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?

# Thevenin Equivalent – Applying Thevenin Equivalent on Diode Problems

circuit analysisdiodesthevenin

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## Best Answer

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\$

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

^{simulate this circuit – Schematic created using CircuitLab}