## Another Power Question from a Newbie

aspro648
Posts: 5
Joined: Sat Aug 30, 2008 12:56 am

### Another Power Question from a Newbie

I'm looking to power a BBB and single 8x8 from a 12V battery as part of a bicycle lighting system (I'm going for a very bright "game of life" tailight). I'm using 20 mA LEDs, so looking a 1.2 A max. Can I replace the BBB voltage regulator with a LM7805 and plug the 12V into the BBB power jack? With a heat sink, can I also use that to power the single 8x8?

Is there a better way to do it? Many thanks.

Cheater
Posts: 9
Joined: Fri Aug 15, 2008 7:15 pm

### Re: Another Power Question from a Newbie

That will make a lot of heat.

Make a switching regulator or buy the Futurlec 5v regulator board.

aspro648
Posts: 5
Joined: Sat Aug 30, 2008 12:56 am

### Re: Another Power Question from a Newbie

Shows what a newbie I am. Now I know the difference between a switching regulator and a linear regulator! No wonder that 7805 was getting so hot.

I'll try the Furtec. Thanks for the info.

paul
Posts: 735
Joined: Mon May 12, 2008 4:19 pm

### Re: Another Power Question from a Newbie

The 7805 is going to get hot if you have all the LED's on for any length of time. If you only light up a few of them at once it might work fine, a better bet might be to use 4 AA, C or D batteries or a 6V lantern battery. The switching regulator will work fine though and is efficient - it's just finding one that does 12V in to 5V out @ 1.5A for a good price that might be a little tricky.

Basically with a linear regulator the voltage represented by the input voltage minus the output voltage @ operating current gets wasted as heat.

Example: 12v input - 5 v output = 7 volts Power is 7v * 1.2A = 8.4 watts This is enough to get a small heatsink very hot and would need a fairly hefty heatsink.

Alternate: 6V - 5V = 1V * 1.2A = 1.2 W = much better. One potential problem with this is that it would require a special regulator called a Low Dropout Regulator.
Another option would be to run the display directly from 6V and dispense with the regulator (on the display only) altogether. The chips in the display are tolerant of 6.5V I think (but check their datasheet).

The switching regulator is also a fine solution - except for the money part.

Paul

Severian
Posts: 14
Joined: Thu May 22, 2008 5:48 pm
Location: Texas

### Re: Another Power Question from a Newbie

Howdy,
I want to control at least 16 high power LEDs from an Arduino. I have a test circuit going now with 4 LEDs and I think I know how to take it to 16 LEDs. I turn the LEDs on and off with a 74HC595. I'll need a second one of those. But, the part I don't understand is the power. I currently drive everything from a 7.5 V DC regulated power supply. I think the LEDs will require the arduino to dump 1.2 watts when all LEDs are on((7.5 V-5 V) times 16 times 30ma). If I ordered a RBBB with the bigger regulator, is that likely to work? I expect to have somewhat random cycling of LEDs and I think it would be rare and brief that all LEDs would be on. I am thinking I might flash them briefly all on when the device powers up.

I see some references in the forums to using a regulated external power supply. I could get a regulated power supply of 5 V to power the device. That would limit my ability to run the device from batteries, which I wanted to do. But, I don't understand how this would even work. I understood I needed to plug in an external power supply of 7V to 15V. Where would the 5V regulated power supply plug in?

And, if that would all work, how about if I doubled the number of LEDs? That is for a different project. I am partially trying to just understand how I figure where the limiting factors on power are.
Good day,
Ralph

paul
Posts: 735
Joined: Mon May 12, 2008 4:19 pm

### Re: Another Power Question from a Newbie

Ralph,
For 16 LED's x .030 A (30mA) current - you get about .5A of current. So the whole thing will run fine on a 1A regulated supply. Since you are using driver chips, only a small amount of the power really is running through the Arduino part, and the real current handling is done by the driver chips, which is one of their jobs.

Here's one clean way to do it. Get a 1A 5V switching regulator wall wart from Digikey (under \$10). Run a wire from the +5V utility hole closest to the power jack on the RBBB to the anodes (positive side) or your LED's, and the +5V pin on the driver chip, run a ground wire from the ground hole to the ground on your chip. Run the stock regulator on the RBBB, and run the serial lines to the driver chip. Plug in your wall wart and life "should" be beautiful. The same setup will work with batteries from 6V to 9V (not a stinking transistor battery though - that will run for 10 minutes maybe), without changing anything, just solder on a 2.1mm power jack on the battery pack. I'd go with six volts anyway though, because the extra volts are just wasted when you're driving LED's that only need 3 volts or so.

Remember - the driver chips are seeing all the current, not the arduino.

Paul

Severian
Posts: 14
Joined: Thu May 22, 2008 5:48 pm
Location: Texas

### Re: Another Power Question from a Newbie

Howdy,
I had been composing a long response and the forum software just threw it all away and made me login again. That is not very nice. I'll try again, but I had more detail before.

The circuit you suggest seems fundamentally different from mine. I am pretty much using the one at:
http://www.arduino.cc/en/Tutorial/ShiftOut. I believe the terminology is that my circuit has the shift register source current and your suggestion has it sink current. Is there any reason to prefer one over the other? It seems like the shift register has as much current flowing in either case.

A better choice might be to find a TLC5940. It seems like a better choice because one chip can drive 16 LEDs and control their brightness. I can control each chip with the shift register, but not the brightness. I can contro the brightness with a Darlington or a MOSFET, but not individually for each LED. But, that is for later. I just mention it here in case someone has used the TLC5940 and wants to comment on it.

I don't understand your wiring for the power completely, anyway. I think you are suggesting the VCC hole that is at about 35 degrees on the image I just uploaded. Is that right? I took your image and added a few numbers for reference. The 35 is labeled and the holes are a little below it.
Good day,
Ralph
Attachments
rbbb_marked.jpg

paul