Sadly passed away in 2015
- Nov 28, 2011
- Nov 28, 2011
OK. Thanks for the link.Yeah.. The problem is the ring indicator pin (RI) is not being used on the breakout board and there are no pins to connect to it. I would have to solder a wire directly to the GSM module. Not a big deal.. but its an SMD module andthe pins are tiny. So Actually I am thinking it might be easier, and definately more durable to not run a wire and just monitor the serial data.
There's not much information on it, is there. I noticed it needs a 5V supply. Are you using a separate battery for that?
I see, RI is only available on pin 4 of the SIM900 IC itself. That's a good reason to use serial data instead.
No, MOSFETs and bipolar transistors operate in different ways.Yeah thats confusing me.. I need to figure out how to visualize it.. breaking it down to how it is in terms of diodes or the actual P and N type materials and whats going on there..
Ive been looking at the diagrams on the datasheet trying to sort it out. Its pretty much the same as a transistor isnt it? just different tolerances?
You just need to understand that the MOSFET is interested in the gate voltage RELATIVE TO THE SOURCE. In that design, the source is connected to the +3.7V rail, so it's always at +3.7V relative to the circuit's 0V rail. So when the PIC's output goes low, i.e. it goes to 0V, from the MOSFET's point of view the gate is 3.7V negative relative to the source.
I kind of get it.. The current can't flow from the drain side (basically the negative side) through to the source (positive side) until there is a 0 or negative voltage on the N type material between the two P type materials.. So I can picture the gate having 0 voltage i guess.. But are you saying that if I were to put my volt meter on the PIC's output pin when I set it high, I will read zero volts?
And the gate going to -3.7v is just throwing me off.. Haha..
In this arrangement, current flows from the source to the drain. That's conventional current, which flows from positive to negative.
Actually it flows from the positive terminal of the battery, to the +3.7V rail on the board, to the source of the MOSFET, through the MOSFET to the drain, then out to the positive supply input of the DC-DC converter.
The MOSFET's source is always at the same potential as the +3.7V rail. They're connected together.
When the PIC's output is high, it is equal to the +3.7V rail, so there is no voltage difference between the MOSFET's gate and source. So it's turned OFF and no current flows out the drain and to the converter.
When the PIC's output is low, it is equal to the 0V rail, which is 3.7V negative with respect to the +3.7V rail. So there is 3.7V on the MOSFET's source, and 0V on its gate, so from the MOSFET's point of view, Vgs is -3.7V. So it's turned ON, and current flows in the source, through the MOSFET, out the drain, and to the converter.
No problem. I hope you understand that last description because it's about as clear as I can make it. Re-read it until it really sinks in. And remind yourself about which pin is the source, and which is the drain. It's not hard to get them confused.Thanks for putting up with me by the way. I kind of feel like a moron right now