Do the statements "circuits/components will only draw the current they need", and "V = IR" contradict? [duplicate]

Question

What I'd like to know is how we can resolve the two statements, "Circuits/linear components will only draw what they need", and "V = IR". They seem to contradict one another, in my view. With a given voltage and a given resistance, current will be set by these two properties in the circuit, regardless of "the current components need". If this could be at all explained mathematically as well, I also think that'd be better... these analogies seem limited after a certain depth...

Answer 1

It is incorrect to say "components only draw what they need to from the circuit." That statement appears to relate to questions like: "I have a power supply that is rated 2 amps and I wast to connect it to a device that only needs 1 amp. Will the power supply give my device too much current?" The answer should be something like: "The power supply determines only the voltage supplied and the current available. The characteristics of the connected device determine how much current it will draw from the power supply." Often, the characteristics of the connected device are well described by ohms law. However there are many devices that have characteristics that are not well described by ohms law. Never-the-less, it is still the characteristics of the device that determine the current in most cases. If the supply has a major influence on the current drawn, there is likely a problem with the connected device or the power supply is the wrong voltage.

Of course, there is such a thing as a current-source power supply. In that case, the power supply delivers whatever voltage is required to force a set current through the load.

Answer 2

draw what they need to from the circuit

I don't think this statement refers to single components.Put too much current through the LED it will be damaged or destroyed.If you put too much voltage across it,the same thing can happen.

An assembly or a system of components,or,to be more specific,the entire circuit of a device like a mobile phone or a laptop has the ability to protect itself,more than a single component has(for example:a transistor).It can shut down charging if it doesn't like the input you give it.Given a correct voltage,it will indeed draw the current it needs.

However,note that no device is invincible:at some point,with too big or too high current or voltage you can damage or destroy them completely.Let's say you have a step up converter which has no feedback from its output and you want charge your phone with it.If the phone doesn't accept it and interrupts charging,the output voltage may jump to a high value and burn it.

All in all,you are right.It can sound ambiguous.

Answer 3

components only "draw what they need to from the circuit"

For simple linear components that is true at the component level.

A better statement would be:

A circuit only draws what it needs from the power supply.

An LED by itself will try and draw what it thinks it needs from the supply and, if that supply voltage is above the forward voltage of the LED then the LED thinks it needs to draw an infinite amount (for an "ideal" LED) of current from the supply. We know better, though, so we include a resistor that will only want a much smaller current, and since the same current flows through all components in series the LED gets its current supply throttled to a level where it is doesn't blow up.

An LED is a non-linear device in that the resistance is not a fixed value, it changes depending on the voltage being applied to it - the higher the voltage the lower the resistance. Get the voltage too high and the resistance will fall too low so the current drawn will be too much for the poor little LED to handle, and it will blow its top (quite literally in some cases).

Answer 4

A properly designed circuit draws what it needs from the power supply. Individual, basic parts used in a design don't necessarily do that. (If they are properly designed circuits in their own right, such as an IC might be, then perhaps they do draw what they need. But those are systems and they are designed, then.) Passive components such as resistors, capacitors, and inductors, are not properly designed circuits. They are circuit elements that may be used to make a properly designed circuit.

An LED is a circuit element, not a properly designed circuit. (Well, usually they aren't, though some may actually come with an IC designed for some purpose built inside them.) It's more complicated, mathematically, than a resistor. But not that much more complex. A diode as a reversed bias behavior and a forward biased behavior, which are different. So right away you have a polarity issue that makes it a little more complex than a resistor.

Beyond that, the forward biased behavior of an LED is itself somewhat complicated. You can take a very simple view for the forward biased case, and just model it as \$V_I \approx V_{fwd} + I \cdot R_{on}\$, where \$R_{on}\$ and \$V_{fwd}\$ are device parameters determined by the specific part you've chosen and \$I\$ is the current through it and \$V_I\$ is the voltage drop across it at that current. It's a simple function. But it is more complicated than the one for a resistor. And besides, it's only good for a relatively small range of currents. And... well, temperature matters a lot to an LED, as well. So it gets even more complicated when you require the LED temperature to be included.

A slightly more complex model of the LED would be something like this: \$V_I = \frac{n\cdot k\cdot T}{q}\cdot\ln(\frac{I}{I_s})+I\cdot R_s\$. \$n\$ is the emission co-efficient, \$k\$ is Boltzmann's constant, \$T\$ is the temperature (usually in Kelvin), \$q\$ is the charge of an electron (in Coulombs), \$R_s\$ is the Ohmic resistance present in the device leads and contacts, and \$I_s\$ is the intrinsic saturation current of the device. Using that, we get the resistance of the LED as something like: \$R = \frac{d V_I}{d I} = \frac{n\cdot k\cdot T}{q\cdot I}+R_s\$. Note here that even this is wrong to use when including temperature, because \$I_s\$ itself is dependent on temperature, too, as a cubed function, and works oppositely. (The full equation wouldn't be possible to even post here -- I've done the calculus before and it's horribly complex.)

And that's just a simple diode/LED.

It's not Ohmic because it doesn't act like a resistor over a broad range of currents and voltages. It's resistance depends on the current flowing through it. That by itself is enough to complicate things a great deal. (If you take the calculus point of view and look at resistance as \$\frac{d V_x}{d I_x}\$, where \$x\$ is a vector of control parameters, then perhaps everything is Ohmic at this infinitesimal perspective. But then the term loses its meaning, too.)

Adding a resistor in series to an LED helps because the voltage across the LED doesn't vary that much based on the current. If you changed the voltage across an LED upward by as little as a 100mV or so, a simple mistake to make, you might see the current in it increase by a factor of 10! But a resistor? If you increased the current by a factor of 10, it would require 10 times the voltage to do that. So the series resistor limits the current in the LED because its voltage drop responds so much more strongly to current increases (or decreases.) That helps to make the current in the LED more predictable, in a way that is as serious of a matter that it means the difference between working and not working. And a resistor is CHEAP, too. Making a circuit work well and doing that cheaply is a good thing, indeed.