lm317 as current regulator

[James Thompson]

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?

 

[William P.N. Smith]

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?

No, it's the differential across it that's limited to 40V.

 

[[email protected]]

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.

 

[James Thompson]

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.

 

[Lostgallifreyan]

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?

How much does your 60V power supply vary its voltage?

Your main need would be to avoid wasting power, if you're considering dropping
20 out of 60 volts, that's one third of your power gone. 20 watts lost at one
amp, when it's easy to limit that to around 3 to 5 watts.

As at least one earlier poster said, you'll want a single current limiter
resistor for each series string of LED's. If you try one LED at the maximum
current you want it to take, measure the voltage. Better to do this empirically
if you want to cut it close to save power.

LED voltage drops will vary, but not much, so each string's resistor should not
need to drop more than maybe a fifth of what any single LED will drop. Your
single LM317 current regulator (and it's control resistor) need only bear the
difference between minimum and maximum expected supply variation, plus up to
one LED's worth of voltage drop, plus its own total voltage drop at 1 amp. (A
low dropout type might save another watt at 1 amp).

Setting the optimum number of LED's is easy, start with one string of a few
more than you need, and cut back one at a time till the string lights up with
the lowest expected supply voltage applied. LED voltage drops are so well
behaved that if your PSU is stable to within less than two LED drops, you can
probably do it right with simple caclulations and not worry about overcurrent
or waste.

 

[Lostgallifreyan]

How much does your 60V power supply vary its voltage?

Your main need would be to avoid wasting power, if you're considering dropping
20 out of 60 volts, that's one third of your power gone. 20 watts lost at one
amp, when it's easy to limit that to around 3 to 5 watts.

As at least one earlier poster said, you'll want a single current limiter
resistor for each series string of LED's. If you try one LED at the maximum
current you want it to take, measure the voltage. Better to do this
empirically
if you want to cut it close to save power.

LED voltage drops will vary, but not much, so each string's resistor should
not
need to drop more than maybe a fifth of what any single LED will drop. Your
single LM317 current regulator (and it's control resistor) need only bear the
difference between minimum and maximum expected supply variation, plus up to
one LED's worth of voltage drop, plus its own total voltage drop at 1 amp. (A
low dropout type might save another watt at 1 amp).

Setting the optimum number of LED's is easy, start with one string of a few
more than you need, and cut back one at a time till the string lights up with
the lowest expected supply voltage applied. LED voltage drops are so well
behaved that if your PSU is stable to within less than two LED drops, you can
probably do it right with simple caclulations and not worry about overcurrent
or waste.

More:
I just saw the post about overcurrent if any series string went opencircuit. I
agree, voltage control is safer. You can still work that inst the context of my
main post though. Why set to exactly 12x40 array? The connection array need not
be the same as the physical one. Any inconvenience might well be more than
offset by the saving in power. You can regulate closer to 60V even though the
LM317 has a differential limit of around 37V, you put a zener of 30V or 40V in
the adjust network's connection to ground to get to a safe operating range on a
60V supply. If you use a chain of 5V1 or 5V6 zeners, you'll eliminate mostof
the change with temperature.

 

[ian field]

No, it's the differential across it that's limited to 40V.

But if you're drawing 1A with a drop of 40V you'd better check if the chip
can handle 40W!

 

[James Thompson]

But if you're drawing 1A with a drop of 40V you'd better check if the chip
can handle 40W!

If the specs say it can handle 1.5 amp with a differential of 37 volts,
would that not mean it can handle 55.5 watts. ??
My circuit will have a differential of 20 volts at 1 amp, which means it
will disapate 20 watt to heat. I have a fairly good sized heatsink for it
so all should be well. My transformer can source around 4 amp, so it also
should stay fairly cool. I am designing it and pcb in a rectangle shaped
caseing with a cooling fan on one end to dirct airflow over it all. Other
than the led's, I have a second winding with 17 volts to run a lm7812 for
running the lcd electronics.

 

[Lostgallifreyan]

If the specs say it can handle 1.5 amp with a differential of 37 volts,
would that not mean it can handle 55.5 watts. ??

No. You might get 1.5A, OR 37V safely handled, butthere are things like heat
dissipation to consider. Even then, I doubt the very best preparations will
keep the junction temperature low enough if you try for both at once.

My circuit will have a differential of 20 volts at 1 amp, which means it
will disapate 20 watt to heat. I have a fairly good sized heatsink for it
so all should be well. My transformer can source around 4 amp, so it also
should stay fairly cool. I am designing it and pcb in a rectangle shaped
caseing with a cooling fan on one end to dirct airflow over it all. Other
than the led's, I have a second winding with 17 volts to run a lm7812 for
running the lcd electronics.

Please read my other posts, I've tried to address this 20 volt drop. It's a
great waste that could be easy to avoid. I've probably made a gaff or two but
nothing uncorrectable. The main weakness of my case might be in assuming
excellent consistency in the voltage drops of your LED's, but if you're wanting
960, I guess you'll be buying 1000 from the same batch, so it's a reasonable
assumption that they're close enough to allow a low series resistance for each
chain of LED's.

If I were making this panel, I'd use a different transformer, or a cheap
effcient power converter if I could get one, to make 140 volts or more. If I
used an old valve transformer which could put out 280 - 300 VAC there need only
be 6 chains of 80 LED's, easy to wire in your array. Each would have a single
LM317T configged as current regulator. This way you need 6 of them, but you
don't need to waste power in series resistors, and most of the current will be
passing usefully through LED's, and each regulator will be running at around
170 mA, so would not need a heatsink at all, and probably no forced air
cooling, either. A further benefit is that you can guarantee equal
brightness from each chain without having large series resistors to even things
out if the LEDs' voltage drops vary too much. Yet another benefit of valve
(vacuum tube) PSU transformers is they have low volt windings as well.

 

[Lostgallifreyan]

Small extra point about safety:

60 VDC is actually dangerous, it only needs to drive 25 mA or so through a
heart to kill. It wouldn't kill everyone who touched it, but the odds are
enough to constitue a good game of Russian Roulette if people try.

My point is, that if you're running 60 VDC you're going to need protection such
that running 300 VDC isn't going to be any worse if you do it safely.

 

[ehsjr]

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

 

[James Thompson]

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

The transformer is from an old hifi receiver which gave 60vdc after the
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.

 

[James Thompson]

Small extra point about safety:

60 VDC is actually dangerous, it only needs to drive 25 mA or so through a
heart to kill. It wouldn't kill everyone who touched it, but the odds are
enough to constitue a good game of Russian Roulette if people try.

My point is, that if you're running 60 VDC you're going to need protection
such
that running 300 VDC isn't going to be any worse if you do it safely.

Your right, and it can sneak up on you. Got zapped good from a meter
calibrator supply once. It was set at 1000 volt too. Worse yet is radio,
had the top of of an old military tube radio and leaned on the top edge and
zap, made me buzz all day.

I will be very carefull with the danger there..

 

[Peter A Forbes]

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?

There is also the LM317HVK which has a 63V or thereabouts input voltage, made in
TO-3 and TO-220 formats.

The 1.5A output quoted for the LM317 is only for the TO-3 version IIRC, the
TO-220 build has a lower output current capacity unless you are well below its
safe operating area.

Peter

 

[Byron A Jeff]

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?
Yes.

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.

Don't white LEDs typically have a 4-5V Vled? You really will need 60V to drive
this panel.

Would the lm317t work in current reg mode since it will only see 20 volt
across it.

Nope. Why do you need an lm317t anyway? Just stick a current limiting resistor
in front of the panel to limit the amount of current through it.

My question is how bright will the LEDs be if you only run 1A through the panel.

Or should I first pre-regulate the 60 volt down to say about 42 volt?

Will you have enough voltage to drive 12 white LEDs at 42V? What's the forward
voltage drop of these LEDs?

Also do you have individual access to each string? The truth of the matter is that
each string really needs its own current limiting resistor.

BAJ

 

[James Thompson]

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?

Update: Advice taken from replies.
Pcb revised to 17 led per string, with a 20 ohm .5 watt resistor per string.
At 3.5 volt per led fv drop that is 59.5 volt drop per led string, with the
remaining .5 volt on the 20 ohm resistor which sets the Current per string
to 25ma and therefore has the resistor dissipating .0125 watt. The resistor
being .5 watt will not even get warm. All power from the supply of 60 volt
is used for producing light instead of just heat. If this is correct, I
will not even need a regulator. Total Current draw for the led's will be
1.25 amp.
Does this seem to be the best configuration?

 

[ehsjr]

The transformer is from an old hifi receiver which gave 60vdc after the
bridge on the caps. It is center tapped, but not using the center.

How do you know the transformer is capable of 3+ amps?
(You need more than 2 amps ac to get 2 amps DC in this
set up.) How much capacitance? Under what load did it
give 60 volts?

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.

You can do that if you want, at higher cost. The LM350K
on a single heatsink is lower cost and takes less space
than 2 LM317K's on two heatsinks.

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.

Yes, and it would screw up the regulation. In this case, with
a fixed load, that's not a problem, provided the +60 is tightly
regulated. But you'd get the same benefit with the resistors
on the input of each LM317, and would not screw up the regulation,
again assuming tight regulation of the +60 volts. If the +60
volts sags appreciably (more than 3 volts) under load the
resistors have to be changed or eliminated. And if it sags
more than 13 volts, you can't set the regulators for 44 volts
output.

It would drop 10 volts at 1 amp on each the resistor and the regulator.

No. The resistor would drop 10 volts at 1 amp.
The regulator would drop Vin-Vout at 1 amp.

If the +60 volts is tightly regulated and you
set Vout to 54 volts, the regulator will drop
6 volts at 1 amp, the 10 ohm resistor will drop
10 volts at 1 amp, each 100 ohm resistor will
drop 4 volts at .025 amps, and each LED string
will drop 40 volts at .025 amps.

By tightly regulated, I mean that there is very
little variation in the +60 volts from no load
to full load, with full load being 2 amps.

There is going to be forced air cooling on the led board as well as this
power supply.

Forced air cooling on the power supply is good.
With the design I posted it will work fine. I
cannot predict with accuracy what will happen
to the LEDs - just applying forced air cooling
may not cool them enough.

Regarding your 60 volt supply: you'll be using
at least 120 watts in your LEDs and regulators.
(2 amps at 60 volts)
Your 60 volts supply has to be capable of at
least that.

Ed

 

[Lostgallifreyan]

Update: Advice taken from replies.
Pcb revised to 17 led per string, with a 20 ohm .5 watt resistor per string.
At 3.5 volt per led fv drop that is 59.5 volt drop per led string, with the
remaining .5 volt on the 20 ohm resistor which sets the Current per string
to 25ma and therefore has the resistor dissipating .0125 watt. The resistor
being .5 watt will not even get warm. All power from the supply of 60 volt
is used for producing light instead of just heat. If this is correct, I
will not even need a regulator. Total Current draw for the led's will be
1.25 amp.
Does this seem to be the best configuration?

It's pretty good You got the idea of getting the LED's to use all available
voltage overhead, reducing the need to get rid of waste heat. If your PSU is
very well behaved, you don't need the regulator. Just bear in mind that your
resistor is only dropping half a volt! While that's efficient, it's a lot
closer than my posts advised, so your supply is going to have to be VERY well
behaved. I think you'll find that the line voltage will vary more. 0.5V in so
is 1 V in 120V, or 2V in 240. You can be damn sure that the average mains
supply is going to drift more than this.

I still advise using s valve (vacuum tube) transformer for around 300V to get
strings as long as 40 LED's, and replacing each resistor with an LM317 current
regulator. You only need 6 of those then, and the performance will be far
better and far easier to control.

I'd been thinking about the heat from the panel itself, btw.. I don't know what
the efficiency is for those LED's. maybe 15% or so? So not all power in will be
heat out, so at least one previous poster gave wrong figures for heat. I guess
it could be 70 watts lost, from a likely panel size of 9x30 cm. The odds are
that you won't need a fan, if you can get decent convection around the LED's.

 

[ehsjr]

It's pretty good

Not really. In fact, it's a disaster. The only regulation
in evidence is the transformer; the only supply in
evidence is the transformer, diodes and capacitor bank.
Look at worst case: Vf 3.3, line voltage 130. Vary the
line voltage up to 130, and his 48v transformer will put
out 52v. Allowing a volt drop for each diode at ~2 amp,
that's 50 * 1.414 or 70.7 volts. At Vf 3.3, 17 LEDs drop
(17*3.3) 56.1 volts. That means 13.9 volts across his 20
ohm resistor, for 695 mA.

Disaster - almost 10 watts for the 1/2 watt resistor and
more than 27 times the target current through the LEDs.
And with 56 strings (960/17)= 56 + the total current would
be (56*.695) ~ 38 amps!

Assume it is not a 48 volt transformer. (Well right away
we're in trouble with "assume".) Lets say its a 44 volt
transformer. At 130 volts, it will put out 47.7V which
after the 1 volt diode drop (2 diodes) becomes ~64.6
(45.7 * 1.414) = ~ 64.6 volts. With Vf = 3.3 the 17 LEDs
still drop 56.1 volts, leaving 8.5 v across the 20 ohm
resistor for 425 mA current. Goodbye to one or more
of powersupply, LED, resistor. The magic smoke will get
out.

If you go with with a higher value resistor to get
..025 mA, that's 8.5/.025 = 340 ohms (best case transformer,
worst case Vf and line voltage). What happens when the
voltage is 120, and the Vf is 3.5 if you use a 340 ohm
resistor? .5/340 = ~1.5 mA and the LEDs are too dim.

He *must* regulate. *If* he posts that the 60V supply
is really a regulated 60 volts, not just a transformer,
bridge and capacitor bank, then a resistor current
limiter would be appropriate. In that case the regulator
is already there, in the supply. He has *got* to have a
regulator in there, somewhere. It's an awful lot of
work to put 960 LEDs in there - it certainly warrants
spending $13.00 on regulation.

You got the idea of getting the LED's to use all available
voltage overhead, reducing the need to get rid of waste heat.

Bad idea. He has *got* to regulate. That means a linear,
which will waste power as heat, or a switcher, which is
not under discussion here. If he is the same person who
has been e-mailing me since 6/13 under a couple of different
id's, then he has a switcher solution that is cheap ($10.00
for the switcher) and solid, but requires work: 3 LEDs per
string. He also has been given a solution in e-mail that
works without a transformer and is easier (40 LEDs per
string) - but is dangerous due to high (up to ~182) voltage.

The waste heat from the regulator is of no concern, other
than cost. It requires about 13 dollars for the LM350K
and heatsink. It is not required to be near the LEDs,
so it won't cook them.

If your PSU is
very well behaved, you don't need the regulator.

Yes, he does. It can be very well behaved, but if
there is no regulator in the supply, line voltage
variation can kill the LEDs, as shown. Varying
the worst case the other way (line voltage = 105,
Vf = 3.7) will just make them dimmer than nominal.

Just bear in mind that your
resistor is only dropping half a volt! While that's efficient, it's a lot
closer than my posts advised, so your supply is going to have to be VERY well
behaved. I think you'll find that the line voltage will vary more. 0.5V in so
is 1 V in 120V, or 2V in 240. You can be damn sure that the average mains
supply is going to drift more than this.
Exactly.


I still advise using s valve (vacuum tube) transformer for around 300V to get
strings as long as 40 LED's, and replacing each resistor with an LM317 current
regulator. You only need 6 of those then, and the performance will be far
better and far easier to control.

That's a non-starter. The differential across the LM317
would be huge. 40 leds at 3.7 = 148 volts, which means
188 volts is the upper limit if the LM317 can handle
40 volts difference. (AIRC, the max difference is 37,
but that's not worth looking up when we're talking
hundreds of volts difference)

I'd been thinking about the heat from the panel itself, btw.. I don't know what
the efficiency is for those LED's. maybe 15% or so? So not all power in will be
heat out, so at least one previous poster gave wrong figures for heat. I guess
it could be 70 watts lost, from a likely panel size of 9x30 cm. The odds are
that you won't need a fan, if you can get decent convection around the LED's.

Yes/no. A total of ~79 watts will be produced. Some of
it will be heat due to inefficiency, the rest (most, I
would think) will convert to heat both in the LED and
external (far away) to it. Light energy becomes heat
energy as it is absorbed in a surface; even a transparent
surface absorbs some and converts it to heat. So he'll
have 15% or whatever due to inefficiency, plus an
additional amount due to absorbtion. The point is not
how much heat exactly - I don't know - but how do you
get rid of whatever heat is there. If there is extremeley
limited convection, I don't know how to do it.
If it's the same person, I told him in e-mail that I was
not qualified to tell him how to deal with the heat in
the LEDs. But I doubt that he'll be able to cram 960
leds into as small an area as possible without a significant
heat problem. In the e-mail he spoke of replacing a projector
bulb with these LEDs. If he can spread his LEDs out so that
decent convection between the LEDs possible, that worry
becomes *much* smaller than what I envisioned.

Ed

 

[Lostgallifreyan]

Not really. In fact, it's a disaster. The only regulation
in evidence is the transformer; the only supply in
evidence is the transformer, diodes and capacitor bank.
Look at worst case: Vf 3.3, line voltage 130. Vary the
line voltage up to 130, and his 48v transformer will put
out 52v. Allowing a volt drop for each diode at ~2 amp,
that's 50 * 1.414 or 70.7 volts. At Vf 3.3, 17 LEDs drop
(17*3.3) 56.1 volts. That means 13.9 volts across his 20
ohm resistor, for 695 mA.

If you're going to tear up a post and say four paragraphs to the extent of
saying that mains power isn't well behaved, you'd do better to quote the
context. I said that in ONE paragraph.

Bad idea. He has *got* to regulate.

Yes, IF the initial supply isn't well behaved, as in regulated to start with. I
said this. You chose to omit that from your quote, but that doesn't make it
unsaid.


That means a linear,

which will waste power as heat, or a switcher, which is
not under discussion here. If he is the same person who
has been e-mailing me since 6/13 under a couple of different
id's, then he has a switcher solution that is cheap ($10.00
for the switcher) and solid, but requires work: 3 LEDs per
string. He also has been given a solution in e-mail that
works without a transformer and is easier (40 LEDs per
string) - but is dangerous due to high (up to ~182) voltage.

Now I get it. You're heading me off at the pass, you've offered a private
solution and you would rather tear my posts up than let them challenge your
efforts, it seems. As I've also mentioned in my posts switching regulators for
efficiency and 40-LED chains at high voltage I can see why it might look like
competition to you. Incidentally, if you also concede that making longer
strings for higher voltages is good, why did you say only a few sentences ago
that my suggestion of making the LED's drop most of the voltage was a bad idea?

Yes, he does. It can be very well behaved, but if
there is no regulator in the supply, line voltage
variation can kill the LEDs, as shown. Varying
the worst case the other way (line voltage = 105,
Vf = 3.7) will just make them dimmer than nominal.

Yes, yes, I DID mention that the mains supply WOULD would vary far more than
the 0.5V he'd allowed across his limiting resistor, but you chose to cut that
out of your quote before proceeding to administer your professorial thrashing.

Exactly.

Ah, so you did quote it. Hell, you even agree. > So why bother with those
4 paragraphs of mathematically inclined invective? It's easier to cut to the
chase. It's easier to advise someone if you keep it short, that way they're
less likely to miss context. Still, my post was short and you still missed it
till now, but I did try...

That's a non-starter. The differential across the LM317
would be huge. 40 leds at 3.7 = 148 volts, which means
188 volts is the upper limit if the LM317 can handle
40 volts difference. (AIRC, the max difference is 37,
but that's not worth looking up when we're talking
hundreds of volts difference)

CURRENT REGULATOR! I guessed someone would see my loose phrasing and make a
more serious mistake, but I didn't expect it to be someone who posted like a
maths professor. Look at the data sheet. The LM317 can be configged as a
current regulator, even though it is a voltage regulator. And as such it does
NOT see a high voltage. It doesn't care about the voltage, all it sees is
current (actually, the small voltage across a resistor), and the high voltage
is mostly dropped across LED's, hence the increased efficiency. And yes, for
the record, the LM317 itself IS limited to 37V, and if you actually read my
posts you'll know that I also know this. You'd also know by now, even from this
quote of my post, that I was insisting on regulation in an ideal solution.

Yes/no. A total of ~79 watts will be produced. Some of
it will be heat due to inefficiency, the rest (most, I
would think) will convert to heat both in the LED and
external (far away) to it. Light energy becomes heat
energy as it is absorbed in a surface; even a transparent
surface absorbs some and converts it to heat. So he'll
have 15% or whatever due to inefficiency, plus an
additional amount due to absorbtion. The point is not
how much heat exactly - I don't know - but how do you
get rid of whatever heat is there. If there is extremeley
limited convection, I don't know how to do it.
If it's the same person, I told him in e-mail that I was
not qualified to tell him how to deal with the heat in
the LEDs. But I doubt that he'll be able to cram 960
leds into as small an area as possible without a significant
heat problem. In the e-mail he spoke of replacing a projector
bulb with these LEDs. If he can spread his LEDs out so that
decent convection between the LEDs possible, that worry
becomes *much* smaller than what I envisioned.

Ed

I'll relax my defence on the heat issue, as I have no idea what actual form
this thing wil take. A fan won't hurt, for sure. As worst case you can assume
heat out locally equals heat in, if you budget for that you won't go wrong, but
in practise this thing would have to be DIRE if that little light got out, and
it certainly wouldn't be competing with a projector lamp. Yor figure of 79W is
the total power drawn by the LED's given 25 mA and 3.3 V dropped for each one.
Again, if 15% of that energy got out as light, that's about 70 W to be removed
locally as heat, as I said.