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
Please help me to understand this.
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:
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.
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.
