Okay, a couple of days ago I smoked a board for my first time ever. it was a Pro Mini 3.3V, and I smoked it when I accidentally and only momentarily touched a 9V battery backwards to leads I had connected to the Raw input and Ground connectors. There was a spark (complete with an audible "crack"), a good amount of magic smoke drifting away, and the input voltage regulator was fried.
The board still works, programs, outputs, reads analog in, etc. when I power it from a FTDI breakout board, so I've just moved it to a different project.
Then yesterday I smoked another (man, stupidity is expensive! I'm glad I'm not always so stupid).
Conclusion: it's time I learned how to build in power supply protection. I THINK I can just add a diode to the +9V in line. I also have a couple of the resettable fuses that Sparkfun sells - I'm thinking about adding one in series next to the diode for additional protection.
I have enough voltage to spare with the 9V so the voltage drop from the diode won't cause problems. I don't think I'll be pulling enough current to need to worry about which diode I use, but I could use advice on that, and also on whether (and how) to use the resettable fuse.
My board is small, and I don't have a lot of real estate, but a couple more components won't be a problem. Oh, and yes, I'm probably going to change out the connector for one that only plugs in one way (so I can't reverse polarity). But I want to learn something from my stupidity as well.
Yes a series diode will prevent a brain fart of hooking the battery up backwards, the cathode side of the diode will wire to the module the anode side to the battery + lead. A 1N4001 diode is fine.
Not sure resettable fuses is really needed as a 9vdc battery really can't supply much current without dropping it's voltage, but a 500ma resettable fuse won't hurt anything.
The problem with a resettable fuse is that it takes time to blow. Much longer than it takes electronics to fry. They are of only limited use is specific applications where the electronics can withstand a short overload.
Thanks, Mike. After I posted, I did see something about 2.5 seconds in the datasheet, which struck me as not very helpful in this situation (the magic smoke was coming out in well under one second).
I've been playing with diodes to see about how to set that up. I had some very small ones on hand: they took my 8.8V battery down to about 8.3V in the forward direction, which was fine, but in the backward direction there was still over 6V flowing, which, I think, would have still fried my Ardiuno. I just bought some larger ones, including 1N4001s - I'm going to play with them in the circuit (again, testing voltages before attaching another Arduino) when I get home tonight.
I wouldn't mind a little advice on how to pick the right diode, though, other than trial-and-error.
The 1N400x series of rectifier diodes (as a previous poster mentioned) will work fine. The x indicates the max voltage rating. The 1N4001 will handle 35 V RMS reverse, which is more than enough to protect a 9V source.
If you put in a reverse protection diode, it's also worth putting in a low resistance (100 ohm) resistor in series. I do this in case the reversed power is applied for a period of time -- the resistor provides a load on the battery, preventing it from being in a shorted state (which could cause it to explode, worst case).
If you put in a reverse protection diode, it's also worth putting in a low resistance (100 ohm) resistor in series. I do this in case the reversed power is applied for a period of time -- the resistor provides a load on the battery, preventing it from being in a shorted state (which could cause it to explode, worst case).
That makes no sense to me for a series connected polarity protection diode. If the protection diode was wired from the +9v terminal to ground (normally reversed biased but conducting if battery is backwards) then a series resistor would be a good safety addition, but for a series connected diode it serves no purpose.
but in the backward direction there was still over 6V flowing,
Voltage doesn't flow current flows. What you are probably seeing is the reverse leakage current of the diode, this is a very small current and your meter is doing well to see any voltage with this, it will not be enough to damage your circuits.
Alternatively the diode could be breaking down or you could have a zener diode but I don't think you have any of these.
I agree with Lefty, you are better off using a series protection diode rather than a shunt diode connection. If you don't like the voltage it drops then try a schlocky diode as these drop less across them.
Mike, it was a zener diode. Be patient - I don't know any better, and it was all I had on hand. Right now I've got a little bag of 1N4001s and I'm going to try one of those.
(headslap) about that "6V flows" comment. I do know better than that.
So to be clear, a 1N4001 in series with the +9V input to the Raw connection should protect my circuit (more specifically, my Arduino Pro Mini). Right?
Again, I'll pretest to make sure the voltage drop is okay. I should be fine - the Arduino is supposed to work right down to 3.3V in, if I remember correctly.
(testing)
Okay, I get 8.9V from the battery with no diode, about 8.6V with the diode there in series, and practically no voltage when I reverse it. Beautiful. Warming up the soldering iron now.
Okay, I learned how to use a diode to protect against reverse voltage. Next related question: the second Arduino Pro Mini that I smoked wasn't from a reverse voltage connection. It was running as expected, and suddenly stopped working and the magic smoke came billowing out. I THINK I now understand what my problem was, and I'd like to check my thinking with you guys.
It was a Pro Mini 3.3V, powered by a 9V battery connected to the ground and RAW input pins. I was using the Pro Mini's 3.3V output pin to power an accelerometer and an XBee Pro, which was transmitting furiously several times per second. I now believe that I might have exceeded the max current output of that 3.3V pin on the Pro Mini, as the XBee Pro draws 215 mA during transmit (per the Digi manual for it). I don't know how much the accelerometer would have been drawing, but I'm suspecting that I smoked the Arduino's 3.3v voltage regulator simply because I tried to draw too much current through it. Is that possible?
A look at the Pro Mini's schematic shows that the onboard regulator is rated for 150 mA, which means that I was exceeding its rated output as soon as I attached the XBee Pro even if nothing else were drawing current. I've got a funny feeling those parts have current ratings for a reason, having something to do with keeping the magic smoke inside where it belongs.
It's kind of important for me to understand this stuff, obviously. I wound up replacing the onboard regulator with an external one to power my devices, and it works so far. However, the regulator I used is an TO-92 LDO regulator that is only capable of putting out 250 mA, so I bet I don't have a lot of headroom on the current even with the current (no pun intended) setup. I could take that one out and replace it with a larger TO-220 sized one that can supply 500 mA or more - should I? This is eventually going to go into a rocket, and I'd rather not have a smoking part next to my ejection charges. :o
do some math 9V - 3.3V = 5.7V * mA = Watt. What is the ratings of your Vreg ? if you exceed it probably will shutdown, even if you exceed the mA output rating it will shutdown.
Even better substitute "will" with "should", because, as you learned the hard way, they rather go belly-up in smoke than properly shutdown as stated in the data sheet.
Lesson learned : NEVER trust data sheet , always be prepared and double check everything !
ON a second tough , way you don't power your external boards/sensors with their own power ? it the same thing that having an extra LDO reg on the ardu and you get more juice to power them .
Ah BTW , when an alkaline battery shows open circuit a voltage under the rated nominal voltage it should be considered dead , by the book ... so if you use a 9V alkaline batt and its say 8.something V don't trust it to stay this Voltage under heavy load, like the ones you have.
So if I added a second 3.3V LDO 250 mA regulator fed from the same battery as the first, I could safely source an additional 250 mA from it? Is that correct?
It'd be pretty simple for me to do that, and wire it up so that the XBee (which, again, requires 215 mA when transmitting) has its own 250 mA regulator and the other regulator provides 250 mA to be shared by the Arduino Pro Mini, the accelerometer, and perhaps a small temperature sensor or two. That'd bring me further away from reaching the current limit of the regulator.
I know you can't treat the battery as an bottomless source of current either, but I think the overhead is a LOT higher there than on the post-regulator side. I don't think I'll tax the battery if I add a second regulator, if that basic idea works okay. It seems to me it would: I'm already running devices that have both a 3.3V and a 5V regulator fed from the same power source.
That would certainly work. However battery capacity (duration) is a limited resource sometimes (think flight time in a R/C aircraft) so one should be aware of the overall efficiency and where the losses are. Running a 3.3 regulator from say a 9 volt battery source wastes 5.7 volts times the current consumed in waste heat. If you can afford that then your good to go. It kind of depends on the MA capacity of your battery and how much time duration you want before having to power off and recharge or replace the battery.
It needs a kind of top down power management design review step.
I'm hoping to use a 3.8V Lipo battery, 1100 mAH. That's partly why I'm using a LDO regulator (okay, also because I had a couple on hand).
It's a rocket flight: duration isn't really a big deal here: I'll turn the device on, it'll sit on the pad for 5 minutes, fly for 2 minutes, and since I'm sending the data down by RF to the ground station, it doesn't matter if the battery runs out after I've landed. I think that 1100 mAH is WAY more than I need anyway. I could almost make do with 1100 milliamp minutes.
In that case I would be temped to just use a single cell Lipo battery without the 3.3v regulator. I would add a single series diode to drop the fully charged cell of 4.2vdc down a little, I bet that would work just fine.
