Confusion about how KCL and KVL are used for diode circuits

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The discussion revolves around the confusion regarding the application of Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL) in diode circuits. The key point is that when a diode is considered "on," it is treated using a constant voltage model rather than as a perfect short circuit. This model accounts for a small voltage drop (VD,on) across the diode, which influences the current in other branches of the circuit, such as R2. Participants clarify that the diode's behavior is modeled as non-ideal, allowing for current flow in parallel branches despite the diode being on. Understanding this concept resolves the initial confusion about current distribution in the circuit.
timnswede
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Homework Statement


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This is just one of the example problems in my book which is already solved for me, but I don't really understand their solution which I'll post here:
They start out with assuming Vin is very negative, which makes D1 turn on and makes Vout=VD,on+Vin. THen they solve for the currents.
JYPhFsY.png

Homework Equations


KCL/KVL

The Attempt at a Solution


What I am confused on is that if the diode is turned on, then the branch with the diode is a short circuit, so why is there current going in the R2 branch?
The book solves all the problems like this one the same way, but I don't really get it. Are we assuming the diode is actually on the brink of turning on/off, so it allows current through it, but it's actually not a short circuit yet?
 
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timnswede said:
What I am confused on is that if the diode is turned on, then the branch with the diode is a short circuit, so why is there current going in the R2 branch?
What does VD.on represent? Is the diode treated as ideal or non-ideal in these problems? :wink:
 
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timnswede said:
if the diode is turned on, then the branch with the diode is a short circuit,
No. You are asked to use the constant voltage model for the diode.
images (4).jpg
 
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Oh wow, not sure why I didn't realize it until you guys said it, but now it makes sense, thanks!
 

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