Hello Eric , Thank you for your reply. You mentioned - " *You should read the voltage across the resistor, not measure on both sides and take the difference." *I did not quite understand this line. Where exactly should I place the resistor ,
Is something wrong with my arrangement - one side of the resistor on the +ve terminal of the battery , the other side on the PCB ? If yes - where do i exactly place my resistor ? As far as the multimeter is concerned , I will try and get hold of a scope by tomorrow. Thanks , Somnath > Yes, it's because you are using a multimeter. With a 1 ohm resistor, you > will be reading about .02 volts difference. You should read the voltage > across the resistor, not measure on both sides and take > the difference. Most multimeters do some autoscaling. > > You may have more success using the ammeter > function if the only instrument you have is a multimeter. We have one > of the better > Fluke hand held multimeters, and it measures the on current fairly > well, but the current > when the mote is sleeping is way off. > > > Eric On Thu, May 29, 2008 at 9:27 PM, Shomnat Mitra <[EMAIL PROTECTED]> wrote: > Hello , > > I have gone through several help archives to find a way to measure the > current draw of the motes. > > All suggest using a 1 ohm resistor in series with the + ve terminal of the > battery and the other end of the resistor attached to the PCB, > and measuring the voltage across this resistor. > > The readings are 1.5V which is the battery voltage. > I have tried several positions on the mote to attach the resistor, but none > show any change. > > The reading also doesn't show any change when I use HPLPowerManagement in my > programs. > > Is it because i am using a Multimeter ? > > Can anybody suggest the exact way to do this ? > > Thanks in advance , > Somnath
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