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One concept of electronics I have a hard time understanding is if things like motors, actuators, solenoids etc. use as much power as they need or what you give them.

If a motor needs 12 volts and 500ma and I supply it with 12 volts and 3000ma will it only consume 500ma? Also, if I supply it with 15volts and 500ma what will happen?

It seems logical that an LED and a DC motor are way different when it comes to requiring/using electric where as an led has to be completely regulated and (I assume) a DC motor doesn't.

Is my understanding wrong?

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  • \$\begingroup\$ Keep in mind that mAh is unit of energy! It's current multiplied by time. Any motor can consume any amount of energy, given enough time. \$\endgroup\$ – AndrejaKo Oct 15 '13 at 14:25
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    \$\begingroup\$ @AndrejaKo it is not. It's a unit of charge. Way off anyways. \$\endgroup\$ – John Dvorak Oct 15 '13 at 14:26
  • \$\begingroup\$ @Jan Dvorak Yes, of course! (C/s)*s=C Internally I was thinking about V*A*(3600s). \$\endgroup\$ – AndrejaKo Oct 15 '13 at 14:38
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    \$\begingroup\$ If your supply is a voltage supply, you supply a specific voltage and the load consumes current in relation to its resistance. If your supply is a current supply, you supply a specific current and depending on the load's resistance the supply sets its voltage accordingly to meet the current flow you specified. If you set your current source to give more current than needed (3A instead of 0.5A) it will output more voltage than needed, and your load will blow up. Remember voltage is the cause and current is the result. \$\endgroup\$ – 1p2r3k4t Oct 15 '13 at 14:39
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    \$\begingroup\$ See electronics.stackexchange.com/a/34746/4512. \$\endgroup\$ – Olin Lathrop Oct 15 '13 at 15:07
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If it needs 500 mA then it will take 500 mA, even when you provide a 3000 mA capacity. If you're standing at the bottom of the Niagara falls with a 10 liter bucket you can fill it till it contains 10 liter, even though the waterfall has the capacity to provide much more.

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    \$\begingroup\$ at the properly rated voltage. \$\endgroup\$ – scld Oct 15 '13 at 14:58
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    \$\begingroup\$ Less like a bucket, more like a balloon or bag with a certain amount of resistive elasticity, where the water pressure changes how much water it can hold. \$\endgroup\$ – Passerby Oct 15 '13 at 15:05
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This is generally true for incandescent lamps, motors, other things made of coils, and most electronics that predates semiconductors. It is also generally true for lots of integrated circuits, which draw from their power rails as needed.

It is specifically false for LEDs and bipolar transistors, both of which can easily draw enough current to self-destruct unless held at a very specific voltage.

Over-voltage is nearly always bad for almost everything. Simple electronics may work when under-volted (motors, lamps). Semiconductors will not.

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Imagine an electrical connection as being a shaft that can rotate, and can connect a machine which would be powered by the shaft to a device that will make it turn. If the driving device is turning the shaft, a real-world machine which doesn't have an energy source will apply at least some torque in the direction opposite rotation (effectively trying to slow it down)--some torque in that direction will come from input-bearing friction if nothing else. The amount of energy transfered via the shaft will be the product of the torque and the rotational speed in radians per second [the units are radians per second because at that speed, the end of a torque arm l distance-units long will move l distance-units per second].

Some types of driving apparatus will "try" to supply a certain amount of torque at just about any speed. Other types of driving apparatus will "try" to turn the shaft at a particular speed, supplying as much torque (up to some limit) as is required to do so. Most types of driving apparatus will spin at some speed without a load, but will turn more slowly under conditions of increasing load torque.

Conversely, some types of driven apparatus will apply an almost constant level of load torque regardless of how fast they are driven, some will apply almost no torque when driven below a certain speed but "try" to prevent the input from spinning faster than that, resisting with as much torque as necessary to do so (up to a certain point). Many types of driven apparatus will resist with some torque almost regardless of speed, but the torque will be greater at higher speeds than lower speeds.

Any time the supplier's torque is higher than the consumer's, the shaft speed will increase; when it's lower, it will decrease. Since increasing speed causes most drivers' torque to drop but will cause most consumers' torque to increase, speed will increase until it reaches a level where the two torque levels are equal.

In some cases, one may think of the rotational speed as being set by the supplier; in some cases it's set by the consumer. In many cases, it's set by an interaction of the two.

In the electrical world, current is largely analogous to rotational speed and voltage is analogous to torque. Just as it's possible to have applied torque without something moving but (absent frictionless bearings) one can't have continuous motion without torque, so likewise one can have applied voltage without current flow, but current flow (except in superconductors) requires voltage. The one weird thing about the analogy is that most motors consume current proportional to mechanical torque, while dropping voltage which is proportional to the sum of their rotational speed (they also drop some additional voltage which is proportional to the applied current).

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Consider Ohm's law:

$$ E = IR $$

Here, we have three variables: voltage, current, resistance. For any resistive load, the three will always be related by this equation.

If that's hard to understand, consider a more observable, familiar three variable equation, Newton's second law:

$$ F = ma $$

Force is the product of mass and acceleration. In a frictionless environment, something that is not accelerating must have no force applied. Accounting for friction, something that is not accelerating must have forces applied that exactly cancel friction, such that there is zero net force. When there is force, a mass will accelerate; and it will accelerate less if it is more massive.

Say you wanted to tow a trailer at a constant speed. Your trailer is going to have some friction from air and the tires, and the towing machine will have to balance that force to maintain your desired speed. If the trailer isn't already moving, the towing machine will have to apply more force to accelerate the trailer. If you are towing uphill, yet more force will be required to overcome gravity. Going downhill you might need to apply a backwards force.

It doesn't matter if you use a bicycle or a locomotive as your towing machine, as long as you can apply enough force to maintain your desired speed. In either case the force is the same, though the range of forces that can be supplied by a bicycle and a locomotive are obviously much different.

You could also have a towing machine which instead of being programmed to travel at a constant speed, is programmed to apply a constant force. In this case, assuming the mass of the trailer is constant, acceleration will be whatever it needs to be to satisfy \$F=ma\$. If we assume a flat highway, probably you will accelerate the trailer to some speed until friction prevents further acceleration, and then your speed will be constant.

Most electrical power supplies are designed to maintain a constant voltage. For most loads, if you want more current to move through them, you must apply a higher voltage to overcome the opposing forces created by the resistance of the load. \$E\$ is set by the power supply, and \$R\$ by the load, thus, there will be only one \$I\$ which will satisfy \$E = IR\$. Not all power supplies will be capable of supplying enough current (and less commonly, a power supply may have a minimum current), but provided the circuit is operating within the supply's specifications, voltage will remain constant, and current will change according to the load.

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Current is pulled, voltage is pushed.

(Simplified explanation) A motor is essentially a large resistor, limiting the current that goes through it. It is a long coil of wire. When given V voltage and Coil Resistance R, with the regular Ohm's Law formula I = V/R you get the current it needs.

An LED is essentially a very very small resistor, like a fuse, as in it lets a very large amount of current through, heating up on the way. Essentially, it is a short circuit. For the useful purpose of emitting light, that current has to be externally controlled. If heat wasn't an issue (heat at the led junction being what kills it), it would simply act like a very very small resistor.

Think of a motor as an led + resistor. That's all it really is in the simplest terms. And as voltage changes, current changes through that led + resistor combo, or the motor.

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    \$\begingroup\$ LEDs are entirely different from resistors to the extent that I think your third paragraph is very misleading. \$\endgroup\$ – pjc50 Oct 15 '13 at 15:07
  • \$\begingroup\$ If current is pulled, how does a constant current power supply work? \$\endgroup\$ – Phil Frost Oct 17 '13 at 18:57
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    \$\begingroup\$ Also, I don't see how motors are like LEDs at all. \$\endgroup\$ – Phil Frost Oct 17 '13 at 18:59
  • \$\begingroup\$ @PhilFrost Because Constant Current supplies use Ohm's Law I = V/R and feedback to self adjust. R is the load (mostly constant), and it adjusts V so that I stays consistent. Current is still pulled. \$\endgroup\$ – Passerby Oct 17 '13 at 19:32
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    \$\begingroup\$ A LED is acts like a diode junction. LED's, diodes and bipolar transistors are current controlled devices, unlike resistors. The voltage is a function of current and not the other way around. There is no R=U/I for diodes at DC, as a V/I curve will clearly show it's non-linear. For AC, there is a model which requires a DC bias to select the slope on the V/I curve, in which an approximation can be made for small AC signals which is 'resistive'. However, that is a theoretical model, in the end it's still a diode. A CC PSU relies on it's feedback loop for constant current. \$\endgroup\$ – Hans Oct 17 '13 at 20:54
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we know the actuators like DC motors,stepper motors,Relay,Solenoid are made by coils(inductors);when supply is given it used to draw much current from the source than its rated;because the back emf of the coil is zero in the start up condition (if we use fast reacting fuses then it may blow) so only they are given with the higher current rating.

another example there is a difference between the batteries used in the automobiles and inverter.when the vehicle is started the battery must supply much current(very high inrush current) for few seconds, then the load current would be very less(light loads,audio systems);the battery used with the inverter must always give a steady state current(inrush current will be less in compare with automobiles).

but loads like LEDs are purely non reactive type so current drawn by them maynot vary so it can be powered by the source with exact current rating.

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  • \$\begingroup\$ The high inrush current of a car battery is due to the use of a Starter Motor which is not on the rest of the time, and then the battery is actually secondary to the ac or dc generator/alternator. This is not a accurate example. \$\endgroup\$ – Passerby Oct 15 '13 at 15:04
  • \$\begingroup\$ @Passerby yes i agree but i meant to say there is an inductive load which requires much start-up current.even incandescent lamps also draw much inrush current when they are turned on \$\endgroup\$ – yogece Oct 15 '13 at 15:07

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