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Let's suppose I want to know the diode bias, how would KVL look when applied on this circuit assuming the diode is off? (Suppose Vin = 10V)

What I would do is: If diode is OFF current should be flowing (or try to flow) from right to left on the diode, so the right side of the diode need to have a higher potential than the left. Applying KVL from the battery counter-clockwise would lead to -Vin-Vd = 0, Vd=-Vin, Vd=-10V and the diode is indeed OFF. But I know that's not the correct answer... What is the mistake I'm making?

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If diode is OFF ...

We don't speak of diodes as "on" or "off". We normally speak of them as conduction or not. To conduct the diode must be "forward biased". That means that the anode (left side on your diagram) must be at a higher potential than the cathode (right side on your diagram).

... current should be flowing (or try to flow) from right to left on the diode,

No. The convention is that current flows from battery + to battery -. This is clockwise in your diagram and left to right in the diode. The diode symbol is actually an arrow showing the direction of current flow. (The convention was established long before the discovery of the electron. We keep with the convention but keep in the back of our minds that electron flow is from negative to positive.)

... so the right side of the diode need to have a higher potential than the left.

No. As above, to get the diode to conduct it must be forward biased. The anode (left) must have higher potential than the cathode (right).

Applying KVL from the battery counter-clockwise would lead to -Vin-Vd = 0, Vd=-Vin, Vd=-10V and the diode is indeed OFF. But I know that's not the correct answer... What is the mistake I'm making?

  • Kirchoffs Voltage Law or KVL, states that “in any closed loop network, the total voltage around the loop is equal to the sum of all the voltage drops within the same loop” which is also equal to zero.

schematic

simulate this circuit – Schematic created using CircuitLab

You want to go counter-clockwise so \$ -V_{IN} -V_R - V_D = 0 \$. Note that arrows point to higher potential so if the sign of Vd is positive the diode will be reverse biased.

We can re-write the above as \$ V_{IN} = -V_R - V_D \$. From this we can see that the \$ V_R \$ and \$ V_D \$ voltage drop is opposite to that of the arrows on the schematic. Therefore D is forward biased, is conducting and current will flow through R.

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What is the mistake I'm making?

When you assume the (ideal) diode is off, you're stipulating that there is zero current through - an ideal diode that is off is an open circuit.

When you solve for the voltage across the diode using KVL, you write:

$$V_d = V_{in} - V_{out} = V_{in} = 10V$$

i.e., the voltage across the resistor must be zero since there is zero current through.

But this is a contradiction since, for an ideal diode, there can be no positive voltage across.

Thus you conclude that the diode isn't off. Solve the circuit for zero diode voltage and then, to be sure, check that the diode current is positive.

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  • \$\begingroup\$ Inside the diode the "triangle" part's potential is always higher than that of | ? See, thats what I'm having problem when we stipulate a current direction and apply KVL on a resistor we consider the higher potential according to the current direction we stipulate. On the diode is the polarities fixed? \$\endgroup\$ Sep 2, 2016 at 22:54
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    \$\begingroup\$ @JoãoPedro, the diode voltage is positive if the anode is more positive than the cathode. For an ideal diode, there can be no positive voltage across, only negative. If you assume the diode is off and find a positive voltage across, your assumption is wrong. Similarly, for an ideal diode there can be no negative current through. If you assume the diode is on and find a negative current through, your assumption is wrong. \$\endgroup\$ Sep 2, 2016 at 22:58

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