When multiple voltage sources push current through the resistances of a circuit, do these sources interact with each other? I understand they superimpose on each other, in such a way that they drop voltages across components as if to be the only voltage source- but when considering the end result, the voltages dropped from the several sources are summed and a net current occurs. Or so i thought, until i made a small circuit to test the idea.

The circuit i made tests the idea that two voltage sources only see a circuit but not the other sources opposing the individual voltages dropped across the resistances. V1 applies 5v through a diode to a 1k resistor, and should see 5ma flow. The other voltage source does the same from its end. This is not the result, and the voltages send a total current as if only one voltage source is present. Obviously there is a foundational misunderstanding, and i am in need of correction. What is happening here? Current from one source is pushed from its own source, and because it is not current to and from the other source- it shouldn't cause an opposition on the other source. The currents should sum with my understanding, but clearly my understanding is wrong.

 

The input signal, a separate source from vcc- in a common collector amplifier allows the two to combine their resulting currents. I suppose my understanding of it isnt correct, and something else is the cause that has a similar result.

 

I guess what i dont understand is, if each source applies a voltage to the same resistor- half the current needed to create a voltage drop equal to the supply (minus the forward drop of the diodes in series to their respective supply) is output from each source. If each source only sees the current sent and received that it is pushing on, how does the source know to apply only half the current? I understand that applying Kirchhoff laws explain what happens from a mathematical stand point but- from a physics stand point how does this happen? How do these sources interact, such that a balance is made where each drop their voltage to pass half the current? I know that parallel voltages are equal, but i am searching for intuition for this situation.

 

If you redraw the circuit this way, is it any more intuitive? What if one battery is higher by one microvolt than the other, and the diodes are ideal?

 

I believe a very large current would flow from one battery, through the smaller potential battery, and back to itself- but at the same time, this action would reverse bias the diode and not allow conduction. Allowing current from one supply and not the other(?) What i dont understand is, if each source pushes current regardless of what another source is doing (and i believe i have a misunderstanding there) what physically decides that its current is cut down dependent on there being another source (of perfectly equal voltage)? Do voltage sources that are completely unrelated, causes forces on each other? Id always thought that batteries, for example, pushed and applied force on current that is leaving and coming back to the battery, regardless of other sources- as their current does not affect the battery. I cant figure out how, if a battery of a voltage parallel to a DC power supply of equal voltage, on a shared resistor- only drop one supplies voltage and each pass half the current. I understand that a law governs that, but how do these two separate and completely unrelated supplies interact to achieve this on a physical level?

 

I know that if two batteries were placed against each other in a loop, a short would occur- so clearly the forces are not felt on each other. How do they interact when in parallel? If they are equal voltage with equal ideal diodes, some how an interaction is made where they source half the current each. This interaction im sure is something simple, and knowing it will tie much of this together for me.

 

How do they interact when in parallel?

There is no interaction at all. For purposes of circuit analysis, two ideal batteries in parallel can be replaced by a single battery.

 

So if two sources were placed over one diode in both directions- How would the diode know to conduct? A force is available to allow it to operate in forward bias, but another force is available to keep it from doing so. What would result? Is this similar to a capacitor, where only charge and force applied to put the charge there matters yielding a force opposition on any sources trying to move current through? I understand the circumstance is much more complex and that circuit analysis is available to simplify the scenario

 

So if two sources were placed over one diode in both directions

I can't visualize what this means. Need a diagram. (It's way past my bedtime.)

 

When multiple voltage sources push current through the resistances of a circuit, do these sources interact with each other? I understand they superimpose on each other, in such a way that they drop voltages across components as if to be the only voltage source- but when considering the end result, the voltages dropped from the several sources are summed and a net current occurs. Or so i thought, until i made a small circuit to test the idea.

The circuit i made tests the idea that two voltage sources only see a circuit but not the other sources opposing the individual voltages dropped across the resistances. V1 applies 5v through a diode to a 1k resistor, and should see 5ma flow. The other voltage source does the same from its end. This is not the result, and the voltages send a total current as if only one voltage source is present. Obviously there is a foundational misunderstanding, and i am in need of correction. What is happening here? Current from one source is pushed from its own source, and because it is not current to and from the other source- it shouldn't cause an opposition on the other source. The currents should sum with my understanding, but clearly my understanding is wrong.

In many circuits especially linear circuits sources can interact wildly but when you place diodes in the circuit that can alter the behavior drastically.

For two voltage sources each with a different voltage level connected together by a lone resistor and with a common ground, the source with the higher voltage will dissipate energy while the one with the lower voltage will absorb energy. If the lower voltage source is a battery, the battery will charge. If the higher voltage source is a battery, the battery will discharge into the lower source.

Now if you add diodes the behavior can change radically because diodes limit the direction in which current can travel. If you have the same circuit as above and a lone diode is placed in series with the resistor, when the diode is connected one way (so that it conducts) the circuit behavior may not change too much (except for the diode drop), but if you then reverse that diode there will be little or no current flow at all. So you see right away that the way we connect the diode has a lot to do with how the current flows.

Diodes can really really complicate a circuit even though they are relatively simple devices. In static circuits the solutions are not usually too hard to find, but in more dynamic circuits (such as AC or transient circuits) the solution can often require a much more complicated math solution that can even involve partial differential equations. rather than simply a set of ODE's. This is a time when we can be glad circuit simulator programs were invented

 

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.

 

... cannot figure out why two unrelated sources with a shared resistor split the current supply.

But this is an unusual situation created with ideal components, so how can you be sure that the two sources split the current supply? What would change at the node if only one of the sources provided all the current? With ideal components the degree of current splitting is indeterminate. There is not something else going on, it's just that it does not matter.

 

Thank you for your consistent response. I ran this through ltspice and it had shown that both sources equally split the current. To me, it appears as if both sources should act as if they are by themselves because of the diodes, and that twice current should appear over the resistor returning to their respective sources. I know i sound like a broken record when i say, but some kind of opposition must occur to each source from the resistor to cause them to apply their voltage over the resistor and result with only half the current from each source.

 

I ran this through ltspice and it had shown that both sources equally split the current.

So the real question is why does ltspice equally split the current? If ltspice is using realistic non-linear models for the diodes, then the answer is obvious. If ltspice is using ideal diode models, then the result is what it was programmed to provide. I'd be interested in knowing the result if there was a small difference (1 micro-volt) in the battery voltages.

 

It works as it should. I soldered this circuit together with two separate batteries. Each battery was 1.55v, and each diode was .55v forward drop. When one battery was put in, .96mA of current made its way from the battery. When i plugged the other battery in, the current had dropped down to .48mA

What does the battery see that forces its 1.55v to only push half the current when the other battery is present even though they do not interact with each other?

 

See next post- sorry for the double post.

 

@Advaka The simple answer to this question:

When multiple voltage sources push current through the resistances of a circuit, do these sources interact with each other?

Yes, if allowed. Your circuit allows them to do so, at the point between the diodes. All the diodes do is drop the voltage and prevent a battery from pushing back against the other battery on the Anode side of the diode. Since the voltage attraction for each battery is seen on the cathode side of each diode, the resistor sees the pushback from both. But because they are forced to equally go through one single resistor, the current value is cut in half for each battery. All the resistor sees is a parallel battery voltage with a larger capacity of current available. Ohm's Law: I = E/R. so I = (5-0.55)/1000 = 0.00445A (or 4.45mA). And this is shown by the simulation (the diodes I use have a slightly large Forward Voltage Drop):

 

What does the battery see that forces its 1.55v to only push half the current

It's not "what does the battery see" but rather why the diodes only allow half the current to flow through each diode. Realize that in this case each diode is behaving as a non-linear resistance, and that each diode must have the same exact voltage drop (~0.7V) and therefore the same current flowing through each battery. If you were to change the circuit by making one diode a germanium diode & the other a silicon diode, then the current splitting between the batteries would be quite different (not equal at all).

 

It's not "what does the battery see" but rather why the diodes only allow half the current to flow through each diode. Realize that in this case each diode is behaving as a non-linear resistance, and that each diode must have the same exact voltage drop (~0.7V) and therefore the same current flowing through each battery. If you were to change the circuit by making one diode a germanium diode & the other a silicon diode, then the current splitting between the batteries would be quite different (not equal at all).

I put in a schottky diode, and now the silcon diode is completely shut off only allowing a handful of nA to pass. Is this because the voltage source with the smaller voltage drop diode is using the imbalance of voltage to keep the diode reverse biased?

 

Is this because the voltage source with the smaller voltage drop diode is using the imbalance of voltage to keep the diode reverse biased?

Rather than being reverse biased, it's more likely that it is just not forward biased enough.