simulate this circuit – Schematic created using CircuitLab

Both diodes are assumed to be ideal. The current source has I Amp current through it .I am not sure about the voltages at the cathode of diodes and thus I am confused when the diodes will be turned ON and OFF. So How we can decide that when the voltage at A and B is Low then both diodes are OFF because it might be case that voltage at cathode is much Lower or negative than anode.I mean how we decide that diodes are ON or OFF when we know the anode voltage only

  • 1
    \$\begingroup\$ The diodes will be "turned on" once the forward voltage drop is exceeding a specific threshold (usually around 0.7V). Take it from here as it is clearly a homework question (BTW, it looks like the information provided is not sufficient for the solution. To know what is going on we need to know what is connected to Y). \$\endgroup\$
    – Eugene Sh.
    May 10, 2017 at 17:10
  • \$\begingroup\$ @EugeneSh.: In an academic question like this, it's safe to assume that the inputs are connected to voltage sources (zero output impedance) and the outputs are connected to voltage "sinks" (infinite input impedance). \$\endgroup\$
    – Dave Tweed
    May 10, 2017 at 17:22
  • \$\begingroup\$ I know silicon diodes are robust, but 1 amp through 1N4148's might be pushing it rather! \$\endgroup\$
    – Neil_UK
    May 10, 2017 at 17:23
  • \$\begingroup\$ This is very similar to the question you asked yesterday. Assume, for A and B, that 5 V is logic 1, and 0 V is logic 0. For each of the four possible input conditions, at least one of the diodes must be on. Take it from there. \$\endgroup\$
    – Chu
    May 10, 2017 at 18:47
  • \$\begingroup\$ @Chu so when the input is 0, then will any diode ON ? \$\endgroup\$ May 10, 2017 at 19:58

3 Answers 3


This seems to be a classic Diode-OR Gate. Both diodes are forward biased, therefore they both serve as a conductive path to Y so long as the input exceeds the Vf of the Diodes. So if Diode A OR Diode B are conducting, A voltage will be present at Y. General Description of Diode Transistor Logic "DTL"

  • \$\begingroup\$ Why can not the voltage at cathode be negative? .In this case the diode will be always turned ON \$\endgroup\$ May 10, 2017 at 17:41
  • \$\begingroup\$ That is true, but I would assume that the schematic you showed implies a single rail judging by the current being sunk to ground. \$\endgroup\$
    – Luke Gary
    May 12, 2017 at 0:03

I assume there's 1N4148 only because the schematic editor insisted that something from the component list must be selected. The question text states the diodes to be ideal.

This question really can't be answered shortly before we have an agreement on what measurables define the logic states and how. An example:

Let the states be defined by the node voltages against GND. Let 0V be the treshold between the states True and False. If positive voltage mean True, then this is OR. If negative voltage means True, this is AND.


It's a "OR" gate made up with diodes. As long as D1 or D2 are the current flow, Y will have a voltage (Va or Vb (-0.7V)).

But there is another scenario : if Va != Vb. In this case, Vy = max(Va, Vb) - 0.7


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