Hi David, I'm not sure I understand your diagram. I would draw the diagram as shown in the attached image file (removed as it is too big for the tinyos mailing list server, even though I made it extra small).
Let's say that the plus terminal of the power supply is connected to one terminal of the resistor. The other terminal of the resistor is connected to the plus terminal of the battery connector of the mote. The minus terminal of the mote is directly connected to the minus terminal of the power supply. The ground connector of the oscilloscope is connected to the plus terminal of the battery connector of the mote (which is also connected to one terminal of the resistor). The measurement terminal of the oscilloscope is connected to the plus terminal of the power supply (which is also connected to the other terminal of the resistor). Thus, the oscilloscope is measuring the voltage drop over the resistor. This voltage drop is proportional to the current being drawn by the mote. In fact, the voltage over the resistor V_R is: V_R = I_M * R = I_M * 10, where I_M is the current drawn by the mote. So if the mote draws 20mA of current (e.g. when the radio is in receive mode), then the voltage drop measured over the resistance is 200mV = 0.2V. So far, I have not used any simulations for power measurements and can't comment on this. Cheers, Urs David Li wrote: > Urs, > > Sorry I forgot you used 3v lab power supply. So I revised my diagram as > following: > > > >> >> +-----------------------------+-----------------------------------+-------------------+ >> | >> | | | >> mote resistor (1 or 10 Ohm) 3v Power >> Oscilloscope or multimeter >> | >> | | | >> >> +-----------------------------+-----------------------------------+-------------------+ >> >> >> >> >> >> David _______________________________________________ Tinyos-help mailing list Tinyos-help@millennium.berkeley.edu https://www.millennium.berkeley.edu/cgi-bin/mailman/listinfo/tinyos-help
