive a TO-220 IRF3205 MOSFET.

Is there any chance you could share this, or even email it to me? Would it be easy to adapt to 500mA (or thereabouts)?

Would it be able to be switched with PWM (I was originally going to use a 5Amp rated transistor for this and use the LM350 as load)? I have a signal from a PIC (16F690) that I'm programming, so speed can be anything required (I'd have liked 500Hz+, as at 10% duty cycle it wouldn't visibly flicker).

It looks like I am going to have to buy alternative or replacements for the LM350, so if there's something that will do a better job and not cost a load more I'd prefer to do it that way.

 

[(*steve*)]

Resistor wise... 5.74V x 0.7mA = 4Watts.... so why a 20W resistor?

Because your resistor is getting the full 12V across it in fault conditions and V^2/R for this is 17.5W (and it's 0.7A, not 0,7mA)

I think you need to send us a sharp photograph of these devices and your wiring.

 

[daddles]

Is there any chance you could share this, or even email it to me? Would it be easy to adapt to 500mA (or thereabouts)?

Here's one description, but there are a number of others out on the web. When I built mine, I could fiddle with the pot adjusting the divided voltage reference and source any current from microamps on up. I have a beautiful old 500 k 10-turn pot with a turns counting dial and one of these days I'm going to turn it into a precision constant current source with full scale ranges by decades from 1 uA up to 1 A. It will get a lot of use on my bench (when I worked in industry in the 1980's, I had a nice HP 6181C precision current source that was very useful).

Oh, a nice thing about this design is that you can switch out the C battery and the circuit turns into a DC load (just stay below the MOSFET's voltage rating).

 

[MagicMatt]

Thanks for the clarification.

I will try and take a close-up shot - tried with the phone but it wasn't up to the job. I do have a better camera, just not to hand. I'll repost here if I can get a good shot.
This is the component, and the place I purchased from...
http://uk.rs-online.com/web/search/searchBrowseAction.html?method=getProduct&R=6710299
I'm normally ok muddling my way through with circuits, but these things just wouldn't behave. I can't think of anything I would have done to destroy them (not all 5 at any rate!).

Thanks very much for the link. It's a bit beyond my understanding, but I can copy the circuit and play around until I get something that does what I need. I had visions of it being a lot more complicated so I might even get to grips with how it works!

 

[TBennettcc]

Matt,

Can you post EXACTLY the way you have everything wired? The schematic diagram? I might be missing something, but I'm not quite sure you have everything wired correctly from your earlier descriptions...


Steve,

In your case, 700mA, 6V... ohms law says 8.57 ohms, use 8.2 ohms (it will drop 5.74V at 700mA). Technically you'd want a 20W resistor, but a 10W resistor should be fine for short periods of time with 12V across it.

...would you mind explaining how you did this? I'm brand new to learning about Ohm's law, and its application. Ohm's Law states that V / I = R, so that means that if there's a 6 volt potential, and you want 0.7 A to flow in the circuit, a resistance of 8.57 ohms is necessary. (Right?) But I don't understand why you used 6 V? Isn't his circuit 12V? And I'm not quite sure what you mean by "it will drop x V at y A"... Isn't the full potential of the circuit dropped across the resistor (and therefore doubling the current)? Or am I missing something?

Thank you so much for your time.

 

[MagicMatt]

Matt,
Can you post EXACTLY the way you have everything wired? The schematic diagram? I might be missing something, but I'm not quite sure you have everything wired correctly from your earlier descriptions...

Sure... here's what I have wired up. Doesn't get much more simple reallly, does it?
(I just got a new electronics program, can you tell?)

Measuring across R1 I get 0V
Measuring across R2 I get about 8V-9V at the max output of the power supply... around 3V less than what goes in (presumably the max it can output as it does state a 3V drop).

 

[ModemHead]

Well there's your problem. The output current must pass through R1, so attach the load to the REF terminal, not the OUT. Take a closer look at the datasheet examples for current regulator configuration.

 

[MagicMatt]

Ohhhhh.... *&$£*£$^^&"£&^!!!

Thank you.

Yes, I can see what I did now. I copied the circuit down wrong to begin with, worked out all the values... so of course when I checked the circuit I'd made against the one I scribbled down it matched...

I feel like such an idiot.

 

[ModemHead]

Thank TB for insisting that you document exactly how you had it wired.

Good news is, maybe some or all of your regulators are not toasted after all. In that configuration, there could only be 0V across the resistor, since ADJ current is negligible. Rewire and check'em out.

 

[MagicMatt]

Well the regs seem fine, but I did fry one of the RGBW Power-LEDs... could have been worse! Mind you, I reflow-soldered them using the cooker hob, so it may not have been full strength to begin with.

 

[(*steve*)]

Firstly, congratulations on everyone who helped work out the solution

In your case, 700mA, 6V... ohms law says 8.57 ohms, use 8.2 ohms (it will drop 5.74V at 700mA). Technically you'd want a 20W resistor, but a 10W resistor should be fine for short periods of time with 12V across it.

...would you mind explaining how you did this?

No problems.

A resistor that dropped a sensible amount of voltage at 700mA was required. I chose 6V because it gave the regulator plenty of headroom, and also because an absence of regulation would be clearly visible. In addition to that, across the full 12V the current would not be so dramatically large as to damage the regulator in the short term.

So having determined that 6V and 700mA were the basic conditions that were wanted, I applied ohms law...

OK, one quick thing. I simply remember V / (I x R) -- if you visualise that then you stick your finger over the thing you're trying to calculate and you'll see the formula. In this case I want R, so the formula is V/I.

Applying R = V/I you get (remembering to convert to amps!) 8.5714285714285714285714285714286 ohms. That many significant figures is rubbish of course, so I say 8.57 -- which is also stretching it, and round off to the closest preferred value. I chose an E12 value because they're pretty much guaranteed to be available. And that value is 8.2 ohms.

Incidentally, you could use 10 ohms and 700mA would drop 7V -- that would be fine too, but (again using ohms law, V = IxR) .7A through 8.2 ohms drops 5.74 volts -- close enough to 6V.

To calculate the power dissipation of the resistor under the fault conditions we use another extrapolation of ohms law P = V^2/R (The derivation of this is a combination of P = V x I and ohms law, using I = V/R and substituting to eliminate I).

At the full voltage (12V) the resistor (8.2 ohms) will dissipate 144/8.2, or 17.5W. For an extended period, you'd want a 20W resistor, but for a short term (long enough to measure the voltage) 10W should be fine.