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So you need +12v signal to turn the led driver on correct? Can u invert the outlet in reef pi so that a low signal is on and a high signal off?
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You can still use it, but you have to invert the outlet in reef pi configuration and probably add a resistor on the high side. I wish i was better at drawing a quick circuit on the computer, maybe someone could chime in and help here.
Ok first off is that a ulq2002a? If so the reef pi wont have enough voltage to run it, it needs 14 volts to turn it on, you need a ulq2003, a uln2003 or a uln2803 to be able to turn on the darlington from logic level
Ok give me a minute and ill make a quick drawing on paper for you.
Heres a pic and a quick explanation.Ok give me a minute and ill make a quick drawing on paper for you.
@Diamond1 that’s fair
Hey let me ask you this question. With the wykat boards. I don’t have u1 yet so that spot is open but the darlington transistors are not putting out the right voltage. I see voltage on the output pins but less than a volt on the 5v darlington that drive my power outlets and just a tad over a volt for the one that is supposed to drive my led drivers.
I cut the trace that originally cut the trace on the board that fed one of the darlingtons 5v and hard wired a 12v line to that pin. Not sure what is going on.
It flows thru the darlington output to ground when the input is on, so the test point where the on/off switch is gounded at zero volts when the darlington is on...the 12 volts is used by the resistor.This is where circuits don't make sense to me. Seems to me that current would flow through the Darlington or the resistor, whichever has least resistance. In my head it would always be on either way? I am in no way saying your wrong.
Not sure what's going on with the 1v output you have. I'm using 2 ULN2803 chips and they both output 5v.
I tried to keep everything as simple as I could since I have a very limited knowledge of electronics.
That's why I used relays for off on. Not the best solution but it has worked flawlessly for the last week while I'm bench testing the new setup.
Ok let us know, do you understand the drawing I made?I have a bad darlington array grr. Replacing that and will see how things go
It flows thru the darlington output to ground when the input is on, so the test point where the on/off switch is gounded at zero volts when the darlington is on...the 12 volts is used by the resistor.
When the input is off, the darlington is no longer on and the output is not "connected" to anything, so the test point is sitting at 12 volts.
Oh..in that case it may not work unless all your drivers grounds are connected to the same ground as the ulq chip, and the ground to your seperate 12v source, so everything has the same reference point. It may be easier to use a 2n3904 transistor for each driver in that case...Ok I am understanding how it is supposed to work. But I am trying to control 4 separate drivers from one darlington array. That is why I wanted to use a separate 12v source for the signal instead of the 12v from the driver.
Oh..in that case it may not work unless all your drivers grounds are connected to the same ground as the ulq chip, and the ground to your seperate 12v source, so everything has the same reference point
Yes it serves to limit current when the darlington is connected to ground...otherwise it becomes a dead short to ground. The resistor should also be rated at 1/4 wattI can connect all the grounds from the driver connector to the same ground as the darlington array. The 12v source is in the box and is connected to all the other grounds.
The ground on the PCA9685 board is connected to the ground of the pi which is connect to the 12v ground.
If I’m not using the 12v from the driver do I still need the 1k resistor?
Yes it serves to limit current when the darlington is connected to ground...otherwise it becomes a dead short to ground. The resistor should also be rated at 1/4 watt

