Problem with serout in BS2p40

ccastro

Hello.
I succeeded communicating a bs2 with a gprs module via serial with serin and serout commands.

But now i need to do the same with a bs2p40 module. I've changed the current baud rate config but the gsm module doesn't answer it.

I entered both signals into an oscilloscope, they both have the same form but 2p40 signal has 0.5 milliseconds longer. i don't know if it's enough to miss the data.

if anyone knows why can be this happening please answer

Thanks!!

Zoot

Post your code.

Mike Green

Use the Go Advanced button in the Quick Reply window and this will get you to the Attachments Manager which you can use to post your source code file with your reply. Please do not cut and paste your source code. We can't tell you what's going on or why without more information about what you're doing and how.

ccastro

Here i go:

This is a very simple example but its enough to watch:
BS2P program:
2p40.bsp
BS2P Signal:

2V x Div / 2mS x Div 1V x Div / 2ms x Div


BS2 program:
bs2.bs2
BS2 Signal:

2V x Div / 2mS x Div 1V x Div / 2ms x Div

When i send
SEROUT 0,813,["AT+SHDN",CR]
with BS2; the gprs module is shutted down like it should be.

but when i send
SEROUT 0,2063,["AT+SHDN",CR]
with BS2P; the module it's not responding.

if there´s anything else that can help you please let me know

ccastro

Hey
I've solved the problem...if you look at the graphics the BS2 sends a 5,2 Volts Pulse Train while the BS2p sends only 4.8V. Those 0.4 Volts were enough for the non responding module.
I've put a 2N2222 Transistor to increase Voltage.

Thanks a lot Mike and Zoot for your concern.