Mains LED with blocking cap

[Dirk Bruere at NeoPax]

Any problems driving an LED from the mains using a blocking cap as a
constant current source?

 

[Gerard Bok]

Any problems driving an LED from the mains using a blocking cap as a
constant current source?

Many
Lack of isolation.
Lack of safety.
No (inrush) current limit.
(You want your LED to emit light, I guess, not to explode?)
Fire in the house after just one failing component

 

[Dirk Bruere at NeoPax]

Many
Lack of isolation.
Lack of safety.
No (inrush) current limit.
(You want your LED to emit light, I guess, not to explode?)
Fire in the house after just one failing component

The inrush can be dealt with using an inductor.
As for safety, a completely sealed and grounded unit although a 1:1
xformer might do both jobs

 

[Raveninghorde]

Any problems driving an LED from the mains using a blocking cap as a
constant current source?

Reverse voltage limit

 

[Dirk Bruere at NeoPax]

Reverse voltage limit

Well, with a diode or two of course.
I was mainly thinking of using it as a cheap constant current source.
Bearing in mind peak currents

 

[Gerard Bok]

The inrush can be dealt with using an inductor.

Neither a capacitor nor an inductor are suitable to create a
constant current source, imho.
In a transformerless mains design a capacitor can only reduce the
amount of power in the real current limiting device, usually a
resistor.

As for safety, a completely sealed and grounded unit although a 1:1
xformer might do both jobs

Hard to tell. With 230 Volt AC mains overhere, I am glad there is
some kind of transformer between mains and a led

 

[Pimpom]

Reverse voltage limit

An ordinary diode in reverse-parallel with the LED can take care
of that, that is if the OP really meant a single LED. Otherwise,
a FW bridge rectifier in series with the cap and the LED. A
resistor can limit the inrush current.

 

[Pimpom]

Neither a capacitor nor an inductor are suitable to create a
constant current source, imho.
In a transformerless mains design a capacitor can only reduce
the
amount of power in the real current limiting device, usually a
resistor.


Hard to tell. With 230 Volt AC mains overhere, I am glad there
is
some kind of transformer between mains and a led

Assuming that the OP meant a current setting device rather than a
true constant-current source, a capacitor can act as one because
it presents a constant reactance at the mains frequency. I've
used caps that way myself and have seen commercial products using
them.

The thing to watch out for is the initial current surge when the
circuit is switched on in mid-cycle. A series resistor can reduce
that inrush current. I've seen resistors as low as 47 ohms in
series with a cap and a string of 20mA LEDs in lamps meant for
230V 50Hz AC. That represents a theoretical worst-case inrush of
more than 6A if the lamp is switched on when the AC is at one
peak. Apparently it doesn't often cause a failure.

 

[John Devereux]

Well, with a diode or two of course.
I was mainly thinking of using it as a cheap constant current source.
Bearing in mind peak currents

Wouldn't a spike blast straight through the capacitor... (I = C.dV/dT)

 

[mike]

Assuming that the OP meant a current setting device rather than a
true constant-current source, a capacitor can act as one because
it presents a constant reactance at the mains frequency. I've
used caps that way myself and have seen commercial products using
them.

The thing to watch out for is the initial current surge when the
circuit is switched on in mid-cycle. A series resistor can reduce
that inrush current. I've seen resistors as low as 47 ohms in
series with a cap and a string of 20mA LEDs in lamps meant for
230V 50Hz AC. That represents a theoretical worst-case inrush of
more than 6A if the lamp is switched on when the AC is at one
peak. Apparently it doesn't often cause a failure.

Works fine until your air conditioner clicks off and the fast-rise
line transient takes out your device. Or lightning, or your neighbor's
air conditioner or welder or the power company flips a switch or or or.

Caps make great AC current limiters if you have FULL control of both
ends of the system.

 

[Pimpom]

Works fine until your air conditioner clicks off and the
fast-rise
line transient takes out your device. Or lightning, or your
neighbor's air conditioner or welder or the power company flips
a
switch or or or.
Caps make great AC current limiters if you have FULL control of
both
ends of the system.

I certainly wouldn't recommend such a system for something that
needs to be highly reliable. But it seems to be good enough for a
quick lash-up that wouldn't cause a disaster if it does break
down.

In fact, during an event that lasted for a few days last year, I
ran some three dozen lamps of the type I mentioned from a
square-wave UPS. The application required switching the lamps on
and off many times in an hour, at times several times a minute.
The switching was done with non-zero crossing solid-state relays.
None failed.

 

[Rich Grise]

Any problems driving an LED from the mains using a blocking cap as a
constant current source?

Yes.

Hope This Helps!
Rich

 

[Rich Grise]

Any problems driving an LED from the mains using a blocking cap as a
constant current source?

It'd be cheaper and safer to go to Radio Schlock and get a switching wall
wart:
http://www.radioshack.com/product/index.jsp?productId=2049709

Good Luck!
Rich

 

[John - KD5YI]

if you used an inductor you would not need to use a capacitor

About 16 henries?

 

[Phil Allison]

"John Fields"

Looking at a single white LED with a 3.5V drop across it and 20 mA
through it makes it look like a 175 ohm resistor:

** Totally irrelevant to the switch on transient situation.

We'll have the configuration which protects against reverse-voltage
failure and also, surprisingly, again inrush surge at turn-on,
regardless of the angle at turn-on.

** Total nonsense.

So why has the apparently nonexistent turn-on surge been considered
the culprit for so many failures for so long?

** Imagine the LED has 200 to 300 volts momentarily imposed on it ?

This is the situation when the switch is closed near a voltage peak.

What resistance does it exhibit then ?

Wot - no maker's data published for that ?

Try your idea out.

Betcha one dead LED - first shot.



..... Phil

 

[Pimpom]

The thing to watch out for is the initial current surge when
the
circuit is switched on in mid-cycle. A series resistor can
reduce
that inrush current. I've seen resistors as low as 47 ohms in
series with a cap and a string of 20mA LEDs in lamps meant for
230V 50Hz AC. That represents a theoretical worst-case inrush
of
more than 6A if the lamp is switched on when the AC is at one
peak. Apparently it doesn't often cause a failure.

---
Looking at a single white LED with a 3.5V drop across it and 20
mA
through it makes it look like a 175 ohm resistor:


E 3.5V
R = --- = ------- = 175 ohms
I 0.02A


If you're running 230V mains and using a capacitor as a
lossless
voltage dropper, then the equivalent circuit is:

175R
230AC>--[C1]--[LED1]--+
|
230AC>----------------+


If the RC is to allow 20mA through it with 230V across it, then
its
impedance must be:

E 230V
Z = --- = ------- = 11500 ohms
I 0.02A

So, since:

Z² = R² + Xc²


we can rearrange and solve for Xc:


Xc = sqrt(Z² - R²)

= sqrt(11500R² - 200R²)

~ 11499 ohms.

At 50Hz that would be a capacitance of:


1 1
C = ---------- = ---------------------- = 2.77 e-7F
2pi f Cx 6.28 * 50Hz * 11498R


or about 0.277µF, so our circuit now looks like this:


0.277µF 175R
230AC>--[C1]---[LED1]--+
|
230AC>-----------------+


If we put in another LED in parallel opposition to LED1:


0.27µF 175R
230AC>--[C1]-+--[LED1>]--+
| |
230AC>-------+--[<LED2]--+

We'll have the configuration which protects against
reverse-voltage
failure and also, surprisingly, again inrush surge at turn-on,
regardless of the angle at turn-on.

Try it:

Version 4
SHEET 1 880 680
WIRE 224 0 192 0
WIRE 336 0 288 0
WIRE -16 128 -64 128
WIRE 112 128 64 128
WIRE 192 128 192 0
WIRE 192 128 176 128
WIRE 224 128 192 128
WIRE 336 128 336 0
WIRE 336 128 288 128
WIRE -64 176 -64 128
WIRE -64 288 -64 256
WIRE 336 288 336 128
WIRE 336 288 -64 288
WIRE -64 336 -64 288
FLAG -64 336 0
SYMBOL cap 176 112 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName C1
SYMATTR Value 3.0e-7
SYMBOL voltage 80 128 R90
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0

So why has the apparently nonexistent turn-on surge been
considered
the culprit for so many failures for so long?

My guess is that it wasn't the turn-on surge at all, but rather
energy
stored in parasitic inductances which generated very high
currents for
very short periods of time during turn-off that did the damage.

Or switch bounce during turn-on doing the same thing.

How to protect against it?

Turn-on debounce and soft turn-off, but that's another post.

I haven't done the sim, but let's look at this qualitatively. The
calculations above treat the LED as a linear resistance which it
is not. If it has a drop of 3.5V at 20mA, at 10mA it may be 3.4V,
giving an equivalent resistance of 340 ohms. At a surge current
of 1A, the drop may be around 4V, giving only 4 ohms. And so on.
The fallacy here is in ignoring the highly non-linear behaviour
of diodes. A fast turn-on will see the capacitor as a very low
series impedance, resulting in a heavy inrush current.

 

[Phil Allison]

"Pimpom"

I haven't done the sim, but let's look at this qualitatively. The
calculations above treat the LED as a linear resistance which it is not.
If it has a drop of 3.5V at 20mA, at 10mA it may be 3.4V, giving an
equivalent resistance of 340 ohms. At a surge current of 1A, the drop may
be around 4V, giving only 4 ohms. And so on. The fallacy here is in
ignoring the highly non-linear behaviour of diodes. A fast turn-on will
see the capacitor as a very low series impedance, resulting in a heavy
inrush current.

** Precisely.

The situation is the same as charging that 0.28uF cap to 200 or 300 volts
and dumping the charge into the LED.

The tiny chip will likely give just one flash of light - and no more.


..... Phil

 

[Grant]

"Pimpom"


** Precisely.

The situation is the same as charging that 0.28uF cap to 200 or 300 volts
and dumping the charge into the LED.

The tiny chip will likely give just one flash of light - and no more.

I made mains LED lamp one a decade ago, with 1K 1W resistor for surges,
220nF cap and inverse parallel series string of 10 LEDs, thing ran
continuously no problem. Others along the same principle still running too.

Only problem I noticed was the older blue LEDs fading away to no light
over several months.

The resistor is required for spike suppression, it was a 1W for the voltage
rating (240V mains here).

Certainly I'd not run a LED from the mains without the resistor, and I'd
use a higher value than the 75R suggested upthread.

Grant.

 

[John Devereux]

The thing to watch out for is the initial current surge when the
circuit is switched on in mid-cycle. A series resistor can reduce
that inrush current. I've seen resistors as low as 47 ohms in
series with a cap and a string of 20mA LEDs in lamps meant for
230V 50Hz AC. That represents a theoretical worst-case inrush of
more than 6A if the lamp is switched on when the AC is at one
peak. Apparently it doesn't often cause a failure.

---
Looking at a single white LED with a 3.5V drop across it and 20 mA
through it makes it look like a 175 ohm resistor:


E 3.5V
R = --- = ------- = 175 ohms
I 0.02A


If you're running 230V mains and using a capacitor as a lossless
voltage dropper, then the equivalent circuit is:

175R
230AC>--[C1]--[LED1]--+
|
230AC>----------------+


If the RC is to allow 20mA through it with 230V across it, then its
impedance must be:

E 230V
Z = --- = ------- = 11500 ohms
I 0.02A

So, since:

Z² = R² + Xc²


we can rearrange and solve for Xc:


Xc = sqrt(Z² - R²)

= sqrt(11500R² - 200R²)

~ 11499 ohms.

At 50Hz that would be a capacitance of:


1 1
C = ---------- = ---------------------- = 2.77 e-7F
2pi f Cx 6.28 * 50Hz * 11498R


or about 0.277µF, so our circuit now looks like this:


0.277µF 175R
230AC>--[C1]---[LED1]--+
|
230AC>-----------------+


If we put in another LED in parallel opposition to LED1:


0.27µF 175R
230AC>--[C1]-+--[LED1>]--+
| |
230AC>-------+--[<LED2]--+

We'll have the configuration which protects against reverse-voltage
failure and also, surprisingly, again inrush surge at turn-on,
regardless of the angle at turn-on.

How does it protect against turn-on surge? (Sorry your netlist did not
work for me...)

[...]

So why has the apparently nonexistent turn-on surge been considered
the culprit for so many failures for so long?

My guess is that it wasn't the turn-on surge at all, but rather energy
stored in parasitic inductances which generated very high currents for
very short periods of time during turn-off that did the damage.

I don't see how stray inductance generates very high currents
(inductance normally works to stabilise current). High voltage,
potentially, but that is what the antiparallel diode is for.

Or switch bounce during turn-on doing the same thing.

How to protect against it?

Turn-on debounce and soft turn-off, but that's another post.

"Turn-on debounce" won't prevent the initial inrush current, and
soft-turn off is not required AFAICS. A simple RC feeding the leds
should work (as Larkin pointed out).

 

[Phil Allison]

"John Fields"
"Phil Allison"

"Pimpom"

Run the simulation.

** Pointless.

The real world always rules.



..... Phil