I have been working with PMDC motors for a while now. In these motors, as the load increases current drawn by the motor also increases. I was just wondering what is the theoretical reason behind this change in current?
Electronic – Current drawn by a permanent magnet DC motor
currentdc motor
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A bunch of issues:
- Are you sure you have a brushed DC motor? This kind has only two connections. You simply apply voltage and it spins, which is what you are assuming in the rest of your question. All you said is that the motor has permanent magnets and is "DC", which could still leave other possibilities, like a brushless DC motor. Those are a lot more complicated to drive. Those have a lot more than two wires coming out, usually 3 for the drive coils and 5 for the Hall sensors.
- $50-$100 sounds like the cheapest way to solve this problem unless you value your time at pennies/hour. If this is a brushless DC motor, then you are in over your head.
- Whether you run the motor from 60 V, 120 V, or something else, that DC power still has to come from somewhere. This is independent of how you might switch it by applying something like PWM, for example. In other words, you still need some kind of power supply.
- PWM does not somehow supply more consistant torque. PWM is merely a technique for modulating the effective motor voltage while starting with the same DC supply and wasting relatively little power in the process.
Systems that drive motors with PWM also often have speed or position feedback. If the feedback is used to maintain constant motor speed, then the motor can appear to have high torque without running away when the load is removed. However, this is due to the ability to modulate the motor drive and a servo loop with feedback. PWM is one means to modulate the motor drive, but by itself doesn't somehow get more torque from the motor.
- Since PWM is a technique to present a lower apparent voltage to the motor than the raw DC power supply it is derived from, you should use the highest voltage supply the motor can handle.
The apparent motor drive voltage is the raw DC voltage times the PWM duty cycle. For example, if you start with a 120 V DC supply but the motor only needs 30 V to turn at the speed you want with the load you are going to present it, then a PWM duty cycle of 30V / 120V = 1/4 will do it, assuming the PWM frequency is high enough so that the motor does not react to individual PWM pulses. Let's say the PWM frequency is 25 kHz, which is a common frequency to run motors at. That means the PWM period is 40 µs. To make effective 30 V with 120 V in, you'd pulse the motor on for 10 µs every 40 µs, or 10 µs on and 30 µs off.
- You say you are concerned about torque, but it appears what you really want is constant speed. If so, a servo loop in the micro that controls the PWM duty cycle to whatever it needs to be to maintain the set speed will give more apparent "torque" than brute force gearing the motor down to the point where whatever load variations you give it won't matter much at the motor.
By adding a battery in parallel, you do not increase the current. You increase the maximum current that the motor can take. Nothing will happen if you add another battery in parallel and the motor isn't suffering from shortage of current.
Keep in mind that than in Ohm's law, you have 3 variables: \$V=RI\$. In this equation, you can affect one variable by changing the other two. For a given motor, R is constant, so that means that one of two possible variables you can change is out.
You can either set the voltage to some level, which you seem to be doing by using the speed controller, and let the current come from the equation or you can use a different type of speed controller which sets the current and lets the voltage come out as a result of the equation.
So how is torque related to this? Well motor has what's called back electromotive force and the equation for the Ohm's law is actually a bit different: $$I=\frac{V_{battery}-V_{back-EMF}}{R}$$ The greater the torque provided by the motor, the lower is the \$V_{back-EMF}\$, resulting in greater current through the motor.
When current is supplied by a battery, the battery's voltage usually drops. The drop depends on the type of battery and the current. If the current is above what battery is expected to provide, you can expect the battery to have lower voltage than expected, to overheat, maybe even explode. If the current provided by the battery is sufficient, the voltage drop isn't going to be as big.
So it's as I said in the first paragraph: If the batteries can provide sufficient current to the motor (and you test this by checking the current when motor should be providing maximum torque), then adding another battery won't affect the current or the torque. If there isn't enough current and you add a battery, you can expect increase in torque because the voltage supplied by the batteries will be higher.
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Best Answer
As the motor spins, the magnetic field induces back EMF in the windings. The induced voltage has opposite polarity than the supply voltage. Theoretically, when no load is present, these voltages cancel out and no current can flow.
When the motor is loaded and slowed down, the BEMF voltage decreases and allows more current to flow. When the motor is stalled, there is no BEMF and current is limited only by winding resistance.