I am making a circuit simulation program, and I am not sure how to simulate the behavior of Diodes.
Take this circuit:
——100 Resistor —–LED——
Positive 5V—–470 Resistor—–| |———Negative 5V
———–200 Resistor———-
How would you calculate the voltage and current through all of these components?
Also, is it true that with a simple circuit like this:
Positive 5V——470 Resistor——LED——–Negative 5V
That you would subtract the forward voltage of the LED from the Battery's voltage, then use Ohm's law to calculate the current and resistance through the circuit?
voltage = 5;
newVoltage = voltage – forwardVoltage;
current = newVoltage / 470;
resistance = voltage / current;
I don't have a background in electrical engineering – so I am really clueless.
Thanks for the help.
Best Answer
Yes, you are correct in your last example, assuming a red LED with 2.0V over it, the current will be \$\frac{(10-2)V}{470Ω} \approx 17mA\$. For your first example, I assume you already know that the total resistance of something in parallel is \$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}\$ and you are struggling of how to calculate the total resistance of 100Ω, 200Ω and the LED, right? I am not sure if it is possible to do it that way. Instead look at each of the current paths. Some current will flow through \$R_1\$ (470Ω) and further through \$R_2\$ (200Ω). Let's call this current \$I_A\$. Then some current will flow through \$R_1\$, through \$R_3\$ and the LED, \$I_B\$.
Let's call the voltages over \$R_1\$, \$R_2\$ and \$R_3\$ for \$U_1\$, \$U_2\$ and \$U_3\$. Then \$U_1 = (I_A + I_B) \times R_1\$, \$U_2 = I_A \times R_2\$ and \$U_3 = I_B \times R_3\$. The voltage over \$R_2\$ will be the same as over \$R_3\$ and the LED combined, i.e. \$U_2 = U_3 + U_{LED}\$. You already know that the total voltage over all components is 10V. So
\$U_1 + U_2 = 10V\$
\$U_1 + U_3 + U_{LED} = 10V\$
and this gives you two equations with two unknowns, \$I_A\$ and \$I_B\$, which is standard math to solve.