# How to determine voltage across LEDs in a circuit?

I simulated a basic circuit in Orcad but I don't know how it found voltages(6.08 V and 3.04 V) on the nodes? How do I calculate those voltages mathematically? I am attaching three image files of schematics and values.

NOTE: Current value in circuit is 23.92mA according to OrCAD calculations.

This is schematic diagram with values

These are LED values

These is the voltage across both LEDs

• Are you interested in how Spice itself does the calculation? Or interested in how it can be approached directly in mathematics? (Spice uses a numerical approach using successive linearizations and matrices. A mathematician would take the quantitative models and solve for a closed equation.) I can answer either, but you need to specify what you want. – jonk Jan 10 '18 at 2:04
• @jonk I am interested in the mathematical approach that we have to do by ourselves to get the values that have been shown by OrCAD in the circuit diagram. Thank You. – Rajan Jan 10 '18 at 2:29
• Your question provides only the simplest possible model for the LED: $V_D=2.96\:\text{V}+2\:\Omega\cdot I_D$ or else $V_D=3\:\text{V}+2\:\Omega\cdot I_D$ depending on how I interpret what you provided. Trivial algebra, either way, but of very limited use. A more general solution would use the Shockley equation of the diode, but it also requires model parameters that cannot be derived from the information you provide. And I don't know which was used by your simulator. Can you add anything here before I try and answer? – jonk Jan 10 '18 at 2:40
• yes. I got the answer but I still not get how the equation formed? I wanted to know about the basic one that you formed! thanks. – Rajan Jan 10 '18 at 3:31

The simplest equation model of an LED is of the following form:

$$V_D=V_{FWD}+R_{ON}\cdot I_D$$

Where $V_{FWD}$ and $R_{ON}$ are what is called "model parameters" that define the specific behavior of a specific diode. You need to provide those in order to get quantitative answers.

The above model is based upon the following idea. Suppose you take three measurements with the LED, getting both the voltage across the LED as well as the current through the LED, and place them onto a chart:

If you then use a ruler and draw a line through the points, you may find that it intersects the y-axis at some value. In the above case, it intersects right at $V_{FWD}=2.96\:\text{V}$. This is the meaning of $V_{FWD}$ in the above model. It's a projected value based upon an extrapolation from a few actual points made elsewhere on the chart. Similarly, the slope of the line, in this case with $R_{ON}=2\:\Omega$, provides the other model parameter. (You can compute these from a collection of raw data points by using a "least squares fitting" algorithm for lines, if needed.)

A circuit with a power supply of $V=30\:\text{V}$, a resistor of $R=1\:\text{k}\Omega$, and two added LEDs (all of these in series), yields the following result (the $2$ comes from the fact that there are two LEDs in series):

$$30\:\text{V} - I_D\cdot R - 2\cdot \left(V_{FWD}+R_{ON}\cdot I_D\right) = 0\:\text{V}$$

That equation is easily solved for $I_D$:

$$I_D=\frac{30\:\text{V} - 2\cdot V_{FWD}}{R+2\cdot R_{ON}}$$

(You should be able to solve and find the above equation.)

Using your table of LED information, I may get either of these two models and their predictions:

1. $V_{FWD}=2.96\:\textrm{V}$ and $R_{ON}=2\:\Omega$, $\therefore I_D=\frac{30\:\text{V} - 2\cdot 2.96\:\textrm{V}}{1\:\text{k}\Omega+2\cdot 2\:\Omega}\approx 23.98\:\text{mA}$
2. $V_{FWD}=3.00\:\textrm{V}$ and $R_{ON}=2\:\Omega$, $\therefore I_D=\frac{30\:\text{V} - 2\cdot 3.00\:\textrm{V}}{1\:\text{k}\Omega+2\cdot 2\:\Omega}\approx 23.90\:\text{mA}$

Neither of those are an exact match with the value you provided. But they are very close. The fact that they are not exact suggests that either I didn't interpret the table you provided correctly, or else a different type of LED model was applied in simulation.

The problem with the above model is that it assumes that the LED behaves linearly, but LEDs don't actually behave that way. And the above model works over only a very small range of actual currents nearby the "standard value." If you stray very far, then the entire model breaks down and the LED no longer matches the expectations.

An improved model of the LED looks like this:

$$V_D=n\cdot V_T\cdot\operatorname{ln}\left(\frac{I_D}{I_{SAT}}+1\right)$$

This is one of the two equivalent ways of writing the widely known Shockley equation, which provides a much better model of the LED over a much wider range of diode currents. (It still fails to take into account things like lead frame, wire bond, and wire resistances. Adding them is easy, though it complicates the solution a little.)

Here, $V_T$ is the thermal temperature and is computed as $V_T=\frac{k\cdot T}{q}$, where $k$ is the Boltzmann constant, $q$ is the charge of an electron, and $T$ is the absolute temperature. At room temperatures, $V_T\approx 26\:\text{mV}$. This is a physics parameter and isn't strictly speaking a model parameter.

$I_{SAT}$ is the saturation current. This is an extrapolated value for the LED (again) where an estimate is made about an axis intercept (as it cannot be directly measured.) This is an LED model parameter.

$n$ is the emission coefficient and is greater than or equal to 1. With LEDs, it will often by closer to 5 or 8 or even more. This is also an LED model parameter.

With this new model, the new equation to solve is:

$$V_{CC} - I_D\cdot R - 2\cdot \left[n\cdot V_T\cdot\operatorname{ln}\left(\frac{I_D}{I_{SAT}}+1\right)\right] = 0\:\text{V}$$

The solution for finding the LED current is:

$$I_D = \frac{2\cdot n\cdot V_T}{R}\cdot\operatorname{LambertW}\left[\frac{I_S\cdot R}{2\cdot n\cdot V_T}\cdot e^{\cfrac{V_{CC}+I_{SAT}\cdot R}{2\cdot n\cdot V_T}}\right]-I_{SAT}$$

(I have provided a methodology towards solving these kinds of equations. Please feel free to examine similar [not the same, though] steps that can be applied to reach the above solution on your own. See: Differential and Multistage Amplifiers(BJT).)

Suppose (and this is just pure supposition since we don't have actual model parameters) that the model parameters are $I_{SAT}=6\:\text{fA}$ and $n=4$. Then at room temperatures we'd find:

$$I_D\approx 23.96\:\text{mA}$$

This isn't necessarily supposed to be close to the values your simulator gave. However, I picked out model parameter values that were designed to get somewhere in the vicinity. That said, remember that Spice uses numerical methods, not closed mathematical solutions. And also remember that I simply don't have any good idea about what model parameter values I should have used.

But it at least provides an idea about the mathematical process involved.

Spice sets up a matrix and uses some internal knowledge about diode models to help it "linearize" the Shockley diode equation around the "current operating point" of the diode. So while Spice does use the Shockley equation, its approach is quite different from the mathematical solution using it that I just gave. Instead, Spice uses a series of tiny time steps together with linear solution methods and incremental adjustments to the linearized models to successively reach an answer.

How do I calculate those voltages mathematically?

To start, research Shockley's diode equation (a.k.a., the "diode law equation" or just "diode equation") and also the "reverse saturation current" equation (a.k.a., "reverse leakage" current). These two equations (among others) are commonly used to model the voltage across, and currents flowing through, a diode. Here's the Shockley equation (without explanation):

$$I_D = I_S \left ( e^{(V_D/\eta V_T)}-1 \right )$$

By the way, the reverse saturation current term $I_{S}$ changes with temperature $T$, and $|\delta I_S/\delta T| > |\delta ()/\delta T|$ where '()' is the full parenthetical term that is multiplied by $I_S$ in the above equation. Therefore, for a given change in diode temperature $\delta T$, the diode's forward current $I_D$ is more sensitive to changes in $I_S$ than it is to changes in the parenthetical term. (Hint: In a real diode, $I_D$ increases as the diode's temperature increases.)

To learn about the mathematical models used in SPICE simulations that include diode parts (including OrCAD PSpice), perform an Internet search using keywords like "spice equation for diode".

Consider not calculating the LED voltage drops. Instead get the forward bias LED voltage drops and current requirements from the LED specifications.

Only then pick the appropriate series resistance based on the needed voltage drop from the power source to the desired voltage at the LED using the specified LED current.

If you put 2 LED in series they should be of the same type (have the same voltage and current specifications). Otherwise one LED may light up the other may not.

All LEDs are diodes with well defined Vf @ If specs.

LEDs, blue and white just happen to be the same diode curve which tend to be 3 to 3.1 Vnom @20 mA. They are not 5V LEDs but 5V is better than 30V to drive the series R chosen to limit current.

• Hi. Can you please explain me about the voltage calculation at the nodes (about 3.04 or )? Can we anyhow calculate it by applying basic ohm's law and other formulas? Thanks for your answer. – Rajan Jan 10 '18 at 2:10
• If you read all my answers on LED ESR Vf and calculations and references to datasheets by using user search engine then I wont need to be repeating myself. Ok? – Tony Stewart Sunnyskyguy EE75 Jan 10 '18 at 3:42

LEDs are current driven devices with a loosely fixed avalanche voltage, some what like a zener diode. That is to say that most of todays LEDs come ON dim at about 2.9 VDC, and are at full brightness by the time the current pushes the voltage up to 3.5 VDC.

Your reading of 23 mA is actually too high for small LEDs, including the 5 mm size. Some 3 mm LED's are specified for normal operation at just 2 mA. As st2000 stated in his answer it is important that all LEDs in a given string are of the same type, color, forward voltage and nominal operating current.

I normally do not drive 5 mm LEDs with more than 15 mA, and they can be blinding bright if they are the high-efficiency type. But they have a narrow beam of about 6 degrees theta, so they are good for spot lights and flashlights only.

The LEDs used in opto-couplers are normally infra-red with a forward voltage of 1.05 VDC. High powered LEDs used on street lights (10 mm rnd/11 mm sq) are epoxied to a metal plate to keep them from overheating, as they are rated as high as 5 watts. These are in the form of LED clusters and the forward voltage maybe 6 to 7 VDC at 500 mA or more.

An LEDs worst enemy in circuit is heat. At room temperature 25 Degrees C or 78 degrees F, they can last for 100's of years. Their MTBF rate is difficult to calculate based on room temperature alone. Many LEDs made today will out-live us.

As for operating parameters, I go implicitly by the data sheets. I have Orcad as well and have little trust in its parameter tables.

• Hi. Thanks for the answer but I am just looking from the math point of view. I know the Vdc applied is very high and so as current but the point here is how did OrCAD come up with the voltage values? I want to know that by some mathematical calculations. – Rajan Jan 10 '18 at 2:06
• I have Orcad as well. Why 30 VDC with only a 1 K resistor? What was your source of parameters? Their are the data sheets, and then there is guessing. Not much in-between. Treat LED's as zener diodes that loosely regulate the forward voltage. – Sparky256 Jan 10 '18 at 2:10
• Ok. The specification of LED which I have attached as the second image is the data and there is no strong reason to use the 1k resistor but just messing with OrCAD and trying to simulate circuits and I stumbled across this question in my mind that how OrCAD determined the value of node voltage (like 3.04V)? Then, I tried it to accomplish with some math but failed because maybe I don't know which equation to use! – Rajan Jan 10 '18 at 2:22
• No single equation would cover all LEDs. Thats why the manufacture publish elaborate data sheets which give you brightness/forward volts/ current graphs. Orcad is using whatever look-up table it has. – Sparky256 Jan 10 '18 at 2:25
1. the number of different number of LED lamps can not be parallel, the number of the same can be.
2. the voltage of the LED is multiplied by the number of series = drive voltage.
3. the driving current for each series of LED should be within 20mA.
4. if the LED light string is parallel, the drive current =20ma is multiplied by the number of parallel connections. Considering the safety of LED, the current is better than the calculated current.
5. it is also necessary to consider the ventilation and heat dissipation of LED.
• Remove the irrelevant spam link, please. – Nick Alexeev Jan 10 '18 at 1:38
• Please learn how to create paragraphs. Hit the Enter button twice to create a gap to the next paragraph. If using numbers then 1), 2) etc is the usual format. – Sparky256 Jan 10 '18 at 1:48
• Also, please improve your grammar and spelling. A verbal lashing for poor grammar and spelling is common around here. – Sparky256 Jan 10 '18 at 1:52