External power supply

I'm trying to understand this:

If i somehow get an external power supply for my sim900 module and plug the sim900 into it, because it needs more power...

How do I power the arduino mega?

Depending on the power supply, you may be able to run the Arduino from the power supply.

Weedpharma

Where could I get power supplies for this and what specs?

I've read the sim900 requires up to 2A.

I could just use my 9v dc for the mega.

Remember to connect all the grounds.... if you don't, the serial signals from the SIM900 will have no meaning at the Arduino side and vice versa.

Data here

http://www.propox.com/download/docs/SIM900.pdf

Says it need 3.4 to 4.5v and is "low power".

The Arduino needs 5v.

Make sure of the power requirements and select your supply from that information. You could use a 5v supply with a switchmode to drop it to 4v for the SIM900.

Weedpharma

Yes I know about that. But I've read a bunch of posts that say that it requires bursts of up to 2A. And that this may cause problems maintaining connections alive.

Right now Im able to connect and everything but I get an error specifically when sending a request. So I'm considering the possibility of it being a power issue at that moment.

Im told I should consider using a power supply capable of delivering up to 2A.

Use a storage capacitor on the power input. The current would be for a short time so be supplied from the C.

Weedpharma

OK. I'll look into that. Know any tutorials online by any chance, where I could look at how that's done?

As the data says "low power", I assume the extra power is for a very short time. I would start with 10 uF an if that works, try 1 uF.

Put it on the closest point to the device that you can easily get at.

Weedpharma

Ok but does that mean I need a breadboard? So I would basically run the positive lead thru a capacitor first and then to the 5v on the SIM900?

What is the recommended way to do so since Im using a battery pack that plugs into the dc port of the MEGA?

The purpose of the C in this case is to store power for the short peak requirement. It goes across the supply.

Show a picture of your setup.

Weedpharma

Well right now its pretty straight forward. Here is the module:

I can take a more complete picture if you'd like. This is one I had from a couple of days ago.

The connection has 4 cables:

SIM900 ------- ARDUINO MEGA
Red-------------5V
Black-----------GND
TX (y)----------Rx(19)
RX (g)----------Tx(18)

So if I have to insert a Capacitor I would need a breadboard between the red cable of the SIM900 and the 5v of the Arduino Mega?

Screenshot 2015-06-01 11.47.34.png

You need to put one lead of the cap to 5V and the other to GND. It will probably be polarised so make sure that you put the lead that has a strip next to it to gnd(check online for polarised caps).
You may get away without breadboard, by just sticking one in the arduino power rail. In the SIM900 datasheet it says that the absolute max voltage is 5.5V so it could withstand the 5V rail, but you should make a stepdown to 4.1V or something close, even an LM317 will do the job. You don't need a breadboard if you have some soldering experience.

Have a nice day,
Andrei.

The data in the link above says the supply is 3.4 to 4.5v. Direct connection to 5v is above its rated voltage. At a minimum you would need a diode in series to drop 0.6v.

Weedpharma

AndreiV:
...you should make a stepdown to 4.1V or something close, even an LM317 will do the job.

No.
The LM317 has ~2volt dropout @500mA.

As weedpharma says, a 1N4004 diode drops ~0.6-0.7volts.
Leo..

So you are saying something like this:

  1. Assuming that red breakout is actually my 5V sim900 and not a 3.3V as the diagram shows (but it was the closest component I could find on Fritzing.

Except change the part marked with 'MHz' for a capacitor. Which you may not need.

The placement of the capacitor is correct but you keep referring to 5v supply to a device rated at 4.5v. Put a diode from 5v to the capacitor across the module power input.

Weedpharma

You mean like this:

That should be correct, yes.