Series capacitor ?

[Edwin2]

Hi All !

I am ashamed to ask, but just to check if I am correct with this.

I wanted to connect a series of LEDs direct to the mains. (230V/50Hz)

Therefore I need 10... 15 kOhm in series.

I calculated 1 / ( 2 x pi x 50 x 220 nF ) and came to 14.468 Ohm.
Is this correct for this situation ?
I will use an extra 1 or 2,2 kOhm resistor in series, and for safety
a zener parallel over the LEDs. Perhaps the 1 or 2,2 kOhm resistor
`below` the zener.

Kind regards,

Edwin

 

[Dean Huster]

Edwin, that should be 14.468K ohms, not 14.468 ohms. However, to set up the entire calculation and judge its accuracy, we need to know the number of LEDs you're planning on placing in series. If you're designing a reactive current limiter, why the resistor?

Dean

 

[Edwin2]

The resistor(s) is (are) used to limit the initial chargecurrent from the capacitor.

In this case it is for 12 white LEDs, each 2,5 Volt, 20 mA. (From datasheet)
Making 30 Volt DC.
These are hanging in a bridgerectifier 4x 1N4007.

About the Xc: My calculator still makes 14 1/2 kOhms.
14 1/2 MOhms, like you said, would be a bit very much, don`t you think ?

Kind regards,

Edwin

 

[silvarblade]

Just asking why dont you build a proper supply (a simple one) with a xformer and then the bridge?that would be lot more efficient and less bulky.

 

[MrAl]

Hi there,


The reason for using a series cap is to keep parts costs down to a minimum.
A transformer obviously would cost more money.
The reason for using a series resistor (with the cap) is to keep the initial
surge (when the circuit is first plugged in) down to some reasonable level,
and this also helps during line transients which look like spikes of voltage.
Without the extra resistor the LEDs could get banged with several amps
during these times.

Anyway, using a full wave bridge rectifier the following values will drive
either 12 x 2.5v LEDs or even 12 x 3.5v LEDs (little less current for 3.5v LEDs):

C=0.33uf (should use a 450v or better rating here however)
R=390 ohms, 1/4 watt

If you feel that R should be higher, 1k, then it must be a 1/2 watt unit (or better).


Using the above component values the average current through the LEDs is
approximately 20ma.

BTW, changing C to 0.22uf the current decreases to about 13ma average,
which still isnt too bad.

Also, adding a 100uf 100v cap to the output of the bridge rectifier will reduce
any blinking effect, if any is noticed.

 

[Valence_4]

Less bulky yes, but much LESS efficient. The series capacitor is a REACTIVE load and it will not absorb any energy. Furthermore, it's capacitive reactance will sightly help compensate for a nearby inductinve (transformer) reactive load.

However, take great care with this circuit. it does NOT provide ANY isolation between the LED's and the mains. Care should be taken to get sure no one will ever be able to touch the LED's legs. To gain isolation, you'll have to use a transformer.

 

[ericgibbs]

The resistor(s) is (are) used to limit the initial chargecurrent from the capacitor.

In this case it is for 12 white LEDs, each 2,5 Volt, 20 mA. (From datasheet)
Making 30 Volt DC.
These are hanging in a bridgerectifier 4x 1N4007.

About the Xc: My calculator still makes 14 1/2 kOhms.
14 1/2 MOhms, like you said, would be a bit very much, don`t you think ?

Kind regards,

Edwin

hi,
As Dean states at 50Hz the Xcap, of a 220nF is 14.468K, which gives a series current of about 16mA when connected to a 230Vac supply.

Xc=1/(2*pi * 50 * .00000022)

This type of circuit is not mains isolated.

 

[MrAl]

Hi there,


The reason for using a series cap is to keep parts costs down to a minimum.
A transformer obviously would cost more money.
The reason for using a series resistor (with the cap) is to keep the initial
surge (when the circuit is first plugged in) down to some reasonable level,
and this also helps during line transients which look like spikes of voltage.
Without the extra resistor the LEDs could get banged with several amps
during these times.

Anyway, using a full wave bridge rectifier the following values will drive
either 12 x 2.5v LEDs or even 12 x 3.5v LEDs (little less current for 3.5v LEDs):

C=0.33uf (should use a 450v or better rating here however)
R=390 ohms, 1/4 watt

If you feel that R should be higher, 1k, then it must be a 1/2 watt unit (or better).


Using the above component values the average current through the LEDs is
approximately 20ma.

BTW, changing C to 0.22uf the current decreases to about 13ma average,
which still isnt too bad.

Also, adding a 100uf 100v cap to the output of the bridge rectifier will reduce
any blinking effect, if any is noticed.

Hi again,


Good points about the mains isolation. These 'offline' power supplies
are not electrically isolated from the mains so much care has to be
used when applying them.
For example, i dont think i would use this circuit in a bathroom or
anywhere else where water could come in contact with the case
and penetrate, which could cause a very bad shock hazard.

 

[Dean Huster]

A reactive divider is more efficient than either a transformer or resistive divider. The cap doesn't dissipate heat and have IR losses like the other devices. Any energy taken in by the cap is released either to the line or the load. And when you figure that most appliances are inductive in nature, maybe a capacitive reactive divider is good for the overall power factor of the house!

I calculated 1 / ( 2 x pi x 50 x 220 nF ) and came to 14.468 Ohm.

That was where I got your original calculation figures. No mention of 14.468M ohms from me, however!

As long as you keep everything within insulated confines, you should be OK electrically. The world is getting away from isolated supplies in many cases because of their lower cost.

Dean

 

[Edwin2]

Hi again all !

about the mains isolation: Do you have all your lightbulbs isolated with a transformer ?
I don`t think so, so why, if one uses LEDs it suddenly should need a transformer ?

OFCOURSE you have to take care, that no-one can touch the wires, but isn`t that the
same with normal lightbulbs ? So, no difference at all.

Again the Xc: Can anyone explain how they come to 14 and a half MegOhms ?
You are writing 14.468 kilo-Ohms.
My calculator always shows 14,468 kilo-Ohms.... (almost 14 1/2 kOhm)

What is going wrong here ?

Kind regards,

Edwin

 

[ericgibbs]

Hi again all !

about the mains isolation: Do you have all your lightbulbs isolated with a transformer ?
I don`t think so, so why, if one uses LEDs it suddenly should need a transformer ?

OFCOURSE you have to take care, that no-one can touch the wires, but isn`t that the
same with normal lightbulbs ? So, no difference at all.

Again the Xc: Can anyone explain how they come to 14 and a half MegOhms ?
You are writing 14.468 kilo-Ohms.
My calculator always shows 14,468 kilo-Ohms.... (almost 14 1/2 kOhm)

What is going wrong here ?

Kind regards,

Edwin

hi,
Xc=1/(2*pi * 50 * .00000022)

Xc= 1/314 * 220^-9) = 1/ 0.000069080 = 14475.969889983 Ohms.


EDIT:

about the mains isolation: Do you have all your lightbulbs isolated with a transformer ?
I don`t think so, so why, if one uses LEDs it suddenly should need a transformer ?

Its because we have no idea of your abilities and where the LED application is going to be used.

 

[MrAl]

Hello again,


Edwin, you dont need a resistor of 14k, or 14.5k, because that would
get very warm if not hot during use, and it would have to be rated for
at least 5 watts.
Instead use 0.22uf or 0.33uf and a 390 ohm, 1/4 watt resistor in series
with the bridge rectifier. That will get you there and wont get hot.

BTW, Dean said 14.5k, that's *k* not *megs* !

 

[ecerfoglio]

BTW, Dean said 14.5k, that's *k* not *megs* !

Edwin2 said 14.468 Ohm.

As he is in the "220 V 50 Hz world" I think that he may use a decimal coma and a dot between thousands, instead of a decimal dot and a coma between thousands

So he intended to say 14 468 Ohm, or 14.5k

(You may also use a blank space to separate thousands, which leaves no doubts. A millon dollars (or pesos) would be $ 1 000 000,00)

 

[Edwin2]

Hi Al, no, I don`t use a 14k resistor, but some 220 nF capacitor, and an extra
resistor, for the peak-currents.

And 2 W resistors are cheap enough, so no problem for the dissipation.

Heah? A decimal DOT ??? Indeed, we use a decimal COMMA, and a dot for
1000-separation.
ok, problem solved, thanks ecerfoglio !


ericgibbs said:
Its because we have no idea of your abilities and where the LED application is going to be used.

ok, indeed, but a small hint could have been my first line in my initial post:
"I am ashamed to ask, but just to check if I am correct with this."

A beginner would not be ashamed; I am already working for over 30 years in electronics,
designing and building. But until now, I haven`t used the seriescapacitor for this purpose
before, and this has to go into a children`s room. But NO problem abt safety, that will be ok!
A decent plastic box completely closed, with only the LEDs shining out, and a normal
mainscable coming out of the top, and will be hanged abt 8 inches from the ceiling.
The stock of this kind of capacitors here, is extreme poor, so hardly possible to try out.
And bcs they cost more than `a few cents` I thought of asking for backup, before I order
the capacitors. (also mind the ordering/shippingcosts)
And if it was for 100s of these things, it would have been worth to order a stock of these,
but sadly, it is for just about 1 (one) lamp... :-(
It must give more light, than a wallpluglamp, how is that called ? but less than a normal
lightbulb. So some twelve hi-bright LEDs, shining onto the ceiling would be an alternative.
If too much light, add an extra capacitor
Maybe some red LEDs shining down from out of the bottom ? To look a bit funny ?


BUT, and that is why the question came up: a seriescapacitor in a.f. situations is also
different calculated:
- 3 dB = 1 / ( 2 x pi x R x C )

If you use Xc and calculating it as voltagedivider you get -1,5 dB at the same frequency.



Kind regards!

Edwin

 

[ericgibbs]

Heah? A decimal DOT ??? Indeed, we use a decimal COMMA, and a dot for
1000-separation.
ok, problem solved, thanks ecerfoglio !

If you use Xc and calculating it as voltagedivider you get -1,5 dB at the same frequency.
Edwin

hi Edwin,
We use a comma for the 1000s but only when a decimal point if the suffix 'K' or 'M' is used.
To use a decimal point as per your your first post is misleading.

As its for the babys room have you considered 'rainbow LED's'.

BTW: I did check your profile for reference, why dont you enter some technical background info.

 

[audioguru]

I don't think the rainbow LEDs can be connected in series. Their current changes as the LEDs are fading so in series their voltage would change and probably be much too high at times.

 

[Edwin2]

Hey Eric,

To use a decimal comma as per other posts is misleading.

Both sides will claim the correctness of their system

Haven`t read the PDF yet, doing so direct.

Byyee!

Edwin

 

[ericgibbs]

I don't think the rainbow LEDs can be connected in series. Their current changes as the LEDs are fading so in series their voltage would change and probably be much too high at times.

Quoting this article for the series reference.

Rainbow glow night light - Patent Application 20020159258

 

[Hero999]

Just because a patent exists, it doesn't mean it'll work.

I wonder what the US patent office is doing.

How can something as obvious and unimaginative as this be patented?

 

[ericgibbs]

Just because a patent exists, it doesn't mean it'll work.

I wonder what the US patent office is doing.

How can something as obvious and unimaginative as this be patented?

I didnt say it would work.

I agree about the trivial patent, but its only a disclosure type patent issued to extract money from someone.