I have a conceptual doubt about torque-speed relation in a DC motors. It's probably a gap in my thinking but I'm posting this question anyway. As per me "in case of dc machine we see that the armature rotates because of the developed torque.so from the physical point of view it leads that if armature gets more torque then it will rotate at high speed; i.e. speed directly proportional to torque." BUT why the reverse thing happens????? why speed is inverse proportional to torque....?/? pls, explain physically......
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1\$\begingroup\$ Speed is not inversely proportional to torque. \$\endgroup\$– TimWescottMar 6, 2019 at 20:55
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\$\begingroup\$ @TimWescott Isn't it? DC machines put out more torque at lower speeds than higher speeds, rather than the most torque at high speeds and the least torque at low speeds. \$\endgroup\$– DKNguyenMar 6, 2019 at 20:58
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\$\begingroup\$ @Toor DC machines CAN put out higher torque at lower speeds, but they can also put out low torque at low speeds, so there is not an inverse relationship between speed and torque unless the requirement is to maintain constant shaft power. \$\endgroup\$– John DMar 6, 2019 at 21:03
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1\$\begingroup\$ \$T = \frac{V_a - k_t \omega_a}{R_a}k_t\$, where \$k_t\$ is the motor torque constant, \$\omega_a\$ is the armature speed, \$V_a\$ is the applied armature voltage, and \$R_a\$ is the armature resistance. Yes, torque goes up with decreasing speed if you're applying a constant voltage (which is not always the case), but it is not an inverse proportional relationship. \$\endgroup\$– TimWescottMar 6, 2019 at 21:03
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\$\begingroup\$ Please keep in mind the person asking the question and the likely context of his question. I think it's pretty clear he is referring to only a single line (aka voltage) on a torque-speed curve, whereas you are referring to every possible line on a torque speed curve. If you want to explain an answer to him on a level that he is obviously not going to get, you might as well vote to delete his question. \$\endgroup\$– DKNguyenMar 6, 2019 at 21:08
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3 Answers
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All of this applies to permanent magnet DC brushed motors -- different kinds of motors behave differently. Be warned.
- Speed is not proportional to torque. Rotational acceleration is proportional to torque over moment of inertia, though: \$\alpha = \frac{T}{J}\$, where \$\alpha\$ is acceleration in \$\frac{\mathrm{radians}}{\mathrm{sec}^2}\$, \$T\$ is torque in N-m/s, and \$J\$ is moment of inertia in \$\frac{\mathrm{kg}}{\mathrm{m^2}}\$.
- Armature current is a function of the motor's back EMF, the applied voltage, and the armature resistance: \$i_a = \frac{V_a - E_{back}}{R_a}k_t\$, where \$i_a\$ is the armature current, \$R_a\$ is the armature resistance, \$V_a\$ is the applied armature voltage, and \$E_{back}\$ is the back EMF
- Back EMF is a function of the motor's speed: \$E_{back} = \omega_a k_t\$, where \$\omega_a\$ is the armature speed and \$k_t\$ is the motor's torque constant.
- Torque is a function of the motor's armature current: \$T_a = i_a k_t\$
If you put that all together, you'll find that a DC motor with a constant voltage on it will "try" to spin up to a speed proportional to the applied voltage, and that its torque (and current) will be proportional to how much slower it is than that ideal unloaded speed.
A DC motor that's driven with a constant current will behave differently than one driven with a constant voltage -- and it's fairly common in servo applications to drive a motor with a current command, rather than a voltage command. But it'll still follow the equations above.
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\$\begingroup\$ Good answer with just enough math to make sense. So torque can be an 'instantaneous' value and a 'integral' value, assuming current control, which is how I control my cooling fans. Due to starting currents and mechanical inertia the PID loops need all 3 elements so the motor does not stall or surge. \$\endgroup\$– user105652Mar 6, 2019 at 22:12
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in case of dc machine we see that the armature rotates because of the developed torque. So from the physical point of view it leads that if armature gets more torque then it will rotate at high speed.
I do not see how the first part leads into the second part at all.
You must remember two things:
the faster the motor spins, the more back-EMF (a type of voltage) is produced.
more current flowing through the motor windings means a stronger magnetic field which means more torque
The speed a motor spins at is the speed where the back EMF perfectly cancels out the power supply voltage. If you apply load torque to the motor to slow it down, the BEMF generated is reduced so there is an excess in supply voltage which pushes more current into the motor. The amount of current pushed into the motor produces an extra voltage drop across the motor's internal resistance. The current stops increasing when this resistive voltage drop plus the new, reduced BEMF equal the power supply voltage so that everything balances out again.
More load torque->slower speed->less BEMF->more current->more output torque (which you would need to spin the higher load torque to begin with).
Therefore, for a given supply voltage, you get more torque at low speeds and less torque at high speeds.
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3\$\begingroup\$ "The speed a motor spins at is the speed where the back EMF perfectly cancels out the power supply voltage." Nope. The speed a motor spins is the speed where the current generates enough torque to balance applied torque on the armature including internal losses. That's why motors need some current to spin, even unloaded, and don't go quite as fast as their torque constants would imply (because the speed constant in radians per second per volt is exactly equal, including units, to the reciprocal of the torque constant in Newton-meters per amp -- I love dimensional analysis). \$\endgroup\$ Mar 6, 2019 at 20:50
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\$\begingroup\$ I'm trying to keep it simple here. I am under the impression that the OP is going to get really confused if we start talking about the motor as anything more than a BEMF generator and a resistor. I wouldn't even mention internal resistance if I thought I could get away with just saying that the increase in current produces another voltage drop inside the motor separate from the BEMF. \$\endgroup\$– DKNguyenMar 6, 2019 at 20:51
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\$\begingroup\$ The OP needs to find a web page or a book on DC motors (specifically, DC permanent magnet motors, because wound-field motors have a whole extra set of behaviors) and study hard. \$\endgroup\$ Mar 6, 2019 at 21:05
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The answers are excellent but if you want a simplified understanding, consider this.
For PMDC motors, Torque is proportional to current to charge the electromagnetic field.
RPM is proportional to the applied voltage above stiction starting voltage with no load.
The drop in RPM as load increases causes current to rise since the RPM drops the internally generated BEMF drops and the difference with the applied voltage draws more current or I= ΔV/DCR, motor coil DCR (Ohms).
Accelerate is just Newton's 2nd Law applied to rotational acceleration. \$a=F/m\$
For Field induced motors the field current also controls torque from the BEMF voltage it creates and thus the difference internal voltage, which has various configurations of series and shunt wound motors.
