Hi everyone, for my project I need more than 1 pin with 3.3v, does anyone of you know a specific type of arduino board that fits this request? Also, I am kind of a noob in elettronics, so if you have an easier solution I would like to hear it. Thanks.
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The .3v pin cannot supply much current. What two things do you want to connect to it?
...R
I solder wire wrap headers to perfboards, wire wrap the end terminals of those headers, solder the pins in the middle, and make Proliferator boards. typically 8: 5V power pairs, 6: I2C data pairs, 6: 3.3V power pairs
solder wires to the PC board side of male headers to feed the modules, plug the header into the Proliferator, no stupid pins breaking off in breadboard holes
this is for the final stages, when you have all the right modules up and running.
is it possibile to use the digital pins as output for voltage? can I just set them on output and have them high for the Whole time?
SVGMatte:
is it possibile to use the digital pins as output for voltage? can I just set them on output and have them high for the Whole time?
What do you mean by this? Do you mean that you want to use an output pin of an unspecified Arduino for powering something else?
An output pin of an ATMega 328P (used in the Uno and some other Arduinos) can power something else
IF you limit the current to about 30 mA (pin current is 40 mA but you probably should have some margin)
and
IF you are not too picky about the voltage.
This will work from the time that the setup() function sets the pin HIGH (and sets the pin as an OUTPUT previously) until the Arduino is powered off. This scheme should only be used for resistive loads. Inductive loads may cause too high a voltage on the pin, and capacitive loads may initially load the pin with too high a current.
I think you need a breadboard
vaj4088:
IF you limit the current to about 30 mA (pin current is 40 mA but you probably should have some margin)
and
40mA ist the ABSOLUTE MAX rating. That is definitely nothing to advertise. The reasonable limit is rather something like 20mA. There is also an absolute max rating of 200mA for all pins combined.
Are you away of the fact that you can just connect multiple devices to the one 5V pin, as long as you find a mechanical solution for that? (A breadboard would be one, but you certainly don't NEED it).
-> Kirchhoff's Voltage law.
Of course, you cannot take arbitrary amounts of current from that either. You might come to a limit with illumination or motors, but probably not with sensors.
ElCaron:
40mA ist the ABSOLUTE MAX rating. That is definitely nothing to advertise. The reasonable limit is rather something like 20mA. There is also an absolute max rating of 200mA for all pins combined.Are you away of the fact that you can just connect multiple devices to the one 5V pin, as long as you find a mechanical solution for that? (A breadboard would be one, but you certainly don't NEED it).
-> Kirchhoff's Voltage law.
Of course, you cannot take arbitrary amounts of current from that either. You might come to a limit with illumination or motors, but probably not with sensors.
so say I set a digital pin as output and set it on high, it will erogate 40mA right?
I could use a breadboard, but I have 2 different components that both need to be powered at 3.3 v and if I connect them both to the 3.3 pin, wouldn't they get less voltage?
I am no good at elettronics so I really appreciate your help. Thanks
Exactly what are you trying to power?
so say I set a digital pin as output and set it on high, it will erogate 40mA right?
The actual current depends on the voltage (normally ~5V) and the load resistance of whatever you are connecting ([u]Ohm's Law[/u]*).
You don't always know the resistance (or impedance) of whatever is connected, but if not there should be a current specification for whatever you are connecting.
Ohm's Law is a law of nature (with man-made units-of-measure) and it's always true. If you directly-connect a motor (or something that takes lots of current) to an I/O pin, the voltage will drop, you won't get the "calculated" (or specified) current, and the Arduino may overheat and burn-up!
If you need to control a motor (or anything that takes more voltage and/or current) you can make a transistor or MOSFET driver circuit, or you can use a relay (although relays usually require a driver too).
I could use a breadboard, but I have 2 different components that both need to be powered at 3.3 v and I connect them both to the 3.3 pin, wouldn't they get less voltage?
The 3.3V output is also very-limited. It's only rated for 50mA.
- Ohm's Law describes the relationship between voltage, resistance, and current and it's the 1st thing you learn when you take an electronics class.
so say I set a digital pin as output and set it on high, it will erogate 40mA right?
Wrong.
If you do manage to draw 40mA out of a pin it will be damaged. The absolute limit is one you should no exceed not one you can't exceed.
I have 2 different components that both need to be powered at 3.3 v and if I connect them both to the 3.3 pin, wouldn't they get less voltage?
Only if the combined current draw of both components exceeds 50mA or what ever the power supply is capable of supplying.
DVDdoug:
Exactly what are you trying to power?
The actual current depends on the voltage (normally ~5V) and the load resistance of whatever you are connecting ([u]Ohm's Law[/u]*).You don't always know the resistance (or impedance) of whatever is connected, but if not there should be a current specification for whatever you are connecting.
Ohm's Law is a law of nature (with man-made units-of-measure) and it's always true. If you directly-connect a motor (or something that takes lots of current) to an I/O pin, the voltage will drop, you won't get the "calculated" (or specified) current, and the Arduino may overheat and burn-up!
If you need to control a motor (or anything that takes more voltage and/or current) you can make a transistor or MOSFET driver circuit, or you can use a relay (although relays usually require a driver too).
The 3.3V output is also very-limited. It's only rated for 50mA.
- Ohm's Law describes the relationship between voltage, resistance, and current and it's the 1st thing you learn when you take an electronics class.
Hi, thanks for helping me.
I need to power a wireless module and an infrared emitter, they both need 3.3 volt power, also the 5v pin is busy in this project too as I need it for something else, so I just need another pin
I need to power a wireless module and an infrared emitter, they both need 3.3 volt power,
An IR emitter does not need a 3V3 supply.
Grumpy_Mike:
An IR emitter does not need a 3V3 supply.
why are you saying so? I've seen a YouTube video where the guy powered it with 3.3v, could it just be model depending?
why are you saying so?
Because an IR emitter is just an IR LED, you can use what ever voltage you like for it.
I've seen a YouTube video where the guy powered it with 3.3v,
I am not saying you can't I am saying you don't have to. It was the NEED word that was wrong.
Grumpy_Mike:
Because an IR emitter is just an IR LED, you can use what ever voltage you like for it.
I am not saying you can't I am saying you don't have to. It was the NEED word that was wrong.
weel if that's the case, my problems are solved, thanks
@SVGMatte, do not cross-post. Threads merged.
spycatcher2k:
Did you know that wire can be connected together? 3.3V out to one wire, one wire connected to two wires, you now have 2 outputs!
yeah but I need to use both components at the same time, so they'd both get half the voltage that comes out of the pin wouldn't they?
thanks for helping
What are the two components? If they are two 20mA LED's and you wired them parallel with the correct dropping resistors. You would get same voltage to each.
Sound like you need help with basic electronics. There are plenty of tutorials online and classes. Please start with series versus parallel concept.
