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Powering the Netduino


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#1 Cabadam

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Posted 22 February 2012 - 03:39 AM

I currently have a 24VAC -> 12VDC (regulated) in place to power my Netduino Plus.

I am about to insert an LCD screen into the mix, which is 3.3V, but theoretically could draw up to 1A, so I do not want to power it through the Netduino.

My intent was to get something like this DC stepdown. Split the 12VDC before it goes into the Netduino and run it through this to step it down to 3.3V.

That got me thinking, if I were to buy 2 of them... I could basically run the Netduino at any of the voltages within it's range.

So question is - given a choice, do I CARE what voltage I provide it with? Will that change the resulting current draw?

Thanks!

#2 hanzibal

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Posted 22 February 2012 - 10:07 AM

I usually use a simpe LD1117V33 LDO regulator for 3.3V:

http://www.sparkfun.com/products/526

It can take a maximum of 15V in and deliver a regulated 3.3V at up to 1.3A. There are simular regulators for other voltages too. From what I know the Netduino will run equally well regardless of what voltage you supply (as long as it is within the correct range) and the current draw will stay the same.

#3 MDS

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Posted 22 February 2012 - 10:52 AM

Do you have the spec sheet on the display. I dont see how the display could draw an AMP.

#4 Cabadam

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Posted 22 February 2012 - 01:51 PM

I usually use a simpe LD1117V33 LDO regulator for 3.3V:

http://www.sparkfun.com/products/526

It can take a maximum of 15V in and deliver a regulated 3.3V at up to 1.3A. There are simular regulators for other voltages too. From what I know the Netduino will run equally well regardless of what voltage you supply (as long as it is within the correct range) and the current draw will stay the same.


Interesting, I'll have to think about that. The 24AC->12VDC converter is rated up to 1500mA, so I was thinking I might actually have two of those (one for Netduino, one for LCD). But then I realized I was comparing 1500mA at 12V to 1A at 3.3V, which I can't do (right??). I'd need to convert to watts/VA first? In which case, I fit with just one AC/DC converter just fine.

Although, a closer look at the datasheet for the regulator you linked says up to 800mA, not 1.3A?

Do you have the spec sheet on the display. I dont see how the display
could draw an AMP.

It's one of the serial-enabled LCD's on sparkfun. Here's it's datasheet. I was basing the power requirement on the statement: "Backlight transistor can handle up to 1A" I see no other indications of what it would need to draw.

#5 Geancarlo2

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Posted 22 February 2012 - 05:19 PM

It's one of the serial-enabled LCD's on sparkfun. Here's it's datasheet. I was basing the power requirement on the statement: "Backlight transistor can handle up to 1A" I see no other indications of what it would need to draw.

The datasheet says the modules uses 3mA with backlight off and about 60mA when it is on. The transistor can handle 1A in case you want to draw current through pin 15-explained on section "Hi-Current Control Pin".

#6 CW2

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Posted 22 February 2012 - 05:29 PM

I see no other indications of what it would need to draw.

SerLCD datasheet page 4 says "The SerLCD uses 3 mA with the backlight turned off and ~60 mA with the backlight activated." No need to use external power supply, Netduino regulator can easily handle that.

#7 Dan Morphis

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Posted 22 February 2012 - 06:36 PM

My intent was to get something like this DC stepdown. Split the 12VDC before it goes into the Netduino and run it through this to step it down to 3.3V.


Why buck it down to then run it through a linear regulator? Very inefficient, and the linear regulator will be running quite warm. I would buck it down to 5v, and then put the power into the Netduino via the 5v pin on the header. Then for your 3.3v needs, you could linear regulate it down to 3.3v with a LD1117. You wouldn't be eating up nearly as much power to regulate 5v down to a 3v3, vs 12v regulated down to 5v (which the Netduino does when you feed it via the barrel connector).

-dan

#8 Cabadam

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Posted 22 February 2012 - 08:40 PM

The datasheet says the modules uses 3mA with backlight off and about 60mA when it is on. The transistor can handle 1A in case you want to draw current through pin 15-explained on section "Hi-Current Control Pin".

SerLCD datasheet page 4 says "The SerLCD uses 3 mA with the backlight turned off and ~60 mA with the backlight activated." No need to use external power supply, Netduino regulator can easily handle that.


Bah. Can't believe I missed that. I think I saw the first part about the 1A and after I saw the part about the serial protocol, assumed I had covered all the important stuff. That certainly changes things.

Attached to the Netduino is going to be a bunch of 5V relay (10 of them in total). Assuming I read THEIR datasheets correctly, I'm looking at ~500mA if they are all switched on at once. An unlikely occurrence, but I'd like it to be able to account for it. ND+ spec says the microcontroller itself could take up to 200mA, which puts me at 700mA total? This wiki page says the output is up to 800mA. Does that include potential 200mA used by the controller itself? If so, that puts me pretty close to the 800mA.So maybe I end up running the LCD through the Netduino (which only needs 60mA) and separately step down to the relays instead.


Why buck it down to then run it through a linear regulator? Very inefficient, and the linear regulator will be running quite warm. I would buck it down to 5v, and then put the power into the Netduino via the 5v pin on the header. Then for your 3.3v needs, you could linear regulate it down to 3.3v with a LD1117. You wouldn't be eating up nearly as much power to regulate 5v down to a 3v3, vs 12v regulated down to 5v (which the Netduino does when you feed it via the barrel connector).

-dan

Why? Well, couple thoughts come to mind:
1) Not really sure what I'm doing :)
2) I was working with what I had and then tried to add in pieces to that. Something else to remember - my first regulator is also converting AC -> DC, not just dropping the voltage.

So that 5V can function either as a 5V out or a 5V in?

Would someone mind also clarifying one of my earlier statements about how to properly add up the power usage of the various components (which are running at different voltages)?

At first I was just adding up the mA, regardless of voltage. But then it occurred to me that it probably doesn't work that way. It seemed to make more sense that I should convert any particular value (such as the 60mA @ 3.3V for the LCD) to VA. I can then add up VA to make sure I don't end up accidentally exceeding the rated value of the AC->DC converter.

I suspect given that the LCD turned out to be much smaller than I initially understood that this won't actually matter - but I would like to know the proper way to calculate that for the future.

Thanks

#9 Dan Morphis

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Posted 23 February 2012 - 06:53 PM

Why? Well, couple thoughts come to mind:
1) Not really sure what I'm doing :)
2) I was working with what I had and then tried to add in pieces to that. Something else to remember - my first regulator is also converting AC -> DC, not just dropping the voltage.

So that 5V can function either as a 5V out or a 5V in?

Would someone mind also clarifying one of my earlier statements about how to properly add up the power usage of the various components (which are running at different voltages)?

At first I was just adding up the mA, regardless of voltage. But then it occurred to me that it probably doesn't work that way. It seemed to make more sense that I should convert any particular value (such as the 60mA @ 3.3V for the LCD) to VA. I can then add up VA to make sure I don't end up accidentally exceeding the rated value of the AC->DC converter.

I suspect given that the LCD turned out to be much smaller than I initially understood that this won't actually matter - but I would like to know the proper way to calculate that for the future.

Thanks


Yeah, it can function as a 5v in, there is no diode on it. So, keep that in mind and be very careful not to swap the positive and negative rails.

The easiest way to figure out your entire load would be to convert your power usage to watts, Watts = Volts * Amps. 1 amp at 5 volts is the same power usage as 0.5 amps at 10 volts. So if you do your calculations purely on the amps used, you would be off.

So for that LCD, which consumes 60mA at 3v3 you have W = .06 * 3v3, or .2 watts, which is a negligible amount of power. For the relay board, you have W = .5A * 5v or 2.5 watts. And the Netduino is W = .2A * 5v, or 1 watt. So total we are looking at 3.7 watts.

Now to take that and figure out how many amps at 5 volts you need. 3.7W = A * 5v, 3.7W / 5v = A, A = .74. So to power what your looking to power, you would need a 5 volt, 740mA power supply. Now, that number doesn't account for any losses that occur with the 3v3 regulator, and it gives you no head room.

So if it was me, I would use a 5v, 1A regulated power supply.

As to your comment about your first regulator doing AC -> DC, you can easily convert AC -> DC by using a bridge rectifier. You can make one yourself with with 4 silicon diodes like the venerable 1N4004, or you can buy one pre-built in a package for < $1.

-dan

#10 Cabadam

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Posted 24 February 2012 - 10:00 PM

Thanks for all the information. I think I've got an idea how I want to lay all this out. Now the hard part... waiting for some of the parts I'm missing to ship :)




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