Led connected to AC power

[2901975]

Hello friends

I have a project in mind time ago and I did it reciently, it is to connect 6 leds (19.6v Dc @ 700mA each) I connected them in series to avoid excessive currect consumption, I'm sending the schematic so you see how exactly I did it.
My question is I think still need something else, it works well but I do not know if it will support the peak currents or overvoltage (even though I put a fuse holder in series of 1 Amp).

Someone expert in this area could tell me what to do next please.

Thank you for your time and consideration.

 

[itsmoked]

For starters remember that rectified 117V with a fullwave bridge results in 117 x 1.4 - 1.4V = 162V

This means 162V - [6 x 19.6V] = 44.4V

44.4V/700mA = 63 ohms

P_res = I² x R
P_res = 0.7² x 63
P_res = 31W

Keith Cress
kcress -

http://www.flaminsystems.com

 

[2901975]

Thank you itsmoked for your time to respond.
I have another question:

(According with the ohms law) if I put a resistor of 63 Ohms this is supouse to have 700mA through it right?
But in testing is not the same, I put 2 resistors in series of 10 ohms giving an equal to 20 ohms and this is giving at least 800ma (giving more brightness), when I put a resistor of 25 ohms the current drops until 560mA (giving less brightness), well if we multiply the voltage for the RMS (root square of 2= 1.4142)less the voltage on the diodes (1.4v) this should be equal to 162v - 165v right? and with this you can obtain the resistor for the current consumed, but again in testing is not the same, I measured the voltage on the socket and is 117vAC.

Am I doing something wrong? why is different according to the ohm's law?

Thank you again for your response.

 

[LionelHutz]

Yes, 162V out of the rectifier is the peak voltage, not the RMS voltage. So, using a 63 ohms resistor would give a peak current of 700mA, not an RMS current of 700mA. The RMS current is what sets the brightness, more or less.

The 20 ohms resistor gives a peak current of about 2.2A. You need to see what the maximum current your LED is rated for from the datasheet.

Keith's answer is correct if you put a capacitor filter after the rectifier.

 

[benta]

Actually, the OSTARs are rated for 700 mA, but tolerate up to 1 A peak current. Forward voltage according to datasheet is 20.8 V, not 19.6

Do the following (just to give yourself an image of what's happening:
Draw a sine half-wave with a peak of 164 V.
Draw a horisontal line at 6 x 20.8 V + 2 V = 126 V (appr.)
You'll see that your peak voltage over the LEDs is 38 V, but ONLY for a very short time!

This means that with this design you can NOT design to the average current without damaging the LEDs, you need to design to the peak current of 1 A. This gives 38 ohms.
Your average current will be well below 700 mA, so you should reconsider your design.

Best Regards,

Benta.

 

[2901975]

ok guys I appreciate your help but I'm afraid to say that I'm a little confused here.

When I tested the led with my regulated power supply it gives me the maximum brightness at 19.6VDC and 700mA, but I will consider the datasheet standard for 20.8VDC.
This leds behave like the fluorescent lights after a few seconds the brightness increase trying to consume more current and obviously I have to limit and I limited to 700mA and gives me 19.6vDC on my power supply.

So do I need consider 1A or 700mA as given in datasheet?
What value of capacitor should I put after the bridge?

Now considering what my friend "benta" said:

(This means that with this design you can NOT design to the average current without damaging the LEDs, you need to design to the peak current of 1 A. This gives 38 ohms.
Your average current will be well below 700 mA, so you should reconsider your design.)

is this means that the peak current of 1A will go through the resistor? how the average current will be below 700mA?

Thank you again for your time

 

[benta]

There is a major difference if you use a storage capacitor or not. In you schematic none is present.

Benta.

 

[burnt2x]

You should design currents below the ratings of the elements! Otherwise this gadget gets toast in minutes! We call it factor of safety. If I wanted bright lights, I would go for higher rating LEDs. Still I will make sure that the amps are lower than the rating of the LEDs!

 

[2901975]

Now, I modified the schematic, could anyone tell me if this is ok?

I believe 470uf is enough for this circuit, still the resistor I do not know, what should I put?

Thanks guys

 

[LionelHutz]

Are you OK with using an ~60W resistor in the circuit?

If you use a capacitor filter then you end up with the 63ohm, 31W resistor requirement as Keith posted.

 

[itsmoked]

What I would do:
Get a 100W rheostat that is maybe 75, or a 100 ohms. Use it in the resistor location.

Then I'd put a 'true RMS' digital multimeter (amps) in series with it all. Turn down the rheostat until you achieve your 700mA. This is with the cap!!

Without the cap you need to instead monitor the current with a correctly utilized scope and a shunt resistor. (Note this can be very hazardous in your application.)

Then again you adjust the rheostat but without the cap you must instead adjust the rheostat down until you reach the peak allowed PLED current. The prudent person would then increase the rheostat position to be perhaps 10% (or more) below the peak allowed PLED current.

The RMS current would then be whatever it is. That is what you'd have to live with.

In all cases you will not have the brightness you could achieve with a proper PLED driver.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[itsmoked]

A rheostat something like this:

http://cgi.ebay.com/OHMITE-150-Ohm-...39:1|66:2|65:12|240:1318&_trksid=p3286.c0.m14

Keith Cress
kcress -

http://www.flaminsystems.com

 

[burnt2x]

I've checked my rectifier math on your first diagram:
Vpeak = sqrt(2) X 117 = 165.46 volts
Vrms = 0.637 X Vpeak = 105.4 volts (full-wave rectifier)
IMO, you won't get your LEDs very bright then!
Suggest to reduce the number of LEDs.

 

[itsmoked]

Hi burnt2x.

Please explain:

Vrms = 0.637 X Vpeak = 105.4 volts (full-wave rectifier)

Keith Cress
kcress -

http://www.flaminsystems.com

 

[2901975]

I was thinking to use a capacitor in series with the bridge instead of the resistor but it turns it will be too expensive for this purpose.

I berely remember that equation burnt2x, we used in college in electromechanic time ago.

could you also explain what "IMO" means burnt2x?

Could anyone clarify and specify more the whole interpretation please?

what about when a rush current or overvoltage ocurr?
the capacitor in parallel will protect the circuit?

Thank you for your time and responses guys it is greatly appreciated.

 

[LionelHutz]

The killer here is that the voltage has to be above 126V before you get any light output. Due to the nature of the sinewave, this only happens for a short time during each period. I haven't figured this out but it's likely amounts to the LED's being on < 50% of the time. Then, above this, the LED's are only on at full brightness for maybe something like 5% of the time, as the sinewave peaks and the current reaches 700mA.

In your first circuit, I was thinking the resistor has to be fairly high power but in reality it's only at full current and power for a short amount of time each cycle so something like a 10W resistor may even work.

If you want to continue persuing this, you need to look up capacitor filter supply theory and properly determine the capacitor size. Pay attention to the ripple current rating of your capacitor.

You are looking at providing approximately 85W of power to these LED's. A PWM switching circuit really the way to go.

 

[2901975]

I'm using a multimeter Fluke III and a Greenlee clamp-ampmeter that measure peak current, min and max and relative current and according with this last diagram that I'm testing right now and the Greenlee multimeter read as follows;

In AC input
Relative current: 0.5 A
Min peak current: -8A
Max peak current: 3A
Maximum current: 1A

In DC output
Relative current: 0.5 A

As you see this diagram includes the capacitor of 47uf/250v.
The resitors are 20 ohms each connected in series in a TO247 package which is convenient for me.
What do you think about this?

I'm concern about the peak current probably my multimeter is bad?
I do not know but everybody is welcome to give an opinion on this project and thank you again.

 

[2901975]

Sorry I forgot to upload the circuit.

And I forgot also to tell you that I put 2 stream of leds, though this will increase the current.
The capacitor according with spec sheet is 210mA rms at 105 degrees 120Hz. with a frequency coefficient of 0.90 @ 50Hz.

Here you go.

 

[burnt2x]

Keith,

"Your wish is my command."
(See attached)

2901975,
I - In
M - My
O - Opinion

 

[itsmoked]

I didn't remember it be so low.

Thanks burnt2x!

Keith Cress
kcress -

http://www.flaminsystems.com