Trying to understand how a load increases current  - Page 1
Forum section: electronics / beginners

author: mart1n
author_rank: Contributor
timestamp: October 28, 2017, 04:53:24 am
content: I'm fairly new to the world of electronics and this question has been bothering me for a awhile.When it comes to simple linear components, everything makes since to me. I connect a voltage source to a circuit, voltage source introduces a potential difference that causes current to flow. The amount of current that flows will be based on the resistance of the circuit and the voltage applied. I've got a grasp on Ohms Law.What is confusing me is when we introduce a non-simple load like a DC brushless motor. If the voltage is constant (lipo battery) and I assume the resistance is the same, how does increasing the speed of the motor cause more current to flow? I understand that doing more work (spinning the motor faster) will require more energy, but from a circuit theory perspective I'm not understanding how it works.As the motor spins faster, does the resistance in the circuit drop? thus causing more current to flow?If the DC brushless motor has a propeller on the end, the larger the propeller the larger the "amp draw" even if the speed stays the same. If it is the resistance lowering that causes more current to flow, does the larger propeller reduce the resistance even more?Thanks for clearing this up

author: ebclr
author_rank: Super Contributor
timestamp: October 28, 2017, 05:01:24 am
content: A motor isn't a resistor is a motor, that is another animal than a resistor.

author: Brumby
author_rank: Unknown
timestamp: October 28, 2017, 12:16:25 pm
content: The right hand side of that equivalence circuit diagram is what holds the key to your question. That, the rotor and the motion.The inductor (L) and the Back EMF are a function of things somewhat more involved than Ohm's Law. The mechanical load has a strong involvement in influencing the dynamics as well.Magnetic fields and motion make for an entertaining night of mathematics.

author: Ice-Tea
author_rank: Super Contributor
timestamp: October 28, 2017, 12:39:39 pm
content: Congratulations!You have now left the world of simple resistance and have entered the realm of impedance. Welcome!

author: mathsquid
author_rank: Frequent Contributor
timestamp: October 28, 2017, 01:10:30 pm
content: I found this book, particularly the explanation of the primitive linear machine, to be very helpful in learning how load and current are related in electric motors.https://books.google.com/books?id=jjuTYtKokc8C&pg=PA24&lpg=PA24&dq=primitive+linear+machine&source=bl&ots=RtPIXDyt4D&sig=mvWfenmodlNtAcJjY0T3EcJkGvs&hl=en&sa=X&ved=0ahUKEwiw6YSSsZPXAhUE9YMKHS-rD2kQ6AEIPTAH#v=onepage&q=primitive%20linear%20machine&f=false

author: Audioguru
author_rank: Super Contributor
timestamp: October 28, 2017, 05:15:18 pm
content: When you increase the speed of a motor then you are also increasing its current. If the motor is working hard then the current also is increased. The motor works as hard as it can when its speed is set to maximum but something keeps its shaft from turning, then its current is maximum and is called the stall current. A motor's inertia causes it to be stalled for a moment when it is first turned on to full speed.

author: rstofer
author_rank: Super Contributor
timestamp: October 28, 2017, 05:20:03 pm
content: Ignore the inductance, concentrate on a) applied voltage, b) internal resistance and c) back EMF.Amps create torque and volts create RPM.Amps are flowing because of the DIFFERENCE in applied voltage and back EMF divided by the internal resistance. If you load down the motor, the RPM will decline, the applied voltage is unchanged and the back EMF will fall. The difference in voltage will increase and the amps will also increase. The increase in torque accelerates the motor.Remember when I said amps create torque? Well that's how it works. The difference between applied voltage and back EMF divided by the resistance.If you want to plot back EMF versus RPM, spin the motor at a known RPM (drive it with a drill motor) and measure the terminal voltage. You will eventually derive Kv for the equations.So, how does volts create RPM? Absent friction, the motor will accelerate until the back EMF is equal to the applied voltage (current falls to zero). If the applied voltage is increased, there will be more current flow which creates more torque which accelerates the motor until the back EMF is equal to the applied voltage and the current flow again drops to zero. The RPM has increased.All numbers approximate because there is internal friction. It should be a small number and you can calculate it by applying a voltage and measuring the current. You know the resistance so current times resistance gives you the difference voltage. Subtract it from the applied voltage to get the back EMF. You can somehow get the torque required to spin an unloaded motor from the measured current and voltage. But you have to know Kv!

author: rstofer
author_rank: Super Contributor
timestamp: October 28, 2017, 05:32:38 pm
content: Back to the inductance I ignored above. All this does is add an exponential curve to current, RPM and torque. It basically makes speed changes mushy. It does help limit the inrush current when the motor is stalled and full voltage is applied. The inductance is only a factor during 'changes' in current and my discussion above was related to steady state RPM and torque, ignoring the mushy changes.On the mechanical side, inertia of the entire system will dominate the effects of inductance. You just can't change RPM in zero time.

author: hermit
author_rank: Frequent Contributor
timestamp: October 28, 2017, 06:13:46 pm
content: Quote from: mart1n on October 28, 2017, 04:53:24 amHow are you increasing the speed of the motor if you are holding the DC voltage constant?

author: IanMacdonald
author_rank: Frequent Contributor
timestamp: October 28, 2017, 06:29:24 pm
content: The free running speed of a simple commutator motor is determined by thedifference betweenthe supply voltage and the back EMF, all divided by the internal resistance. If the motor is forced to slow, the back emf reduces. Thus the difference between it and the supply voltage is greater so more current flows. To a certain extent this cancels out the loss of speed.If you were to force the motor to turn faster than its free-running speed, then the back EMF will exceed the battery voltage. In that case it will drive current back in the opposite direction, charging the battery. So a motor and a generator are essentially the same thing. The generator being forced to run faster than its natural running speed as a motor.Incidentally, there are no DC motors or generators, just AC ones that pretend to be DC, thanks to the commutator acting as a crude mechanical rectifier. As far as the motor's armature is concerned, everything is AC.DC brushless motors are actually 3-phase synchronous AC motors, but with some clever electronics to generate a 3-phase supply that matches the rotational speed. This type may not rely on back the EMF to control the speed, it may be controlled by the electronics. It is still the case though, that it cannot run faster than the speed where the (average) back EMF of its windings equals the supply.3-phase supplies are actually much better at driving motors (Tesla realised that) because they effectively create a rotating magnetic field that pulls the rotor around continuously instead of just giving it a push now and then. That is why the brushless type can be surprisingly powerful for their size.HTH.

author: rstofer
author_rank: Super Contributor
timestamp: October 28, 2017, 06:59:40 pm
content: Quote from: hermit on October 28, 2017, 06:13:46 pmReduce the load which reduces the requirement for torque to the load which will then be applied to accelerating the motor until the difference in voltage reduces the current flow to some new equilibrium point.

author: hermit
author_rank: Frequent Contributor
timestamp: October 29, 2017, 12:26:15 am
content: Quote from: rstofer on October 28, 2017, 06:59:40 pmI had a different take on the original question. It had nothing to do with loading. A constant voltage motor drawing more current as it sped up. What causes the speed up? I think the OP is putting some things together in their mind that don't belong together. That is why I asked, what causes the motor speed to increase? Loading it will slow it down. At least it did in the labs experiments I did in school.

author: KL27x
author_rank: Super Contributor
timestamp: October 29, 2017, 01:57:29 am
content: ^Exactly. How many answers we got to a question that doesn't even make sense.

author: IanB
author_rank: Super Contributor
timestamp: October 29, 2017, 03:03:42 am
content: Quote from: mart1n on October 28, 2017, 04:53:24 amIn general the opposite is true. If the motor spinsfaster less currentflows and if the motor spins moreslowly more currentflows. At the limit if you stall the motor (make it stop turning) the maximum current flows.You have to ask yourself, what changes the speed of the motor, if the voltage is constant? The answer is the load on the motor. If you put more resistance on the motor (if you fit a bigger fan), if you make it work harder, then it pulls more current to overcome the resistance. If you take all load off the motor, if you let it spin freely without doing any work, then it will spin as fast as possible and will draw minimum current.

author: Brumby
author_rank: Unknown
timestamp: October 29, 2017, 04:23:27 am
content: We need the OP to clarify how he came up with this question:Quote from: mart1n on October 28, 2017, 04:53:24 am

author: hermit
author_rank: Frequent Contributor
timestamp: October 29, 2017, 04:45:21 am
content: Sounds like someone in the early learning stages that is kind of bombarded with data and put two pieces together that didn't fit and was asking why. That's why I asked them to think about what would cause the motor to go faster if they hold the voltage steady. I think that is the part they were missing.

author: mart1n
author_rank: Contributor
timestamp: October 30, 2017, 10:58:49 pm
content: Wow, thanks for all of the replies. I wasn't expecting so much help, what a great community.Yes, it's very possible that I'm confused and my question doesn't make sense as asked. I probably shouldn't have used a DC brushless motor as the example either, as my confusion is more general.The way that I think about current flow is that the potential difference causes the electrons to start moving, thus current flowing. The amount of current flowing depends on the voltage and resistance in the circuit.In my mind it's kind of like the voltage is "pushing" the current, the more resistance the more pressure required to push the electrons. So when I hear most folks say things like, "The load is drawing X amount of amps", I get confused. Why should the load determine how much current is flowing, isn't it the voltage and resistance that determines the current flow? The way most folks refer to "current draw", it would seem the load is "pulling" the current flow.I feel like my understand works with simple circuits with a battery and a resistor and maybe an LED. Because if you don't provide a big enough resistor, the voltage will "push" too much current to the LED and fry it. But I feel like my understand breaks down with non-passive loads.I build and fly a lot of FPV racing multirotors (quad copters). And I know that when I throttle up the AMP draw of my quad can hit 100A or 25A per motor.With more active or complicated loads, my understand breaks down I feel. But please don't worry about explaining how DC brushless motors and ESCs work to draw current in a multirotor if you don't want to, as a more basic answer of explain how non-passive loads can "draw" more amps the harder they work even though the voltage and resistance seem constant (The resistance and voltage must be changing).Thanks!

author: Audioguru
author_rank: Super Contributor
timestamp: October 30, 2017, 11:42:40 pm
content: You mentionedspeed controlof the DC motors in a quad copter. Do you know how the electronics varies the speed? It uses Pulse Width Modulation (look it up in Google). The pulses of high current are at a high frequency and wide pulses produce a lot of average current in the motor so it spins faster. Narrow pulses are on for a short duration for each pulse and are off for most of the time for each pulse so the average current in the motor is low so it spins slower. So it is the speed controller that increases the current, not the motor.The brushless DC motors in my radio controlled model airplanes also use PWM to control the speed of the motor, play a little tune with the motor when the airplane is powered and is setting up its gyros and causes a loud scream when the motor is stalled (stuck on something). When the motor is stalled then its current is very high and the electronics detects the very high current and causes the scream then it shuts down the motor to avoid damage.

author: IanB
author_rank: Super Contributor
timestamp: October 31, 2017, 12:20:21 am
content: Quote from: mart1n on October 30, 2017, 10:58:49 pmThe answer is that loads have two ways of reducing or increasing the current that flows through them. One way is to change their resistance. The other way is to apply a reverse voltage that opposes the applied voltage. So for a load to "draw" more amps it can either reduce its internal resistance, or it can reduce the opposing voltage that it generates. Electric motors tend to work the second way, by reducing the opposing voltage when the load is increased.

author: rstofer
author_rank: Super Contributor
timestamp: October 31, 2017, 12:46:16 am
content: Quote from: mart1n on October 30, 2017, 10:58:49 pmYou aren't having a problem with Ohm's Law, your problem is with the way motors work. Particularly the back EMF.QuoteThis is exactly right. But in the case of a DC motor, the potential difference is NOT the applied voltage, it is the applied voltage minus the back EMF. When the motor is running unloaded at some speed (with no friction), the back EMF will be the same as the applied voltage and the current will be zero. In the real world, there will be some current flow caused by friction (bearings).QuoteGenerally, when this comment is made, the term 'load' and 'resistance' are the same. Except as I pointed out for motors and back EMF. If I have a 12V battery and a 1 Ohm load, I get 12 Amps of current flow. If I change the load (the resistance), the current will change.QuoteIt's the fact that the motor is another source of voltage that is complicating the discussion. Think of an electric motor driving a helicopter. As you increase the pitch of the rotors, they try to move more air and this increases the requirement for torque. As a result, the motor needs to draw more current. The only way it can draw more current is for the back EMF to decrease which happens when the motor slows down. This snippet applies to a constant applied voltage. When you get to PWMing, all you are doing is changing the desired RPM by changing the applied voltage (average).QuoteNo, the resistance isn't changing. The voltage is changing and, specifically, it is the voltage difference between what is applied to the terminals and what is being generated internally.Look at the attached Motor.pdf. I'm going to jump to Kirchoff's Voltage Law and you can look it up later. If we write the loop equation starting in the lower left corner and running all the way back, we go through V1 and gain 10V (moving - to +), the current in the loop (call it I1) causes a voltage drop across R1 and we will take it as a loss of I1 * R1 so the equation, up to this point is +V1 -I1*R1 to which we have to subtract V2 because we are going through it from + to -. The final equation is:V1 - I1 * R1 - V2 = 0 Kirchoff's Voltage Law says the sum of the voltages around a loop total zero.But we know that V1=V2 so we cancel them out and we have -I1 * R1 = 0. Well, R1 isn't zero so I1 must be.So, if the voltages are equal in both batteries, no current flows through the resistor.However, if the back EMF (V2) drops to 5V, we get 10 - I * R1 - 5 = 0 or I1 * R1 = 5 and, since R1 = 100, I1 = 0.05 Amps. We have a positive voltage drop so the current is flowing from left to right. That is pretty intuitive because V1 is larger than V2. You would expect V1 to try to charge V2.Ohm's Law is important but Kirchoff's Laws are probably more important because they allow you to analyze circuits in a methodical manner. The voltage law says the sum of the voltages around a loop total to 0 - voltage can't just pile up. The current law says the sum if the currents entering and leaving a node (junction) total zero. Current can't pile up either.If you want to up your knowledge, you should look into these topics.Try this program:https://learn.digilentinc.com/classroom/realanalog/

author: Brumby
author_rank: Unknown
timestamp: October 31, 2017, 01:44:31 am
content: As others have said, your basic understanding - which is described by Ohm's Law - is quite correct.To include motors in your understanding, you will need to ADD to that.You have to consider what else is happening - and with a motor, the most important factor is one you are probably already aware of: You can use a motor as a generator. Apply rotation to the shaft of a motor and you will create a voltage at its input terminals. (You will need to have a magnetic field around, but for motors with permanent magnets, that is already taken care of.)What is interesting is that this voltage isalwayscreatedwhenever the rotor is spinning. It doesn't matter whether the motor is being used as a motor or as a generator. It is simple, basic physics that applies to moving conductors in a magnetic field - and the faster the movement, the greater the voltage. Add this fact to your knowledge base ... and you are ready to understand about motors.So .... you apply a voltage to the motor and it spins - butbecauseit is spinning, it creates a voltage itself. This voltage is called the "Back EMF" (where EMF = electromotive force, which is just a fancy name for voltage ... but it gets used in this situation all the time). As a result, the effective voltage that is getting applied across the motor's windings is thedifferencebetween the applied voltage and the back EMF.If you apply 12V to a motor and the back EMF is 8V, then the motor will be working on the equivalent of 4V. You can plug this 4V into your Ohm's Law calculation to get a current.Increase the load on the motor and it will slow down. This will reduce the Back EMF. If the Back EMF is now 5V, then the motor will be working on the equivalent of 7V - which will mean your Ohm's Law calculation will give you an increased current.Your understanding of Ohm's Law is safe - and you knowledge has increased to the point where you can get a handle on what is happening with motors, loads and currents.The next step is understanding how multirotors run their electronics which drive the motors. That is a little more complex - and builds on the above even more. Learning electronics is like eating an elephant ... take it one bite at a time.