I have an Imp as part of a project containing several parts. The Imp is on a SparkFun breakout board. One part of my project in particular, an LED strip, runs at 12V. The 12V is in the breakout board’s V(in) range. Can I solder the wires from a wall wart providing 12V to the breakout board to power the Imp, then turn around and continue to run the 12V from the breakout board to the LED strip?
This probably seems like a basic electronics question - and it probably is - but I just want to be sure before I end up letting the magic smoke out of the Imp. Thanks for any guidance!
Wow - schematic in text! That looks about like what I want. The LED strip is actually common anode, so I don’t need ground on that side.
From what it looks like here, you’ve split the incoming power. The SparkFun breakout board has solder pads for incoming power, and then two holes right above those solder parts. I assume this is some fashion of connector setup, but what I’m thinking of doing is using those holes to run the power back out to the LED strip. Effectively using that little part of the breakout board as the split. I suppose I could just use a stripboard, but I’d like to avoid the additional part if possible. Does that seem feasible?
If the imp is going to be turning the led strip on and off, you can use a general purpose transistor to do the power switching. Better yet, an optic isolation component. Don’t let that 12v get on the imp pin directly.
I noticed that voltage is referenced in a lot of different places. The TI chip on the SparkFun breakout for example shows support for up to 17V. But there’s not any mention of current. So if I get a 12V/2A power supply, which is supported by the LED strip, does that fry my Imp?
The TI’s regulator is the one that we all use for the breakouts, as is the one suggested by Electric Imp and is one of the best step down converter you can find around:
The max output rating is 0.5A but to power up the imp001 you will need at least 400mA available all the time, even if the device can drain so much current only in extreme situations with WiFi in operation.
You can use the 12V directly but please consider that any kind of regulator needs to “put” somewhere else the electricity that you are not taking on the outputs. I mean that if you supply a 5V as input and you require a 3.3V as output, the 1.7V difference will be dissipated by the device, mostly as heat. 1.7V x 0.5A = almost 1W that’s easy to handle for the very small package.
Quite a difference is to supply 12V for an output request of 3.3V , with more than 4W left to the incredible small regulator for dissipation.
At page 20 of the (linked) datasheet you can see the suggested design for the best thermal dissipation and low noise, that’s exactly the design we have used for the breakout&shield:
Even with this design of the PCB, I still would suggest to avoid to send the 12V directly to the regulator. I believe it’s much better to bring down the voltage to 5V somewhere else, something that you can easily do with a simple LM7805 or with a *1117DT5 and a couple of capacitors.
This will leave to the bigger regulator 70% of the heavy duty and will keep healthy for long time the regulator on the breakout board.
Also to properly choose the size of your power supply, you should calculate the max consumption of the LEDs, considering that you will need to reserve 0.5A for the Electric Imp.
