digital-logicdiodes
Consider the circuit below:
I'm assuming ideal diode model(\$V_{ON} = 0V\$). If \$V\$, \$A\$ and \$B\$ are all 5V, the diodes will be reversed biased and no current will flow through the circuit. Therefore, \$V_{OUT}\$ = 5V.
However, for either of the diodes, the voltages across them will be
\$V_{A} + V_{D} = V_{OUT}\$.
Since \$V_{A} = V_{OUT}\$, \$V_{D}\$ will 0. Since we're considering an ideal diode, wouldn't that mean that the diode is actually forward biased? What am I missing here?
I think the easiest method to solve such problems is to assume that the diodes are off (both, and then one of the two), compute the voltages across the diodes and see if there's a contradiction with your assumption. Let's call the top left diode \$D_1\$ and the diode in the middle \$D_2\$.
Case 1: \$D_1\$ off, \$D_2\$ off: Since \$D_1\$ is off there is no current through the top 5k resistor, and since \$D_2\$ is off, there is also no current through the bottom left 10k resistor. So \$V=0\$ and the voltage at the anode of \$D_1\$ is 15 Volts. Contradiction! (\$D_1\$ should be on).
Case 2: \$D_1\$ off, \$D_2\$ on: again no current through top 5k resistor. Voltage \$V\$ is
$$V=\frac{15V\cdot(5k||10k)}{10k+(5k||10k)}=3.75V$$
Contradiction! (Because the voltage across \$D_1\$ would be \$15V-3.75V=11.25V\$ and it should be on.)
Case 3: \$D_1\$ on, \$D_2\$ off: Voltage \$V\$ is
$$V=\frac{15V\cdot 10k}{5k+10k}=10V$$
The voltage at the anode of \$D_2\$ is \$15V\cdot 5k/15k=5V\$. This agrees with our assumption, because with these voltages \$D_2\$ must be off. So your solution is
$$I=0A,\quad V=10V$$
From the sounds of it, the diode model you are using is the simple "ideal diode" with a fixed forward voltage. This model is an open circuit when \$V_{\textrm{Anode}} - V_{\textrm{Cathode}} < V_D\$ (reverse biased), and a fixed \$V_D\$ voltage supply otherwise (forward biased).
Start by making assumptions about the state of D1 and D2 (for example, D1 is forward biased, and D2 is reverse biased).
Your circuit would then look like this:
simulate this circuit – Schematic created using CircuitLab
What is \$V_o\$ here? The last step is to check your assumptions on each diode. If the assumptions are correct, the model is applicable. If not, permute your assumptions (ex.: what if D1 is reverse biased, and D2 is forward biased? or D1 and D2 are reverse biased, etc.) Only one of the 4 possible permutations will have a consistent answer.
As a side note, the answer you get will not match what the circuit simulator gives you (though it will be close). This is because the circuit simulator uses a more advanced diode model.
Best Answer
If the voltage across the diode is equal to its \$V_f\$ (what you call \$V_{ON}\$) and no current is flowing through it then the diode is on the knife edge between forward and reverse bias. For forward bias we usually assume a positive current flowing through the diode and for reverse bias we assume that \$V_D < V_f\$.
Since the initial conditions (\$V_A = V_{SOURCE}\$ and \$V_{OUT} = 0V\$) have the diode reverse biased I would probably call it reverse biased even when the output voltage rises to the source voltage.