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I've always wondered why is it valid to assume some voltage drop (different from the junction voltage drop) on diodes or LEDs from the power up of a circuit. It simplifies the analysis, but it's unclear to me why it is a valid approach.

For example, in the question Capacitors charging in a DC circuit, the assumption is a drop of \$2V\$ across the LED.

What is the explanation?

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It's a valid approach just for general cases simply because most LEDs happen to have a forward voltage drop near that value.

Here is a table take from wikipedia on the voltage-drop of different LEDs:

Voltage drop

You can see that for some reds, oranges, yellows, and some greens, 2V is close to the value. I suppose if you were making a circuit with ultraviolet LEDs you could assume 3.5, or 4V. Now if you actually know the forward voltage drop on the LEDs you are using, it would make a lot more sense to use that value.

Going off of what echad said, the constant voltage drop model is the simplest one, and speeds up analysis. In reality, voltage drop on diodes have an exponential relationship.

Also, there are several different models for analyzing circuits that contain diodes.

Taken from a textbook I use at school, Microelectronic Circuits 6th Ed, by Sedra and Smith:

Graphical Analysis of the Exponential Model, using a load line

Graph Analysis Exponential Model

Constant Voltage Drop Model

Constant Voltage Drop Model

Now this is for plain silicon diodes, but the same math holds true for all diodes, just the parameters are slightly different and the drop for LEDs comes out different based on how they are manufactured.

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Diodes have very steep output characteristics - there are almost no current below some voltage value, and after this voltage is reached the current increase is very steep:

enter image description here

This steep increase in current at \$V_d\$ essentially means that no matter what current you'll force through the diode, the voltage will not vary much from \$V_d\$.

You can argue that for very large currents the voltage can increase by a significant amount, but in practice this will not happen. What will happen is that when the current increases over a nominal diode's current, the diode will be permanently damaged (short-circuit, open-circuit or some other bad effect).

Due to the above reasons, if the diode is present and is conducting - it will have a voltage drop of magnitude \$\approx V_d\$ across it.

Now, you may ask: "But how do we know a priori that the diode is conducting?". Good question. We really can't know this a priori, but we can ASSUME that this is the case and perform an analysis of the circuit. After the analysis is done we must ensure that all the voltages and the currents in our circuit are consistent. If they are - our assumption was correct. If, on the other hand, we find that the voltage drop (as follows from the analysis) on the diode is below its \$V_d\$, then it is a contradiction and our assumption was wrong - the diode is in cut-off region of operation and behaves as an open circuit.

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There are several ways to model the diode forward characterstics, one of the simplest forms is the Constant Voltage Drop Model

Other than that, there's also

  • The Exponential Model
  • Piecewise-Linear Model

What makes the constant-voltage-drop model useful is it allows speeding up the analysis of circuits. However you are exchanging quality for time. The exponential model gets you the most accurate answer out of the these three models.

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  • \$\begingroup\$ So basically, if by simple observation of the circuit I conclude it's on forward mode I may use any of those models depending on how accurate I need to be, right? What about the transient analysis? \$\endgroup\$ – user2553780 Jul 30 '13 at 13:04
  • \$\begingroup\$ Yes. But with transient and steady-state analysis you're getting into control system analysis, and modeling transfer functions \$\endgroup\$ – Iancovici Jul 30 '13 at 13:05
  • \$\begingroup\$ For a transient analysis, you're best of using a circuit simulator with an accurate model of the diode you're using. For an LED, though, whose purpose is to generate light, unless you're talking about something like optoisolation there's eyes and a brain somewhere in the loop, with delays of tens to hundreds of milliseconds, and you generally don't care. \$\endgroup\$ – Scott Seidman Jul 30 '13 at 14:03
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It's simply a rule of thumb. Whether the approximation is good enough or not is a function of the question you're asking, why you're asking it, and how accurate do you need the answer to be. For example, often when approximating for LED's, its because you're trying to limit the current so you don't damage the LED, and you're almost shooting for an order of magnitude estimate (well, a tad better than that).

Indeed, if you need better, you can parametrize the diodes I-V curve, or do a load line, but these aren't necessarily numbers that you know very precisely (sort of like beta for a transistor, so good electronic designs work even though such parameters can vary substantially.

As others have pointed out, the rule-of-thumb voltage drop differs for different colors.

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It is a valid assumption because of how, or more precisely what, LEDs are composed of. From decades of manufacturing LEDs in a common way, with a common material that we have full understanding of, simplifications and assumptions of how they work are good enough for most Practical use.

Your average Red led has a 1.8V to 2.0V drop, based on the semiconductor material used to construct it. Blues tend to be 3.2V.

This is the same reason Ohm's Law is used, even though it doesn't take into account Johnson–Nyquist noise or Temperature. This is why Newtonian Physics are still used, even though they are slightly flawed and are largely superseded by Relativistic and Quantum Physics.

For most practical, simple uses, approximations are fine. Using a diode or led with a 1.8v drop when you calculate for 2.0v drop won't affect most circuits, unless that 0.2v difference is so important. Same reason me pouring in 2.2oz of butter for my pancake mix instead of exactly 2.0oz won't turn my pancakes into mush. It's close enough that the slight variation won't matter.

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It's a valid assumption because it's close enough to true that the difference is negligible in many cases. See for example this graph from LTL-307EE:

led voltage vs current

Notice that the voltage axis starts at \$1.2V\$, and over the typical operating range of the diode, the voltage only varies about \$0.6V\$. This is mostly due to the internal resistance of the diode, which in this case is about \$13\Omega\$. If you are going to put this diode in series with a \$4.7k\Omega \pm 1\%\$ resistor, then the \$13\Omega\$ resistance of the LED is quite insignificant compared to resistor you've added, which might deviate \$\pm47\Omega\$ from the nominal value of \$4700\Omega\$. Put another way, that \$13\Omega\$ of resistance from the LED represents a \$0.28\%\$ error in your calculations.

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