Electronic – Voltage drop – Diode reverse biased

A diode has a limited bandwidth, reversing voltage will take time to react. However it is not that big current is flowing in the opposite direction, it is more like the capacitance of the diode is charging in this time, limiting the bandwidth.

The story about the DC motor switching is just that the charge within the inductor of the motor needs to go somewhere, the current can not be suddenly stopped. If you do not use the diode the voltage will peak because of the inductance of the motor possibly blowing the transistor. Because of the diode the charge of the motor inductance will be "shortcircuited" over the motor-coil itself and the voltage will just ripple a little.

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.