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I have studied the diode model which treats diode like elements with constant voltage drop and depending on the bias of the diode the voltage drop varies. Now I'm starting to learn about the correct model, the diode has a variable voltage drop depending on the voltage, resistance.If I have a diode in series with a resistor how do I find the current through the diode so according to the IV curve I will be able to find the voltage and how does the resistor affect it?

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  • \$\begingroup\$ See if lednique.com/current-voltage-relationships/… helps for a simple graphical calculation. It's for LEDs but the same principle applies if you can draw the diode IV curve. \$\endgroup\$ – Transistor Mar 26 at 15:23
  • \$\begingroup\$ I used to solve it graphically. The intersection of VI curve and the load line. \$\endgroup\$ – Mitu Raj Mar 26 at 17:50
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If all you can see is a diode and a resistor, with no other information, then you can not find the current through the diode.

You must look at the entire circuit. Substitute an ideal voltage source for a forward-biased diode and calculate the current. Use whatever exponential model you like to calculate the actual forward voltage of the diode at that specific current level. Change your ideal voltage source voltage to the calculated diode voltage. Repeat until the values of diode voltage and current converge to your satisfaction.

Or, run a SPICE simulation.

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If the diode model is given by the function: $$I_d=f(V_d)$$

we can write the following equation for the series circuit:

$$V_d + V_R = V_s$$ $$ I_d=I_R=I $$

substituting with Ohms law and the first equation (here the subscripts \$s\$, \$R\$ and \$d\$ are for the source, resistor and the diode respectively: $$ I=f(V_d) = \frac{V_R}{R}=\frac{V_s-V_d}{R}$$

So now you have a function of \$V_d\$ on both sides, meaning that you can solve it graphically by plotting both \$f(V_d)\$ and the \$\frac{V_s-V_d}{R}\$ on the same plane and finding the intersection point. Or solve it analytically if \$f\$ is known and is of a solvable form.

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  • \$\begingroup\$ is the voltage function the IV curve? \$\endgroup\$ – Miss Mulan Mar 26 at 18:03
  • \$\begingroup\$ Yes. You can see that it is the relation between I and V \$\endgroup\$ – Eugene Sh. Mar 26 at 18:21
  • \$\begingroup\$ Can we use this process to solve more complex diode circuits? \$\endgroup\$ – Miss Mulan Mar 26 at 19:53
  • \$\begingroup\$ @MissMulan Depends what you mean by process :) But yeah, by writing equations for KVL and KCL (which I basically did here in the second and third equations) you can solve more complex circuits. \$\endgroup\$ – Eugene Sh. Mar 26 at 19:56
  • \$\begingroup\$ Ah okay thank you. \$\endgroup\$ – Miss Mulan Mar 26 at 19:57
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By hand you need to use the simplified model you learned.

Graphically you may find the bias point of the circuit.

Mathematically that circuit has no closed form solution.

Use LT Spice.

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  • \$\begingroup\$ Depending on the IV model it may very well have an analytical solution. Using LTSPICE is not an answer when someone is asking to calculate something. \$\endgroup\$ – Eugene Sh. Mar 26 at 15:03

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