In my microelectronics course we have just begun learning about diodes and semiconductor physics. We have been analyzing simple circuits with diodes, mostly solving for component values. The approach we have been taught is to simply "guess" how the circuit is behaving, running the math to see if it works out, and adjusting if necessary. For example, consider this circuit:

enter image description here

NOTE: I have already solved this question (I am not asking for homework help).

It turns out D1 is reverse biased and D2 is forward. I only found this out after testing two other assumptions (D1 and D2 both on, and D1 on/D2 off).

Are there any clues that indicate a certain assumption is more likely to be correct? I ask this because we are expected to solve these rather quickly - I know practice is the best, but I am hoping for some tips.

  • 1
    \$\begingroup\$ Well, you can tell that the potential at the node B is positive because of the voltage divider formed by the 5k and 10k resistors, but lower than the voltage indicated by V+-. Therefore, D2 being forward and D1 being reverse is the solution. This is a simple circuit however, there is generally no trick to be sure which diode is conducting or not. \$\endgroup\$
    – lucas92
    Jan 29 '16 at 2:18

Here's your circuit, redrawn a bit with more labels.


simulate this circuit – Schematic created using CircuitLab

You can determine the state of the diodes without any math, just reasoning about the circuit. In these situations I find it helpful to imagine that there's no voltage across any of the resistors (through some magic mechanism), and then consider how the circuit will behave as it approaches equilibrium.

Initially the voltage across R2 is 0, so D1 and D2 are forward. This permits more current to flow through R2, pushing up nodes B and C with it.

Once C is at 0V, both diodes are forward, but only just. If C creeps above 0V, D1 will be reverse biased.

Will it? I can see that R1 and R2 make a voltage divider, and the voltage across R2 must be twice that of R1. With C at 0V, that's not the case: we have 10V across R1 and 9.3V across R2.

So the voltage at C will continue to rise, and D1 will be reverse biased. The voltage at C will continue rising until the voltage divider is in equilibrium, and D2 will remain forward biased.


to do this exhaustively, you have to test all 4 possible hypothetical ON/OFF combinations for D1 and D2.

so you make an assumption regarding the ON-ness of D1 and D2. every ON diode is replaced with a 0.7 v source (with the + side the same as the triangle side of the diode) but the current flow must be positive going into the triangle side of the diode. every OFF diode is an open circuit, but then confirm that the voltage on the bar side of the diode comes out to be more positive than the voltage on the triangle side.

when i look at the circuit, i can say right away that D2 must be ON (no reason for it not to be). so then the only two cases are, as you have mentioned, D1 ON and D1 OFF. and D1 OFF is confirmed when you solve for the voltage at point B.


It's about spotting stuff and the thing with spotting stuff is that you really only get there with practice.

In this case we spot that 10V is the highest voltage in the system and the only way current flowing in the 5K resister can flow is through D2. Therefore D2 must be in forward bias.

How about D1, well lets think about what would happen if D1 wasn't there. We would essentially have a voltage divider and since the top resistor is smaller than the bottom one it would have a positive voltage on the tapoff point. Therefore D1 must be reverse biased.


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