I am just trying to learn about wiring, specifically how a POT works in conjunction with the dim function on a driver, and how I could wire in an additional on/off switch.
Here’s what I want to achieve: one SPDT switch, where one side gives full power from the driver, and the other side wired into a POT so I can trim the brightness to desired levels. In addition I’d like to wire in an on/off switch that will work independently of the SPDT switch.
Going by some diagrams of different config options provided in the datasheet (it’s a LuxDrive A009 BuckBlock) I thought this might be a good starting point:
Would that work how I expect? I would be expecting the on/off switch to work independently, then I would be able to switch between two brightness levels, one full power and one set by the POT?
I also have another question based on a diagram in the A009 datasheet: if the switch is placed at the end/outside of the circuit as shown below, couldn’t current still flow to the POT as highlighted in red?
I’m guessing it’s actually correct since it’s in their documentation, I just can’t get my head around it… I mean there is no break in the circuit between DIM +/- as far as the POT is concerned? If anyone can explain that I would appreciate it!
I don't know the driver well enough to say about the function of the pot, but I do know switches...
In your first pic: What about instead of two switches, an on-off-on rocker switch? Combine both functions into one. It'd be wired as mode1=dimmer, middle=off, mode2=full on.
In the spec sheet diagram next, I think they just don't show the main power switch, which cuts either/both of the Vin leads. The switch shown just enables/disables the dimmer pot. edit: And the dimmer override switch shown doesn't need to remove the pot from the circuit to do its job, the switch will have lower resistance than any setting the pot's capable of, so it acts like a shunt and all the current goes thru the switch and none thru the higher resistance leg that includes the pot.
The answers are all in the data sheet. You must read the section labeled “External on/off control” and it says:
“Where a manual on/off control is desired, the potentiometer in Figure 14 may be replaced by a
pushbutton or toggle switch. The output current will be zero and the input current will drop to the
quiescent level when the switch is closed. Figures 16 and 17 show external dimming control combined
with on/off control. ”
The dim pin provides a reference voltage to the dim/ground pin. The drawing provided by luxdrive, figure 14 that you have posted, when the switch is open the reference voltage flows to the dim/ground (or control pin), via the potentiometer which limits the voltage/current. When the switch is closed, the current takes the path of least resistance and pretty much ignores the potentiometer.
I expect your circuit would give you two modes of operation, 100% bright and dimmed via the pot, but I am not entirely sure without having the driver in hand.
Yours appears to connect the reference voltage directly to ground. I don’t know how the driver would handle that.
Thanks a bunch for your replies - I get it now! Well, mostly…
So it seems like Fig. 16 shows exactly what I want, but I would just have to add an on/off switch on the Vin side? Or should it really be wired into the DIM circuit to do it “properly”? That’s the only thing I’m not sure of - it seems like the data sheet only describes an on/off for the POT function, not for the light output as a whole?
For example Fig 16 & 17 according to the data sheet should show external dimming combined with on/off control, but I’m still not seeing it. Unless their idea of on/off is actually just to control the POT?
Anyway - seeing as the documentation only speaks about the on/off switch being wired to the DIM circuit, I suppose the below would be the “correct” option rather than putting a switch on the Vin side?
Or is having it on the Vin or DIM not such a big issue? There wouldn’t be any lag if the switch was on Vin, would there?
The main reason to use the dim switch as your on/off switch or for PWM is it draws much less current, so can easily be switched via a microcontroller, arduino, or just a smaller switch. While switching v in will require a switch capable of carrying the entire load. Also if doing high freq switching like I said, PWM etc you should use the dim pin.
To add more to this, take your example from figure 16 and ignore the pot/fixed resistor for a moment. If you have a switch across Dim/Dim ground then when the switch is closed, the driver will essentially be off (well, it will have a small quiescent amount of draw, but the output will be off) when the switch is open, and if you have NO resistance, then the driver will operate at full power.
Since you will have the pot and or a fixed resistor (did you know you could do both, in series?) then when the switch is closed, the driver will be off and when the switch is open the driver will output a current level in relation to the resistance between dim/dim ground.
As stated in figure 17 “output is approximately %IOUT=R1/50. Or, to reverse the formula, I*50=R1
Lets assume your desired current level is 700mA. .7*50= 35. So if you put a 35 ohm resistor across the two pins, and your switch is open, then your output will be 700 milliamps, as soon as you close the switch, the resistor will be bypassed as the current takes the path of least resistance. This shorts out the two pins and causes the driver to switch off.
As I said, you could instead use a switch on the input of the driver, functionally it would be reversed. Open switch = off, closed switch = on but you would have to drawbacks I posted above.
