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I am trying to understand the way constant current LED drivers work. Lets say I have a driver with these specs:

Description:

Power: 18W

Output Voltage: DC 36-63V

Output Current; 300mA

Input Voltage: AC 85 - 265V

I can attach one LED light (which accepts 300 mA current) of 18 watts or 3 LED lights (each of which accept 300 mA current) of 6 watts each (in series) to the driver and it should work. In both these cases, the output voltage will be 60 V so that power demands are being met.

Q1 - Am I right up to this point?

Now suppose, I want to use the same driver to power up two 6 watts LED lights. I can still do so by wiring them in series. Output voltage in that case will be 40 V.

Q2 - Am I right?

Now lets say I want to light up a single 6 W bulb which requires 300 mA.

Q3 - What will happen in this case? In order to drive 6 W LED, the output voltage needs to be 20 V but the driver can give output in the range of 36 V to 63 V.

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  • \$\begingroup\$ It is impossible to accurately model your answer without a model of the lamp's VI characteristics, so do not be misled. \$\endgroup\$ – Tony Stewart EE75 Aug 20 '18 at 1:24
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The specifications for this LED driver (300 mA, 36-63V) means the following:

  1. You must use LEDs (or LED clusters, whatever) that are specified for 300 mA nominal current;

and

  1. You must use a number of LEDs in series, such that their total forward voltage (Vf) is above 36 V, but less than 63 V worst case (usually at cold).

Then the LED driver will drive the chain of LEDs to their sum of Vfs, which should be between 36 and 63 V, with constant current of 300 mA.

If Vf of your chain of LEDs sums up to 60V, this LED driver will work to its maximum capacity of 18 W.

In short, "wattage" of LEDs has little meaning without knowing their Vf. If you say you have a LED lamp of 6 W at 300 mA, it means that nominal Vf is 6/0.3 = 20 V (internally there are several LED chips in series). So yes, connecting 3 of these LED bulbs in series will sum up to 60 V and be about optimal. This answers Q1.

For Q2, yes, two bulbs make Vf at 40 V, which also meets the driver specification. The driver will drop its output to 40 V while maintaining 300 mA drive current. They will work just fine, and the light fixture will consume 12 W (plus losses).

For Q3, it depends on driver design. The 20V is below the CC (constant current) capability of this driver. The result will be either intermittent flickering, or reduced (but steady) current. Here is the load characteristic of some MeanWell driver:

enter image description here

Red region shows unpredictable (design-specific) behavior of the driver, but the current shouldn't exceed 300 mA in any case.

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  • \$\begingroup\$ I am very curious to ask two follow up questions: Lets say I have two LEDs which require: (a) 40 V and 200 mA and (b) 40 V and 400 mA. What will happen to these two if connected to this driver? My guess - (a) LED will burn due to overcurrent and (b) LED will glow at a reduced brightness. What do you say? \$\endgroup\$ – Whiskeyjack Aug 15 '18 at 17:19
  • \$\begingroup\$ @Whiskeyjack, I would agree with you. \$\endgroup\$ – Ale..chenski Aug 15 '18 at 17:22
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The Watts of an LED is the wrong criteria to select an LED driver.

LEDs for lighting applications typically have a wide range of operating current. Therefore they also have a wide range of wattage. Meaning LEDs do not have a set wattage. The actual wattage depends on temperature, current and resultant forward voltage Vf where the Vf is a function of temperature, current, and manufacturing characteristics.

Step 1 is to determine the current requirement for your LED(s).

Step 2 is to calculate (min-max) or measure the forward voltage Vf of the LED or sting of LEDs at the target current.

Below is an example IV curve for a Cree XP-G3 LED. Its wattage, as shown on this IV curve, ranges from 200 mW to 6 watts. But is generally referred to as a 1 watt LED (@ 350 mA "test current").

enter image description here


An AC to DC constant current driver is a two stage circuit. The first stage is to convert the AC to a fixed DC voltage. The second stage converts the fixed DC voltage to a constant current. If the Vf and DC supply to the constant current stage are too far apart efficiency will suffer. So LED drivers usually are sold with a range of voltages and currents.

Below is the voltage and current ranges for a popular LED driver. This particular driver (Mean Well HLG-40H) can supply up to 40 Watts.

Constant current range for LED driver

Generally speaking as the voltage of the LED driver increases from 12V to 54V the efficiency improves. So for white lighting applications I use 16 sets of 3 parallel LEDs (3P16S) in series.

enter image description here


I use Samsung LM301B LEDs.
which have a forward voltage from 2.5V to 2.9V so the total Vf ranges from 40V to 46.4V.

enter image description here

The maximum power requirement is 28 Watts per strip of 48 LEDs (2.9V x 200 mA x 48 LEDs).

So I select an HLG-40H-48B to power a single strip. This 48V driver has a range of 28.8V to 48V which will work with the min-max (40V-46.4V) range of possible forward voltages. 40 Watt is the smallest wattage driver in the HLG series.

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Vout min=36V so 36/Vf@0.3A defines the minimum string xS of LEDs required. If they are a ~12V array then a minimum of 3 series arrays are needed.

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  • \$\begingroup\$ Tony, in my opinion, it works more like this: Max wattage will be received at max output voltage. So you can have a bulb rated for lesser wattage and same current as the output and you can still attach to the driver. PROVIDED the voltage requirements are in limit as well. So, the minimum wattage that I can drive is 36V x 0.3A = 10.8 W. \$\endgroup\$ – Whiskeyjack Aug 15 '18 at 17:10
  • \$\begingroup\$ Yes of course this is minimum and Max is 18W =60V which may depend on tolerances and cooling so 62V is possible. \$\endgroup\$ – Tony Stewart EE75 Aug 15 '18 at 17:12
  • \$\begingroup\$ After reading you answer again, I figure that I commented exactly what you wrote in the answer. Just the other way around. My bad. Thanks. \$\endgroup\$ – Whiskeyjack Aug 15 '18 at 17:15
  • \$\begingroup\$ I have two follow up questions: Lets say I have two LEDs which require: (a) 40 V and 200 mA and (b) 40 V and 400 mA. What will happen to these two if connected to this driver? My guess - (a) LED will burn due to overcurrent and (b) LED will glow at a reduced brightness. What do you say? \$\endgroup\$ – Whiskeyjack Aug 15 '18 at 17:19
  • \$\begingroup\$ It depends if any additional R is inside the string. The total current will be 0.3A so the 400mA string will drop in Voltage the 200mA string with has higher ESR will rise until the current is shared but not equally nor perfectly depending on ESR of each string or the voltage drop with 20mA thru each, then I can compute it. \$\endgroup\$ – Tony Stewart EE75 Aug 15 '18 at 18:08

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