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I've looked and looked but i can't find it anywhere. I remember reading once upon a time that it wasn't wise to power the my n+2 through the 5v pin no matter how smooth the supply was. Now i can't find where i read it and i'm confused because I've found a few projects that do. I've got a project that requires lots of 5 and 12 volt devices and i'd hate to need to add 9v just so i could use Vin.
Please tell me I'm wrong and that it's perfectly fine to bypass Vin if you've got good 5v supply?
Maybe was me, but perhaps someone else found the same issue.
In the first Netduino edition, the +5V pin is a simple hook for that power rail. The pin is directly connected to the regulator output, and also to the +3.3V regulator input.
In the Netudino 2, the +5V pin is controlled by a mosfet, which is controlled via software, so by the MCU. When the MCU is powered off, there's none being able to drive the mosfet gate and it appears impossible to power the board through the +5V pin.
BTW, the mosfet embeds a small diode which is able to yield the current flowing *from* the +5V header pin *to* the internal +5V power rail. When you apply +5V on that pin, the diode will let the current energize the whole board for a brief, until the MCU wakes up and actually drive the mosfet.
In my (very honest) opinion, I wouldn't have added such a mosfet, because the external power control (e.g. shutdown extra logic) is rather hard to generalize. A simple mosfet would become effectively useful in a very small number of cases.
Why?
Suppose having a famous 74HC595 shift-register SPI-driven with a bunch of leds connected, and the goal is to power down this logic in order to save energy.
If you cut off the +5V power from the 595+leds using the mosfet trick, you only detach the rail, but *NOT* the other wires of the SPI. Because of that, when the 595 power is down, most of the energy will flow through the MOSI and SCLK lines. This condition is even worse than leaving the power always connected.
Solution?
Unless you wire a short across the mosfet using an iron, you'll have to accept the risk of leverage the diode as temporary way for the power up.
Cheers
Biggest fault of Netduino? It runs by electricity.
Thanks for a terrific reply. Forgive me if this is a silly question.
Would connecting the Vin to the 12v line with a resister that only allowed 150ma, solve the problem of assuring it booted up. The only problem I could see is that the on board 5v reg would be potentially competing with the other 5v supply. And if it was a fraction higher voltage wouldn't it all try and come through that tiny reg on the netduino.
Without blowing my brains out with info overload. How would you approach this.
Do I man up and add a 9v reg in there some where. Or will powering it with 12v through Vin be OK because all the other devices would draw there power off the other 5v rail. All though there is quite a bit of 3.3 volt stuff I'm using so that's probably not a good idea either. Since I'm planning to use the netduino reg for that.
Or do i take the risk. What are the odds of that diode biting the dust?? BTW what does "[color=rgb(40,40,40);font-family:helvetica, arial, sans-serif;]risk of leverage" mean.[/color]
I am not sure to understand your second option: is it to power the board through the regular Vin at +12V?
If so, I'd prefer this way. Although the board is rated for up to 9VDC, you can power even at +12V until the current drawn is limited (e.g. the board itself). In such a case, the onboard +5V regulator keeps cold and the reliability won't be an issue.
There's another issue regarding the automatic USB-power switch, that draws a weak current through the Op-amp when the Vin is greater of +10V approx. However, that current is very weak and the Op-amp stress is low.
The counterpart is powering the board via the +5V header.
The mosfet diode is rated for up to 150mA (if I remember well), but that diode's purpose is rather a protection for spurious inverse currents/voltages. Now, when you apply the external power at the very first time, all the capacitors are totally discharged, of course. However, in order to rise the rail voltage, the caps requires *a lot* of current for a brief time.
It's not easy to say what's the reliability of that diode: perhaps you can use without any risk for years, or even get the mosfet's life short. It depends on several factors, and the simplest way to test is...to test!
Again, my choice would be using the regular Vin.
Cheers
Biggest fault of Netduino? It runs by electricity.
Well that does it then. I'll add a 9v rail of sorts. Just another question about that then. Since a 7809 just bleeds heat anyway. Would a simple voltage divider do the trick. Any other advice?
Well that does it then. I'll add a 9v rail of sorts. Just another question about that then. Since a 7809 just bleeds heat anyway. Would a simple voltage divider do the trick. Any other advice?
Thanks
IMHO another regulator is a waste because useless. Moreover, you have an source input of +12V and a supposed output of 9, that is 3V of drop across the 7809. Bear in mind that such a regulator works fine when the drop is not-so-close to the minimum, which is 2V (according to the specs).
Please, don't be afraid by the heat: the Netduino itself requires around 100mA. To get just 1W of wasted power, you should have 10V of drop across the regulator! For a 7809, 1W in open air yields about 90°C of the case, but any metallic shield would sink the heat well!
A resistor divider is the worst thing you could do. The power waste is much bigger than a regulator, so you must use big resistors. Moreover, a divider won't give you any regulation, just a waster eating a bite of energy from the original source.
Voltage divider are good just for very small currents (< 1mA), or when you have no other way to solve.
If you feel better adding an extra regulator on the Vin, just use a LM7808 than a '09. However, as far the source +12V are pretty stable, I wouldn't use any regulation for the Vin. As "stable", I mean a battery pack or another regulator PSU (e.g. wall), but *NOT* a car supply, which rises up to 16V and it's full of noise.
Cheers
Biggest fault of Netduino? It runs by electricity.
Hi Grant,
Mario's advice is good
Netduino Plus 2 is really designed to be powered via USB or VIN (power barrel or VIN header). You can drive power to the 5V header, but it's really designed to provide 5V power rather than be powered via 5V.
The FET on the 3V3/5V headers was added to help address USB enumeration issues on certain computers (i.e. enforce <=100mA during USB enumeration) and to allow developers to power down/up expansion shields.
Chris
Do not understand why not just use a usb cable to connect to to your 5v filtered supply, or tap the netduino usb 5v in point. Adding another supply seems ....bulky.
. .....I've got a project that requires lots of 5 and 12 volt devices and i'd hate to need to add 9v just so i could use Vin.
Please tell me I'm wrong and that it's perfectly fine to bypass Vin if you've got good 5v supply?
Hi RAG,
The 5V and 3V3 headers are just there for your convenience, as an easy way to supply power to components in your project. You can certainly tap into other power sources for your project.
Chris
The bigger question that would determine my advice would be this. How serious is your project, are you selling it and what would be the pain in downtime if the FET diode blows out. If your not selling it and it's not mission critical then by all means, power it through the 5v header. If you answered yes to the other two questions then like me. You'll probably end up adding another reg to be on the safe side.
Netduino Plus 2 is really designed to be powered via USB or VIN (power barrel or VIN header). You can drive power to the 5V header, but it's really designed to provide 5V power rather than be powered via 5V.
The FET on the 3V3/5V headers was added to help address USB enumeration issues on certain computers (i.e. enforce <=100mA during USB enumeration) and to allow developers to power down/up expansion shields.
Chris
Answered
