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I am trying to control a dc motor using potentiometer(armature control). Increase in voltage leads to increase in speed of motor. Even though this seems in accordance with general understanding, but i am confused on seeing the speed - Armature current characteristics of DC motor which suggest that increase in armature current leads to decrease in speed as shown

speed vs armature current for dc series motor

So why does speed increase on increasing the voltage? since motor resistance is constant, current must increase and hence speed should decrease. Is this graph valid for certain cases or always true? If invalid, when is it valid? If always true, how can we explain my observation?

I seem to be missing some fundamental concept here. Please help. Thank you

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  • \$\begingroup\$ IMPE, controlling a dc motor directly with a potentiometer is not a particularly good control scheme. Using a potentiometer to control something else that controls the motor is generally better... \$\endgroup\$ – Ecnerwal May 31 '16 at 14:59
  • \$\begingroup\$ It's the other way round - at a specific voltage, decrease in speed thanks to increased load, means an increase in armature current as the motor works harder. \$\endgroup\$ – Brian Drummond May 31 '16 at 15:52
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Is this graph valid for certain cases or always true? If invalid, when is it valid?

The shape of the curve and the word "series" on the curve indicate that the curve is for a motor with the armature and field wired in series with each other. Permanent magnet motors, motors with the armature and field connected in parallel (shunt), and motor with a separate power supply for the armature and field, have curves that are more like straight lines as shown below. The curved red lines are for a series motor and the straight green lines are for a permanent magnet motor. The dotted lines show the effect of voltage reduction.

The location location of the operating point is the point where the load line crosses the motor curve. The load line is the characteristic of the load that shows how much torque is required to drive it at a given speed. The applied voltage adjusts the speed by moving the motor curve and the operating point. The shape of the curve as determined by the motor design, determines how much influence the applied voltage and the motor current (torque) each have on the operating speed.

Note that motors normally operate on the small portion of the curve at or below the rated current (100%). In that range, the load has much less influence on the speed than does voltage, for motors with a permanent magnet, shunt or separately powered field.

enter image description here

Load Line

The load line is the steady-state speed vs. torque characteristic of the load driven by the motor. The % current axis for the speed vs current motor curves can be considered to represent % torque for the purpose of illustration. The equation for the load line is a summation of friction, fluid drag, gravitational forces and any other forces that oppose the rotation of the drive shaft.

For most loads, torque increases as speed increases. For centrifugal pumps and fans, load increases in proportion to the square of the relative speed increase. For most other loads, torque is relatively constant to slightly increasing with speed. Torque decreases with speed increase only when there is an active control mechanism that reduces the torque, such as increasing the speed and reducing the force on the bit when drilling with a smaller bit. The load line is normally determined only by the characteristics of the load and completely independent from the motor characteristics.

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  • \$\begingroup\$ Thanks! Could you also tell me the equation of load line for a motor? \$\endgroup\$ – user5089054 Jun 1 '16 at 4:23
  • \$\begingroup\$ See "Load Line" addition to my response. \$\endgroup\$ – Charles Cowie Jun 1 '16 at 14:53
  • \$\begingroup\$ Hi , i did. however the load line according to my equation seems to have negative slope ( same as the equation from answer by abdo mohamed) .The observations correspond to your representation of the load line (positive slope); Could you help me out please? \$\endgroup\$ – user5089054 Jun 6 '16 at 10:25
  • \$\begingroup\$ I corrected the load line on the drawing and added a paragraph. When I originally added the load line, I drew it for a horizontal speed axis and vertical torque axis. I don't see where the question or @Abdo Mohamed's answer have any information related to the load line. \$\endgroup\$ – Charles Cowie Jun 6 '16 at 13:14
  • \$\begingroup\$ Oh so, the equation of the load line would'nt be determined from the circuit it is in? What i mean to say is, in my circuit, which resembles the one given in @Abdo's answer (where V is the applied potential for controlling the speed), the equation given by abdo must hold true. right? So, if i plot that equation on the speed- Ia curve, the point of intersection of that line and the characteristic at given V must give me the point of operation. And, if this is so, my line has a negative slope (unlike the one you have drawn) \$\endgroup\$ – user5089054 Jun 7 '16 at 4:32
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It's hard to tell what exactly you are asking, so a brief introduction to how motors work is the best answer.

Think of a motor as a resistor in series with a voltage source. The resistor is the resistance of the windings, and the voltage source opposes the applied voltage and represent the motor acting like a generator. Mechanical output torque is simply proportional to current.

When you apply some fixed voltage with the motor initially still, the internal voltage source is 0 and all the applied voltage is across the resistor. This allows a large current to flow, called the stall current. As the motor spins up, the internal voltage source opposes the applied voltage, leaving less voltage across the resistor. This means less current, which means less torque. Eventually the motor gets to the speed where just enough voltage is left across the resistor to allow just enough current to flow to provide the torque to keep the motor going at that speed.

Hopefully you can see that a voltage increase at any particular speed appears completely across the resistor, which means more torque. Conversely, the steady state speed with the same load is higher with higher voltage.

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Assuming your DC motor is a Separately excited DC machine such that the magnetic flux "Φ" is constant and does not change as the armature current changes, like the case in the Shunt or Series DC motors.

An equivalent circuit of your motor could look like this

enter image description here

V: the voltage of your battery

E: the voltage across your armature

Ra: the armature resistance

Ia: armature current

ω: motor speed

T: torque developed

K: a geometric constant that depends on geometry of the coils inside the motor

E=K x Φ x ω

T=K x Φ x Ia

Solving the motor equivalent circuit you will find that V= E + (Ia x Ra)

from this relation we can get a relation between ω and the circuit current Ia

enter image description here

So as Ia increase the voltage drop in Ra increase and the voltage across the armature decrease so you get less speed but more developed torque as the torque equation is a function in Ia

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It's got things to do with the magnetic field strength and the back emf, every motor has a back emf which helps to limit the supply current to a sane value (motors draw less current than their DC resistance would imply), a smaller armature current needs a faster rate of rotation to generate the same back emf as the emf is proportional to the rate of change in the magnetic field (because it's spinning, rate-of-change here means rotational speed). This only applies to motors with both armature and stator windings, permanent magnet motrs get a bit more complicated

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