How to test total power used by led + driver?
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I don't have a bench power supply but how would that work? The constant current driver I am using is 700mA and accepts up to 24 volts DC. So if I hook it up shouldn't it put out the 700mA regardless of the voltage I set it at? I just want to confirm that it's really putting out 700mA and how many total watts the 3 LED's + the driver are actually using??
Is there any way to do this with a multi-meter?
Thanks!
Is there any way to do this with a multi-meter?
Thanks!
Genes
Newly Enlightened
The 700 ma to the leds is only partly related to the total power consumption. You have to measure the input voltage to the driver and the current to the driver, not the leds.
Christexan
Enlightened
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- Sep 29, 2006
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Won't be perfect, but it'll be ballpark....
you'll need 4 measurements, first, voltage at the input to the controller (+/- from the battery at the input point preferably to negate wire resistance).
Then, switch the meter to current mode (10A typical setting), and test "in-line" on either the plus, or minus side (connect one lead of the meter to the battery, and the other lead to the controller where the battery would normally connect (disconnect the normal connection wire while doing this test).
That is the total power used in the system (power in watts = volts*amps). Now, do the same measurement, but from the controller to the LED (test voltage at the +/- as close to the ends of the LED string as possible (last connection at the LED PCB/star, for example). Then in current mode, disconnect one lead from the LED string to the controller, and put the meter in it's place and measure the current. This is the power getting to the LEDs out of the controller.
Divide those and you get a rough efficiency number, however realize a LOT of factors play in this, including switching frequency for switching controllers (the readings you see may be an average or a peak per unit time instead of RMS, etc)... anyhow, it's good enough for rough guesstimating.
you'll need 4 measurements, first, voltage at the input to the controller (+/- from the battery at the input point preferably to negate wire resistance).
Then, switch the meter to current mode (10A typical setting), and test "in-line" on either the plus, or minus side (connect one lead of the meter to the battery, and the other lead to the controller where the battery would normally connect (disconnect the normal connection wire while doing this test).
That is the total power used in the system (power in watts = volts*amps). Now, do the same measurement, but from the controller to the LED (test voltage at the +/- as close to the ends of the LED string as possible (last connection at the LED PCB/star, for example). Then in current mode, disconnect one lead from the LED string to the controller, and put the meter in it's place and measure the current. This is the power getting to the LEDs out of the controller.
Divide those and you get a rough efficiency number, however realize a LOT of factors play in this, including switching frequency for switching controllers (the readings you see may be an average or a peak per unit time instead of RMS, etc)... anyhow, it's good enough for rough guesstimating.
Illum
Flashaholic
if a driver is really putting out 700ma?
um...disconnect one wire from the string and measure the current with your multimeter. It is not recommend to test without some type of load like the LED as some drivers fail when powered on without the output connected to anything. The way current is measured without an appropriate load what the driver sees is an short circuit and it may fry both your converter and your multimeter
CREE XREs consume 3.5V at 700ma
[voltage x current] x quantity = total wattage, or power dissipation
3.5 x 0.7 x 3 = 7.35W at a junction temperature [Tj] of 25C, or about 77F. as the junction temperature rises, power consumption will increase slightly. 7.35 here represents the minimum power output
if your wondering how much current is being drawn from the source you can measure it in two ways, one being more accurate than the other
method 1: determining your converter's efficiency, then using that efficiency value to solve for input current
this translates to this. Solve the missing variable
If your efficiency is 80%, then output power 7.35W divided by 80% of 12V equals ~0.7656, so at 80% efficiency the converter is drawing 766ma at 12V to output 3.5x3V at 700ma.
Method 2: grab your Fluke, set it to Amps, unplug the probe from the ma socket, stuff it in the A socket then run your multimeter in linear path to the converter from the battery. record the current input [if its set at amps, then it should be a positive decimal] take that decimal and multiply it with voltage yields the total wattage input.
um...disconnect one wire from the string and measure the current with your multimeter. It is not recommend to test without some type of load like the LED as some drivers fail when powered on without the output connected to anything. The way current is measured without an appropriate load what the driver sees is an short circuit and it may fry both your converter and your multimeter
CREE XREs consume 3.5V at 700ma
[voltage x current] x quantity = total wattage, or power dissipation
3.5 x 0.7 x 3 = 7.35W at a junction temperature [Tj] of 25C, or about 77F. as the junction temperature rises, power consumption will increase slightly. 7.35 here represents the minimum power output
if your wondering how much current is being drawn from the source you can measure it in two ways, one being more accurate than the other
method 1: determining your converter's efficiency, then using that efficiency value to solve for input current
Code:
Useful power output = Total power input x Efficiency
Code:
Iout x Vout = Iin x Vin x E
0.70 x 3.5(3) = Iin x 12 x EMethod 2: grab your Fluke, set it to Amps, unplug the probe from the ma socket, stuff it in the A socket then run your multimeter in linear path to the converter from the battery. record the current input [if its set at amps, then it should be a positive decimal] take that decimal and multiply it with voltage yields the total wattage input.
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