Enlighten Me, LED me to Knowledge! Watts Going On? - Page 1
Forum section: electronics / projects

author: SuzyC
author_rank: Frequent Contributor
timestamp: January 25, 2015, 02:06:27 pm
content: I have taken apart a 5W 240VAC .043A LED bulb (lights up like a 25W incandescent bulb). Schematic shown below.How do they get .043 Amp at 240 volts 50Hz (thats whats written on the bulb) to be 5W? I calculate twice as much.How does around 10Watts become 5W

author: mij59
author_rank: Frequent Contributor
timestamp: January 25, 2015, 02:43:52 pm
content: Hi,The apparent power is 10 VA, the real power is 5W, you have to take in account the power factor.

author: SuzyC
author_rank: Frequent Contributor
timestamp: January 25, 2015, 02:49:01 pm
content: How do I do that? Ok so I guess that the answer is Power Factor =.5, but how do I do the math to get this answer?

author: SeanB
author_rank: Super Contributor
timestamp: January 25, 2015, 03:00:20 pm
content: Power factor, and that the light only uses the top half of the waveform to generate light. Add to that it is very likely that the claims are somewhat optimistic to say the least for power and current, and there is likely to be very little relation in the values used in production and what was originally specified. There would have been a lot of "substitution" of values depending on what was cheap that day in the market when they went out to buy. But those values in the circuit are about right for a nominal 3-6W light from what I have seen.

author: SuzyC
author_rank: Frequent Contributor
timestamp: January 25, 2015, 03:15:30 pm
content: SeanB, thanks for your reply, but the light uses both halves of the waveform to generate light, there is a full-wave bridge rectifier used.

author: mij59
author_rank: Frequent Contributor
timestamp: January 25, 2015, 03:22:58 pm
content: The led's and the bridge rectifier are a non linear element, so you end up using differential equations.An approximation would be to assume a sine wave voltage at the bridge rectifier with an amplitude of 106 V, giving an effective voltage of 74 V.The current in the circuit is would be (240-74)/Xc , about 46 mA.Total power of the led's 46 mA* 106 V.Sorry for the bad math.

author: Zero999
author_rank: Super Contributor
timestamp: January 25, 2015, 03:25:17 pm
content: Quote from: SuzyC on January 25, 2015, 03:15:30 pmThat's true, both the positive and negative half of the waveform are used but power is only drawn when the waveform exceeds the total forward voltage of the LEDs, plus that of the bridge rectifier.

author: SuzyC
author_rank: Frequent Contributor
timestamp: January 25, 2015, 03:38:58 pm
content: If I go to Farnell.com I can see that the closest LED that identically match the ones being used in shape, have amax operating current of 30mAand are tested at 20mA..the current is shown on the lamp as 43mA.The LEDs at Farnell also cost about (for 26 of them) in price, even if buying at quan. 30000, is more than the price of the LED lamp I bouight, even without the consumer tax, packaging, shipping, etc, plus the light's lens, PCB, housing, alumina heatsink and of course, the power supply components used in the circuit posted.Thanks to you Hero999, your answer seems to make sense.So, when I need them, I get the high efficiency white LEDs I use in projects by removing them from a store bought LED lamp.A clue to do the math?, to calculate PF, I know use the cosine, turns out that cos(60) = .5 .. a 60-Degress phase angle?In a capacitive reactance circuit, does not the current lead the voltage by 90-degress???I certainly see that current flows only when the rectified input voltage exceeds the on-voltage of the series-connected LEDs,Does everyone agree the effective power is really calculated this way?

author: mij59
author_rank: Frequent Contributor
timestamp: January 25, 2015, 03:48:07 pm
content: Quote from: SuzyC on January 25, 2015, 03:38:58 pmI guess its their secret, better led it be.If there was only a capacitor the real power would be zero, so somewhere is a resistor is present.

author: SeanB
author_rank: Super Contributor
timestamp: January 25, 2015, 03:51:28 pm
content: AC with a full wave rectifier uses magnitude of the voltage, not the polarity, in this case. there is little current drawn up till the voltage on the capacitor is above the forward voltage of the LED's, when the mains capacitor starts to pass charge through to light the leds and charge the main capacitor. Thus it only really is using the top portion of the mains voltage, and drawing current then. This leads to the power factor being around 0.5 ( even though there is a capacitor across the mains in this case the current will be roughly in phase as there is energy transferred to the led load) in the best case, and will show on a measurement of the current as a pulse being drawn only during the peak part of the waveform. Current drawn in pulses is 40 odd mA, but the integration of the capacitor reduces the average current to a lower value. Might still not be good for LED life, but they probably will last out the warranty period, so the manufacturer does not care any more and might get replacement sales.As to cost of the parts you are comparing the overhead of an entire organisation that ships it from China in bulk, then stores it and packs it for you, and sends by courier locally. Local people, local business costs, local wages and local delivery costs and taxes on all thatCompare that to a factory in China whose overheads are so much lower, including salaries. Plus he buys parts cheaply locally, and can bargain for a better price as well, so might be paying less than 10% of the cost of those LEDS delivered to a EU supplier. He pays less for shipping as well, and then the buyers that he supplies pay for shipping of a container load of finished product by sea, at a low cost for a 6-8 week lead time, as opposed to courier by air for small quantities.All in all he makes less per unit, but his costs are so much lower. Likely as well the price you paid for the lamp had more of the total price as markup, agent commissions, warehousing and shipping costs locally than the cost of the unit FOB from China. On a 10 euro unit it might only be 1 euro for the item.

author: SuzyC
author_rank: Frequent Contributor
timestamp: January 25, 2015, 04:00:56 pm
content: LED it be!Thanks SeanB,BTW the life of the lamp is shown on the packaging to be 15000 hours!

author: SeanB
author_rank: Super Contributor
timestamp: January 25, 2015, 04:39:49 pm
content: Good luck if it reaches that, the LED's might, but the cheap electronics definitely will not do that. The capacitor will likely self heal to nothing from the normal mains spikes before that, even with the VDR inside the lamp.