Re: lm317 as current regulator
- From: "James Thompson" <Jamesthompson2002@xxxxxxxxxxx>
- Date: Sun, 25 Jun 2006 20:43:21 -0400
"ehsjr" <ehsjr@xxxxxxxxxxxxxxxx> wrote in message
news:YyFng.9160$U%1.5651@xxxxxxxxxxx
James Thompson wrote:The transformer is from an old hifi receiver which gave 60vdc after the
<kellrobinson@xxxxxxxxxxxx> wrote in message
news:1151252918.638378.162040@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
James Thompson wrote:
When using the lm317t voltage regulator with a 1.2 ohm 5 watt resistor
for 1
amp limiting, my question is: Is the input limit on the ic still
limited to
about 40 volts?
What I am doing is, I have a supply voltage of 60 volt dc and I want to
limit the current draw to 1 amp for an led board I made. the leds will
be a
series link of 12 with 40 of these in parallel. By limiting the current
available to this panel, it should put the voltage of each led at 3.33
volt.
White leds here.
Would the lm317t work in current reg mode since it will only see 20 volt
across it.
Or should I first pre-regulate the 60 volt down to say about 42 volt?
You probably need a resistor for each string, to equalize currents.
And there's another reason you need extra resistance in the circuit;
your regulator will get too hot.
Even assuming all the led strings have equal current and voltage, look
at the circuit --
you drop 40 volts across the led's (12 X 3.3) and 1.2 volts across the
resistor,
leaving 18.8 nominal across the terminals of the lm317t. At one amp
that's 18.8 watts, way to much power
unless you have one hell of a heatsink. So use a resistor in each led
string.
It will equalize currents among the strings and get rid of extra heat.
You want to drop about 15 volts.
Exactly how much depends on how stiff your power supply is. But for now
we'll say 15 volts.
Now since you want 25 mA in each string, you will get a 15 volt drop
with
about 600 ohms (or nearby standard value) resistor in each string. Use
a half watt or one watt resistor.
If your power supply sags with an amp load, you could start with a 560
ohm resistor and see how that works.
If your supply doesn't droop you can go for more than 600 ohms. The
next standard value is 630.
The higher you go with the equalizing resistors, the cooler your
regulator chip will run, so try different values until
you find the right one.
Thanks kell, I have decided to abandon the current mode regulator
function and go with the lm317hvt regulator in voltage normal regulation.
The regulators can handle 1.5 amp and my led board I designed is split in
half, as to one side of 12 X 40 string then the other mirror image to
that. So a total of 960 leds.
The reason I abandoned the current mode is after reading what I typed in
I realized that by doing current reg mode, If 1 or 2 strings of 12
opened, It would over power the others and the whole board would burn
out. So Voltage mode it is. I am going to run the lm317hvt (2 x
parallel) to supply the 2 amp desired current. This project is for an
led light source to make an lcd projector. Started out making a board to
run at 102 vdc from mains direct regulation, and decided for safety
reasons to shy away from that foolishness. This seems as far as voltage
and current demand to be more logical. Lower voltage - higher current ,
that is why firstly looking at the 102 volt, as I could run 1024 leds
with only 800 ma. So this is middle of the road on voltage versus
current.
If anyone wants my pcb designs, I will send to then once I get them
tweaked just right. The led board is 8" X 8". JTT.
Oh boy. You really need to settle on a design.
Don't ever use two LM317's in parallel, without equalizing
resistors. In this case, do not parallel the LM317's
at all. Use a 3 amp LM350 - one heatsink, simpler design.
Assuming your 60 volt DC supply is tightly regulated and
will maintain 60 volts under a 2 amp load, use an LM350K
mounted on a good heatsink - Mouser part # 567-641-A will
work - and two 25 watt resistors in series between the +60
and the input pin on the LM350. One is 3 ohms, the other
is 3.3. Each one is 25 watts. Don't change this - it is
to spread the heat.
That will drop the voltage at the input to the LM350
to 60 - 12.6 or 47.4 volts with a 2 amp load. Keeps
some of the heat out of the LM350.
Set the LM350 for 44 volts out, with 240 ohms for R1 and
8.2K. 1/2 watt for R2. Use a 100 ohm resistor in series
with each of the 40 strings.
You *still* have a bucket of heat in a small area with the
LEDs: 960*.025*3.3 = 79.2 watts
Finally, what are you using for your 60 volts DC source?
The design above assumes a well regulated 60 volt DC
source capable of at leats 2 amps. You know what
"assumes" means.
Ed
bridge on the caps. It is center tapped, but not using the center. My led
board was designed as 2 banks sharing a center conductor, so I could feed
each side separately. This let me have 1 amp per side, and that is why the
lm317 being 2 of them sharing a single adj connection, but feeding each side
independently. I worded it wrong last post.
If I were to add a 10 ohm 10 watt resistor from the regulator output to feed
the led string, that would help on the dissipation of the regulator- right.
It would drop 10 volts at 1 amp on each the resistor and the regulator.
There is going to be forced air cooling on the led board as well as this
power supply.
.
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