diodesparallelresistorsseries
how do you calculate V\$_{R3}\$ in the circuit below?
V\$_{in}\$, all the resistor values and the type of the diode are known.
By simulating this circuit one can see that V\$_{D}\$
is increasing with higher values for R\$_{3}\$. How is that even possible?
I would not use one resistor for all LED branches. Use one resistor for each LED branch.
simulate this circuit – Schematic created using CircuitLab
Uled=3V
Iled=300mA
Uledbranch=9V
Iledbranch=300mA
Vcc=12V
R= 12-9 / 3e-1 = 10 Ohms
Pr= 12-9 * 3e-1 = 0.9W
Your 12V 2A switching power supply has required current output for driving 3x300mA branches. It might flicker or get hot.
You should also try experiment with connecting 4 LEDs in series and limit the current with resistor (it will be much cheaper and smaller).
The diode is there to discharge the capacitor quickly.
If the output of U3A is high the diode is reversed polarized, so you can ignore it then. Then C7 gets charged via R22 with a time constant of 4 ms. That means it takes about 20 ms to charge completely.
If the output of U3A is low the diode conducts and discharges C7 very quickly, which you can see as the sharp falling edge on the scope.
Best Answer
If you are using a diode that follows the Shockley equation, you can iterate to find a solution. Assume a plausible Vf for the diode (say 0.7V for a silicon diode), calculate the currents and voltages, then find Vf for that current and substitute it back it. Repeat until the answer has converged to the desired accuracy. In most cases the diode voltage won't change much with R3 so it will converge rather quickly.
For rough calculation and relatively high voltage \$V_{IN} (\frac{R_3}{R_1+R_3})\$ compared to \$V_F\$ you may get a pretty accurate answer without any iterations.
Before you do this, convert the resistors and voltage source into their Thevenin equivalent - one resistor and a voltage source.
If you are required to come up with a closed-form solution, there is a method described in Wikipedia using the Lambert W function.