I have a few simple projects under my belt. They were low power so didn’t run across this before. But now I have a project using a lot of WS2811b LEDs (168 of them,) and I can easily get to 2.2A (and beyond if I run them at full white.) Controlled by a knock off Arduino nano. So everything runs on 5v – nano and the LEDs. Battery I’m using is sealed lead acid and says 12 Ah on it, so should run a few hours at 2A.

Short story/question: how to run off a 12v (or a 6v) battery and drop down to 5v - to power everything. Right now I'm set up to run right at 2 amps by limiting the brightness. But future revisions it could bump up the amps. Maybe to 3 or 4 amps.

Long way around...Bear with me, a lot of description of my process to get to the questions.

I am using a small cheap voltage regulator (http://www.readytoflyquads.com/rtf-1v-17v-adjustable-voltage-regulator). No data sheet is given, but a few specs from the above page state:

Non-isolated step-down module (BUCK)
Input 4.75-23V DC
Output 1-17 V (adjustable with small screw on board)
Output current: Rated current 2A (3A MAX)

I’m assuming the 3A max is a short burst at a time. So for sake of this conversation, I’ll say the 2A is the max.

My assumption – before I knew about watts and power, was I could take a battery say a 12v which is under the maximum input, and use this regulator to go down to 5v. After tying this, and running at 2A you could imagine my surprise when I started smelling something hot. Way to hot for my comfort. Too hot to touch. Of course it was the regulator. So I unplugged and started researching. I found a good article (don’t have the link) that basically said, “Yup this gets every noob.” And I learned you need to check power. So the change in volts across the regulator, or other device, (12v down to 5v = 7v) x amps (2A) = 14 Watts. The description on the regulator page doesn’t have anything about watts. I assume I’m going beyond what this little guy can handle.

QUESTION 1:
So this is a lesson of cheap parts and not knowing full specs of it then? Just because something says it can handle 2A doesn’t mean you can run it at 2A since you probably will exceed the watts? And this is the difference of having a data sheet and not having one? Maybe it's a lesson of "with the right heat sink you can run at full amps given on the specs."

My solution – I switched to a 6v battery so now my power is (6-5v) x (2A) = 2 Watts. And it doesn’t get too hot.

QUESTION 2:
What is a good way to power projects from a 12v battery that needs to step down to 5v? For low amp projects the regulators mentioned above are just fine. Maybe better to ask this question: What is a good way to step down 12v (or even 6v) to 5v when you are running a circuit at 2A or even more?

What I’ve read is a linear regulator dissipates as heat and they get hot if running high amps – so I’m assuming the one I used is a linear regulator. And I’ve read a little on switching regulators are better since they don’t dissipate as heat as much as linear regulators. Which I assume would increase the life of the battery as well? I am interested in extending the battery life.

Any good examples of some products or solutions anyone uses to go from 12v down to 5v and run at 2 or 3 or even more amps?

Looking for “off the shelf” part, not make your own with parts on PCB (but would be curious how that is done as well or if it's even worth it.) No need for adjustable, a non-adjustable 5v output would be fine. I've seen LM-xyz parts mentioned but a few posts I've read said they really are for 1A or less. And is that a linear regulator not the switching type?

 

If you can use 6V directly it will be the best option, otherwise use a buck regulator like Lm2576 or similar, less heat waste .

 

If it were a properly operating 2A switching regulator it should not have gotten hot with 12V in and 5V@2A out.
But for$.99 it apparently is not a properly operating one.

A linear regulator would, of course, get hot since it would be dissipating 7V x 2A = 14W.
The one you are using states it's a switching regulator, apparently just one not able to handle its rated current and voltage.
You can determine what type it is by measuring the input current and the output current.
If the currents are the same, it's a linear regulator.
A switching buck regulator will have a lower input current, than output current.

So you either can buy a better/larger switching regulator, or use several of the cheap ones to power just part of the lights each, at a lower current that it can handle as determined by experiment (but don't try to run them with the outputs in parallel).
You definitely want a switching regulator to avoid having to dissipate all the heat generated by a linear regulator.

 

Thanks for the advice. Thinking back, I think I had my multimeter between the battery and the input side of the regulator and got 2A. So I might have just been measuring the "input current"? This part is new to me - input/output current. I was thinking it didn't matter where the multimeter was set up in the circuit. So my "output current" could have been much higher than the 2A reading I got? I will measure on the output side later tonight to see the difference.

 

Thinking back, I think I had my multimeter between the battery and the input side of the regulator and got 2A. So I might have just been measuring the "input current"? This part is new to me - input/output current. I was thinking it didn't matter where the multimeter was set up in the circuit. So my "output current" could have been much higher than the 2A reading I got?

Yes, and this was the problem. A switching regulator does not have a traditional power transformer, but it stores up and plays out energy in an inductor and this opens up possibilities. In a small linear regulator, the output current equals the input current. Not so with switchers.

Lets start with a synchronous buck converter like yours, only 100% efficient. Watts in = watts out, zero heat loss. If you have it adjusted to 5 V out, and connect a 2 A load, that is 10 W. If the input is 10 V, the circuit will draw only 1 A (again, 10 W). If the input is 12 V, the circuit will draw 0.833 A. Etc.

In your case, you connected a 12 V source, and adjusted the 5 V load until the *input* drew 2 A. That means the output was delivering 4.8 A.

Now the fun starts. If the regulator is 80% efficient, then for every 1 watt that goes out to the load, 1.25 W comes in the front and 0.25 W is dissipated as heat in the inductor and the switching transistor. Turning that around for your case, for every watt that comes in the front, 0.8 W goes to the load and 0.2 W goes up as heat. You had 24 W at the input, so 19.2 W (3.84 A at 5 V) went to the load and 4.8 W was dissipated in the regulator components. This is *way* more than it can handle without a heatsink, or fan, or both. If you left it running very long, the regulator chip would get hot enough to unsolder itself (yes, that's a thing), if it didn't die first.

Switching regulators are way more efficient than linears, but they are not magic. Watts is watts.

ak

 

Some follow up information...
I finally got back and checked the current again. I placed the multimeter between the battery and the regulator input and turned everything on, but reduced the brightness of all the LEDs, and it read 0.4 A. I unplugged and moved the meter to the output side (between the regulator output and all my other components) and it jumped around a little, but read 0.34 to 0.39 A (some LEDs flickered a bit so wasn't surprised I got a reading that moved a little. Just as a test, I reconnected the meter between the battery and the regulator and it read 0.411 A. I was curious if I bumped anything just to see if I got the same reading as before. So that's roughly 82% - 0.411 A (input current) x 0.820 = 0.340 A (output current.)

If that held true for my original setup where I measured 2A between the battery and regulator then it would have been 2 A (input) x 0.82 = 1.64 A on the output side of the regulator? And I set the output to 5v so W=V(output) * A(output) = 5v x 1.64A = 8.2 watts. Is that thinking correct? And then input watts = 12v x 2A = 24 watts. So was I dissipating 24 - 8.2 = 15.8 watts in heat? I don't think that's correct since AnalogKid did some math and says 4.8W was dissipated. Don't think I'm fully understanding everything yet.

@AnalogKid I read your reply a few times to try and understand it better.

In your case, you connected a 12 V source, and adjusted the 5 V load until the *input* drew 2 A. That means the output was delivering 4.8 A.

So you are saying I measured 2A on the input side, and I know I was using 12V source so watts (input) = 12v*2A = 24 w. So then on the output side, I am using 5v and the watts need to equal, so W=v x A...A = W/v = 24/5 = 4.8 A. So this part is confusing me if I now measured input current vs output current and the output is lower, not higher. I guess I still don't know if the regulator I have is linear or switching? I think linear since you mentioned

In a small linear regulator, the output current equals the input current.

And I measured just about the same current on input and output side this time around.

Let me know if a schematic would help here. I can get one made and uploaded.

 

I measured just about the same current on input and output side this time around..

But not for the reason you think. A switching regulator capable of a large output current that is asked for a small output current is not nearly as efficient as when running fully loaded. Two things. The regulator itself consumes some power. Also, some of the losses in the regulator circuit are relatively independent of the power passing through it. Your efficiency calculation is incorrect. Comparing input and output currents doesn't tell you what you want to know. Efficiency is about total input power and total output power, not just voltages nor just the currents. Redo the trial but measure both the voltage and current, calculate the two powers, and calculate the efficiency based on that.

ak

 

So with everything turned on, I measure 5.38v on the pins at the regulator input side and 4.24v on the pins at the regulator output side. Then for amps, I measure on the input side (meter between battery and input + pin of regulator) 0.376 A and on the output side (meter between output + pin of regulator and + rail on breadboard) 0.394 A. So input watts = 5.38v x 0.376 A = 2.02 watts, and on the output side = 4.24v x 0.394 A = 1.67 watts.

So does this tell me the regulator is dissipating 2.02 - 1.67 = 0.35 watts as heat? And efficiency is 1.76 / 2.02 = 87.1%?

 

Yup.

ak

 

Thanks for all the help everyone. I now know a little more about regulators! I really appreciate this site and the community. I might stay with the regulators I have and limit the brightness, or as @crutschow mentioned maybe use more to power different parts of the LED display to spread the load out. That would be relatively easy to use (2) of them in my project. I also found the Hobbywing UBEC which is a switching regulator. The instructions say output 5v/3A so I should be able to run higher brightness with this instead of the regulator mentioned in my first post. The hobbywing one is a little bigger. And it looks like there is a heat sink over almost the entire part. Need some time to experiment more with these parts.

If anyone was interested in the project, I've attached a couple photos. It is for a scoreboard display. It uses an Arduino Nano to run WS2812b LEDs. I also learned a LOT about soldering and layout parts on a PCB as well! You can see a ton of wires. I already have the next version somewhat planned out in my head that will eliminate a lot of the wires on the PCB (and save a ton of time!)

 

This series of lessons may help you.
http://www.learnabout-electronics.org/PSU/psu31.php
You could probably benifit from starting at the beginning. I have found this site pretty good.

 

So with everything turned on, I measure 5.38v on the pins at the regulator input side and 4.24v on the pins at the regulator output side. Then for amps, I measure on the input side (meter between battery and input + pin of regulator) 0.376 A and on the output side (meter between output + pin of regulator and + rail on breadboard) 0.394 A. So input watts = 5.38v x 0.376 A = 2.02 watts, and on the output side = 4.24v x 0.394 A = 1.67 watts.

So does this tell me the regulator is dissipating 2.02 - 1.67 = 0.35 watts as heat? And efficiency is 1.76 / 2.02 = 87.1%?

You are using a 12V battery and measuring 5.38V at the input? That cannot be right.

Bob

 

When selecting the DC DC check the datasheet look for the "Efficiency Curve" the higher the Efficiency the better you can pick up many off the shelf Buck's that have >92% Efficiency at 2A like the LM53635 from TI as an example LTC does some rally nice stuff also

 

You are using a 12V battery and measuring 5.38V at the input? That cannot be right.

Bob

Yea, that doesn't seem right does it!?!? I'll have to revisit that when I get time to go back to this project. Thanks for the comment.