|Dec 04, 2004, 10:55 PM|
How do you measure AMPS with a multimeter?
My CC P-10 is getting warm while testing a motor i built. I can't seem to figure out how to use my multimeter to measure amps.
Please be descriptive.
|Dec 04, 2004, 11:03 PM|
If your meter has a special hole for the red lead to check amps...put it in there...then hook up everything as normal. Then take the red lead from the batter and hook the red lead from the meter to the battery lead....and the black lead from the meter to the remaining red wire going to the controller. The best way someone described it is pretend everything is normla...then a kid (meter) breaks in the middle of the circle (red battery lead) and holds hands with the two kids (both ends of the red battery lead.
run the motr and you should read amps. If you have a buddy with the clamp on amp meter it's much easier...you don't have to cut wires.....
I made up a harness out of two connectors....and wired it so one end goes to my battery...the other end goes to the controller.....and I just left the red side so I could insert my meter...and taped the plack side together..
hope that helps....
|Dec 04, 2004, 11:23 PM|
You can use a shunt resistor to measure current using a common voltmeter.
As Ohm's law states:
I = V/R
I = Current
V = Voltage
R = Resistance
To implement the formula, simply put a low-value resistor in series with the power or ground connection to the ESC, then measure the voltage difference between each lead of the resistor. The difference can be plugged into the formula to yield a current.
For example, a .05 ohm shunt resistor reads "0.17V" across the two terminals. We simply replace the variables with our known values and solve the equation
1: I = 0.17/.05
2: I = 3.4A
When using this though, take care that the shunt resistor is of the lowest possible value to minimize voltage losses, and by minimizing voltage losses, you are also minimizing power dissipation, as:
W = VI
W = I x I x R
Using this formula we can calculate power dissipation of the given resistor. Suppose the conditions of the above example, 3.4A and 0.05 ohms.
1: W = 3.4 x 3.4 x 0.05
2: W = .578W
Keep in mind that the resistor power rating of your shunt must be a higher number than your calculated dissipation at max continuous current draw you plan to measure.
Hope this helps, it will allow you to easily measure current using just a DC voltmeter.
|Dec 05, 2004, 12:52 AM|
If you are using Deans Ultras you can flip one plug over and plug in the negatives leaving the positives open. Now plug one probe into the pack's + lead and push the other in between the exposed rx + lead and the connector. They stay put like this and it frees up both hands.
|Dec 05, 2004, 07:55 AM|
Post # 7 here:
The resistence of 12" of #10 solid copper wire is .001 ohms.
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