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I'm wondering if I'm thinking right when wanting to use the Thévenin equiv. on this diode problem? I'm thinking that in order for me to calculate \$I_D\$ then I have to "open circuit" the diode? Am I thinking wrong?

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  • \$\begingroup\$ Please clarify what you are trying to do. Are you talking about replacing everything except the diode with its Thevenin equivalent? When you say "in order to calculate Id" do you mean after creating the Thevenin equivalent or during the process of creating it? If you open circuit the diode, doesn't that mean that Id is 0 by definition? \$\endgroup\$ – Elliot Alderson Apr 6 at 12:33
  • \$\begingroup\$ The question says determine \$I_D\$ \$\endgroup\$ – Andy aka Apr 6 at 12:35
  • \$\begingroup\$ You could use Thévenin to make a simplified equivalent circuit of \$I_A, V_B, R_1\$ and \$R_2\$. However do you really need that? Since the \$V_T\$ of the diode is given we know the voltage across the diode (assuming it is in forward mode). So then why not simply replace the diode with a voltage source, determine the currents and check if that solution would also work if the voltage source was actually a diode. \$\endgroup\$ – Bimpelrekkie Apr 6 at 12:42
  • \$\begingroup\$ You know the current in \$R_2\$ is imposed by \$I_A\$ and the one in \$R_1\$ flows towards \$V_B\$ as 0.7 V is greater than 0.5 V. Calculate \$IR_1\$ and subtract it from \$I_A\$, this is the diode current you want (800 µA). \$\endgroup\$ – Verbal Kint Apr 6 at 12:49
  • \$\begingroup\$ Clearly 200 uA has to flow into \$V_B\$'s positive terminal leaving 800 uA to flow into the diode. R2 is irrelevant. Superposition is not required. \$\endgroup\$ – Andy aka Apr 6 at 12:51
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I think you can solve this problem using different ways.

And yes you are right, in order to calculate the Thevenin equiv. looking from diode terminals, you have to open the diode, calculate the \$R_{th}\$ and \$V_{th}\$ and then put the diode back.

Using nodal analysis to calculate \$V_{th}\$ across diode terminals

$$ \frac{V_B-V_{th}}{R1} +I_A=0 $$ $$ \therefore V_B-V_{th}=-I_AR_1 $$ $$ \therefore V_{th}=V_B+I_AR_1=0.5+0.001\times1000=1.5volts $$ Then we calculate \$R_{th}\$ By replacing every source with its internal resistance (making current sources open and voltage source close) $$ R_{th}=1k \Omega $$

Since \$V_T\$ of the diode equals \$0.7\$

$$ I_D=I_{total}=\frac{1.5-0.7}{1000}=800\mu A $$

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Please don't give out homework solutions. We expect the OP to show a significant amount of effort and ask a specific question. \$\endgroup\$ – Elliot Alderson Apr 6 at 14:27
  • \$\begingroup\$ @ElliotAlderson I gave him procedures for solving Thevenin equivalent for this circuit. \$\endgroup\$ – NAND Apr 6 at 14:41
  • \$\begingroup\$ No, you gave them the complete answer before they had done any work themselves. \$\endgroup\$ – Elliot Alderson Apr 6 at 14:46
  • \$\begingroup\$ @ElliotAlderson Is this site suggests not to answer the questions completely? \$\endgroup\$ – NAND Apr 6 at 15:09
  • \$\begingroup\$ @AbdulrhmanAboghanima thank you for your answer. I just need to ask, Rth, how come it becomes 1kohm? \$\endgroup\$ – Vetenskap Apr 6 at 15:11

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