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Current is the rate at which electrons flow through a circuit, and voltage is the potential "drop" between two points in a circuit.

Do voltage sources output an unknown current, and vice versa? Why aren't they the same, if V=IR when there is no resistance? This question made me think deeper: It's intuitive (for me) that a current supply would provide a flow of electrons around a circuit, giving each element a charge and voltage. But how can a voltage source provide a constant output of a "drop" around a circuit? It doesn't seem to make sense. How does a voltage source work, compared to a current source?

These questions have come up for me several times throughout my circuits 101 course.

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  • \$\begingroup\$ Those words are shortcuts we use. Their full names are Constant Voltage Power Supply and Constant Current Power Supply. Since it is understood that we want to regulate our power supply to a specific constant output (eg 5V, 12V etc) we simply ignore saying the word "constant" \$\endgroup\$
    – slebetman
    Commented Mar 18 at 7:20
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    \$\begingroup\$ V=IR applies to a resistor. If you don't have one, it simply won't apply. \$\endgroup\$
    – winny
    Commented Mar 18 at 10:37
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    \$\begingroup\$ @slebetman, "constant" does not mean not to regulate the source voltage. You can adjust it and yet it is constant. "Constant" here means that the voltage does not depend on changes in load, input voltage and other disturbances. \$\endgroup\$ Commented Mar 18 at 11:07

13 Answers 13

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Voltage is the cause, current is the effect.

So both sources provide a voltage on their terminals.

In the voltage source the voltage on the output terminals has, or is intended to have, a constant value. Sometimes it's left to the internal resistance of the voltage source to determine what the final voltage will be, and sometimes the voltage at the terminals is measured and fed back to a controller which then compensates for the voltage drop across the internal resistance.

In the current source, there is a measurement of the current, and through a controller the voltage is adjusted such that the current will have the desired strength.

The question

How does a voltage source work, compared to a current source?

were better asked as follows: "How does a current source work, compared to a voltage source?" with the answer being: A current source is just a voltage source, but with a (different) controller.

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    \$\begingroup\$ If you're actually moving electrons, like a Van de Graaff generator, then current is the cause and voltage the effect. \$\endgroup\$
    – mow
    Commented Mar 17 at 19:39
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    \$\begingroup\$ I'm also not convinced that voltage is the cause. Voltage and current are two sides of the same coin, and the relationship between them is not one of cause-and-effect. In the lumped element model, the relationship between the two is simultaneous, which is why we are able to derive and solve simultaneous equations for the instantaneous state of a circuit. \$\endgroup\$ Commented Mar 18 at 3:04
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    \$\begingroup\$ @HarryH The voltage produced in the kick is determined by the current that needs to flow. \$\endgroup\$
    – DKNguyen
    Commented Mar 18 at 14:58
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    \$\begingroup\$ @KarlKnechtel "is there any reasonable approximation to a current source that works passively?" A solar cell (albeit a poor one). It's output current is (roughly) linearly dependent on the amount of light, while the output voltage is set externally and does not vary (much) with the amount of light. It is usually modeled as a current source with a parallel diode. \$\endgroup\$
    – BrtH
    Commented Mar 18 at 15:14
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    \$\begingroup\$ Another approximation to a current source is a bike "dynamo" (really an alternator). \$\endgroup\$
    – Chris H
    Commented Mar 18 at 16:27
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There are only two quantities that can be considered fundamental, or "real", in the sense that they actually exist, and those are voltage and current. Resistance is artificial, a contrived quantity that merely describes the relationship between voltage and current.

Voltage across some component, and current through it, are usually related somehow. For instance, for a resistor we find that the two values are directly proportional to to each other. That is, if you double the voltage \$V\$ across a resistor, the current \$I\$ through it also doubles, or if you halve current \$I\$, voltage \$V\$ is halved too. That particular relationship is called "Ohm's law":

$$ V = R \times I $$

The constant of proportionality in this relationship is resistance \$R\$. This formula only applies to resistors. For other elements, such as voltage sources, current sources, capacitors, diodes and so on, the relationship between current and voltage is not directly proportional. For instance, the current \$I\$ through a silicon diode, as a function of the voltage \$V\$ across it is:

$$ I = I_S\left(e^{\frac{qV}{kT}}-1 \right) $$

It will be helpful to draw graphs of voltage vs. current for three components, a 10V voltage source (blue), a 1A current source (red), and a 2.5Ω resistor (green):

enter image description here

The key features of these graphs are:

  • The voltage source (blue) has a constant voltage across it, regardless of the current through it. It is a straight, horizontal line, with gradient (slope) \$\frac{dV}{dI} = 0\$. Notably, current and voltage are independent of each other, when current changes, voltage does not.

  • The current through the current source (red) is fixed at 1A, regardless of the voltage across it. It is also a straight line, but this time it's vertical, having slope \$\frac{dV}{dI} = \infty\$. Again, current and voltage are independent; when voltage changes, current does not.

  • The resistor (green) is not vertical or horizontal. Its slope is non-zero and finite: \$\frac{dV}{dI} = 2.5\$. The line passes through the origin \$(I=0, V=0)\$. This is a graph showing direct proportionality between \$V\$ and \$I\$. When current changes, voltage does too, in proportion to the change in current, and vice versa.

That facts that neither voltage source nor current source have any relationship at all between voltage and current (they have infinite/zero V-I gradients), and that their V-I curves do not pass through the origin, are clear indicators that the concept of resistance is not applicable to sources. Resistance implies direct proportionality between voltage and current, a behaviour which sources simply do not exhibit.

The short answer to your question is that a voltage source "outputs" only a voltage, and doesn't impose any constraint on the current through it. A current source "outputs" only a current, imposing no constraint on voltage across it.


Perhaps the simplest demonstration of a source's independence of voltage and current is when we connect a current source in parallel with a voltage source:

schematic

simulate this circuit – Schematic created using CircuitLab

You might think that there's a conflict here, but there isn't. The only job of voltage source V1 is to apply a fixed potential difference of 10V between A and B. The only job of current source I1 is to ensure that 1A flows around the loop. Neither of those roles are in conflict with the other. Source V1 has no problem with I1's insistence that current be 1A, and source I1 has no problem with V1's insistence that the voltage across it be 10V.

There's no ambiguity about the current, here, or the voltage, and either source can choose the conditions that it "imposes" without the other so much as batting an eyelid. In other words, where does resistance play a role here? No equations that you derive for the behaviour of this circuit make any mention of "resistance" at all. There are two equations:

$$ V = 10V $$ $$ I = 1A $$

That's it. Where's the R?


If you insist on assigning some kind of resistance value to a source, then you must first define what that resistance means. A practical definition of resistance in these circumstances goes as follows: resistance is the ratio of change in voltage to change in current:

$$ r = \frac{\Delta V}{\Delta I} $$

In mathematical treatments you will normally see the above relationship written:

$$ r = \frac{dV}{dI} $$

If you paid attention in math class, then you'll understand that this is referring to the gradient (slope) of the graph of \$V\$ vs. \$I\$, at some particular point \$(I, V)\$ on that graph. I've used a small \$r\$ to indicate that this is dynamic resistance, as distinct from ohmic, which applies only to resistors.

By that new definition, the dynamic resistance of a voltage source will be the gradient of its \$V\$ vs. \$I\$ plot, which is zero:

$$ r_{V1} = 0\Omega $$

The dynamic resistance of a current source is also the gradient of it's \$V\$ vs. \$I\$ plot:

$$ r_{I1} = \infty\Omega $$

The dynamic resistance of a resistor is the same as its ohmic resistance, because the gradient is constant, and proportionality is direct. For a resistor, Ohm's law is always true, for all values of \$V\$ and \$I\$, and both \$V\$ and \$I\$ are able to change:

$$ R = r = \frac{dV}{dI} = \frac{V}{I} $$

Dynamic resistance is a very useful concept. For any component that does not have a linear (straight line) \$V\$ vs. \$I\$ curve, the gradient of the curve changes as you vary either voltage or current. Take the silicon diode, for example. Its V-I curve is the blue line here:

enter image description here

The response of the diode to very small changes in current is different depending on where you "reside" on this graph, since the gradient of the curve is different everywhere. In other words, by passing an average (DC) current of 18mA through the diode, and making small ±100μA fluctuations (the "AC signal") , the resulting voltage changes across the diode will be much smaller than the changes you can expect if the average DC current through it were 3mA. This permits you to control gain.


Cells and batteries of cells are voltage sources. They maintain a constant potential difference chemically (at least until they are depleted of energy, or over-charged to destruction). Their chemical and physical construction are such that charges at one end tend to have a fixed amount of potential energy with respect to charges at the other terminal. How that happens is for the chemists to know.

Other ways exist to produce a fixed potential difference, such as charge pumps, and it's sufficient to design a circuit that constantly monitors some potential difference, and if it changes, take some action to restore it. "Pump faster or pump slower", in the same way you occasionally pump air into a tyre that has a slow leak, to maintain constant pressure.

At the moment I can't think of any current sources that are able to self-regulate like chemical cells, although probably some such thing exists. However, it's possible to build regulators that maintain a constant current flowing through them, in the same way that voltage regulators do; monitor and adjust.

At the risk of confusing voltage sources with current sources, in practice it's probably easiest to think of a current source as something that varies its own potential difference to whatever value is required to obtain the requisite current through whatever's connected to it (and, consequently, itself). The result is the behaviour you would expect from a current source; current through it remains fixed while the voltage across may vary. As long as there's no functional difference, there is no distinction between a current source and a voltage source that varies voltage to maintain constant current.

By that same argument, you might consider a constant voltage source to be something that varies current, to whatever amount produces exactly the correct fixed potential difference across it. Again, a variable current source that maintains a constant potential difference is functionally identical to a fixed voltage source. Both practically and mathematically there's no distinction.

However you envisage these sources, they each only represent a single quantity in the simultaneous equations that you derive for any system. When you see a voltage source in a schematic, all you know about it, until you solve for all the other variables, is the voltage across it. For a current source in a schematic, all you know (until the system is solved) is the current through it. How they work is actually irrelevant in the analysis.

Cells aside, to build practical, active voltage and current sources is not trivial. When you see a source in a schematic, it represents an abstraction of something that, in practice, may have substantial complexity, consisting of transistors and/or op-amps, and other elements, whose role is to make continuous adjustments to maintain some fixed voltage or current.

In the analysis, though, their behaviour is truly trivial. When you see a voltage source in a schematic, you know the potential difference between two points. That's all. When you see a current source, you know the current in that path. That is all.

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    \$\begingroup\$ "mow" mentioned the Van de Graaff generator as a constant current source. It physically moves a certain number of electrons per second, at least until it reaches some very high maximum voltage. \$\endgroup\$ Commented Mar 18 at 17:27
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Do voltage sources output an unknown current, and vice versa?

Yes, a constant voltage source puts out a variable current (into different resistances).
A constant current source puts out a variable voltage (into different load resistances).

Why aren't they the same, if V=IR when there is no resistance?

A constant voltage source is quite content to put out no current at all, if you connect no path into a load resistance of infinity. This voltage source works harder when a current path is provided into a load. A load resistance of zero makes it fail, because current would be extremely high.

A constant current source is quite content to put out no voltage - this requires a zero-ohm load - a short circuit. This current source works harder when load resistance is larger than zero. A load resistance of infinity (an open circuit) makes it fail, because voltage required would be extremely high.

But how can a voltage source provide a constant output of a "drop" around a circuit?

Consider a constant voltage 1.5V battery...
A fresh battery, before inserting into a device has a potential difference between terminals of 1.5V...no current flows, because load resistance is infinite. Its chemical engine need not cause current to flow.

A lamp is connected across its terminals. It still supplies 1.5V, but now current must flow through the lamp...

Another lamp is added to its terminals. It still supplies 1.5V, but now current must flow to two lamps, so the battery must put out 2X current. When this lamp is added, it attains the same brightness as the first lamp.


When a constant current source is connected to one lamp, it attains a certain brightness.

Adding a second lamp to the constant current terminals results in both lamps being equally bright, but dimmer than the one-lamp case. The lamps must share the constant current. Voltage across the two lamps is half of the one-lamp case.

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All other answer are good to perfect; I just like to share my thinking model for beginners. It might give something imaginable.

Compare electric voltage with the pressure of water due to level difference.

Compare electric current with the water current through the pipe.

A constant voltage source is like an endless ocean of water behind the wall. Independently of the amount of water current you let flow, the pressure stays the same.

A constant current source is like a pump of endless power. Independently of the pressure needed, it maintains the amount of water current it sources.

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A voltage source is a power supply that will sink or source current into a load until its predefined voltage is reached. An ‘ideal’ voltage source can supply infinite current in either direction, regardless of load (except into zero ohms, see below). Real-world voltage sources have a current compliance range.

Note: an ideal voltage source driving a zero-ohm load is an undefined condition. This makes sense, since I = E/R, so for any nonzero E, the limit as R approaches zero doesn’t exist. Simulators will catch this condition and flag it as an error.

A current source is a power supply that will vary its voltage to sink or source its defined current into a load, regardless of the voltage or load it is driving into (except for an open circuit, see below.) Whether the voltage changes or not depends on the load. An ‘ideal’ current source can support an infinite voltage range. Real current sources have a voltage compliance range.

Unlike a voltage source, a zero ohm load poses no problems for a current source, as we have E = IR. Zero-ohm R merely results in the source forcing a current into zero volts.

You might ask, do you need voltage to have current? Not necessarily.

A real-world example of a current flowing without any voltage potential is a superconducting coil. We force current into the coil, building up to its target field strength. The current source is then removed and the coil ends connected together, forming a loop where the current keeps circulating. Once the coil is loaded up and connected in a loop, there's no further voltage or current needed to keep things moving.

Moving on, an ideal current source driving an open circuit (E = IR, where R = infinity) could result in an infinite voltage for any nonzero current source value. However, this is ok math-wise as multiplying infinity x any non-zero value of current is also infinity. There's no dividing by zero happening and thus no undefined condition.

For an open circuit we can further conclude that current source current is zero, and simulators will model this behavior and in addition, helpfully set the open-circuit voltage to zero as well.

In the real world an unconnected current source will be at one extreme of its voltage compliance range.

Wrapping things up:

  • A voltage source outputs variable current to force a fixed voltage
  • A current source outputs variable voltage to force a fixed current

And, the two corner cases to consider:

  • Voltage source behavior is undefined when driving into zero resistance (I = E/R, where E \$\neq 0\$ and R \$= 0\$).
  • Current source output is assumed to be zero when driving infinite resistance, (E = IR, where R \$=\infty\$.) Since I is zero by definition, E is also zero since zero (I) x \$\infty\$ (R) is still zero.
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It's basically about output impedance and how they form lop-sided voltage dividers or current dividers.

Remember that in real power sources, there is output impedance which means the output is not truly constant with variations in load. However, you can design things so that the output, either voltage or current (but not both at the same time), remains relatively unchanged over a wide range of load impedances.

VOLTAGE SOURCES

A constant voltage source can be interpreted in the following TWO ways:

schematic

simulate this circuit – Schematic created using CircuitLab

In the series output impedance version of a voltage source, an ideal voltage source is used. Proper design dictates that in order for the entire circuit to behave as a voltage source, the series output impedance must be low relative to the load impedance. It forms a lop-sided voltage divider where the load impedance dominates. As long as this condition is met, a wide range of load impedances, more or less, experience the same voltage since most of the voltage is dropped across the load impedance which results in a voltage source. This should be straightforward and obvious.

In the parallel output impedance version of a voltage source, an ideal current source is used. Proper design here dictates that the parallel output impedance be low relative to the load impedance. The way this ends up being a voltage source even though the circuit has an ideal current source is that lop-sided current divider in which the parallel output impedance dominates which results in the vast majority of the the current produced by the current source flows through the parallel output impedance rather than the load. As long as this condition is met, the total impedance seen by the ideal current source remains relatively constant even if the load impedance changes and only a small amount gets diverted through the load. This results in the voltage drop produced being the constant current mostly being determined by the parallel output impedance rather than the load impedance and since the parallel output impedance is fixed, you get a voltage source even though the circuit is being driven by an ideal current source.

Why might you want to turn a current source into a voltage source? One example is the ballast resistor on a current transformer.


CURRENT SOURCES

Similarly, a current source can be interpreted in the following TWO ways. If you understood what I wrote about the voltage source above, you should be able to work out how the current sources below operate. Give it a try before reading the explanation below.

schematic

simulate this circuit

The parallel output impedance version of a current source requires that the output impedance be high relative to the load impedance. You should be catching on now that this is required in order to form a lop-sided current divider so most of the current from the ideal current source flows through the load. As long as the load impedance remains relatively low compared to the parallel output impedance, the current in the load remains relatively fixed.

The series output impedance version of a current source forms a lop-sided voltage divider where most of the voltage from the ideal voltage source is dropped across the output impedance. This may seem strange, but remember that we are trying to make a current source, not a voltage source. If the series output impedance is very high it means that the load impedance can vary quite a bit without changing the overall impedance seen by the ideal voltage source very much. That means the voltage source supplies the same amount of current, more or less, resulting in a constant current source.

Why might you want to turn a voltage source into a current source? In transistor differential amplifiers, current sources are needed a lot of the time but are rather expensive to implement discretely. So what they often do instead is use a very high value resistor at the top or the bottom of the differential pair before it splits. As long as this resistor is relatively high compared to the impedance of the transistors and other components in the differential legs, it approximates a current source very cheaply.


You may have noticed that it requires an unusually powerful current source if you want to convert it to a voltage source (since you need to drive a lot of current through a low parallel impedance which seemingly wastes a lot of current). Or you may have noticed that it requires an unusually powerful voltage source if you want to convert it to a current source since you need to drop a lot of voltage across a high value series resistor.

However, sometimes that is not a problem as is the case in the current transformer. There, you have excessive current and power available since you are tapping current off a power circuit but only need to provide a weak signal for measurement).

In the case of the differential pair you are amplifying a signal which means the input voltages are low relative to the output voltages. Since transistors can only reduce voltage, that means the voltage supply needs to be at least as high as the highest output signal desired which is significantly larger than the input voltages (and usually the transistor bias voltages as well), so there is plenty of excess voltage to work with.

A regulator changes this impedance dynamically to actively keep the output relatively constant within limits, but you don't necessarily need regulation to have a voltage or current source. Changing the impedance dynamically allows you to reduce variation under load and in some cases also allows you to increase efficiency since you aren't always burning off so much energy through a high impedance (the aformentioned point of requiring an unusually powerful current or voltage source if you want to change it to the other kind).

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  • \$\begingroup\$ I decided to express my opinion (although it is not well accepted here) for your very good overview of the two ubiquitous circuit configurations - series and parallel (I have re-read it several times with pleasure). This duality is very beautiful and can be seen in many practical circuit implementations (I think you haven't shown the most typical ones). Just to add that for beginners it is very strange to "spoil" ideal sources by connecting additional resistors; it should be clarified why this is done... \$\endgroup\$ Commented Mar 22 at 12:42
  • \$\begingroup\$ ... It would be good to say that nowadays these configurations are implemented by elements with dynamic resistance which allows to make almost "ideal" sources with moderate resistances. \$\endgroup\$ Commented Mar 22 at 12:43
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    \$\begingroup\$ @Circuitfantasist Well I think that's more about regulation which is somewhat distinct from an actual current or voltage source. \$\endgroup\$
    – DKNguyen
    Commented Mar 22 at 18:03
  • \$\begingroup\$ DKNguyen, I agree. These are really the two basic regulation configurations. But when I think about it, it turns out that everything is regulation under different names - amplification, stabilization, etc. \$\endgroup\$ Commented Mar 22 at 18:10
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Basic idea

In electronics, it is accepted with "voltage source" and "current source" to call devices that have the property of keeping the voltage and current constant when the load and other disturbances vary. Usually they do that by varying their resistance. So, this kind of voltage and current sources are dynamic resistors that compensate for the disturbance by changing their resistance.

From this perspective, the definitions "Voltage source has low resistance" and "Current source has high resistance" are misleading in electronics; it is quite possible for the voltage source to have a high (static) resistance and the current source to have a low (static) resistance. It should be specified that they apply to varying resistance when defining the term "differential resistance".

The simple "resistor type" voltage and current sources with constant resistance can be used when the load resistance is accordingly too high and too low.

Implementation

The best way to understand how sources do it is to put ourselves in their shoes. For this purpose, let's represent them with a 5 kΩ variable resistor and start changing its resistance accordingly.

Voltage source

R1 = 5 kΩ, R2 = 5 kΩ: Let's connect two resistors (R1 and R2) in series and make one of them (R2) a constant 5 V voltage "source". This network is known as a voltage divider.

schematic

simulate this circuit – Schematic created using CircuitLab

R1 = 10 kΩ, R2 = 10 kΩ: If we double the R1 resistance, the current through and accordingly, the voltage across R2 will decrease. But R2 is a (constant-voltage) "dynamic resistor" that also increases two times its resistance so the divider ratio and accordingly, Vconst does not change.

schematic

simulate this circuit

R1 = 20 kΩ, R2 = 20 kΩ: If we repeat this experiment and double R1 once more, R2 will also double, and the voltage Vconst will remain the same.

schematic

simulate this circuit

True voltage source: In the circuit above, there is an actual source of power, but it is implied. If we want to show that there is a voltage source, we need to attach it to R2.

schematic

simulate this circuit

Current source

R1 = 5 kΩ, R2 = 5 kΩ: We can use the same arrangement of two resistors (R1 and R2) in series to make one of them (R2) a constant 1 mA current "source". R1 acts here as a load.

schematic

simulate this circuit

R1 = 7 kΩ, R2 = 3 kΩ: If we increase the R1 resistance with 2 kΩ, the total circuit resistance will increase and the current through R1 and R2 will decrease. But R2 is another type of "dynamic resistor" (constant-current) that decreases its resistance with the same 2 kΩ so the total resistance and accordingly, the current, does not change.

schematic

simulate this circuit

R1 = 9 kΩ, R2 = 1 kΩ: If we continue and increase the R1 resistance with more 2 kΩ, the total circuit resistance will increase more, and the current through R1 and R2 will decrease again. But R2 will decrease its resistance with new 2 kΩ so the total resistance and accordingly, the current, does not change.

schematic

simulate this circuit

True current source: As above, there is an actual source of power, but it is implied. If we want to show that there is a real current source, we need to attach it to R2.

schematic

simulate this circuit

Conclusions

  • Voltage source and current source have the property of keeping the voltage and current constant when the load and other disturbances vary; hence the name "constant source".

  • They can do this in two ways - by changing the resistance or by changing the voltage.

  • The first is more common; then they are dynamic resistors.

  • By changing its resistance, the dynamic resistor compensates for the disturbance.

  • From this viewpoint, the voltage source is a constant-voltage dynamic resistor, and the current source is a constant-current dynamic resistor.

  • These sources can be variable (in the sense that we can adjust their voltage or current), and at the same time they are constant (in the sense that their voltage or current is not affected by the load).

  • The humble static "resistor type" voltage and current sources make sense to use when the load resistance is accordingly too high and too low compared to the source resistance.


See also another related answer of mine.

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voltage is the potential "drop" between two points in a circuit.

Well it can also be interpreted as a potential increase depending on how it is measured. \$V_{ba}=-V_{ab}\$

Do voltage sources output an unknown current, and vice versa?

Voltages sources output voltage. Current sources output current. When the load is applied to the source then charge flows, or voltage appears.

An "unloaded" voltage source has an infinite load resistance applied to it. So by Ohm's law, current is \$0A\$. There is current only if the resistance is finite. The load (current) increases as the resistance decreases.

An "unloaded" current source has a load resistance of \$0 \Omega\$ applied to it. So by Ohm's law, voltage across the current source is \$0V\$. There is voltage across it only if the resistance is greater than zero. The load (current) increases as the resistance decreases.

Why aren't they the same, if V=IR when there is no resistance?

Sorry but I don't know what this means. Be careful with the phrase "no resistance". Some mean it to be zero ohms, others use it to refer to a resistor that has been removed from the circuit meaning an infinite resistance. Two very different meanings. I can't understand the they reference.

It's intuitive (for me) that a current supply would provide a flow of electrons around a circuit, giving each element a charge and voltage

Current Source

Current will not happen without a closed path. So the elements are connected with wires to the current source at one place in the loop. The electrons (charge) already exists at all points along the path: in the copper wires and in the electronic elements. Copper for example has one electron per atom that is free to move. The current source does not provide this charge. It comes with the wire.

To personify, the job of a current source is to suck electrons into its positive terminal, and push electrons out of its negative terminal. It does not care about the voltage unless it is too high, called the voltage compliance limit. This voltage is not created internal to the source, but is a reaction voltage by the external elements (not necessarily in the sense of reactive elements, but including them) that apply the voltage to the source.

Voltage source

Voltage does not require a closed circuit to exist. It comes as a field surrounding every charged particle, negative or positive. It also occurs in the region of a changing magnetic field. This effect is used in all the ac power generation throughout the world.

To personify, the task of a voltage source is to make one terminal more positive than the other. It does not care how much current it must supply, up to a limit.

But how can a voltage source provide a constant output of a "drop" around a circuit?

It doesn't. The voltage source knows nothing of the external circuit. Applying a voltage source to a string of external series elements, so that there is a closed path, allows the voltage to be distributed along the path as an electric field. This happens at the speed of light in the materials. The electric field is a voltage gradient. and since voltage is an energy, charged particles within the field accelerate resulting in electric current. If the resistance is zero along the path, then the current will continue to increase without bound. If there is a resistance then a reaction voltage will develop until the applied voltage from the source equals the sum of the reaction voltages from the elements.

So the voltages in the external circuit are created in place, by the current through the reacting material. As well, the current "demanded' by the external circuit is what the source must supply.

If the sources have an internal resistance then aspects of voltage source and current source occur at the same time. Whether to consider as a current source or voltage source depends on the external circuit. For example a voltage source in series with a resistor can be analyzed as a current source with a parallel resistor under restricted loading.

Note: The terms reacting and reaction are used in the sense of responding to an action, not electronic reactance.

The roots of this discussion lay in Newton's action-reaction and "the Principle of Locality".

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    \$\begingroup\$ RussellH, Very interesting and originally written answer... I have only a few remarks: 1) Regardless of the measured polarity, the essential thing about the voltage drop is that it is subtracted from the source voltage (that it is a loss). 2) It is better to say that in a current source, the load (voltage) increases as the resistance increases. 3) In a true current source supplying resistor load, the voltage is created by the very source, as a “reaction” to the external load resistance. 4) The configuration “current source with a parallel resistor” is not so useful and rarely used. \$\endgroup\$ Commented Apr 1 at 15:52
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    \$\begingroup\$ Not really so original @Circuitfantasist. The roots lay in Newton's action-reaction and the Principle of Locality.Remark 1) Certainly there is power loss, but the voltage drop still exists. There is no voltage loss. Otherwise it could not be measured. Without the reaction voltage across the resistor, the current would increase forever. Remark 2) It is an alternate way, but better depends on circumstance. Remark 3) The source has no knowledge of the resistor, only the voltage and current at its terminals.(continued) \$\endgroup\$
    – RussellH
    Commented Apr 1 at 20:17
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    \$\begingroup\$ @Circuitfantasist The resistor could be a kilometer away behind all sorts of circuitry. As long as the source and resistor are in series, the voltage across the resistor is determined by the resistance only in response to the current through it. So yes, a voltage across the current source appears as a reaction to the voltage created in the external circuit. Remark 4) The entire sentence is:"...a voltage source in series with a resistor can be analyzed as a current source with a parallel resistor...". This "source transformation is used analytically and for me is very useful depending on circum \$\endgroup\$
    – RussellH
    Commented Apr 1 at 20:35
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    \$\begingroup\$ @Circuitfantasist: Thank you for your thoughtful comment. \$\endgroup\$
    – RussellH
    Commented Apr 1 at 20:37
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    \$\begingroup\$ RussellH, Thanks for the comprehensive answer! Just to clarify that by "voltage loss" I mean that the voltage drop is something that is subtracted from the input voltage and the rest is applied to the load. Obviously, the latter is less than the input with this "loss". Thus we apply the input electromotive voltage to the entire circuit, part of it is "lost" in the resistance and the rest reaches the load. Sounds a bit textbook indeed :-) Cheers! \$\endgroup\$ Commented Apr 2 at 6:13
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V=IR only applies to loads and devices that are ohmic(i.e. a resistor). Current and voltage sources are non-ohmic, and do not follow Ohms' Law.

Voltage sources source a constant voltage, no matter how much current is being drawn by the load(hence the name voltage source). Current sources are the opposite; they source a constant current no matter how much voltage is at its input, and how much current the load wants to draw.

In reality, perfect voltage and current sources are impossible(an ideal voltage/current source works up to infinite voltages and currents). Voltage sources will always have a maximum amount of current that can be passed through before the voltage begins to droop, and current sources will always have a maximum amount of voltage at its input.

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  • \$\begingroup\$ You ,ay (or may not) wish to change "would be extremely high" to eg "would need to be infinite". Nit picking only. \$\endgroup\$
    – Russell McMahon
    Commented Mar 20 at 3:53
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It's about their output impedance. These sources are pretty much the same thing, but we call low impedance sources "voltage sources" and we call high impedance ones "current sources".

Example

Take a 12 V car battery. By itself, it is a reasonably good voltage source, because its internal series impedance is low for most purposes. Its output voltage would be always about 12 V, regardless of the load.

Now add a 1 MΩ resistor in series and suddenly it is a reasonably good current source, which will output a current of 1.2 µA through most loads. But it's output voltage would now be highly load dependent.

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Current source outputs a flux of charge. \$1.6 × 10^{19}\$ electrons per second = 1 Amp. Through your wire, lets say. (Or through some arbitrary cross-section area you define, generally).

That's it. Easy.

If you could somehow build a current source with a single terminal, there wouldn't be any mathematical problems. But conservation-of-charge would stop you.


A voltage source outputs energy per unit charge, onto whatever amount of charge is going through it. 1 Joule for each 1 Coulomb = 1 Volt.

As an abstract thing, the voltage source makes more sense with two terminals. Because potential is always relative to something else.

Also, with a voltage source, it sortof begs the question of what the heck is energy, and potential, and how do force fields define them, and physics and math which are ultimately tangential to circuit design 101. So IMO voltage is a more elaborate concept.


Next, Which came first, the chicken or the egg?

The egg comes first, obviously. That's charge. But it wouldn't move - you'd get no current flowing through your circuit - without having a potential.

Either a current-source and a voltage-source, could be used to generate that potential. In reality, the current-source is also giving you voltage; the voltage-source is also giving you current.

It's pretty arbitrary which you pick as causing the other. Voltage-first is typically easier to think about, but there are definitely exceptions to that. Stick a resistor on it, and it becomes a mix of the two anyway.


How does a voltage source work?

You need a nonlinear circuit element.

For example, you could build a voltage source from a crappy (poorly defined) current source, plus a nonlinear element like a diode. The VI curve of the diode would result in a roughly constant voltage across the diode, and those two terminals could become your output terminals giving roughly constant voltage. If that's not good enough, you can make fancier semiconductor devices, whose VI curve is flatter. If you haven't unlocked semiconductor diodes yet, you can use a vacuum tube - but the concept is similar.

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Negative, Voltage is the potential to push current through the load. Having voltage across something just means there is potential for current to flow.

When a power supply used as a voltage source is powered on without a load, it will use energy to charge up the output filter capacitors and remain in regulation. The only energy used are losses. If this was a perfect system. The regulators would stop consuming energy the moment the capacitors were charged to the regulated voltage.

When a power supply used as a current source is powered on without a load, it will use energy to charge up the output filter capacitors to the maximum output voltage that was set when the supply was configured (or which is specified in the datasheet, it might be the input voltage!). It operates much like the voltage source does with no load attached. Only leakage currents while maintaining the voltage regulation are consuming energy.

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Do voltage sources output an unknown current, and vice versa? Why aren't they the same, if V=IR when there is no resistance?

Your math is broken. When there is no resistance, R=0, not R=1. That resolves to V=0 while I can take arbitrary values. Which means a shortcircuit will make any current source happy while it will not cooperate with any voltage source other than a 0V voltage source.

In a similar vein, an infinite resistance (open circuit) will be perfectly acceptable for any voltage source while it will only be compatible with a 0A current source.

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