Help regarding driver efficiency calculations

Could some one check these calculations, as Ive bought yet another driver, and apparently its doing a resistor job and dissipating the surplus current.

Pre driver - 3.69v and 150ma giving 55.4 watts


Led side - 2.77v and 143ma giving 39.6 watts

Efficiency of 71 percent, which is basically a resistor

Now is it me or i it the driver, ?



All help welcomed

Pete

I think your misplacing the decimal point for amperage? 143mah should be calculated as .143 amps using a common online calculator. Your wattage would then be .3961 At least that's how I see it

Probably just a cheap driver. As far as the error in calculating wattage it doesn't effect his efficiency calculation as the error cancels itself out. The only 2 ways I know of to drop power to an LED that are more efficient is either using PWM or a DC/DC convertor circuit (buck in your case) and buck circuits require headroom (minimal voltage difference between in/out) to operate effectively which you don't have. Personally at those low of current levels and voltage to the LED I wouldn't worry about it as you are only losing 0.15watts in the driver and 7ma lower output to the led vs from the battery. If the driver has other higher modes they may be more efficient as the voltage drop to the LED is less.

It's actually less efficient that a resistor. With a resistor you'd have only 143 mA input, and 75% efficiency.

This suggests it may be a linear currrent regulator. If so, the input current, output current, and output voltage should stay the same while the input voltage varies. The efficiency would be higher as the voltage drops, but lower if the voltage were higher.

These are not impressive numbers for such a driver. The difference in input and output current is the 'quiescent current' consumed by the driver. In this case that's 7 mA. The ubiquitous AMC7135 which has more than twice the output current has only 0.2 mA quiescent current. I've built drivers with much less than that.

Does it have a coil of any sort on it? Or a large lump of dark grey ceramic? If so it's more likely a switchmode driver, and not a good one.

DIWdiver said:

It's actually less efficient that a resistor. With a resistor you'd have only 143 mA input, and 75% efficiency.

This suggests it may be a linear currrent regulator. If so, the input current, output current, and output voltage should stay the same while the input voltage varies. The efficiency would be higher as the voltage drops, but lower if the voltage were higher.

These are not impressive numbers for such a driver. The difference in input and output current is the 'quiescent current' consumed by the driver. In this case that's 7 mA. The ubiquitous AMC7135 which has more than twice the output current has only 0.2 mA quiescent current. I've built drivers with much less than that.

Does it have a coil of any sort on it? Or a large lump of dark grey ceramic? If so it's more likely a switchmode driver, and not a good one.

Click to expand...

Yep its got the coil, (-nduction coil)which i thought was needed to convert using the switching methOd. Its got the chip on it for the switching too. It is a buck by name.

I did wonder whether the metre was reading peak current and voltage on the way in ? Rather than the average.

Yep the sums are wrong but no they do not make a difference to theefficiency. Its power is 0.5535 watts

Thanks i will have a look at the driver you name.

I put a thread asking for driver reccomendations a while ago to no sucess. What do you think to these ?

http://m.ebay.com/itm/3w-Constant-C...mA-with-2pcs-7135-IC-/111855113500?nav=SEARCH

These look like you select your resistors yourself

And

http://m.ebay.com/itm/LED-Flashligh...L-XM-L2-CC-XP-G-7135-/191062917478?nav=SEARCH

Fixed output.

They have no coil on them ?

The '7135 is a linear current regulator, all self-contained in one tidy little package. Each one contributes 350 mA to the output. Want 700 mA? Use two. Want 2.1A? Use 6.

There's typically a capacitor or two on the board to support the '7135s. If there are any other electronics, it's a microprocessor to provide modes, and some support stuff for the micro.

Those boards in your first link are set up to accept up to four, with only two installed. It has a single mode - full power all the time. The other one has eight installed, and a micro for modes.

If that's a buck-type switching regulator you have (which is consistent with the coil being there), and only 71% efficient, it's a poorly designed one, or perhaps a failed one.

Most meters set to measure DC will measure average. Some have peak-capture feature, but they have limitations on how fast a signal they can capture. Few if any could capture the peak of a regulator's switching signal. Set to AC they will try to measure RMS values, but again the signal is too high frequency for most to measure.

There is one thing that may partly explain how you got a lower than expected efficiency. When you insert the meter in the input line to measure the current, you introduced additional resistance in the circuit. This would cause a voltage drop at the input of the driver, which (because it is a switcher) causes the input current to rise. Unless your meter has unusually high resistance, this likely doesn't account for more than a few percent, given the numbers you posted.

Thanks all, a great help.