Could you explain how that is so? Does this occur because the other source has more voltage available to do so?

 

From https://en.wikipedia.org/wiki/Schottky_diode

"When sufficient forward voltage is applied, a current flows in the forward direction. A silicon p–n diode has a typical forward voltage of 600–700 mV, while the Schottky's forward voltage is 150–450 mV. This lower forward voltage requirement allows higher switching speeds and better system efficiency. "

 

It sounds as if the other source having more voltage available at the node opposes the other source keeping the diode from the other source out of conduction (with the source containing the schottky conducting) but that is like i said. These two voltages interact at the diode (that isnt the schottky) and keeps it from conducting?

 

But it is conducting. Determine the forward voltage across diode in the circuit, plug that into the I(V) equation for a pn junction and find the current flowing through the diode. It won't be much, but it will be more than zero.

 

But it is conducting. Determine the forward voltage across diode in the circuit, plug that into the I(V) equation for a pn junction and find the current flowing through the diode. It won't be much, but it will be more than zero.

I certainly will. I understand that with superposition, voltages in a way super impose, and i guess dont oppose each other except for when the net current of two sources going opposite directions over a resistor times the resistance yields the resultant voltage drop. If two currents flow through a resistor- tho opposite directions, the power is found as the net current squared times the resistance for which it flows through. Unless im wrong, why isnt the total of both currents considered? They are flowing through the resistance in the end anyway arent they?

 

I mean, hypothetically. If two sources opposed on both sides of a resistor sharing the same ground pushed current through this resistor, would each source flow as if the other source wasnt there? Wouldnt there be opposition of any sort because of the electrons flowing in opposite directions? I understand if there were equal voltages over this resistance- the net current would be zero, but surely alot of current would flow, and alot of power would be consumed in the resistor. I understand that circuits like this dont typically occur, but the results are something id certainly learn from

 

Wouldnt there be opposition of any sort because of the electrons flowing in opposite directions?

It seems you need to get a better handle on the fictitious currents used with superposition & mesh current analysis versus the real currents that actually flow in the circuit. One of the reasons that I only use nodal analysis is that there are no fictitious currents to keep track of.

 

Thank you for the thought out response. Fortunately, this is a simple DC circuit with a simple steady state. My issue, is that I cannot figure out why two unrelated sources with a shared resistor split the current supply. From the sources point of view, all it sees is a diode and a resistor with the other diode acting as an open circuit- clearly something else is going on that i do not understand.

Well there are finer points to consider in this circuit depending on what model you are using for the two diodes.

If you are using the ideal diode model with fixed forward voltage drop, then one diode will be completely OFF if one source is of higher voltage than the other.
Now if you introduce leakage current to that model then the diode that was OFF before will conduct a very small amount of current, depending on the level of the reverse voltage across the diode. If there is a high reverse voltage it may conduct a lot.

If you go to a more advanced model like an exponential model, then if the two voltage source levels are close to each other even though not exactly the same, then both diodes may conduct with one conducting more than the other.

So maybe you should mention what kind of model you want to use, and if you are doing this in real life then you have to say what the voltage levels actually are so we can use a better model to explain what you are seeing. In a real life situation you would have to use a more advanced model if the two source voltage levels are somewhat close to each other. For example, for 5v and 100v you would only have to consider leakage current, but for 5v and 5.01v you would have to use a more advanced model to show what each current is.

Also, you have to say what voltage levels you are using for the tests and also what currents you are measuring and how you are measuring them. I would suggest measuring the currents with fixed resistors of somewhat small values in series with each diode and measure the voltage across each resistor with a high impedance DC volt meter, or use two quality current meters in series with each diode.
If using the two resistor method, keep both resistors in the circuit at all times and make sure they have the same value for both resistors, we can add that resistance to the model which will not affect the results too much at low current levels.
It may be hard to measure leakage current though you may have to use a higher value resistor in series with the diode or a good current meter and accept some error in measurement.