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I often see the same method for computing voltage distribution in electronic circuits implying components whose V-I curves are not trivial.

For example, in a circuit implying a resistor and a LED having 2V / 20mA caracteristics, we substract 2V from the supply voltage to determine the remaining resistor's voltage, and we assume current in the whole circuit is 20mA to determine resistor's value using Ohm's law. But what about a direct computation? If the resistor's value is too low, the LED voltage would be higher than this ; on the other way, if the resistor's value is too high, the LED voltage will be lower.

Is there a standard way to compute voltage distribution directly, based on chosen components (i.e. the inverse of the method that determines other components)? And more generally, is the voltage distribution always unique?

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  • \$\begingroup\$ yes you could develop a mathematical model based on the exponential model,en.wikipedia.org/wiki/Diode_modelling But there are varying parameters one LED to another. If you want to force something to an operating point, there are other methods in electronics to do that. What you asking? \$\endgroup\$ – Standard Sandun Mar 23 '13 at 11:56
  • \$\begingroup\$ What is "voltage repartition"? I thought you might mean repitition, but that doesn't make any sense. \$\endgroup\$ – Olin Lathrop Mar 23 '13 at 13:26
  • \$\begingroup\$ I wonder if the OP means repitition as in repeating a solution until it merges on a answer that is more accurate? Like successive approximation? \$\endgroup\$ – Andy aka Mar 23 '13 at 13:30
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    \$\begingroup\$ Or you draw the resitance in the V-I graph (I=0 to V=0) in the datasheet and see where it crosses the diode's curve. But in practice it'll come down to the same values. Regular resistors values are separated at a 1.2'ish factor each ( 10, 12, 15, 18 ) so you have to round up or down anyway as your calculation only needs 10% accuracy. \$\endgroup\$ – jippie Mar 23 '13 at 13:54
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    \$\begingroup\$ Seems the repartition term was inadequate. I replaced it with distribution, I hope it makes more sense :) \$\endgroup\$ – lledr Mar 23 '13 at 14:07
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Yes there is a general method, but its generally not useful.

If you go back to first principles, there is a equation for each component relating voltage and current. Let's simplify the problem to where one is a direct function of the other, like with your LED and resistor example. Put another way, there is only one degree of freedom to decide what each component is doing.

The problem then is a set of simultaneous equations. The general answer is to solve these simultaneous equations. Usually this is rather complicated and the extra level of accuracy is not necessary and even misleading because it exceeds the tolerance to which other values can be known.

In your specific LED and resistor example, you could solve this simple set of simultaneous equations visually with a graph. Plot the resistor and LED current separately, each as a function of the voltage of the node between them. You know one additional constraint, which is that the current thru both is the same. Visually, this means you look for the place the two curves intersect, because that is the point at which both have the same current.

Hopefully you can see that it gets a lot more complicated for anything but a simple contrived example.

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  • \$\begingroup\$ Great. But what if the intersection (for some components) gives multiple matches? Is it impossible? If not, how to interpret it? \$\endgroup\$ – lledr Mar 23 '13 at 17:34
  • \$\begingroup\$ @ysomane: If there are multiple intersection points, then first it means the components aren't monotonic. However, at a higher level it means there are multiple solutions. If those multiple points are stable solutions (not a given), then it means the circuit could settle at any of those points. If there are two stable points, then it can be called a "bi-stable" circuit, which is basically one that can be made to switch (due to some external influence) between two states it can otherwise hold indefinitely. Two transistors arranged as a flip-flop is one such example. \$\endgroup\$ – Olin Lathrop Mar 23 '13 at 18:24
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Since an LED is a current driven device, if you are using a resistor you won't ever get a very predictable or accurate current draw (variations in process, operating temperature, resistor tolerance, supply variation, etc).
This doesn't matter though as in most cases a +-10% variation won't be a problem.

If you look in the datasheet for any decent LED you will notice the Vf is given at a nominal current (e.g. 20mA) and (as Jippie mentions) there will also be an I-V curve you can use to graphically predict the Vf at other currents.

So you could develop a mathematical model and try and calculate the resistor value this way, but I wouldn't bother for two main reasons:

  • There are already SPICE models that will do this for you, created by the manufacturers themselves so likely to be more accurate than anything you can come up with easily.

  • If you want to drive an LED precisely, then use a constant current source. There are hundreds of cheap and accurate driver ICs out there (or an opamp with precision reference can be used) For a random example, the SP7618 has a typical regulation tolerance of 1.5% and a maximum tolerance of 6% from -40°C to +85°C.

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