I have knowledge about the definitions about both speed as well as of the torque. But I am unable to think dc motors in terms of speed and torque. When the armature of a dc motor rotates , we call it torque because of rotational movement about the fixed axis. So where is speed?
1\$\begingroup\$ Andy aka beat me to it, but your assumption about torque is wrong. Torque is not a measure of speed, it is the rotating force. \$\endgroup\$– I. WolfeMar 13, 2015 at 15:43
\$\begingroup\$ I wrote an answer a while back that has several related formulas that may be helpful: electronics.stackexchange.com/questions/43066/… \$\endgroup\$– embedded.kyleMar 13, 2015 at 17:55
\$\begingroup\$ @embedded.kyle - Yes, your answer is helpful too. Thank you for suggesting it. \$\endgroup\$– VibhuMar 13, 2015 at 18:09
When the armature of a dc motor rotates , we call it torque because of rotational movement about the fixed axis. So where is speed?
We don't call it torque, we call it speed aka rpm or rotational speed.
The force that it takes to spin at a certain speed we call torque and usually, for a dc motor high speed means low torque and high torque means low speed. Look at this formula for the mechanical power output of a motor: -
Power out = \$2\pi n T\$
Where n is speed in revolutions per second or, if you wish \$2\pi n\$ is radians of rotation per second. and T is torque.
For a constant power out n.T must be constant.
Torque is how hard you push, and speed is how fast it goes.
In a electric motor, torque is proportional to current, regardless of the applied voltage.
Think of a motor as the coils that cause the torque in series with a voltage source. The coils are always the same, and the torque produced by the magnetic interactions inside the motor are always proportional to the current thru these coils. Here is a useful first order simplified model of a motor:
L1 and R1 represent the part of the motor that produces torque proportional to current. V1 is the motor acting as a generator. When the motor is spinning due to applied voltage, then V1 opposes that voltage. The voltage applied across the torque-producing part of the motor is therefore V+ - V1. This means the faster the motor goes in the direction you are trying to spin it, the less voltage is avaiable to drive the motor, therefore less current is drawn, and less torque developed. In steady state, the motor goes fast enough so that V1 cancels out just enough applied voltage to leave the exact amount applied to L1 and R1 to produce the torque to spin it at the speed it is going at.
For example, let's say you have a small DC motor that is not mechanically connected to anything (unloaded). Let's say the motor is rated for 10 V. When you first apply 10 V, the motor is not moving, so V1 is 0, and all of that 10 V is applied across L1-R1. This causes current of 2 A thru the motor, which causes torque, which makes the motor speed up. As it speeds up, V1 increases. After a few seconds the motor speed stabalizes and stops increasing. You now measure only 200 mA with the same 10 V applied, and the motor is running at 80 Hz (4800 RPM).
Since the current is now 1/10 what it was originally when the motor was not spinning, called the stall current, you know that 1/10 the original voltage is applied across L1-R1, which is 1 V. That means the motor is internally generating 9 V. From that you can compute the generator constant, which is (9 V)/(80 Hz) = 113 mV/Hz. You also now know that 1 V across L1-R1 is what is needed to just counter friction and other losses in the motor at 80 Hz.
If you were to externally spin the motor at 80 Hz you would measure a open-circuit voltage of 9 V. You would need to spin it at 89 Hz to get 10 V out. If you applied the 10 V while spinning it at 89 Hz, no current would flow. If you then disconnected the shaft from whatever was driving it, the motor would start slowing down. This is because is it has some friction. As it slows down, the V1 voltage goes down, so some voltage is applied across L1-R1, which causes current, which produces torque to keep the motor going. Initially that current is too small to produce enough torque to keep the motor going at the existing speed, so the motor slows down more. That reduces V1, which increases the voltage on L1-R1, which increases the current, which increases the torque. Eventually it will reach equillibrium at 80 Hz and 200 mA, just like before. If the motor were to go slower, V1 would drop, the voltage on L1-R1 would increase, current would increase, more torque would be generated, and the motor would speed up. The reverse happens if the motor were to go faster.
\$\begingroup\$ Torque is directly proportional to current in a DC brush motor, but I don't think that's true for universal motors. For steppers and brushless motors, I think torque is directly proportional to current for any given phase angle, but as phase angle varies so will the relationship between current and torque. \$\endgroup\$– supercatMar 13, 2015 at 20:03
\$\begingroup\$ @supe: Even DC brushed motors have torque ripple. Basically, torque is proportional to curent averaged over full commutation cycles. Since for ordinary spinning motor applications it's that average that matters, this wasn't a detail worthing adding confusion over. \$\endgroup\$ Mar 14, 2015 at 13:52
\$\begingroup\$ While practical DC brush motors have torque ripple, even a three-pole motor will always have a phase angle that will let most of the current contribute usefully to torque. Steppers by contrast will often have a phase angle such that current contributes almost nothing to useful torque. \$\endgroup\$– supercatMar 14, 2015 at 14:15