Voltage rpm relation

I measure the voltage ESC drawing while increasing the dc motor speed. Multimeter shows that as long as the speed increases the voltage value decreases. Can anybody explain why this is happening?

You can't understand this because you don't know what you observe. The electromotive force (EMF) of your motor is proportional to its speed (the ratio is called Kv). So, when the speed of your motor rises, the EMF rises too.

What does an ESC ? Basically, it hashes the input voltage to generate the expected output voltage.

So, what is this decreasing voltage value you observe ? It won't be the output voltage of the ESC, because it's rising, not falling, and, as it is hashed, you probably can't measure it with your multimeter (basic multimeters only measure continuous or sinusoidal signals).

I expect the decreasing voltage to be the input of your ESC. Why would it drop ? Because your driving a motor with a load that rises with speed (typically a propeller), and this requires current. The wires between your battery and your ESC are resistors. Let say you have 50mOhms of resistivity. If you uses 10A, the input voltage of your ESC is 0.5V less than your battery output. Moreover, your battery also have internal resistivity. If you drive more current, the output voltage of your battery drops too.

• Many thanx ....now I got it.. Thanx a lot again
– lsn
Jul 17 '15 at 9:25

I think the answer here should be insightful. But basically when the motor is running, you're working under the equation

P=V*I

This equation has to stay balanced, so assuming your (P)ower remains a fixed value, then if (V) changes, current (I) has to make and opposite change in order to keep the equation balanced. In this instance, current (I) increases as speed increases, thus you get a related decrease in (V)oltage in order to maintain the balance.

EDIT: My answer is flawed because as pointed out in the comments, the assumption of power being fixed is wrong and doesn't make any sense.

• Im gonna use this meuserment for guessing the motor rpm from Kv constant. my motors is 1350 Kv(rpm/V) and it has a propeller I try find out rpm-thrust diagram. According to my measurment result, the rpm value will decrease (though it increases) as a result of decrement in V values and I couldn't understand. I think Im missing a point for rpm calculation with Kv.
– lsn
Jul 17 '15 at 8:13
• This is an incorrect way to evaluate the problem because this assumption means that the motor spinning at low speed and at high speed both draw the same power. Even assuming the motor wasn't loaded, and ignoring viscous friction effects, there is still dynamic friction between the shaft and whatever bearing/bushings are in the motor. Assuming this is only due to the rotor weight, you get a constant frictional torque (friction force applied at $r_{rotor}$), and since $P = \tau w$, where $w$ is motor speed, you have a higher frictional losses at higher speed - they can't be the same power. Jul 17 '15 at 12:24
• Also, current doesn't increase as speed increases, current increases as torque increases; that is, $\tau_{motor} = K_T I$. The reason you may see current increase as speed increases is precisely because of frictional losses as described above - frictional power losses increase at higher speed, requiring higher torque from the motor to maintain the given speed. Higher torque requires higher current, so again, a real motor with the same load draws higher power, not the same, at higher speeds. Jul 17 '15 at 12:32