Calculating Total Amp Draw

We need to know exactly what they are and how they are connected.
1watt led bulb is pretty meaningless, what voltage rating are they?
They could be 12v DC jobs wired in series and strung across rectified mains for all we know

Or maybe he has his meter on the wrong range or the wire in the wrong socket and it's 2amps!

Del

The bulbs are 1 watt 45 milliamp led bulbs. I have them installed on a casino sign my shop manufactured. It's controlled by a Rocox flasher to make the bulbs scintillate. 120 v feed.

1W at 45milliamps implies about 22Vacross each bulb.
How are these bulbs connected? Series strings? What is the flasher output?
It sounds like maybe they are configured incorrectly for the flasher.
All sounds weird to me.
Does the bulb have a voltage rating on it?
Del

Hello Dholsomback.

Given the numbers you have given we achieve 20A when 432 leds, each consuming 0.045 A, are connected in parallel to the voltage = 1W/0.045 A = 22.22V. Then, it turns out that consecutively (in a line) you may switch only 5 or 6 leds to 120V (the more leds the poorer will be light). You need leds for a much less rated voltage (to make a longer line).

Thanks. I am currently talking with the manufacture of the bulbs. I will pass this info on to them.

No matter how they are connected, series, parrallel, parrallel+series, series+ parallel, the current will ALWAYS be 432Watts / Supply voltage. In this case : 432/120 = 3.6 Amps

If the LED's are connected in another way so that they need a supply voltage of, let's say, 24V then this 24V power supply will have to give 432/24= 18 Amps... Suppose you put all the LED's in parrallel, and you feed 'em with 22V then this "string" would draw 432 / 22 = 19.63 Amps what equals 432 LEDs x 0.045 Amp each

Of course when you gonna use extra resistors, then you have to recalculate. Those extra resistors consume too ...

Actually it is so easy ... No matter what voltage, current, frequency, whatever ...

P (watts) = U (volt) x I (amps)

So tell your field guy that following his calculations your LED string would consume
20 x 120 = 2400 Watts !!!

No matter how they are connected,
Oh the naiveity of youth<whistful sigh>

That assumes they are connected correctly.
If you put a 24volt device across 120V it will probably draw more than it's rated current.
Del

Of course, Dell ... I assume that all the LED's get their proper 22V... If you give 'em less volts, they wont be as bright as they should be (but use less current/power) and if they're connected to a higher voltage, they'll give a lot of light, but not for long ...

What you prolly didn't understand is that 432 Watts stays 432 Watts ... If you put the LED's in parallel or serial, or mixed (AS LONG EVERY LED GETs its 22V !!!) the total power consumption stays the same (432W). If you connect the LEDS so, that you can use a higher voltage, then the current will accordingly gets smaller.

If we go to the extremes:
we put all the LEDs in parallel and use a supply voltage of 22 Volts : 432 x 0.045A = 19 Amps
When we put all the LEDs in series, we'll need 432 x 22V = 9500 V and a current of 0,045A
In both calculations the total power is 22V x 19A = 9500V x 0,045A = yes yes : 432Watts !!!

But practically : I would advice our friend to make 86 strings of each 5 LEDs in series, and put those 86 strings in parallel (for the 2 remaining LED's he'll has to build a separate PSU of eiter 22V/90 mA or 44V/45mA)
So : there will be 86 strings that use 45mA each, that is
86 x 0.045 = 3.8 A, and they'll be fed by a 110 V supply ... and guess what ... 110V x 3.8A = 432 Watts !!!

or he makes 5 strings of each 86 LED's in parallel and put those 5 strings in series.

Again he'll need 110V / 3.8 Amps ... Again all together : 432Watts !!!

You see, Mr.Dell : no matter how you wire them : the power consumption always stays 432 W

Concluding : he needs a power supply of 120Volts that can deliver 4 Amps to light all the 432 LEDs ... (practically : I would suggest a supply of 120V / 6A)

(Please notice : all numbers are rounded !)

Is the flasher a Rocox 120V 20Amp for incandescent lamps or the LED flasher rated at 60 Watts? How many are being used? Just because the LED say they are 1 Watt does not mean they can't draw more than that.

Don't forget, that you are measuring the amount of energy it takes, not the amount of energy that's used by the Flasher and the LEDs. Their efficiency is less than 80%. Is there a transformer or switching power supply in the circuit?

Hi dholsomback,

Tell us your make and model of LED bulb (so we can find a data sheet) and how they are powered.

It looks to me as if there are some pretty shady practices going on in the "green" lighting market place.

Look at this light bulb:-

You can see the "standard LEDs" mounted to look a bit like an old valve.

I'll bet these LEDs are the "common or garden" 20 mA 3.5V forward voltage variety.

The blurb says:-

Power consumption is only 0.5W compared to 11 or 15 watts for the standard incandescent lamp.

Hmmm 6 LEDs * 0.02 Amps * 3.5 V = 0.42 Watts . . . . . .Seems reasonable if that base contains any kind of a power supply. BUT, now look at the "technical data" at the bottom right of the page:- "Electrical Data: 120VAC, 150ma": that's a whopping 18 Watts. It looks as though they've used a separate series resistor for each of the LEDS.

I guess they might be able to justify their claim of half a Watt by saying that's what the LEDs themselves use (and we've neglected to tell you that we're wasting most of the power by heating up the base with some resistors).

Am I missing something here?

Anyone want to go into business with me making LED replacement light bulbs?

Thanks for the reply. The common answer seems to be that this company has mislabled the bulbs. That's got to be the answer.

Probably 120mA (and 18 W) are the apparent current (and power), the active power - for which the consumer pays - would still be the 0.5 W I believe, although, the resultant p.f. unbelievably small.

Guys

re LED lamps, bulbs or whatever.

An LED is a diode, it will have the classic VI curve of a diode with it's knee presumably being determined by the doping being used to create the colour/luminosity of the output.

LED lamps are generally driven by Current drivers which, of course need power to work. Trying to calculate the required Voltage, current, power is major guesswork without precise knowledge of the particular bulbs charcteristic, including, see below , the spread of same.

An alternative is to provide the current of a string of LEDs in series through a high voltage input and a series resistor. Ok but what is the spread of the VI chractersirtic of the LEDS, you want to share the Power and luminosity of each bulb/lamp to be as near as possible to each other, I assume.

Running LED s in parallel is not to be recommended as the spread of charateristics works against you.

Donald the Lighting man could help here??

Sleepy

The leds power consumption is 432W. Now 432 W/120V=3.6A That's the current draw on the 120 VAC side so, you have to add 3.6A + current drawn by the controller itself; it shouldn't consume more than a couple of Amps unless it is an electromechano-archaic device based on drum switches and many motors.

I don't believe it will sink 20 Amps on the supply lines.

Yahlasit

The ohm's Law formulae helps you to calculate the currents draws in a resistive load circuit at 432watts. Hence Power(watts) = Voltage(V) x Current (I)

I= P/V = 432watts /120volt = 3.6 ampere.(current draw)

_________

REY KSA

Guest,

This is an AC system, and this is not purely a resistive cicrcuit.

While Ohms will put him in the ball park, it will not give him\her an acurate answer.

Regard,
Sapper.

Mr. Sapper,

Yes you are right buddy, the question is not clear whether it is incandescent bulb(considered as resistive) or fluorescent bulb(considered as inductive due to transformer ballast) plus may be the capacitive reactance... it said only a bulb hook up in AC circuit system, then i agree your brilliant idea that there should be many factors to be considered in getting the precise current flows/draws in a circuit.

It's an LED bulb.

Actually the questions does specify that the bulbs are LED.

The problem is that LEDs do not work in an AC circuit the way that they are being describing.

The LEDs will require a DC power supply, which today is more than likely to be a Switch Mode Power Supply, whick is going to include a Transformer (inductive) to get the voltage to the correct level, a rectifier cirtcuit to convertr AC to DC (Capacitive and Resistive) and probably a regulator to minimise variations.

Now you have mixed in all the variables and you add losses, the numbers will be out of wack (yes, it's technical term )

Regards,
Sapper