Hello Everyone,
I come from more of a programmer background then electronics, so forgive me if I mence terminology. I am currently building an arduino powered gps device that will transmit geo data every 1-5 minutes (interval undetermined) over the cellular network via the SM5100B shield. This device needs to be battery powered and ideally last all day on a single charge. The shield specs state that it can draw 2 amps of current at peak transmit, which I assume will happen when polling the gprs network. To accomplish the needs of the module I am looking at a few different options but wanted to get some input on what method would be best.
option a)
6v battery rated at 3.5mah
1 diode to drop the voltage to 5v
option b)
12v battery
LM2576 switched voltage regulator to drop to 5v
And I am open to better suggestions if anyone has one. ideally I would like to power the device with a 6v, as I found one in a local store that is perfect size for this application. Going to a 12v battery will make the device a little bulkier then I really want. But using a voltage regulator seems like it might provide a more reliable circuit, however I was unable to find any voltage regulators that worked on less then 7v. Another concern with a voltage regulator would be power consumption. i am worried that going that route could waste alot of power from the battery.
Ok, so after further reading I realize that I need to connect 7-12v to the vIn pin or power supply jack to power up the arduino via a battery, so that helps answer some of my questions. Knowing this, can I pull 2 amps of current via the vIn pin or is that going to blow the arduino?
The diode would work, but batteries are not like gas tanks... The voltage drops as the battery discharges. The might be OK, since you only need the 2 amps intermittantly...
Switching regulators are very energy-efficient (usually in the ballpark of 90%). With a 12V battery, you will find that you are only drawing about 1A from the battery when you are supplying 5V @ 2A to the load! And with a 12V battery, the battery could discharge "badly" to about 7V, and the thing would still work. (This would make your battery work more-like a gas tank.)
Also, try to find the specs (discharge curves) for the batteries you are considering. The mA-Hr rating is a rough starting-point, but doesn't tell you how much peak or average current the battery can supply... Along with the voltage rating, it tells you how much energy the battery stores.
I have a feeling that you can use a smaller battery, depending on how long your system needs to run between charges.
So I started reading the data sheet and realized that the chip has a max voltage of 4.2v and this got me wondering how its powered. Thus I dug into the schematic for the shield and low and behold it is already equipped with an spx29302 voltage regulator. So it looks like I am covered and pulls its power from the vIn pin on the arduino. Looking at the arduino schematic it looks like the vin should have the exact same voltage and current available to it as the power source plugged into the dc socket. Am I correct in that assumption.
cangeceiro:
Oh wow, thanks for the quick responses.
So I started reading the data sheet and realized that the chip has a max voltage of 4.2v and this got me wondering how its powered. Thus I dug into the schematic for the shield and low and behold it is already equipped with an spx29302 voltage regulator. So it looks like I am covered and pulls its power from the vIn pin on the arduino. **Looking at the arduino schematic it looks like the vin should have the exact same voltage and current available to it as the power source plugged into the dc socket. Am I correct in that assumption. **
So now its on to find an appropriate battery.
Between the arduino external power connector input and the Vin pin is a series polarity protection diode, rated at a nominal 1 amp maximum current flow and around .6vdc voltage drop. Be sure to include this information in your project planning.
6 good quality NiMH AA cells will give 7.5V or so, and support high current (I said good quality).
2 LiPo cells give 7.4V (and definitely support very high current). LiPo cells have issues though (need protection circuit, proper charger, should not be charged unattended).
When you said "3.5mah" did you mean 3.5Ah ? If so then that would imply NiMH C-cells which go up to about 4 to 5Ah.
"Don't charge LiPo cells unattented" is a bit of mysticism to me. If you have a correct charger that cuts off at the right time, it's not needed. Consider that some fancy pieces of computer hardware (such as large DRAM disk caches) have on-board LiPo cells for memory protection, managed by a proper charger, and those computers are stored unattended for months and years in server closets and data centers around the world. (This is a major use case for 18650 size cells and on-PCB holders for those cells)
RC enthusiasts, that use "raw" LiPo cells without designed-in circuitry and with perhaps un-matched chargers, do well to not leave their cells unattended while charging. But charging your cell phone, or laptop, or some other well-designed LiPo powered device is just as safe as a NiCd or Lead-Acid or whatever other battery chemistry you want -- except you get more power for less weight
If you really worry, use LeFePO4 instead. Almost same energy-per-weight ratio, with much safer battery chemistry. Basically impossible to "charge until they explode."
sorry to bring up this topic, but based on this recommendation...
how would you connect the 5Vout to the arduino?
I am reading threads like this one and i am a bit confused on the best way to power the arduino
sorry to bring up this topic, but based on this recommendation...
how would you connect the 5Vout to the arduino?
I am reading threads like this one and i am a bit confused on the best way to power the arduino http://arduino.cc/forum/index.php/topic,54334.msg409987.html#msg409987
You apply the +5V to the +5V pin on the board, bypassing the linear regulator. You don't feed it in thru the power coaxial connector of the board. I believe you will still be powering the onboard 3.3V linear regulator though.
sorry to bring up this topic, but based on this recommendation...
how would you connect the 5Vout to the arduino?
I am reading threads like this one and i am a bit confused on the best way to power the arduino http://arduino.cc/forum/index.php/topic,54334.msg409987.html#msg409987
You apply the +5V to the +5V pin on the board, bypassing the linear regulator. You don't feed it in thru the power coaxial connector of the board. I believe you will still be powering the onboard 3.3V linear regulator though.
That does however conflict with the Arduino's warning about power the board via it's 5V pin:
5V.This pin outputs a regulated 5V from the regulator on the board. The board can be supplied with power either from the DC power jack (7 - 12V), the USB connector (5V), or the VIN pin of the board (7-12V). Supplying voltage via the 5V or 3.3V pins bypasses the regulator, and can damage your board. We don't advise it.
thats why i was asking CrossRoads, cause I just learned today that its not advice to do so. I have been doing this for ages, and today I learn in openenergymonitor forum that its a bad practice.
CrossRoads, could you explain your suggestion?
afremont:
But that's precisely why you would use that pin, to bypass the linear regulators.
Several posters have reported damage to their on-board 5vdc regulator by applying a regulated +5vdc to the 5V pin. Arduino folks seemed to have changed the brand/module of regulator on the rev3 boards. CrossRoad member feels the regulator doesn't like having a external voltage applied to it's output with no input voltage on it's input. The regulator's datasheet suggests adding a diode across the regulator's input and output terminals to protect the regulator.
And that is where the topic has been for some time now. The Arduino folk's warning has not come with a technical description of the why for their warning.
I was wondering if that might be the issue. I didn't look at the datasheet for the regulator being used now, but with something like a 7805 you're not supposed to allow the output pin to be higher than the input. A diode is often used to route voltage back to the input pin. I would have hoped that a modern part would automatically include this trivial feature to protect the regulator. Applying +5V to Vin and the +5V pin should eliminate the problem by prevent the reverse bias of the regulator.
I have seen quite a few recommendations to do what I suggested, but hadn't read yet of anyone experiencing any problem. I didn't search for it though.
afremont:
I was wondering if that might be the issue. I didn't look at the datasheet for the regulator being used now, but with something like a 7805 you're not supposed to allow the output pin to be higher than the input. A diode is often used to route voltage back to the input pin. I would have hoped that a modern part would automatically include this trivial feature to protect the regulator. Applying +5V to Vin and the +5V pin should eliminate the problem by prevent the reverse bias of the regulator.
I have seen quite a few recommendations to do what I suggested, but hadn't read yet of anyone experiencing any problem. I didn't search for it though.
And the thing about it that puzzles me is that when one is powering a board with just USB power the on-board regulator also then has +5vdc applied to it's output with nothing on it's input, so what is the difference in that case as when one applies a regulated +5vdc to the 5V pin? The only think I can think of is that there is some current limiting when on USB power via the 500ma thermofuse, so maybe that limits a possible peak current spike/path going backwards through the regulator? Who knows, but the diode addition sounds like cheap insurance if one is going to power via the 5V pin.
I didn't look at the datasheet for the regulator being used now, but with something like a 7805 you're not supposed to allow the output pin to be higher than the input. A diode is often used to route voltage back to the input pin.
With a voltage drop across the diode, how would that work? The voltage regulator on my old arduino put the arduino +5v pin at 8v+ when externally powering using 12v in on the barrel jack. I then used a 7805 chip from the 12v to the +5v arduino pin and the arduino worked without issue.
Interesting. I looked at the datasheet for the regulator and it does indeed have an internal protection diode connecting Vout to Vin. It is capable of handling a 50A surge which is far in excess of what might be encountered in normal cycling of the device. In case the Vin is shorted to ground and Vout contains more than about 50uF of capacitance could this limit be exceeded requiring another external diode. I think reports of connecting a truly regulated 5V supply to the +5V pin causing damage are highly suspect.
I didn't look at the datasheet for the regulator being used now, but with something like a 7805 you're not supposed to allow the output pin to be higher than the input. A diode is often used to route voltage back to the input pin.
With a voltage drop across the diode, how would that work? The voltage regulator on my old arduino put the arduino +5v pin at 8v+ when externally powering using 12v in on the barrel jack. I then used a 7805 chip from the 12v to the +5v arduino pin and the arduino worked without issue.
Obviously the last diode drops worth of voltage won't be routed back, but will not cause a problem. If you measured +8V on the +5V pin with +12V connected to the barrel connector then something wasn't measured right. +8V would easily destroy the CPU.
