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Protecting LEDs from exceeding max current

S

sinthreck

Jan 1, 1970
0
I'm using a step down converter to step down 5V to 3.3V

The load is: 3 Lumileds (350mA * 3) = 1.05 A

I'm worried that in the event of the PWM being locked high the max
current of the LEDs will be exceeded and they will all burn out, a
very expensive anomoly.

I'm wondering what cheap and efficient solution is available to me.

Cheers,

James A
 
L

Leon Heller

Jan 1, 1970
0
sinthreck said:
I'm using a step down converter to step down 5V to 3.3V

The load is: 3 Lumileds (350mA * 3) = 1.05 A

I'm worried that in the event of the PWM being locked high the max
current of the LEDs will be exceeded and they will all burn out, a
very expensive anomoly.

I'm wondering what cheap and efficient solution is available to me.


If you have an MCU in the system you could detect the PWM output and
shut it down if it malfunctions.

Leon
 
S

Spehro Pefhany

Jan 1, 1970
0
If you have an MCU in the system you could detect the PWM output and
shut it down if it malfunctions.

How about detecting overvoltage across the LEDs and shutting the PWM
down? Maybe with an automatic re-try, depending on the consequences
of losing the light due to electrical noise or whatever.

Best regards,
Spehro Pefhany
 
G

Geir Klemetsen

Jan 1, 1970
0
sinthreck said:
I'm using a step down converter to step down 5V to 3.3V

The load is: 3 Lumileds (350mA * 3) = 1.05 A

I'm worried that in the event of the PWM being locked high the max
current of the LEDs will be exceeded and they will all burn out, a
very expensive anomoly.

I'm wondering what cheap and efficient solution is available to me.

Use an constant-current generator. The scheme I have contains only two NPN
transistors and two resistors.
 
D

Dana Raymond

Jan 1, 1970
0
Well, if the supply goes bad then something needs to be repaired, right?
Why do you need fault tolerance for the leds?

Does the converter have an output from the error amp, or depending on type,
some other type of voltage detector? If so, integrate that (resistor, cap,
time delay) and use it to shutdown the fets. Or, if voltage sensing is the
only issue (control shutdown is easy) then use a dallas or likewise 5V
voltage detector chip.

You can also protect at the load via a raychem polyfuse, or a series of
diodes/zener diode to limit the voltage.

There are so many ways to skin this cat. Care to share a schematic?

BTW is 3V3 supply being used by other circuitry? If not, then consider
turning it into a constant current source instead and eliminate the
resistor, which will improve efficiency. A raychem can protect in that case.

Dana Frank Raymond
 
I

Ian Stirling

Jan 1, 1970
0
sinthreck said:
I'm using a step down converter to step down 5V to 3.3V

The load is: 3 Lumileds (350mA * 3) = 1.05 A

I'm worried that in the event of the PWM being locked high the max
current of the LEDs will be exceeded and they will all burn out, a
very expensive anomoly.

I'm wondering what cheap and efficient solution is available to me.

I think I'd go with a quick-blow fuse.
Put an amp fuse in there, and it should work.
The LEDs should be able to take a little surge.
For extra protection, add a 5W 3.7V diode across the LEDs.

--
http://inquisitor.i.am/ | mailto:[email protected] | Ian Stirling.
---------------------------+-------------------------+--------------------------
"Melchett : Unhappily Blackadder, the Lord High Executioner is dead
Blackadder : Oh woe! Murdered of course.
Melchett : No, oddly enough no. They usually are but this one just got
careless one night and signed his name on the wrong dotted line.
They came for him while he slept." - Blackadder II
 
I

Ian Stirling

Jan 1, 1970
0
sinthreck said:
I'm using a step down converter to step down 5V to 3.3V

The load is: 3 Lumileds (350mA * 3) = 1.05 A

I'm worried that in the event of the PWM being locked high the max
current of the LEDs will be exceeded and they will all burn out, a
very expensive anomoly.

I'm wondering what cheap and efficient solution is available to me.

Alternative to my other suggestion, you might try a 0.75A fuse in the
5V line.
I would consider reconsidering the 3.3V drive.
There are various voltage bins that may draw significantly more than
nominal current at 3.3V, unless you select the LEDs, they will be
unevenly driven.
And the tempco (-.175V/100C) will cause the current to vary by around
a third over that voltage.
 
S

Spehro Pefhany

Jan 1, 1970
0
Alternative to my other suggestion, you might try a 0.75A fuse in the
5V line.

With fuses, watch the voltage drop. Beware that fuse data sheets
typically only specify the "cold" resistance.
I would consider reconsidering the 3.3V drive.
There are various voltage bins that may draw significantly more than
nominal current at 3.3V, unless you select the LEDs, they will be
unevenly driven.
And the tempco (-.175V/100C) will cause the current to vary by around
a third over that voltage.

There are also diminishing returns on something like this. As someone
has already pointed out, if it fails something needs to be fixed. In
many businesses that represents a minimum of $100 or $200 worth of
cost, and $50 or $75 worth of LEDs in some fraction of the failures
may not really influence it that much. OTOH, if the added protection
components *cause* slightly more field failures, you've saved the LEDs
but lost any cost advantage. This is not really like a safety-critical
system where the cost of a failure might be many orders of magnitude
higher, but you can use some of those techniques- redundancy, watchdog
circuitry, avoiding or tightly controlling the use of microprocessors
and other unduly complex logic in the critical bits.

Best regards,
Spehro Pefhany
 
K

Kevin McMurtrie

Jan 1, 1970
0
sinthreck said:
I'm using a step down converter to step down 5V to 3.3V

The load is: 3 Lumileds (350mA * 3) = 1.05 A

I'm worried that in the event of the PWM being locked high the max
current of the LEDs will be exceeded and they will all burn out, a
very expensive anomoly.

I'm wondering what cheap and efficient solution is available to me.

Cheers,

James A

Use a crowbar to blow a fuse. It's typically an SCR with a zener diode
or some other overvoltage trigger.
 
W

Watson A.Name - Watt Sun

Jan 1, 1970
0
I'm using a step down converter to step down 5V to 3.3V

The load is: 3 Lumileds (350mA * 3) = 1.05 A

I'm worried that in the event of the PWM being locked high the max
current of the LEDs will be exceeded and they will all burn out, a
very expensive anomoly.

I'm wondering what cheap and efficient solution is available to me.

You can play the overvoltage crowbar game that the SMPS makers use.
You use a current sense resistor in series with one of the LEDs. This
goes into a comparator, and the comparator output pulls the gate of a
SCR high, which turns on the SCR and shorts out the output of the PS.


--
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W

Watson A.Name - Watt Sun

Jan 1, 1970
0
I think I'd go with a quick-blow fuse.
Put an amp fuse in there, and it should work.
The LEDs should be able to take a little surge.
For extra protection, add a 5W 3.7V diode across the LEDs.

Did you mean a 3.7V Zener diode? In that case, it had better be a
heavy wattage one. It will have to handle more than an amp or more
than 3.7 Watts. But the problem is that it will have some dynamic
resistance, and as the current rises, the current may not all go thru
the zener, and still overload the LEDs. Or would that be DEDs?


--
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###Got a Question about ELECTRONICS? Check HERE First:###
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My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
 
W

Watson A.Name - Watt Sun

Jan 1, 1970
0
Alternative to my other suggestion, you might try a 0.75A fuse in the
5V line.
I would consider reconsidering the 3.3V drive.

I think I'd reconsider replacing it with a constant current source.
There are various voltage bins that may draw significantly more than
nominal current at 3.3V, unless you select the LEDs, they will be
unevenly driven.
And the tempco (-.175V/100C) will cause the current to vary by around
a third over that voltage.

Exactly.

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
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My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
 
T

Tim Shoppa

Jan 1, 1970
0
sinthreck said:
I'm using a step down converter to step down 5V to 3.3V

The load is: 3 Lumileds (350mA * 3) = 1.05 A

I'm worried that in the event of the PWM being locked high the max
current of the LEDs will be exceeded and they will all burn out, a
very expensive anomoly.

I'm wondering what cheap and efficient solution is available to me.

It's straightforward to convert a "simple switcher" style supply
to constant-current use. Just use the -ADJ part and size the current
sense resistor so that at the rated current, the voltage drop is 1.2V.
The sense resistor has to go on the negative side of the load unless you
want to play some games.

If you go outside the "simple switcher" realm the current sense resistor
can be made even smaller.

You may want to look at other power supply/load topologies in the
context of your application. Putting the three LED's in series is
an obvious step... that way you end up with a 350mA supply, not a
1.05A supply. Step-up from 5V is quite feasible, and you end up
dissipating less power in the sense resistor.

The best device for protecting against overcurrent in case of
catastrophic failure is a fuse. A fast-blow will protect those
expensive LED's quite nicely.

Tim.
 
I

Ian Stirling

Jan 1, 1970
0
Watson A.Name - "Watt Sun said:
Did you mean a 3.7V Zener diode? In that case, it had better be a
heavy wattage one. It will have to handle more than an amp or more
than 3.7 Watts. But the problem is that it will have some dynamic
resistance, and as the current rises, the current may not all go thru
the zener, and still overload the LEDs. Or would that be DEDs?

The zener would only cope with that that the LEDs won't take, it'll
actually be conducting somewhere in the range of 1-300ma, at its nominal
voltage.
It's just to ensure the fuse blows promptly.

The LEDs (assuming they have a similar internal resistance of a couple
of ohms each that my luxeon star has (when I reconnect the bondwires to
something).
They won't go up to double nominal current until around .6V over nominal
voltage.
Looking at the derating curves, they seem to imply a maximum die temp of
125C.
A luxeon star (hanging in mid-air) that I measured got around 40C/W.
Doubling the current for a short period (under the second or so
a quickblow fuse would take to blow) should not
cause any harm, even at high ambient temperatures.
 
J

John Fields

Jan 1, 1970
0
I'm using a step down converter to step down 5V to 3.3V

The load is: 3 Lumileds (350mA * 3) = 1.05 A

I'm worried that in the event of the PWM being locked high the max
current of the LEDs will be exceeded and they will all burn out, a
very expensive anomoly.

I'm wondering what cheap and efficient solution is available to me.

---
Cheap and efficient don't necessarily go hand in hand, but if
reliability is what you're looking for it's hard to ignore a KISS
solution!

What I'd do would be to forget the 5V to 3.3V converter and power the
LEDs with PWM'ed 5V and use a series current limiting resistor on each
LED. Assuming full brightness with 3.3V across, and 350mA through each
LED, the resistor comes out to:

R = (Vsupply-Vled)/Iled = (5V - 3.5V) / 0.35A ~ 4.3 ohms.

The power dissipation will be:

P = (Vsupply-Vled)*Iled = (5V - 3.5V) * 0.35A ~ 0.53 watts,

which is not bad, and on top of that 4.3 ohms is a standard 5% value, so
if you wanted to make up a one watt resistor out of cheap quarter
watters you could do something like this:




+5V>-+-[4R3]-+-[4R3]-+-[LED>]---+
| | |
+-[4R3]-+-[4R3]-+ |
C
PWM>--------------------[R]---B
E
|
GND>----------------------------+
 
L

Leeper

Jan 1, 1970
0
Well, LumiLEDs actually warned about too much current though the Luxeon
devices,
during a seminar, they said there was a lack of carriers at higher currents
and
it causes tunneling in the device, that damages the die, which leads to
early failures.

Although, they did not indicate at which current levels this starts to
begin.

There is an application note on their website now, about how to "safely"
pulse their
devices...
 
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