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Increasing voltage increases the RPM of the motor, but what does increasing current do?

I can't seem to get a straight answer. Some googling has led me to believe that torque increases while rpm stays the same.

I currently have two 24V batteries wired in parallel, and they are connected to a DC motor through a speed controller. I understand that the speed controller works by turning the circuit on and off very quickly to vary the speed of the motor.

I just bought a third 24V battery that I plan to connect in parallel to the other two, and was wondering what I should expect when I hook it up.

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  • \$\begingroup\$ Related: Choosing power supply, how to get the voltage and current ratings? \$\endgroup\$ – Phil Frost Jan 29 '14 at 0:05
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    \$\begingroup\$ This question is not going to get great answers because it's based on a false premise: that adding another battery will increase the current to the motor. Each answer will probably mention that, and then each go in a different direction. \$\endgroup\$ – Phil Frost Jan 29 '14 at 0:06
  • \$\begingroup\$ @PhilFrost yes, I was definitely misunderstanding the problem. The batteries I had are each rated for 10 amps, while the motor is rated for 30 amps. I had a feeling that the batteries were limiting the performance of the motor because they couldn't deliver that much current. \$\endgroup\$ – Ben Jan 31 '14 at 7:04
  • \$\begingroup\$ Batteries are normally rated in Ampere-hours (Ah), not in Amperes. An Ampere-hour is a measure of the energy stored in the battery, and is not directly related to the current that the battery can deliver. \$\endgroup\$ – Peter Bennett Jun 10 '18 at 18:07
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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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  • \$\begingroup\$ This was a fantastic answer. Thank you for correcting my misconception -- I did not know that the resistance of a motor is constant. \$\endgroup\$ – Ben Jan 31 '14 at 7:03
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Increasing voltage increases the RPM of the motor, but what does increasing current do?

Increasing Voltage, Increases the Current Pulled, which Increases the Strength of the coil, which increases the RPM AND Torque of the motor. They are all connected.

I currently have two 24V batteries wired in parallel, and they are connected to a DC motor through a speed controller. I understand that the speed controller works by turning the circuit on and off very quickly to vary the speed of the motor.

This is called PWM. By turning the circuit on and off rapidly in a given period (ex: 25% on time 75% off time every 10 milliseconds) will average out the voltage to that amount (25%). The motor sees about 25% the voltage, so it draws about 25% the current, the torque is decreased, the rpms are decreased, etc.

As a basic comparison, a motor is simply a resistor. Ohm's law applies to it. I = V/R. Current and Voltage are related, especially with a fix Resistance like the motor. Except that the motor is only a fixed resistance depending on it's load. A motor with no load has lower resistance than a motor with a load, and a motor that is overloaded will stall out, having the highest resistance, drawing a stall current that (rule of thumb) is between 2 and 2.5 times the no-load current.

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Adding the third battery will give you a 50% longer run time before the batteries need recharging, but shouldn't change the motor's performance.

Adding more batteries in parallel won't cause the motor to draw more current because the voltage remains the same as before.

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  • \$\begingroup\$ 50%, or 33%? Why would 1 extra battery add 50% runtime? \$\endgroup\$ – Passerby Jan 29 '14 at 3:02
  • \$\begingroup\$ @passerby: He already had two batteries, and is adding a third. If each battery is 100 AH, he had 200 AH, but now has 300 AH - a 50% increase (but if he takes the third battery away, he'll have a 33% decrease.)(of course, this assumes that all three batteries are the same capacity). \$\endgroup\$ – Peter Bennett Jan 29 '14 at 3:15
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Adding another battery in parallel would allow the resulting battery to provide more current as the load increases. It can allow the motor to produce a higher torque if the engine's winding resistance is much lower than the internal resistance of the battery set.

It also has been mentioned that adding a battery will allow a longer run without battery replacement under the same load.

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