# Ohms law and LED driver efficiency? How can this possible be true?

I am having a meltdown with a Chinese supplier at the moment and hope dearly that someone can help.

I have an LED driver that is in my mind 28.8W. The output of the driver is 48 V @ 600 mA (48 V x 0.6 A = 28.8 W).

I was however talking to the Chinese supplier who tells me this is a 32.72W driver.

They are justifying this by quoting me P=U·I/efficiency formula (48 · 0.6 / 0.88 = 32.72).

Now what I can't understand is how a driver that puts out only 28.8 W by definition of Ohms law can this be then considered 32.72 W by terms of this formula regarding efficiency? Surely this is simply not possible?

I have gone around in circles trying to get my head around it but can't seem to get past the fact this won't power 30W of LEDs.

Can anyone explain this to me or are they trying to pull the wool over my eyes?

Thanks Guys really impressed with you responses and you have confirmed what I suspected. Essentially he trying to sell the input power as output power, which is what I thought. Absolutely ludicrous.

@ Marcus I think you are absolutely right need to nail down specs before I even have a discussion with these jokers. Unfortunately I buying off the shelf and not having anything specifically tailored. I am committed to the purchase unfortunately luckily it doesn't actually matter for the project but lesson learnt.

Out of interest are there any EU or US conventions of directives for the labelling of Power supplies for output power rather than consumption?

• I'm not sure if Ohm's law is relevant here - switched mode power supplies aren't resistors, and neither are LEDs. – Simon B Nov 12 '18 at 11:38
• Wouldn't the formula rather be P = U * I * efficiency? As in, to determine how much power you actually get after taking the loss of the converter in account, as a percentage. – Lundin Nov 12 '18 at 11:56
• Chinese suppliers have varying habits. I have a couple of times got seriously overspecified stuff. When I made a reclamation, the answer was "Dear valued customer, we sell only highest quality products with exact specs". A photo of a miniscule battery + a big piece of styrox foam in a big box around the battery changed nothing - except I saw my next email rejected by R.O.C foreign spam email blocker automate. – user287001 Nov 12 '18 at 12:14

He's trying to sell you something that consumes > 30 W as a "30 W driver", whereas it actually can only supply less than 30 W. The ratio between power that goes in and goes out is called the efficiency. The difference in power in and out is simply converted to heat.

You can upgrade your LED driver to a 2500W LED driver simply by putting it in parallel with a water heater of roughly 2470W, or any other resistor, for that matter. You can simply waste more energy.

So, that's what we call a scam and that's why we only buy from reputable sources in jurisdictions where one can ultimately sue someone.

As a personal note: Don't argue with someone who'll be getting your money. Simply make "fulfill this or don't get the deal" statements. Make the requirements verifiable, as in "The driver has to deliver $$\x\$$ A at $$\y\$$ V with a ripple of less than $$\\delta\$$ V and without getting hotter than $$\T\$$ °C in continuous operation. These are my requirements, and I can only pay for equipment that fulfills these."

An LED driver is essentially a power supply. The rating of a power supply is given as the Output Power.

So your thinking is correct, the LED driver you are talking about is a 28.8W LED driver.
Regarding efficiency, this 28.8W LED driver would require 32.72W Input Power if it is 88% efficient as the seller is claiming.

$$\Efficiency = \frac{Output Power}{Input Power}\$$

However, claiming that the input power is the same as the power rating is preposterous. As Marcus stated in his answer, you could just make the device require more power and fudge the power rating.