11 replies to this topic

[r007rbk]

Hi everybody, Netduino plus technical specs (http://www.netduino....oplus/specs.htm) rates the input voltage from 7.5 to 12 VDC, and schematic has a note saying "9V preferred"), while the LDO used in the board MC33269DT-5.0G is rated at 20v max (http://www.onsemi.co...l/MC33269-D.PDF) Would it be risky to power the board at 13 volts? and at 15v? The only thing I think might be a problem is the voltage R4/R10 voltaje divisor to VIN_COMP net because if not, why you specified "9V preferred"? (apart of minimizing the LDO thermal overheating) Thanks in advance Rebeca

[Chris Walker]

Hi Rebeca, In a word, heat. We could only fit so much heatsinking on the board...and pushing 800mA at 20V (dropping to 5V) through the 5V LDO regulator is going to be hot. The higher voltages aren't as much of a concern when current is low. But if you're driving quite a bit of current then I'd recommend providing a 5V regulated power source to the 5V header. Welcome to the Netduino community, Chris

[Mario Vernari]

Chris, when the Vin rises over 10-11V, the inverting input of the comparator (VIN_COMP) is getting higher than +5V. BTW the comp specs recommend to limit the input pins within the V+..V- range. Did you perform any test? Is it worthwhile to add a 4.7V Zener? Cheers

[Chris Walker]

Chris, when the Vin rises over 10-11V, the inverting input of the comparator (VIN_COMP) is getting higher than +5V. BTW the comp specs recommend to limit the input pins within the V+..V- range.
Did you perform any test?
Is it worthwhile to add a 4.7V Zener?
Cheers

Hmm, good thought. We've been looking at this circuitry recently, so I'll chat with our electric engineers about it. We haven't seen any issues when testing with 12VDC power supplies--but you are correct that the 7.5VDC-10VDC range is best (especially since most AC-DC power supplies are in a word, unregulated).

Chris

[Mario Vernari]

I would perform the test up to +15V, which is above the car supply range. It would be useful to connect directly to the car outlet. Cheers

[Inquisitor]

I would perform the test up to +15V, which is above the car supply range. It would be useful to connect directly to the car outlet.
Cheers

I second this! I was wondering the same thing... when it says limit to 12 Volts, but cars routinely show 14+ volts when charging.

IF I understand you Chris... The issue is not the voltage (up to 20 V), but a matter of what I'm driving off the Netduino pins and thus trying to pull through the voltage regulator(s) on the Netduino board. As long as I'm using small, low current stuff, I should be ok on a car's voltage system?????? Or am I overly simplifying things too much?

If so, I'm guessing there should be some upper limit of current... would that be the 200 mA on the specs... or something less? And No! I wouldn't hold you to it... if I fry a Netduino by putting it on a car, its totally my fault.

[Kenny Spade]

I second this! I was wondering the same thing... when it says limit to 12 Volts, but cars routinely show 14+ volts when charging.

IF I understand you Chris... The issue is not the voltage (up to 20 V), but a matter of what I'm driving off the Netduino pins and thus trying to pull through the voltage regulator(s) on the Netduino board. As long as I'm using small, low current stuff, I should be ok on a car's voltage system?????? Or am I overly simplifying things too much?

If so, I'm guessing there should be some upper limit of current... would that be the 200 mA on the specs... or something less? And No! I wouldn't hold you to it... if I fry a Netduino by putting it on a car, its totally my fault.

I've powered it through a car USB adapter that was rated at 5V 2.1A for iPad charging. The Netduino worked great, but it would be interesting to know if I was courting disaster. I know that I have experienced some instability in power coming from my car outlets before, though the car I am driving now seems to have better isolation than others I have tried. I can think of a few neat ideas for using a Netduino in my car, so it would be good to know that it would be safe to do so.

So far, this thing has been incredibly tough. I stayed away from hardware because I always ended up frying electronics when I touched them in the past, but the Netduino has performed admirably considering how many times I thought I must have killed it.

[Mario Vernari]

The power wasted in heat by the regulator is:

P = (Vin - 5) * I

where I the current flowing. Assuming I=80mA (Netduino Plus without anything else), when Vin is 15V, then power is:

P = 0.8W

The regulator specs say that the thermal-resistance (junction-ambient) is Zja=92'C/W. Since the regulator is solded on the PCB, we may assume that all the heat will be dissipated by the PCB itself.
What is the "thermal-resistance"? Is the "ability" to conduct the heating from the junction (where the heat is originated), to the ambient (heating sink).
At this point:

Tj = Ta + Zja*P = 25 + 92*0.8 = 98.6 'C

where: Ta=25'C is the ambient (PCB) temperature, and Tj is the internal regulator (junction) temperature.
Since the reg specs say that the max Tj is 150'C, the chip is still within the safe range.
However, that's much like an optimistic depict. The PCB will be warmer than 25'C, and the ambiental temperature could be higher as well.

Old school: an easy and good test is to put your finger on the regulator: if you can leave it without frying it, the chip is *surely* safe!
Cheers

PS: I am sorry for using the temp in 'C instead of 'F.

[pa-1000]

hi there, I'm also interested in using my netduino plus in my car. Does anyone use a netduino plus in his car? If yes is your netduino still a live? cheers

[mohammad]

Hi all, Is there any way for Netduino to understand its input voltage value? Cheers, Mohammad

[Chris Walker]

Hi Mohammad, Sure! If you take the voltage from VIN (or 5V0 if using USB power), divide it with a resistor divider circuit, and then plug the result of that circuit into one of the analog inputs...you can read the resulting analog voltage. You'll need to multiply that analog value by the same amount that you divided the original voltage by...and you'll have your input voltage. Chris

[mohammad]

Hi Chris, Thanks for your useful reply. Cheers, Mohammad