Does anyone know what the modems power consumtion is at 12V. The documentation says 1.7A at 5.5V. But what is it at 12V?
And can I measure this “peak” current? (Connecting an ammetre in series when powering the modem at 5.5V or 12V gives an answer in mA around 60mA and 90mA respectively but the modem cannot RX (sms’es). When the ammetre is removed the modem can receive SMS’es just fine.)
it is very difficult to measure the peak current with a normal ammetre.
Because the peaks are very short, so you need a quick peak detector to meassure it.
The reason why you can’t receive SMS (also a voice call) is, that the impedance on your power supply is high. You should try to reduce it.
And when you connnect an additional ammetre, the impedance is too high.
You should use low ESR buffer caps to increase the performance.
The Fastrack Modem M1306B User Guide Revision 002 of 5 April 2005 has a table of peak & average current consumption figures on p48.
At 13.2V, the peak is 1.4A, and average is 500mA during communication.
But I would also like some more details of the current consumption over the full input voltage range:
wavecom.com/modules/movie/sc … .php?t=813
As skos says, you can’t measure this with an ordinary ammeter - ordinary ammeters will read average current.
You will either need specialist equipment, or you could try putting a small resistor in series in the supply line, and observe the voltage drop on an oscilloscope…