I have a bare-bones Uno using a L7805CV regulator. It is for a remote battery powered device. The battery is a 12VDC car battery.
I realize the device will shed a lot of excess power as heat due to the 12-5 V conversion. Is there a better / more power friendly voltage regulator I can get? The battery is a deep cycle one so the voltage will drop as it "ages".
Aside from the board it will have 3 PIR sensors and one LED. This is the minimum configuration for my project.
Thank you all!
Specifically: I have noticed my full Uno boards can be fed 4.85 VDC from an old lantern battery and still power up (with no peripherials). The LEDs work and so on. My bare-bones 329P-PU will take the same battery input and put out 0.14 VDC at the 7805 Vcc.
So it seems the Chinese Uno can scavenge theb power left in the battery where as the 7805 is useless when the battery drops below 5VDC.
Hi.
Consider what's known as a 'Step Down Module'.
Those are a lot more efficient than any 78xx solution, but they're a bit more expensive.
If you mean the same thing as i understand when reading 'car battery', then by the time it has 5 volts left, you'll have other problems than keeping your 7805 working.
MAS3:
Thanks. Do you have a wiring schematic for one of these "step down modules" so i can ditch the 7805?
I would consider something like this:
Mini-DC-DC-Converter-Step-Down-buck-Power-Supply-Module
If you search eBay you will find a lot of this type of converter.
Good luck.
JohnRob: Yep... already found them!
Are they really more efficent than the 7805 when used with a 12VDC battery? I have read some of the ads on eBay and it does not tell how the two compare.
They are.
The difference between the two of them is that the 78xx has to burn the excess away, where the module switches until the correct voltage is reached (something like that).
The 1st one means a lot of heat production, the 2nd means some heat production (of course they aren't 100 % efficient).
Expect a number like 88 % efficiency for the 2nd. one for a conversion of 12 - 15 volts to 5 volts.
Efficiency will go up a bit if the output is closer to the input by the way, but that's not the case here.
Choose a seller who offers all details, including the numbers and how to connect and adjust.
These modules are just a few bucks each so for the first project make sure you know what you'll get.
MAS3:
Wow! I did not realize the difference. Thank you for the advice and input!
One more question if you don't mind:
If I use a deep cycle marine 6VDC battery instead of a 12VDC car battery... will the buck / step down converter still be more efficent than the L7805 and if so by how much? I just want to power my Uno board with 3 PIR sensors.
Thank you again!
I'm not sure about that.
You might get too close to the lowest input voltage for the module.
Rule of thumb for a 78xx: input voltage needs to be about 3 volts over output voltage.
So that one will for sure not be suitable for a 6 volt to 5 volts conversion.
2:32 AM on a working day: I'm off to bed...
OK... thank you for all your help!
Switching regulators are not a magic bullet that will solve all problems. It is true that they are more efficient at converting power than a linear regulator, but they also have higher amounts of quiescent current. In a battery powered operation where the system is sleeping most of the time, the power saved by the better conversion efficiency could be dwarfed by the extra power the regulator itself wastes while continuously operating.
Your average switching regulator is probably going to take a few hundred microamps to a couple milliamps of current even with no load. It's not impossible to find micropower switching regulators with much lower operating current, but you're not going to find them ready-made on ebay.
Linear regulators can go far, far lower, like the MCP1703 with a nominal 2 uA of operating current (0.62 USD each direct from the manufacturer for the non-automotive version). This thing would be golden for a very low power project, as long as you can deal with the fact that it's only in SMD packages SOT-223 is pretty big though, that's the package of the AMS1117 regulator right next to the barrel jack on an Uno. Breakout boards are pretty easy to get on ebay, you just have to wait a really long time for the slow boat from China.
Deciding which is best requires a detailed analysis of how power is used in your system, called its power budget. Which parts of the system are using power? How much? For how long? How often? The answers to those questions can be used to figure out the normalized average power usage of each part of the system, so that you know (instead of just guessing) which parts of your system use the most power.
Answering those questions also requires analyzing the behavior of your system. You mentioned an LED, 3 PIR sensors, and the microcontroller. What kind of LED is it? A tiny little 5mm gumdrop, or a 100W beast of a floodlight? How often and for how long is it going to be turned on? How much current do you need it to use? How often do you need to power the PIR sensors? How much current to they use while operating?
It's not out of the question to use a hybrid strategy. A low power linear regulator can power the digital circuitry, and a switching regulator that you can switch on and off can power the stuff that needs a lot of current less frequently. That all depends on where things fall in your power budget.