I am trying to repurpose a cheap Chinese LED driver. The particular unit takes 110V AC and outputs 90V, 420mA DC AND has a second smaller wires that puts out 12V ,250mA DC. I am using the 12V curcuit to power a cooling fan and water pump and the larger circuit to power a few LEDs.
The issue is, that when there is no load on the main circuit from the LEDs, then the secondary 12V circuit is not powered and the cooling shuts off. This is an issue for me, because if I lose one of the LEDs in that particular circuit then I loose cooling for all the LEDs (including all the ones that are not in that circuit).
I want to install a resistor to maintain a very small load on the main circuit to keep the smaller 12V circuit on in case I lose the main circuit. However, I do not want to waste a bunch of power in the process. I was thinking about bridging the +/- of the main circuit with a resistor but cannot decide on what to use or if this will work. I was hoping to create a 90V , 0.25A load with a ~333 ohm resistor. Is there a way for most of the current to flow through the LEDs except if the circuit is broken, at which time the resistor would maintain a load. In that case it will need to dissapate the full 37W. I am not sure how to set this up such that is does not steal half the power from the LEDs...
Yes, it is 250mA, not 25mA for the fans and pump, and I have 2 of these drivers. I am using one for pump and one for fan, while the main outputs power LEDS. Need to keep both on.
There are 2 pairs of wires exiting the drivers. One for LEDs and one for fans. LEDs are powered without fan plugged in but fan is not powered without LEDs plugged in. I assume this means the driver is designed not to power the fans if there isn't a draw on the main circuit (LEDs). I am hoping to bridge the main circuit to create a dummy load so that the secondary circuit stays on incase the LEDs fail and that circuit breaks. As long as there is load on the main circuit then the fan circuit gets a constant 12V 420mA. I am almost certain a draw of less that 0.1A would be more than enough. It is OK if the resistor is a little big (within reason).. I am just trying to create this small dummy load and I am worried that when the main circuit is on (powered at 90V, 420mA) the current will be split between resistor and LEDs, which would waste half the power. I am thinking about just bridging + and- of main circuit with resistor, thus it would be in paralell to the LEDs. Is there a better way to set this up? Am I thinkaing about this right?
What you need is a constant current sink. One that has a dynamic voltage range that encompasses the whole 90V span. This current sink would consistently draw the required minimum current to keep the fan running.
The way a constant current sink operates is, it modulates it's output voltage, to keep the target current flowing. It will have a maximum voltage, and a minimum voltage.
Both Bi-Polar, and MOSFET transistors can function as Constant Current sinks [and sources], but a bi-polar tends to have a better characteristic for this. And, since, in this application, the current regulating function can be a bit "sloppy" [that's a technical, term, right?!], you won't have to design the hell out of it. But, it will have to handle a bit of heat -- i.e. beefy transistor with a beefy heat-sink.
Something like this [untested, set to around 250mA]:
You can set the approximate current by changing R4, using the following formula:
** **R4 = 0.6/I[sub]S[/sub]** **
Where: [b]I[sub]S[/sub][/b] is the Sink Current.
I wouldn't go much higher than 250mA. But, you can go lower.
And, actually, this one might perform a bit better -- i.e. greater dynamic voltage range -- but, again, I must emphasize "sloppy" as the key behavior, here. The current regulation could dip down as far as 200mA [from 250 mA] at around 30V across the sink. So, perhaps a usable range from 90 to 30V [or more if it can be even more sloppy].
And the formula becomes more like:
** **R4 = 0.68/I[sub]S[/sub]** **
Where: [b]I[sub]S[/sub][/b] is the Sink Current.
