Set your electronic load to resistance mode and adjust the synthetic resistance to get the desired voltage drop.
I think this is probably best general answer for almost all of those with commercial electronic loads. All I have used have constant current, constant resistance and constant power modes. That's the whole point of such devices- to act as flexible loads- most will also allow you to simulate changing loads etc. If yours does not for some reason, keep reading:
If yours does not, you can use a shunt regulator based on an op-amp with a PNP darlington (and a reference voltage) or use the ciruit in the TL431 data sheet, substituting a PNP darlington (eg. TIP125) for the PNP transistor and choosing the B-E resistor so that the TL431 always conducts at least 1mA.
The resistor in series with Vi is not required in this application- the series impedance of the current source takes its place (and should be quite high in dynamic resistance if it is a good constant current source). Vref for the TL431 is about 2.5V (2.495V nominally). Depending on the discrepancy between the load and source the transistor could see a lot of dissipation (18V * 0.8A = 14.4W if the load was completely disconnected). That would require a large heatsink. If the load is set to a higher current than the source, the output voltage will collapse which may cause your source to misbehave, or worse.
If your "load" is just a current sink (CC mode only) it is not very appropriate for this purpose). You might as well just use the above circuit alone with a suitable heatsink.
Your driver has an output of 416 mA @ 12V and is constant current (CC).
You need to know the voltage input to the LED strip.
Based on this picture it appears there is one current limiting resistor for each three LEDS. What make this more difficult is, in the photo, the ends of the strips are cut off, you say there are 32 LEDs, and the PCB says 36 LED. Three LEDs per resistor makes more sense if there are 36 LEDs (12 resistors) than 32 (10.666 resistors). I cannot see how the strips are wired.
The five needed unknowns
- What is the voltage of the driver output when connected to a strip?
- What are the values of the resistors?
- How many resistors?
- Are there 32 or 36 LEDs per strip?
- Are the strips wired to the driver in series or parallel?
I am going to assume the power supply should be 12V constant voltage. Although a CC driver may (or may not, depends on the value of the resistors) work with one strip. If the strips are wired in series, the pictured driver would not work becasue you would then need 24V.
If the driver does not light up one strip by itself then I would guess the board needs more than 416 mA. 5 watts seems to be very inadequate for even one strip with 36 LEDs. They appear to be high power LEDs like a Cree Xlamp with a 350 mA test current. So the strip may be trying to draw more the 416 mA which drops the driver voltage below 12V and the LEDs do not have enough voltage to light up.
I do not know what RS means.
UPDATE
The 241 on the resistor mean 24 x 101 or 240Ω
So the current flowing through each resistor is about 37 mA
Which means the 3 LEDs and resistor = 444 mW
.444 W x 36 resistors = 5.328 W x 2 strips = 10.656 W
You need a plain old power supply, not an LED driver.
Power Supply Needed: 12VDC out and at least 1 Amp or 12 Watts.
RS 12V Power supplies
Best Answer
LEDs work best if you drive them with a current.
When you would apply a voltage to a LED or a string of LEDs in series, the current can vary a lot over production lots and temperature. It is very well possible that too much current will flow through the LEDs which makes them hot and limit their lifetime.
So it is better to drive LEDs with a current.
Such a LED driver module does just that. It adapts the voltage such that the specified 300mA current flows. It can adjust the voltage between 60 and 90 V DC in order to achieve that.
So as long as you would use LEDs in a configuration (series/parallel) that they should be powered at between 60 and 90 V at 300 mA, (20 - 30 W) then this module is suitable.